Theories in Ecological Risk Assessment 9819903084, 9789819903085

Table of contents :
Preface
Acknowledgments
Contents
Part I Linking Ecology and Ecotoxicology
1 Basic Concepts of Ecological Risk Assessment
1.1 What Is Risk?
1.2 What Is Risk Assessment?
1.3 Ecological Risk Assessment
1.4 Are Different NOEC Values Comparable Indices for Ecological Risk Assessment?
References
2 Population-level Assessment
2.1 Introduction
2.2 Differential Equation Model
2.2.1 Logistic Growth
2.2.2 Equilibrium and Stability
2.2.3 Multiple Species
2.2.4 Including Toxic Effects
2.3 Difference Equation Model
2.3.1 Increase or Decrease
2.3.2 Sensitivity Analysis
2.4 Case Studies
2.4.1 The Effect of Zinc on an Algal Population
Zinc Toxicity on Fathead Minnow
The Life-History Parameters of the Fathead Minnow and Sensitivities
Combining the Toxic Effect and Life-History Matrix
2.4.2 Population-level Species Sensitivity Distribution
2.4.3 Comparison of the NOEC and PTC
2.5 Toxic Effects on r- vs. K-Selected Species
2.5.1 r vs. K Selection
2.5.2 Model
2.5.3 Optimum Allocation Rates in a Crowded Population
2.5.4 Optimum Allocation Rates in a Sparse Population
2.5.5 Toxic Effects: r vs. K
References
3 Population Models of Extinction
3.1 Introduction
3.2 A Model of the Birth and Death Process
3.3 Models of Stochastic Processes: Necessary Elements
3.3.1 The Random Walk
3.3.2 Mean and Variance in Location
3.3.3 The Forward (Fokker–Planck) Equation
3.3.4 Backward Equation
Solving the Backward Equation: Obtaining MTE
3.4 A Model for Exponential Growth and EnvironmentalStochasticity
3.5 Logistic Equation with Environmental and Demographic Stochasticity
3.5.1 Mean Time to Extinction
The Numerical Solution for Eq.3.47
3.5.2 Time Evolution of the Probability Density and the Probability of Extinction
References
4 Population-Level Assessment Using the Canonical Model
4.1 The Canonical Model and the Adverse Effect of Chemicals
4.2 Parameter Estimation
4.2.1 Intrinsic Rate of Reproduction
4.2.2 Carrying Capacity and Environmental StochasticityIntensity
4.3 The Effect of DDT Exposure on a Herring Gull Population
4.3.1 Parameters for Natural (DDT-Free) Population
4.3.2 The Effects of DDT on a Herring Gull Population
4.3.3 Mean Time to Extinction (MTE)
4.3.4 Effect of DDT on MTE
4.4 Risk Equivalence
4.4.1 DDT Exposure vs. Habitat Loss
4.4.2 DDT Exposure vs. Environmental Stochasticity
4.4.3 How Long Do We Have for Habitat Restoration?
4.5 Effect of Nonylphenol Exposure on Medaka Fish
4.5.1 Medaka Life History
4.5.2 NP Toxicity for the Medaka Fish
4.6 Conclusion
References
Part II Models for Ecotoxicology
5 Species Sensitivity Distribution in Ecological Risk Assessment
5.1 Introduction
5.2 Basic SSD Elements
5.3 Criticisms of SSDs
5.4 The Probability that We Achieve the Protection Goal
5.4.1 The Minimum NOEC Approach
5.4.2 The SSD Approach
5.4.3 The t-Distribution
5.4.4 The Noncentral t-Distribution and the ExtrapolationFactor
5.4.5 The Distribution of the Estimated HC5
5.5 The Relationship Between , Sample Size, and the AF
5.5.1 Reduction of the Protection Goal
5.5.2 How Far from HC5
References
6 BLM: A Model for Predicting Metal Toxicities
6.1 Introduction
6.2 Estimation of Metal Toxicities Using the BLM
6.2.1 Metal Speciation: From Total Concentration to Bioavailability (i.e., Free-Ion Concentration)
6.2.2 The Concentration of Free Ions that Bind to the Biotic Ligands
6.3 Relationship Among Metals
6.4 The Toxic Effect of Metal Mixtures
References
7 Mathematical Models for Chemical Mixtures
7.1 Introduction
7.2 Two Classical Models for Predicting the Effects of Chemical Mixtures
7.2.1 Bliss's Independent Action
7.2.2 Loewe's Additivity Model
7.3 Statistical Test for Additivity
7.4 The Conditions for Additivity and Non-additivity
7.4.1 Case 1: Single Enzyme-Substrate Reaction
7.4.2 Case 2: Sequential Enzyme-Substrates Reactions
7.5 The Funnel Hypothesis
7.6 Metal Mixture
7.6.1 The Outline for the Model Predicting the Toxicity of Metal Mixture
7.6.2 The Total Metal Concentration Resulting in 50% Effect
7.6.3 Binary Mixture of Two Metals
7.6.4 The Reason for Non-additivity with Identical Toxic Mechanisms
References
8 Statistics and Related Topics
8.1 Introduction
8.2 Hypothesis Testing
8.3 Regression Analysis
8.3.1 Regression Analysis Based on the Maximum Likelihood
8.4 The Confidence Interval
8.5 AIC and Model Selection
8.6 SSDs, Hypothesis Testing, and Model Selection
References

Citation preview

Theoretical Biology

Masashi Kamo

Theories in Ecological Risk Assessment

Theoretical Biology Series Editor Yoh Iwasa, Kyushu University, Fukuoka, Japan

The “Theoretical Biology” series publishes volumes on all aspects of life sciences research for which a mathematical or computational approach can offer the appropriate methods to deepen our knowledge and insight. Topics covered include: cell and molecular biology, genetics, developmental biology, evolutionary biology, behavior sciences, species diversity, population ecology, chronobiology, bioinformatics, immunology, neuroscience, agricultural science, and medicine. The main focus of the series is on the biological phenomena whereas mathematics or informatics contribute the adequate tools. Target audience is researchers and graduate students in biology and other related areas who are interested in using mathematical techniques or computer simulations to understand biological processes and mathematicians who want to learn what are the questions biologists like to know using diverse mathematical tools.

Masashi Kamo

Theories in Ecological Risk Assessment

Masashi Kamo Research Institute of Science for Safety and Sustainability National Institute of Advanced Industrial Science and Technology Tsukuba, Ibaraki, Japan

ISSN 2522-0438 ISSN 2522-0446 (electronic) Theoretical Biology ISBN 978-981-99-0308-5 ISBN 978-981-99-0309-2 (eBook) https://doi.org/10.1007/978-981-99-0309-2 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

Some years ago, I had the opportunity to write a book, in Japanese, on ecological risk assessment. I have been pleased, but somewhat surprised given its multitude of mathematical formulas, to learn that the book has been reasonably well received. Happily, Springer gave me the opportunity to publish an English version of the work. I initially thought this would be easy, as I was simply translating a Japanese book into English, but having set about the task, I saw the need to address a number of other topics not discussed in the Japanese version, and the English book ended up taking much more time and effort than I had originally expected. I was originally a theoretical biologist. In recent years, I have been working exclusively on ecological risk assessment, but have never changed my original colors; and like other works written by theoretical biologists, this book contains many mathematical formulas. Some may ask why. The reason is simple: they provide the easiest and most powerful tools for argument. In such discussions, we make a few assumptions, design a model, and analyze it, confident in the accuracy of our results, however strange they may appear (of course, when mistakes are made in the analysis, the results cannot be trusted no matter how persuasive they may appear). It is impossible to pursue such logical development as easily using words alone. In many cases, unfounded beliefs (e.g., it can’t be!, or it must be!) get in the way of logical thinking. In environmental science, mathematical models are often used as predictive devices, and I sometimes use models in this way. However, one of the messages of this book is that mathematical models should not be seen as merely predictive devices, but also, like words, as tools for developing arguments. Some of the most important research I have been pursuing is introduced in the book. The first part describes the population-level assessments I conducted at the beginning of my ecological risk assessment. Chapter 1 begins with a description of the current methodology for ecological risk assessment, a methodology sometimes referred to as individual-level assessment, and one in common use worldwide. However, I believe that ecological risk must be assessed not only at the individual level but also at the population level, and this constitutes the main theme of the chapter.

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Chapter 2 describes the population dynamics models, developed in ecology, which are necessary to conduct population-level assessment. Such models are an essential element of theoretical ecology, and should be well understood. To that end, they are described in considerable detail in this chapter, along with a number of case studies employing the models. I believe that understanding the chapter will enable us to make initial population-level assessments, as well as more complex modeling. At the end of the chapter, an evolutionary biology approach is used to show how chemical sensitivity appears to differ for different life-history traits. In ecological risk assessment, it is important to find chemical-sensitive species; thus, I also discuss how our understanding of sensitivity depends on what level we focus on. Overall, the chapter serves as an example of using a mathematical model as a tool for logical development. Chapter 3, which represents the culmination of the first half of the book, introduces mathematical models for dealing with population extinction. Although, in general, the book is an introduction to my research, the derivation of mean time to extinction (MTE) presented in this chapter, and the case study of herring gulls using MTE, discussed in Chap. 4, are not parts of my research. I have long wished to pen a chapter like this, and am happy to have finally done it. What I consider most important in conducting a risk assessment study is to be able to compare risks of various qualities. Using the concept of risk equivalent derived from MTE, the chapter helps us understand how risks of different qualities, such as habitat destruction, chemical effects, and the magnitude of environmental fluctuation, can be rigorously compared. Chapter 4 discusses case studies employing MTE, and one way to approach the book might be to read Chap. 4 first and then return to Chap. 3 if the derivation of MTE looks interesting. The second part of the book discusses mathematical models involved in hazard assessments. Chapter 5 introduces the method of risk assessment using species sensitivity distributions (SSDs). SSDs are statistical distributions of toxicity data, and their use in ecological risk assessment has been studied for a long time; however, they have also been a target of criticism, and SSD-based risk assessment has yet to be widely practiced. Given the ubiquitous use of statistical methods in scientific activities, this reluctance has long seemed strange to me. In general, however, when using statistical methods, we need a great deal of data, and it is rare that we can obtain an abundance of toxicity data, which is likely one of the reasons why we hesitate to use SSDs in ecological risk assessment. Uncertainty can be compensated for through the use of uncertainty factors (AFs). However, if the uncertainty is not quantitatively known, the magnitude of the AF cannot be determined. The motivation for this research was the idea that this problem could be solved if we could quantitatively identify the level of uncertainty when data is scarce. Sample size is frequently considered in assessments of uncertainty, but I wish to show that other factors are equally important. Chapter 6 is concerned with the hazard assessment of metals. The toxicity of metals is known to vary depending on the water quality, and metals have long been considered difficult for risk assessment. This, however, changed dramatically with the appearance of the biotic ligand model (BLM) for predicting the toxicity of

Preface

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metals. I began my career as an environmental scientist, when this model was being rapidly developed, and it was exciting to witness the enthusiasm of the researchers who were involved in shaping a new era. As a theorist, I initially thought the BLM would not be of great use as theoretical issues, because of its focus on metal toxicity prediction, but a figure that appeared in a BLM review article completely changed my mind. It merely showed the toxicity values of some metals, but to me it seemed highly meaningful, and led me, eventually, to conclude that the toxic effects of metal mixtures could be additive. This sixth chapter, then, explores how consideration of the toxicity of individual metals has significant implications for our understanding of metal mixture toxicities. In Chap. 7, I discuss chemical mixtures using the Michaelis-Menten equation, a simple chemical reaction model which, along with its slight modifications, is probably familiar to everyone. I show that even such a simple model can provide insights into mixture effects. For example, the so-called funnel hypothesis states that the mixture effects become additive when many chemicals are contained in the mixture. I now believe this hypothesis to be correct, but my reasons are somewhat different from most others. This chapter provides a further example of the use of mathematical models as logical tools. Chapter 8 considers the basic elements of statistics and related topics. In recent years, there have been calls to stop using hypothesis testing, because, it is argued, the results of such testing are often misinterpreted. While I think that a ban on hypothesis testing would be too extreme, I agree that we must interpret the results with great care. In ecological risk assessment, for example, the no observed effect concentrations (NOECs) obtained by toxicity testing play an important role, and these are determined through hypothesis testing. Those who think that NOECs refer to concentrations representing “no effect” are encouraged to read this chapter. Throughout the book, I describe the relevant calculations in some detail, so that readers can arrive at the final equations without having to consult other reference books. Those with a mathematical background, who may find the calculations somewhat facile, are welcomed to skip the derivations and proceed directly to the relevant case studies. I detailed the derivations because one of the working principles of the work was that even those without mathematical training should be able to reach the final equations with some effort. Even in theoretical studies, good expository works rarely use equations (as Stephen Hawking said in A Brief History of Time (1988), “Someone told me that each equation I included in the book would halve the sales”), while math-intensive books are typically forbidding to beginners. In shaping the book, I targeted the middle ground between these two extremes. Beyond these considerations, however, I think solving equations is inherently interesting. When, for example, I myself arrived at the formula for mean time to extinction discussed in Chap. 3, I almost soared with joy. It is exciting even to arrive at a formula that someone else has already derived, when one arrives on one’s own steam as it were. Imagine, then, how exciting it is to build your own model, do the relevant analysis, and arrive at an equation that no one else has yet discovered. As the author of this work, I would be delighted if it helped someone to someday have this experience themselves.

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The codes (Mathematica, C, and possibly R) used to generate the figures in the book, along with related commentaries, will (I hope soon) be found at https://github. com/masashi0209/Theories_ERA; however, the repository is currently empty. I welcome those interested to contact me if they should require a particular code, in which case I will generate it as soon as possible. Tsukuba, Japan

Masashi Kamo

Acknowledgments

Research cannot be done alone. Especially for someone like me, who lacks selfconfidence, it is essential to have mentors or collaborators who guide and correct them. This book has described the research I have done, and I am sure that I could not have been able to do this work without great support from these people. It is very fortunate for me to have received the guidance of two great researchers: Professor Yoh Iwasa, a giant of mathematical biology, and Professor Junko Nakanishi, a giant of risk assessment. It is already a rare event to meet and be mentored by a single, highly accomplished researcher; few have the good fortune of receiving direct instruction from two. It is truly fortunate for me to have had such a chance, the probability of whose occurrence may be considered to be zero by linear approximation. It was Dr. Wataru Naito who gave me much advice when I still knew nothing about ecological risk assessment. Some of my early work on population-level assessments was done in collaboration with him. It was Dr. Robert Pastorok who gave me high marks for my first paper on the population-level species sensitivity distribution in ecological risk assessment. Without these, I would not have had such confidence in my work, and would probably be doing something else by now. In my research on population-level assessment, the contribution from Dr. Takehiko I. Hayashi cannot be neglected. He and I had many discussions about how to incorporate ecological risk assessment into ecology. Dr. Takashi Nagai gave me advice on the toxicity of metals. I could not have proceeded without his contribution on the combined effects of metals. Dr. Hiroyuki Yokomizo made a great contribution to my theoretical studies on chemical mixtures; Chapter 6 is the result of my collaboration with him. I hope to continue collaborating with him in the future (if he agrees, of course, and if he doesn’t, I may quit my research in despair). Dr. Yuichi Iwasaki made a significant contribution to the overall content in this book. In risk assessment, in addition to purely scientific research, the perspective of how to apply such research in assessment practice is important. As a theorist, I am not skilled in such applications; but he has skillfully translated my theoretical terms into risk assessment terms. Without his advice, my publications in risk assessment research would be much fewer. Although he sometimes says something strange, he ix

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is my most reliable research friend. Dr. Junichi Takeshita is the person I go to when I have a problem with mathematics. This book would have been much weaker without his support. Professor Hiroshi Hakoyama, who derived the mean time to extinction in Chapter 3, gave me a great deal of feedback on this book. His comments dramatically improved the work’s accuracy, especially in Chapter 3. Nevertheless, if there are still mistakes, they are all my fault. He was my senpai (senior student) and we spent many fruitful hours in the same lab. I fondly remember the days when he was struggling with the mean time to extinction. As I have no systematic experience in risk assessment, chatting with my colleagues at my institute has also been indispensable. In risk assessment, the meaning of the results can change with one’s perspective, and I have learned the folly of stubbornly insisting on one’s perspective, and the importance of tolerance for different perspectives, during lunch-time chats. Finally, my thanks to all the readers of the book. It will be my great pleasure if you find that the mathematical model is useful and fun.

Contents

Part I Linking Ecology and Ecotoxicology 1

2

Basic Concepts of Ecological Risk Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 What Is Risk? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 What Is Risk Assessment? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Ecological Risk Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Are Different NOEC Values Comparable Indices for Ecological Risk Assessment?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 10

Population-level Assessment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Differential Equation Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Logistic Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Equilibrium and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Multiple Species . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4 Including Toxic Effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Difference Equation Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Increase or Decrease . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Sensitivity Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Case Studies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 The Effect of Zinc on an Algal Population . . . . . . . . . . . . . . . . . . . . 2.4.2 Population-level Species Sensitivity Distribution . . . . . . . . . . . . . 2.4.3 Comparison of the NOEC and PTC. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Toxic Effects on r- vs. K-Selected Species . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 r vs. K Selection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.3 Optimum Allocation Rates in a Crowded Population . . . . . . . . . 2.5.4 Optimum Allocation Rates in a Sparse Population . . . . . . . . . . . 2.5.5 Toxic Effects: r vs. K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 11 13 16 17 19 22 23 25 28 31 31 37 38 40 40 40 41 43 44 46

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Contents

Population Models of Extinction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 A Model of the Birth and Death Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Models of Stochastic Processes: Necessary Elements . . . . . . . . . . . . . . . . 3.3.1 The Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Mean and Variance in Location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 The Forward (Fokker–Planck) Equation. . . . . . . . . . . . . . . . . . . . . . . 3.3.4 Backward Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 A Model for Exponential Growth and Environmental Stochasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Logistic Equation with Environmental and Demographic Stochasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Mean Time to Extinction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Time Evolution of the Probability Density and the Probability of Extinction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 49 50 53 56 58 61 62

Population-Level Assessment Using the Canonical Model . . . . . . . . . . . . . . 4.1 The Canonical Model and the Adverse Effect of Chemicals. . . . . . . . . . 4.2 Parameter Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Intrinsic Rate of Reproduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Carrying Capacity and Environmental Stochasticity Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The Effect of DDT Exposure on a Herring Gull Population . . . . . . . . . . 4.3.1 Parameters for Natural (DDT-Free) Population . . . . . . . . . . . . . . . 4.3.2 The Effects of DDT on a Herring Gull Population . . . . . . . . . . . . 4.3.3 Mean Time to Extinction (MTE). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.4 Effect of DDT on MTE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Risk Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 DDT Exposure vs. Habitat Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 DDT Exposure vs. Environmental Stochasticity . . . . . . . . . . . . . . 4.4.3 How Long Do We Have for Habitat Restoration? . . . . . . . . . . . . . 4.5 Effect of Nonylphenol Exposure on Medaka Fish. . . . . . . . . . . . . . . . . . . . . 4.5.1 Medaka Life History. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 NP Toxicity for the Medaka Fish . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

81 81 83 83

66 70 74 77 79

84 85 85 86 88 89 91 91 92 93 94 94 95 98 99

Part II Models for Ecotoxicology 5

Species Sensitivity Distribution in Ecological Risk Assessment. . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Basic SSD Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Criticisms of SSDs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The Probability that We Achieve the Protection Goal. . . . . . . . . . . . . . . . . 5.4.1 The Minimum NOEC Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

103 103 104 107 110 111

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5.4.2 The SSD Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.3 The t-Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.4 The Noncentral t-Distribution and the Extrapolation Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.5 The Distribution of the Estimated HC5 . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Relationship Between σˆ , Sample Size, and the AF . . . . . . . . . . . . . . 5.5.1 Reduction of the Protection Goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 How Far from HC5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

113 114

BLM: A Model for Predicting Metal Toxicities . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Estimation of Metal Toxicities Using the BLM . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Metal Speciation: From Total Concentration to Bioavailability (i.e., Free-Ion Concentration) . . . . . . . . . . . . . . . . . 6.2.2 The Concentration of Free Ions that Bind to the Biotic Ligands . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Relationship Among Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Toxic Effect of Metal Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

135 135 139

Mathematical Models for Chemical Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Two Classical Models for Predicting the Effects of Chemical Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Bliss’s Independent Action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.2 Loewe’s Additivity Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Statistical Test for Additivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Conditions for Additivity and Non-additivity . . . . . . . . . . . . . . . . . . . . 7.4.1 Case 1: Single Enzyme-Substrate Reaction . . . . . . . . . . . . . . . . . . . 7.4.2 Case 2: Sequential Enzyme-Substrates Reactions . . . . . . . . . . . . 7.5 The Funnel Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Metal Mixture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 The Outline for the Model Predicting the Toxicity of Metal Mixture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 The Total Metal Concentration Resulting in 50% Effect . . . . . 7.6.3 Binary Mixture of Two Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.4 The Reason for Non-additivity with Identical Toxic Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

151 151

Statistics and Related Topics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Hypothesis Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Regression Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Regression Analysis Based on the Maximum Likelihood . . . .

183 183 184 186 189

119 123 126 128 130 133

139 140 145 148 149

153 153 154 157 162 163 167 171 174 175 176 177 178 182

xiv

Contents

8.4 The Confidence Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 AIC and Model Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 SSDs, Hypothesis Testing, and Model Selection . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

191 196 201 202

Part I

Linking Ecology and Ecotoxicology

Chapter 1

Basic Concepts of Ecological Risk Assessment

Abstract This work considers various mathematical models used in ecological risk assessment, primarily discussing models used in hazard assessment. This chapter provides a very brief introduction to the basics of risk assessment in general and ecological risk assessment in particular, including explanation of some terminology used in the latter, and illustration of risk assessment currently being done. Lastly, population-level assessment, which has received increasing attention in recent years, will be described, and the reason why we need such assessment will be discussed. Keywords Risk · Risk assessment · Ecological risk assessment · Population-level effect

1.1 What Is Risk? This work focuses mainly on ecological risk assessment, but first must briefly consider risk assessment in general. The apparent simplicity of the title of this section conceals a great deal of complexity. The word “risk” is frequently used in our daily lives, but often with different meanings, depending on the speaker and the situation. A common definition of the term is Risk = impact of the given event .× probability that the given event occurs;

however, I do not think it wise to limit the concept to a single abstract definition. I think we should be tolerant of varying definitions of the term, and when it is used, in argument for example, we should be careful to ascertain its precise sense in the respective context.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_1

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1 Basic Concepts of Ecological Risk Assessment

1.2 What Is Risk Assessment? Risk assessment is an integral part of decision-making. The need for decisionmaking is ubiquitous in our daily lives, and most of us perform risk assessments every day; for example, the decision whether or not to take an umbrella with you before you set off for work. In terms of the definition above, the “magnitude” of the event’s impact is the fact of getting wet in the rain. Let us say, then, that you do not want to take an umbrella with you, if at all possible, because umbrellas get in the way and you may lose it somewhere. Now, these two “events”—that you get wet and that you take an umbrella—are different and thus may not be directly compared; hence, we unify the unit of assessment, in the form of the degree of dislike, and the rest is easy. You turn on the TV and consult the weather forecast, which tells you that the probability of rain is p. You then compare “.p× degree of dislike of getting wet” and “.(1 − p)× degree of dislike of taking an umbrella” and decide whether to take an umbrella with you. This is a formal risk assessment and comparatively easy to perform. But in this example, the probability of rain is supplied by the weather forecast. Can you determine it yourself? Possibly, if you are skilled in reading weather maps but can you generate such maps yourself? The weather is, in fact, a highly complex phenomenon, and proper forecasting requires an extensive observational network, including weather satellites and powerful computers, far beyond the reach of common individuals. When dealing with such complex events, then, experts and expert equipment are typically needed. One of the important features of risk assessment is that it cannot be paused or postponed. This is a major difference from typical scientific activities. Acting the know-it-all is the greatest sin for scientists, who will—or certainly should—make no conclusions until they have conclusive evidence. For them, the right course of action is to wait until they are sufficiently certain of their results to make them public. As far as I know, there is only one reliable way to conclusively know the evening weather, and that is to wait until evening. Can we behave like good scientists on this issue? Should we call our boss and say that we will not come to work until the evening because we are not sure if it will rain? In the absence of a remote-work option, the answer would likely be “Fine, don’t come. Ever.” The fact is that we are not typically allowed to postpone our decisions—in this case about the umbrella— until the ideal moment, but here we need to make our risk assessment in the morning. Risk assessment, then, is the study of directing one’s current behavior based on the possibility of future events [2]. Put simply, the aim of risk assessment is to provide information for decisionmaking, and the most important element in such assessment is the ability to compare qualitatively different events using a common evaluation unit. Thus, in the risk assessment of “whether to take an umbrella in the morning,” the qualitatively different events of “getting wet in the rain” and “taking an umbrella” are first converted into a common unit of “degree of dislike,” and the two events are then compared, relative to each other, using this common unit. The reason why such

1.3 Ecological Risk Assessment

5

relative comparison is necessary is that understanding the absolute amount of risk is typically very difficult for us. Furthermore, in practice, most people are more concerned with relative vs. absolute values. For example, say the local risk of getting cancer from exposure to the chemicals of a waste treatment plant is .10−20 , but in a town somewhat farther from the plant the risk is .10−19 . If we asked a risk assessor which town we should live in, they would say “either” because the risk is negligible in each case; the absolute amount of risk is not worth serious consideration. But if we were free to choose, who would choose to live in the town nearer to the plant? Or say that a plant is to be built in the town where we live. The risk assessor says that the current cancer risk is .10−20 , and after the plant is built, it will be .10−19 , but the absolute amount of risk is still low enough that we should not worry. Do we agree to have the plant built in our town? More likely we would say “I understand that the risk is low, but not in my backyard.” In these examples, the absolute amount of risk is ignored and only the relative risk focused on. Humans tend to judge things in dualistic terms. Whether such thinking is right or wrong is not at issue here. It is simply a fact that we tend to think this way, partly perhaps because we come across so many instances in life where minor differences eventually lead to major results; hence, we care more about the difference than the absolute amount of risk. An anecdote may serve as illustration. Two explorers are walking through the jungle. Suddenly they hear a tiger roar. One explorer sits down and takes a pair of running shoes out of his backpack. “You’re crazy, you’ll never out-run a tiger,” says the other explorer. “I don’t have to out-run the tiger,” he replies. “I just have to out-run you.” [6]

1.3 Ecological Risk Assessment As there are numerous detailed works on ecological risk assessment, I will here only focus on some key terminologies for our present purposes. Ecological risk assessment is also a type of risk assessment and requires determining the event to be avoided. This event is called the endpoint. In ecological risk assessment, the endpoint is the toxic effect of a chemical substance, and the ecotoxicity test is used to examine this effect. A conceptual diagram of the test is drawn in Fig. 1.1. Toxicity tests are divided into acute and chronic toxicity tests, with the former conducted over a relatively short period of time and the latter over a much longer period (for example, the lifetime of a given species). In acute toxicity tests, the half-effect (EC50) and half-lethal (LC50) concentrations are reported, whereas, in chronic toxicity tests, the no observed effect concentration (NOEC) is often reported. In each case, the testing protocol is strictly regulated by the OECD test guidelines [4].

6

1 Basic Concepts of Ecological Risk Assessment NOEC

0

2

EC50

4

8

Chemical concentration Fig. 1.1 Test schematic. The zero concentration of a chemical substance is called the control. The maximum concentration at which the effect is not significantly different from the control is called the NOEC (no observed effect concentration) and is often reported in chronic toxicity tests. In the case of the crustacean, Daphnia magna, such testing is typically conducted for 21 days. The concentration that kills half of the individuals is called the half-lethal concentration (LC50), the concentration that affects half of the individuals is called the half-effect concentration (EC50), and these are often reported for acute toxicity tests lasting for 48 or 96 h

In ecological risk assessment, multiple species must be examined because species are typically diverse and have different levels of sensitivity to chemical substances, one respect in which species risk assessment is more difficult than human health risk assessment. Basically, at least one species from each of the three taxonomic groups (algae, crustaceans, and fish) must be examined, and since a large number of individuals are required for toxicity testing, species with established breeding systems are often used in such testing. In the case of crustaceans, Daphnia magna is often used, but this does not mean that D. magna alone must be tested; other crustaceans are often used. The difference between chronic and acute toxicity depends on the taxonomic group. For D. magna, a 48-h test is used to determine acute toxicity and a 21-day test for chronic toxicity. However, in tests for rare species, there may be disagreement as to whether the test is for acute or chronic toxicity. Typically, the results of chronic toxicity tests are more important than those for acute toxicity tests, which are conducted over a much shorter period. The endpoints also vary with the taxonomic group. For crustaceans, immobilization is the endpoint; however, lethality and other effects are sometimes used. If there are toxicity values based on different endpoints, the minimum toxicity value or the geometric mean is used for risk assessment. Ecological risk assessment consists of two parts: hazard assessment and exposure assessment. As different knowledge is required for each, they are often assessed separately by different experts and finally integrated in a composite risk assessment [7]. Exposure assessment will not be considered here. In order to conduct a risk assessment, toxicity data, such as the NOEC, LC50, EC50, etc., are first collected, and the predicted no effect concentration (PNEC) is derived based on these values. An important concept in deriving the PNEC is the uncertainty (or assessment) factor (UF or AF), whose magnitude depends on the abundance of the data. Tables 1.1

1.4 Are Different NOEC Values Comparable Indices for Ecological Risk. . .

7

Table 1.1 The magnitude of the uncertainty factor (UF) in ecological risk assessment for deriving the PNEC. The results of chronic toxicity tests are considered more important (hence the larger UF in the absence of chronic toxicity studies). ACR stands for the acute chronic ratio (see Table 1.2) Adopted value Minimum NOEC by chronic test from 3 trophic groups Minimum NOEC by chronic test from 2 trophic groups Minimum NOEC by chronic test from 1 trophic group Minimum L(E)C50 by acute test from 3 trophic groups Minimum L(E)C50 by acute test from less than 3 trophic groups Table 1.2 The magnitude of the ACR. The ACR for daphnids varies depending on whether the chemical is an amine or not

Interspecific extrapolation

Lab to field extrapolation 10

Total UFs 10

5

10

100

10

10

100

ACR

10

10 .× ACR

ACR

10

100 .× ACR

10

Acute chronic extrapolation

Trophics Algae Daphnids

Amins Non amins

Fish

ACR 20 100 10 100

and 1.2 show the magnitude of the uncertainty factor for deriving the PNEC, as specified in the Chemical Substances Control Law of Japan. When the acute toxicity test results for fish are the only data available, the toxicity value is divided by 10,000 to calculate the PNEC. If the LC50 for fish is 1 mg/L and this is the only toxicity data available, the PNEC would be 0.1 .μg/L. In exposure assessment, the predicted environmental concentration (PEC) is derived, and in the simplest risk assessment, the PEC is compared with the PNEC. If the PEC is less than the PNEC (PEC/PNEC .< 1), then the environmental concentration is less than the concentration that causes an effect, and there is no risk, and if the PEC is greater than the PNEC, there is risk, and this is the simplest ecological risk assessment.

1.4 Are Different NOEC Values Comparable Indices for Ecological Risk Assessment? I will now discuss some of the problems with current ecological risk assessment methods and some related issues from an ecological perspective. In practice, two NOEC endpoints are considered: reproduction (NOEC.r ) and survival (NOEC.s ). These two NOEC values are compared, and the lower value is

8

1 Basic Concepts of Ecological Risk Assessment Ambient concentration NOECr

NOECs

Chemical concentration

Fig. 1.2 Conceptual diagram of ecological risk assessment. If the environmental concentration of a chemical is lower than the lower NOEC value, no adverse effect on any endpoint is expected

Ecological impact

NOECs

NOECr

Chemical concentration

Fig. 1.3 Two axes that define the ecological impact of chemicals. The actual impacts are indicated by the black dots; however, in current ecological risk assessment, the ecological impact axis is ignored, and only the NOEC values on the horizontal (chemical concentration) axis, indicated by the gray dots, are considered. The ecological impacts involve multidimensional indicators, representable by numerous evaluation axes, but they are typically integrated as one indicator on the vertical axis

selected for PNEC derivation. The selection is based on two related assumptions: that the NOEC values are comparable indices, and that, if the environmental concentration is maintained below the PNEC derived from the lower NOEC value, there should be no effect from the higher NOEC endpoint (Fig. 1.2). Ecologists consider reproduction and survival to be different traits, and thus, as the respective ecological impacts of adverse effects on these traits should also be different, the two NOEC endpoint values cannot be easily compared. As shown in Fig. 1.3, ecologists often utilize two axes to evaluate the toxicity of a given chemical. However, there are cases where a chemical-sensitive trait (NOEC.r in this example) has minor ecological impact and cases where a chemical-insensitive trait has major ecological impact, and if there are several indicators representing adverse effects, with different evaluation units that cannot be easily compared, assessment is difficult. Therefore, such indicators must somehow be integrated into a single indicator. One form of assessment that integrates the effects on reproduction and survival into a collective growth rate (called the intrinsic growth rate) is populationlevel risk assessment. A concern arises, however, when the NOECs obtained for individual traits are integrated as a single index. As the respective effects on various traits are integrated into a single index in population-level risk assessment, the small effects on individual traits, which cannot be statistically detected below the NOEC level, are

1.4 Are Different NOEC Values Comparable Indices for Ecological Risk. . .

9

also integrated and finally expressed in terms of population-level risk. The NOEC is often considered as an index indicating “no effect,” but its proper meaning is the concentration (the maximum value among the concentrations) where the effect is not statistically different from the control; thus, the NOEC does not guarantee that there is no impact. We may express this concern in greater detail using equations. Let .st , then, be the survival rate of an individual at age t and .mt be the number of new births by an individual of age t. We may then express the survival rate of the individual, from age 0 to age t, as .lt , where .lt is the cumulative probability of the survival rate at each age, or lt = s0 s1 . . . st .

.

(1.1)

The number of offspring (R) produced by this individual during its lifetime is R=

T 

.

lt m t ,

(1.2)

t=0

where T is the maximum age of the individual. When the ambient chemical concentration is less than the NOEC, the adverse effect is, statistically, negligibly small but not zero. Let us denote this small effect by .. The survivability at age t (.st ) is then st = st− − ,

.

(1.3)

where .st− is the survivability with no chemical present, and the reproduction level is mt = m− t − ,

.

(1.4)

where .m− t is the reproduction level with no chemical present. By substituting these into Eqs. (1.2) and (1.1), we have R = R − − ω1  + ω2  2 − ω3  3 . . . ,

.

(1.5)

where .R − is the reproduction lifetime in the absence of the chemical, and .ωi is a coefficient for . i . Since .ε is small, . 2 (and higher orders of .) can be approximated as zero. This type of result, known as linear approximation, will frequently appear in this work. The remaining term, .ω1 , may be expressed as ω1 = m0 + s0 + (s0 + s1 )m1 + (s0 s1 + s0 s2 + s1 s2 )m2 + l1 + l2 ,

.

(1.6)

when the maximum age (T ) is 2. If T is larger than 2, a greater number of terms are included in .ω1 , and since many terms are summed in this variable, there is a concern that .ω1  may no longer be negligibly small. The analysis indicates that the

10

1 Basic Concepts of Ecological Risk Assessment

environmental concentrations are lower than the minimum NOEC, and therefore no adverse effects are observed for any trait, but effects may be observed, at the population level, at the given environmental concentration. Whether this can actually occur will be discussed in Chap. 2. The importance of risk assessment at the population level has been repeatedly noted. However, support for the practice has been notably uneven; when I commenced my research, for example, it was in the middle of a revival [1, 5]. Nonetheless, a few research groups have made considerable and sustained efforts in this regard, and such assessment is now a very active area of research, with applications to practical risk assessment also having seen good progress [3]. I have been away from population-level research for some time (the research I was doing over this time will be discussed later in the book), and it is a personal regret of mine that I have contributed little to the development of such research in recent years. The first half of this work provides an introduction to model construction for population dynamics and the means by which we include chemical effects in the model, with a focus on the basic requirements for using such models. We know that the maximum eigenvalue determines the population growth rate but often do not know why it does so, and though such knowledge may not be needed for practical risk assessment, it is necessary for proper analysis of the mathematical models. The various chapters of the book are mutually related but can be read independently; thus, one may begin wherever one’s interest lies.

References 1. Akcakaya HR, Stark JD, Bridges TS (eds) (2008) Demographic toxicity: methods in ecological risk assessment. Oxford University Press, Oxford 2. Bernstein PL (1998) Against the gods: the remarkable story of risk. Wiley, New York 3. Grimm V, Johnston ASA, Thulke HH, Forbes VE, Thorbek P (2020) Three questions to ask before using model outputs for decision support. Nat Commun 4959. https://doi.org/10.1038/ s41467-020-17785-2 4. OECE. OECD Guidelines for the testing of chemicals. Organisation for Economic Cooperation and Development. https://www.oecd.org/chemicalsafety/testing/oecdguidelinesfor thetestingofchemicals.htm (last access 2023/03/15) 5. Pastorok RA, Bartell SM, Ferson S, Ginzburg LR (eds) (2019) Ecological modeling in risk assessment: chemical effects on populations, ecosystems, and landscapes. CRC Press, Boca Raton 6. Trott D (2013) Predatory thinking: a masterclass in out-thinking the competition. Pan Macmillan, London 7. USEPA (1998) Guidelines for ecological risk assessment. US EPA, epa/630/r-95/002f edn.

Chapter 2

Population-level Assessment

Abstract In this chapter, we will discuss mathematical models of population dynamics. First, we will consider differential equation models, using a well-known logistic equation as a primary tool both for the modeling itself and for analysis of such models. Next, we will look at difference equation models (often referred to as matrix models), and how they are analyzed. We will then consider how the toxicological effects of chemicals are input into these models and evaluate population-level effects in terms of several examples. The models used in this chapter are relatively simple. In recent years, population-level assessments have become popular, and relatively complex models are sometimes used in such assessment. However, in order to create such complex models, it is necessary to first acquire the basic knowledge of simple models. For example, in the case of matrix models, we know that the maximum eigenvalue of the matrix indicates the change in the population size. One aim of the present chapter is to understand why. At the end of the chapter, we will focus on evolutionary dynamics and discuss differences in toxin response in terms of life-history traits. Keywords Population dynamics · Logistic model · Matrix model · Eigenvalues · Sensitivity analysis · .r/K-selection

2.1 Introduction There are numerous valuable works on population dynamics models,1 and I encourage the interested reader to consult them for a more rapid and accurate understanding than can be gained from the present work. The use of population dynamics models to understand the effects of chemicals, now known as population-level risk assessment, has a rather long history, and a number of works may now be read as classics in the field. Suppose, however,

1 The

books that I consulted heavily in writing this book are Gurney and Nisbet [6] and Caswell

[3]. © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_2

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2 Population-level Assessment

that all the knowledge of ecological risk assessment has been lost, and we must create anew a methodology for such assessment. I believe that it would turn out to be a population-level assessment methodology. From an ecologist’s point of view, then, it is rather strange that ecological risk assessment does not take into account population dynamics. Ecosystems are complex, with many species in mutual interaction. Even with the most powerful computers, it is quite impossible to reproduce such complex ecosystems in any detail. We can never fully describe all the interactions of all the species, and there may and likely will always be further interactions unknown to us. A well-known large-scale ecosystem simulator is the CASM (Comprehensive Aquatic Systems Model) [4], but this model still generates only a very limited snapshot of the entire ecosystem. The key to creating a useful model is to be clear about what it is, precisely, that we wish to know, and then to construct the simplest possible model necessary for achieving this goal. To this end, we must sometimes be bold and ignore known systems if they are not necessary (though we must explain why they are unnecessary). Typically, the more variables we include in our model, the more complex the population dynamics will become, and the more difficult will be the analysis; the greater the number of parameters, the more difficult it becomes to estimate them. If we try to incorporate all the known information into our model, in order to make it more realistic, we may end up with a model that is applicable to the target field but inappropriate for other fields (that is, ungeneralizable). It is important, then, to include in the model only what is absolutely necessary. My impression is that risk assessment at the population level has gained increasing attention in recent years. There has, for example, been a great deal of discussion about how to use population models for risk assessment and decision making [5, 18], and assessment tools such as the Dynamic Energy Budget (DEB) [10, 11] have been developed. However, I think it is dangerous to merely understand how to use such tools without understanding the details involved in their design and proper application, if for no other reason than to know if you are using them incorrectly, because sufficiently advanced technology is often indistinguishable from magic. In this chapter, then, we begin with the basics, showing how to build models and how to incorporate the effects of chemicals into such models. Readability being one of the aims of this work, however, mathematical rigor has, in some cases, been sacrificed to some extent, and it is hoped that the book will be read with a spirit of tolerance.

Box: Ecologists do not Know Why Biodiversity Is Stably Maintained In ecosystems, chemicals do not have unique adverse effects. Such effects form a mere part of the cumulative effects that may threaten biological populations and biodiversity. The advantage of population-level risk assessment is that the effects of chemicals can be discussed using a single composite indica(continued)

2.2 Differential Equation Model

13

tor, along with other effects such as habitat loss. In ecological risk assessment, as in conservation biology, the key is to avoid biodiversity loss and species extinction. I often hear people say that we should examine the effects of chemical substances because it is important to maintain biodiversity; but such a claim feels weak to me, because it assumes that biodiversity is maintained in a stable manner, whereas, in fact, we have a limited understanding of why ecosystems are stable. Intuitively, even if one prey organism becomes extinct, as long as there are other prey organisms, the predator can survive. The richer the diversity of species, the greater the likelihood that such prey will exist, and hence the stability of biodiversity is maintained by the diversity itself. However, a theoretical study using the Lotka-Volterra competition model of multispecies systems have shown the exact opposite result: as the number of species increases, the likelihood of instability increases [14]. This counterintuitive result is called May’s Paradox, and since the appearance of the study, the study of the mechanisms that actually sustain biodiversity has remained one of the major research topics in ecology.

2.2 Differential Equation Model As differential equations will frequently appear in this work, let us briefly consider their nature and use. Suppose an individual produces two offspring in a unit of time (say, a day). How many offspring will it produce in two such periods of time? The answer is simple: 2(offspring) × 2(days) = 4.

.

The point is that changes in population size are proportional to both the number of offspring and the unit time. Suppose, then, that there are .Nt individuals at time t. Let r be the number of offspring that each individual produces over a certain period of time (.t). Since the increase in the number of individuals is proportional to both the number of individuals and the unit time, the increase in the total number of offspring is rNt t.

.

At time t, there are .Nt individuals, and the increment is .rNt t; hence, the total number of individuals at the next time step .(t + t) is Nt+t = Nt + rNt t.

.

(2.1)

14

2 Population-level Assessment

By defining that N ≡ Nt+t − Nt ,

.

Equation 2.1 becomes .

N = rNt . t

If we now take the limit of .t → 0, we derive a differential equation for population growth: .

dN = rN. dt

(2.2)

This equation can then be solved analytically for .N > 0: 

1 dN = N

.



d log N =

 rdt,  rdt,

log N = rt + C, N = ert+C = N0 ert , where C is a constant and .N0 = eC is the initial value of N (at time 0). With the exception of special cases, we cannot solve the equation analytically, and in such the cases we solve the equation numerically. To find the numerical solution, we reverse the reasoning thus far. First, we discretize Eq. 2.2 as .

Nt+t − Nt = rNt . t

If we arrange this equation, it reverts to Eq. 2.1. Obtaining the value at the next time step from .N0 (as the initial value), N1 = N0 + rN0 t,

.

N2 = N1 + rN1 t = N0 + rN0 t + rN1 t, N3 = N2 + rN2 t = N0 + rN0 t + rN1 t + rN2 t, .. . NT = N0 + r

T −1 t=0

Nt t,

(2.3)

2.2 Differential Equation Model

15

Fig. 2.1 The concept of integration

N N4 N3 N2 N1 N0 N0 Δt

N1 Δt

Δt

N2 Δt

N3 Δt

N4 Δt

t

we obtain the number of individuals (.Nt ) at time T . The numerical method of solving the equation is equivalent to filling the function to be integrated with rectangles of width .t and height .Nt (Fig. 2.1). While .t is large, the area under the function differs from the sum of the areas of all the rectangles, but as .t approaches zero, the areas of the two converge. When solving differential equations numerically, the magnitude of .t remains a finite value because it cannot be set at 0 in the computer. In my experience, if the integral function is not very complicated, .t = 0.001 is small enough. If the function we want to integrate is rapidly changing, we must make .t even smaller, or utilize a more sophisticated approach, such as the Runge-Kutta method. If, however, we perform such numerical integration with a very small .t, it will take a long time to get the answer, whereas if .t is made large, the calculation time decreases, but the accuracy of the calculation is also reduced. Thus, there is an inherent trade-off between the calculation accuracy and calculation time, and deciding what size of .t to use in the numerical approach requires a risk assessment. The solution of 2.2 is N(t) = N0 ert .

.

(2.4)

The dynamic here is referred to as exponential growth, where the number of individuals increases at an accelerating rate as time goes on. However, it is rarely the case that a biological population will or can increase at an accelerated rate forever. In the next section, then, we will consider a term that expresses the limitation on such increase.

16

2 Population-level Assessment

Quiz: Exponential Growth Assume that one cell divides into two cells once per minute, and if we start with one cell, it will take 60 min to fill a Petri dish. If the cell division starts with two cells, how many minutes will it take to fill the Petri dish?2 The typical (and incorrect) answer of “30 min” derives from our intuitive assumption that population growth is linear. Non-linear population growth, such as exponential growth, is not as easy to understand.

2.2.1 Logistic Growth In Eq. 2.2, the growth rate (r) is constant, but as the population size increases, the habitat will tend to deteriorate due to waste accumulation and food depletion, so that a constant growth rate is no longer maintained. The simplest assumption about the effect of such population-crowding is that the growth rate decreases linearly with population size, or r(N) = r0 (1 − qN),

.

where .r0 > 0 is the growth rate at .N = 0 (often termed the intrinsic growth rate), and q is a constant that regulates the intensity of the adverse effect of populationcrowding. By combining this equation with Eq. 2.2, we have .

dN = r0 (1 − qN)N, dt

(2.5)

and this is an equation for logistic growth. The constant, q, is often defined as .q ≡ 1/K, and K is termed the carrying capacity. It is important to estimate the .r0 of the specific species which is the focus of our risk assessment; however, the estimation of .r0 is not easy in practice, because it is the value at .N = 0 (absolutely nothing!).3 Thus, the value is typically estimated based on the observed population growth data, but this data is always subject to the impact of population-crowding, as discussed above.

2

Correct answer: 59 min.

3 If we have time series data of

N near the carrying capacity, there is a way to estimate the intrinsic growth rate of .r0 . See Chap. 4 for details.

2.2 Differential Equation Model

17

Equation 2.5 has a second-order term of N and is non-linear in nature. It is typically difficult to solve such non-linear equations analytically, but the related logistic equation can be solved as:  .

1 dN = (1 − qN)N

 r0 dt.

(2.6)

By arranging the left side of the equation, we have   .

q 1 + 1 − qN N



 dN =

r0 dt,

which may be solved as follows:  .

q dN + 1 − qN



1 dN = N

 r0 dt,

− log(1 − qN) + log N = r0 t + C,   N log = r0 t + C, 1 − qN N = er0 t+C , 1 − qN

(2.7)

where C is an integration constant, and it is implicitly assumed that .N > 0 and 1 − qN > 0. By arranging the final equation above, we have

.

N(t) =

1 . q + e−r0 t−C

N(t) =

K 1 + C0 e−r0 t

.

Because .q = 1/K, we have .

and this is a familiar form of the logistic equation, where .C0 = Ke−C .

2.2.2 Equilibrium and Stability Equilibrium values (.N ∗ ) for Eq. 2.5 are easily obtained by setting .dN/dt = 0: N∗ =

.



0, 1/q = K.

(2.8)

18

2 Population-level Assessment

Assume, then, that N is at equilibrium, and that a perturbation slightly dislodges it from equilibrium. If N returns to equilibrium, we say that the state of equilibrium is stable. To know if the equilibrium is stable or not, stability analysis is conducted. Consider, then, a population at equilibrium, and a small perturbation of the population occurs, changing the population size to N = N ∗ + n,

.

(2.9)

where n is a very small value. Substituting this equation into Eq. 2.5 yields .

d(N ∗ + n) = r0 (1 − q(N ∗ + n))(N ∗ + n), dt dn dN ∗ + = r0 (1 − qN ∗ )N ∗ + r0 (n − 2qN ∗ n − qn2 ). dt dt

(2.10)

Now, equilibrium means that the population size does not change over time, which yields .

dN ∗ = r0 (1 − qN ∗ )N ∗ = 0. dt

As n is assumed to be very small, it can be approximated (i.e., linear approximation) as n2 ≈ 0.

.

By applying the approximation, Eq. 2.10 becomes .

dn = r0 (1 − 2qN ∗ )n, dt

(2.11)

yielding a linear equation for dynamics with small perturbation (n). Let us now consider the case where .N ∗ = 0. Substituting .N ∗ = 0 into Eq. 2.11 yields .

dn = r0 n. dt

When the perturbation is positive (.n > 0), since .dn/dt > 0, the magnitude of the perturbation increases, and hence .N(= N ∗ + n) is dislodged from its equilibrium (.N ∗ = 0). When n is negative, N is also dislodged from equilibrium, but .N < 0 is biologically meaningless. In the case of the other equilibrium (.N ∗ = 1/q), we have .

dn = −r0 n, dt

2.2 Differential Equation Model

19

which means that, when .n > 0, n decreases, and when .n < 0, n increases. In both cases, n returns to 0 as times goes on, and thus N returns to .N ∗ = 1/q, implying that the equilibrium is stable. Intuitively, the coefficient of n determines whether it is stable or unstable. If the coefficient is negative, then when n is positive, .dn/dt is negative, so n will decrease and thus approach 0. When n is negative, .dn/dt is positive, so n will increase and thus still approach 0. In either case, since n tends toward 0, the equilibrium is stable. Conversely, if the coefficient is positive, the deviation increases, and the equilibrium is unstable. Stability analysis is performed for small perturbations, and the dynamics of the perturbation is restricted to a region very close to the equilibrium. The stability expressed in terms of such analysis is often termed local stability. Whether or not the equilibrium is globally stable involves a separate discussion (see, for example, Murray [16]).

2.2.3 Multiple Species It is easy to extend the logistic equation to multiple species. For two species, such as hares (X) and wolves (Y ), the population dynamics may be expressed as .

dX = aX − bXY, dt dY = cXY − dY dt

(2.12)

which are the well-known Lotka-Volterra equations. X increases at rate a and Y dies at rate d. XY is often referred to as a term of mass action and is widely used in mathematical modeling of physical, biological, and numerous other types of system (e.g., nuclear reactions, infectious diseases, etc.). In our example above, the term represents “wolves eat hares.” The reduction in the hare population due to predation is proportional to the number of both hares and wolves, and hence is described as XY . The rate of predation is regulated by a parameter, b. The wolf population increases only when the wolves eat hares, and c regulates the rate of increase. As equilibrium states are determined by solving .dX/dt = 0, dY /dt = 0, we have   d a ∗ ∗ , . .(X , Y ) = (0, 0) and c b To know the stability of these equilibrium states, we consider small perturbations from (.X∗ , Y ∗ ): X = X∗ + x,

.

Y = Y ∗ + y.

(2.13)

20

2 Population-level Assessment

Substituting these equations into Eq. 2.12 yields dx = ax − b(X∗ y + xY ∗ ), dt dy = c(X∗ y + xY ∗ ) − dy, dt

.

(2.14)

and again the xy term is approximated to be 0. For .(X∗ , Y ∗ ) = (0, 0), we have .

dx = ax, dt dy = −dy. dt

(2.15)

These equations mean that a perturbation of y from .Y ∗ = 0 returns to .y = 0, but a perturbation x from .X∗ = 0 does not return to .X∗ = 0, and eventually the equilibrium is unstable. Equation 2.14 is often described using the matrix: .

d dt

     x x a − bY ∗ −bX∗ = . cY ∗ cX∗ − d y y

(2.16)

The matrix  J=

.

a − bY ∗ −bX∗ cY ∗ cX∗ − d

 (2.17)

is often termed Jacobian. The stability of the equilibrium is known by determining the eigenvalues of the matrix. Eigenvalues can be computed by finding the .λ that satisfies |J − λI| = 0.

.

When .(X∗ , Y ∗ ) = (0, 0), we have .

  a − λ 0    0 −d − λ  = (a − λ)(−d − λ) − 0 × 0 = 0,

and to satisfy the equation, obviously λ = a and − d.

.

2.2 Differential Equation Model

21

One of the eigenvalues, a, represents a change in x over time, while the other eigenvalue, .−d, represents a change in y (see Eq. 2.15) near the equilibrium ∗ ∗ .(X , Y ) = (0, 0). The judgment of stability or instability is the same as in the case of the logistic equation. But in the two-species model, the system has two dimensions, and thus there are two corresponding coefficients (eigenvalues) for small perturbations. In the example, one of the coefficients is negative (.−d), and the equilibrium is stable in this dimension, but since the other coefficient is positive (a), the equilibrium is unstable in this dimension, and overall the equilibrium is unstable. In some conditions, the eigenvalues for the 2.×2 matrix are complex conjugates of the form √ λ = α ± β −1,

.

and .α is a real part of the complex conjugate and represents a change in the variables. The remainder is an imaginary part and represents the period for the internal oscillation of the dynamics. When .β is non-zero, the dynamics shows oscillatory behavior, with the period of the oscillation being .2π/β (near equilibrium). This can be confirmed by looking at the eigenvalues at .(X∗ , Y ∗ ) = (d/c, a/b). An example of population dynamics is shown in Fig. 2.2. We must construct a Jacobian matrix at the equilibrium, and then derive eigenvalues, but we will leave this as an exercise.

y 2

1.5

1

x

1.0

-2

-1

1

2

x

0.5 0.5 -0.5 -1.0

-1

1

2

1.5

Rad.

y

-1.5

-2

Fig. 2.2 Examples of .x, y dynamics when the eigenvalues are pure imaginary numbers (the real part is zero). In the left panel, the horizontal axis is x and the vertical axis y. In the right panel, the horizontal axis is time (in radians) and the vertical axis is x and y. x and y continue to oscillate around the same value

22

2 Population-level Assessment

2.2.4 Including Toxic Effects Now let us return to Eq. 2.5, and consider the situation where the number of individuals is reduced by the toxic effects of chemicals. How we incorporate the toxic effects depends on the chemical type. If the chemical does not kill the organisms but reduces the reproduction rate, then the equation is r = r0 f (c),

.

where c is the chemical concentration and .f (c) is a function representing the dose response. The shape of the function is determined by the toxicity data. A widely used model focusing on mortality is .

  N dN = r 1− N − αN, dt K   N = r 1 − α/r − N, K   N N, = (r − α) 1 − K(1 − α/r)

(2.18)

where .α is a toxic effect. If we arrange this equation, and substitute the parameters r  ≡ r − α,

.

K  ≡ K(1 − α/r), the equation reverts to Eq. 2.5. As the population level cannot be sustained if the growth rate becomes negative .(r  = r − α < 0), the maximum toxic effect is α = r.

.

Note that the equation does not take into account population fluctuations due to environmental or demographic stochasticity. How we incorporate such stochasticities is the subject of the next chapter. In a previous work, we termed the concentration of chemicals leading to .α = r as the population threshold concentration (PTC) [8]. If .α = r holds, then the population size does not decrease or increase; however, the adverse effects of chemicals that result in zero population growth may be too extreme as a form of ecosystem management, and the optimal magnitude of such effects, with the goal of system protection, is often difficult to determine. As explained in the previous chapter, current ecological risk assessment focuses on the maximum concentration with no statistically observable effect (NOEC), compared to the control (or the PNEC derived by dividing the NOEC by an assessment factor), as the goal of protection. However, this approach seems to me to grant statistics too great and exclusive a role in determining what the goal of

2.3 Difference Equation Model

23

protection should be. In any risk assessment, we must first determine the endpoints that we wish to avoid. The endpoint of an ecological risk assessment should be the ecosystem we do not want to lose, and this we must decide for ourselves in a holistic manner. I believe that statistical significance should not be simply equated with the environment we want to protect and, further, that there will and should be impacts that are statistically significant but acceptable, and conversely, impacts that are not statistically significant but unacceptable. Let us now consider a rather strange problem. Imagine there is a certain pest that damages crops. The government wants to get rid of the pest and decides to offer a reward for its extermination, buying each insect for one dollar. A coalition of volunteers elects a bounty hunter for the pest, but the hunter realizes that if they catch all the pests this year, they would receive no bounty next year, and they wish to maximize their bounty. How many pests do they need to catch this year, then, in order to catch as many as possible next year (and so on), assuming that the pest population has logistic growth with K as the carrying capacity? If they decide that they will catch the pests at the rate of .α each year, the population size at equilibrium (Eq. 2.18) is N ∗ = K(1 − α/r).

.

The number of pests they catch is .αN ∗ , and hence αK(1 − α/r).

.

(2.19)

This is optimized at .α = r/2. For the bounty hunter, then, in order for the business to be sustainable and the bounty maximized, it is best to continue catching pests in such a way that the intrinsic growth rate is reduced by half each year. If we consider the pest to be fish, then, and the extermination rate to be the amount of fish caught, this yearly catch is termed the maximum sustainable yield (MSY) [19]. The assumption at work here should, I think, be criticized for equating chemical management with the management of fisheries, which are operated as means of using resources; however, until an appropriate theoretical tool, analogous to the MSY, emerges in ecological risk assessment, I would support the use of the MSY as one of the tools in the management of chemical risk.

2.3 Difference Equation Model If we let .Xt , now, be the number of individuals in a given population this year, and Xt+1 be the number of individuals next year, we obtain the time-discrete version of the logistic equation:

.

Xt+1 = rXt (1 − qXt ).

.

24

2 Population-level Assessment

This model shows chaotic behavior when r becomes large, but such high growth rates are not typically considered in chemical risk assessment, because high r inherently means low risk; thus, we will not consider this situation in detail. A typical example of a discrete model used in the risk assessment of chemicals is the age-structured model or Leslie matrix, which is constructed as follows. We assume that an organism starts breeding in early summer. If the census is carried out in spring, then we count the one-year-old or older individuals that survived the winter. Let the number of individuals whose age is a at time t be .na (t). When the age-based survivability of each individual from time .t − 1 to t is .sa , then the number of two-year-old individuals is n2 (t) = s1 n1 (t − 1);

.

(2.20)

that is, the number of two-year-olds is equal to the number of one-year-old survivors. We must, however, be careful in the case of the final age group. If, for example, all individuals die at the end of 3 years, then n4 (t) = s3 n3 (t − 1) = 0

.

and there are no four-year-old individuals; thus, it is defined that .s3 = 0. The number of one-year-olds is equal to the number of zero-year-old survivors, and therefore, n1 (t) = s0 n0 (t − 1).

.

The number of zero-year-old individuals (.n0 (t − 1)) equals the total number of new individuals which were born last summer. When an a-year-old individual produces .ba new individuals, we have n0 (t − 1) = b1 n1 (t − 1) + b2 n2 (t − 1) + b3 n3 (t − 1),

.

and the number of one-year-olds at time t is n1 (t) = s0 b1 n1 (t − 1) + s0 b2 n2 (t − 1) + s0 b3 n3 (t − 1).

.

(2.21)

As the .s0 ba terms are unwieldy, for simplicity’s sake we drop the .s0 component and simply use .ba . Now we will summarize all the equations. We do not observe .n0 (t) (observation precedes recruitment) or .n4 (t) (all die at age four). We need one equation for each of ages one, two, and three (based on Eqs. 2.20 and 2.21). These equations may be expressed in matrix form as ⎛

⎞ ⎛ ⎞⎛ ⎞ n1 (t) b1 b2 b3 n1 (t − 1) . ⎝ n2 (t) ⎠ = ⎝ s1 0 0 ⎠ ⎝ n2 (t − 1) ⎠ . n3 (t) 0 s2 0 n3 (t − 1)

(2.22)

2.3 Difference Equation Model

25

It is often defined that ⎛

⎞ b1 b2 b3 .M = ⎝ s1 0 0 ⎠ 0 s2 0

(2.23)

and this is the age-structured matrix (also termed the transition matrix).

Box: Census in Autumn If we carry out the census in autumn, what does the age-structured matrix become?4

2.3.1 Increase or Decrease As a simple case, let us assume that all individuals die when they are 3 years old; thus, .s2 = 0. Eq. 2.23 is now  M=

.

b1 b2 s1 0

 .

(2.24)

The reader likely knows that the maximum value of the real part of the eigenvalues of this matrix will decide if the population is increasing or decreasing. If the maximum value of the real part of the eigenvalues is greater than one, the total population size is increasing, and if less than one, decreasing. However, let us consider why such a determination can be made. To know the reason why the eigenvalues are important, we must first understand the meaning of the eigenvectors of the matrix. The eigenvalue indicates the rate of increase in the total population, and the eigenvector indicates the direction of movement of the numbers. The matrix in Eq 2.24 is a 2.×2 matrix, with two eigenvalues and two eigenvectors corresponding to these eigenvalues. The two eigenvalues are described as .λ1 and .λ2 (where .|λ1 | > |λ2 |), and the corresponding eigenvalues as .v1 and .v2 , respectively. In the transmission matrix, .M, the eigenvalues and eigenvectors have the following relationship: Mvi = λi vi ,

.

4

Hint: we must take into account the number of zero-year-old individuals (.n0 (t)).

(2.25)

26

2 Population-level Assessment

Fig. 2.3 An example of population dynamics, where t represents time, the initial value is .(n1 , n2 ) = (0.15, 0.2), and the dashed lines represent eigenvectors. Since eigenvectors only have a directional meaning, we can draw the lines anywhere we want as long as they indicate the proper direction

0.30

t=19

0.25

n2

t=0

0.20

v2

0.15 0.10

t=2 t=3

t=10

t=1

v1 0.15

0.20

0.25

0.30

n1 where .i = 1, 2. Eigenvectors are special vectors that, when multiplied by .M from the right, do not change direction but increase in magnitude by a factor of .λ (they are referred to as right eigenvectors). An example of such population dynamics may be seen in Fig. 2.3. The value of .(n1 , n2 ) initially moves along .v2 , but from around .t = 10, it starts to move along .v1 . Each value on .v1 then continues to increase at a rate .λ1 . The same result is obtained with any arbitrary initial values (of course, for .n1 > 0, n2 > 0). Some may be curious why .n1 , n2 moves along .v1 and not .v2 . We first, then, describe .n1 , n2  n(t) =

.

 n1 (t) , n2 (t)

(2.26)

as a vector. If the initial value is .n(0), then the value of the next time step (.n(1)) is n(1) = Mn(0).

.

Repeating this t times, the number at time t is n(t) = Mn(t − 1) = M2 n(t − 2) = · · · = Mt n(0).

.

(2.27)

The initial vector (.n(0)) can be rewritten as n(0) = c1 v1 + c2 v2 ,

.

by using two eigenvectors where .c1 , c2 are constants. Substituting the above equation into Eq. 2.27 yields n(t) = Mt (c1 v1 + c2 v2 ) = c1 Mt v1 + c2 Mt v2 ,

.

(2.28)

2.3 Difference Equation Model

27

and by using the relationship in Eq. 2.25, we have Mt v1 = Mt−1 λ1 v1 = Mt−2 λ21 v1 = · · · = λt1 v1

.

and therefore Eq. 2.28 becomes n(t) = c1 λt1 v1 + c2 λt2 v2 ,   λt = λt1 c1 v1 + c2 2t v2 . λ1

.

(2.29)

Because we assume that .|λ1 | > |λ2 |, we have .−1 < λ2 /λ1 < 1, and therefore  .

λ2 λ1

t ≈0

as t goes large, and eventually Eq. 2.29 becomes n(t) = c1 λt1 v1 ,

.

(2.30)

for large t. This is why the population size is on the axis determined by .v1 , and why only the maximum eigenvalue determines the increase (or decrease) in the population size. We now return to Eq. 2.22, which includes individuals up to age three. We denote the transition matrix by .M, the eigenvector which corresponds to the eigenvalue (.λ) by .v, and the elements in the eigenvalue by ⎛

⎞ v1 .v = ⎝ v2 ⎠ . v3 After long time-steps, we have Mv = λv.

.

(2.31)

If we apply this to each element, we have b1 v1 + b2 v2 + b3 v3 = λv1 ,

.

s1 v1 = λv2 , s2 v2 = λv3 .

(2.32)

28

2 Population-level Assessment

From the second and the third equations here, we have s1 v1 , λ s2 s1 s2 v3 = v2 = 2 v1 . λ λ v2 =

.

These equations show that the ratios of the respective numbers at each age, after long time-steps, are v1 : v2 : v3 = 1 : s1 /λ : s1 s2 /λ2

.

(2.33)

and the distribution of .vi is termed a stable age distribution. After the number of individuals assumes this type of distribution, the number increases at the rate of .λ, while maintaining the shape of the distribution. By substituting these ratios into the first equation in Eq. 2.32, we have b1 v1 +

.

s1 s1 s2 b2 v1 + 2 b3 v1 = λv1 , λ λ

(2.34)

and by arrangement,  1 s1 s1 s2 λ−x lx bx , b1 + 2 b2 + 3 b3 = λ λ λ 3

1=

.

(2.35)

x=1

and this is the characteristic equation, often termed the Euler–Lotka equation, where l1 = 1,

.

lx = s1 s2 . . . sx−1 .

(2.36)

Box: Census in Autumn—Revisited What would the Euler–Lotka equation look like in a model where the census is taken in autumn?

2.3.2 Sensitivity Analysis We have shown that the population growth rate is equal to the maximum (real part) eigenvalues. Our next task is to determine how the eigenvalues change when we change the elements of the matrix. This is termed sensitivity analysis. Before we consider the derivation of the change, however, we must define the left eigenvectors.

2.3 Difference Equation Model

29

The eigenvector we considered in Eq. 2.31 is a special vector that, when multiplied by the matrix .M from the right, becomes .λ times greater. Usually, the term eigenvector is used to refer to this vector. However, there is another special vector that, when multiplied by the matrix .M from the left, becomes .λ times greater. We term the vector .w, and it satisfies wM = λw,

.

(2.37)

and w = (w1 , w2 , . . . , wn ).

.

When we wish to distinguish these two vectors, the vector, .v, in Eq. 2.31 is termed the right eigenvector, and .w the left eigenvector. Suppose, now, that small changes occur in the elements in .M, and .M changes to .M + M, the right eigenvector changes to .v + v, the left eigenvector changes to .w + w, and the eigenvalue changes to .λ + λ. Since the relationship between the eigenvectors and eigenvalues still holds, we have (M + M)(v + v) = (λ + λ)(v + v),

.

Mv + Mv + Mv + Mv = λv + λv + λv + λv. Since the change is small, we can ignore the numerous . terms, and since .Mv = λv, we have Mv + Mv = λv + λv.

.

By multiplying the left eigenvector by both sides of the equation from the left, we have (w + w)(Mv + Mv) = λ(w + w)v + λ(w + w)v,

.

wMv + wMv = λwv + λwv. Again we ignore the numerous . terms, and since .wMv = λwv, we have wMv = λwv.

.

Now, since wv =



.

i

wi vi

30

2 Population-level Assessment

is an inner product of .w and .v, and is a scalar, we have λ =

.

wMv wv

as the change in .λ. When the change occurs only in an element in the ith row and j th column (.aij ), the change in .λ may be expressed as λ =

.

wi aij vj . wv

By dividing both sides by .aij , we have .

wi vj λ = aij wv

and rewrite the equation in a continuous form, we obtain the sensitivity .sij as sij =

.

wi vj ∂λ . = ∂aij wv

(2.38)

Sensitivity is a measure of the absolute amount of change. It represents the change in .λ when the rate of reproduction increases, by 1, for example. However, some parameters, such as survivability, cannot increase by 1 (survivability must be between 0 and 1). In such cases, we want to know the relative amount of change. To convert an absolute change to a relative change, we simply divide the amount of change by the original value.5 We again write Eq. 2.38 in a discrete form as aij λ sij = . λ λ



aij aij

 ,

and we obtain the elasticity .eij as eij =

.

aij aij ∂λ sij = . λ λ ∂aij

(2.39)

The knowledge we have thus far gained is sufficient to do population-level risk assessment using population dynamics and is sufficient also, I believe, for constructing some advanced models. The question that may arise, however, is whether to use a continuous-time or discrete-time model for population-level assessment. In ecology, discrete-time models are used for annual plants, and differential equations are used for organisms

5 If money we have increases from 100 to 101 dollars, we calculate the rate of change as 1/100 .= 0.01.

2.4 Case Studies

31

with overlapping generations, but there is no clear distinction between the two in usage. Rather, I feel that the distinction is often based on what we want to know, and which model is more serviceable for analyzing the system that we want to understand. In the case of chemicals, what we want to know is the effect of the chemical(s), but in my experience, it is the data we have obtained that determines the model.

2.4 Case Studies 2.4.1 The Effect of Zinc on an Algal Population My first task in ecological risk assessment was to assess the ecological risk of zinc, and I reviewed more than 1000 papers (there are a staggering number of papers on the toxicity of metals). After reading them all, I realized that ecological risk assessment does not really involve ecology. Most of the studies reported NOEC values or half lethal concentrations (LC50s). With these values, an ecological risk assessment could be made, but the significance of the results was not clear to me. When we live in a foreign country, we may be bothered by language differences. Yet in this literature survey, though solely in Japanese, I met with the same difficulties, trying to puzzle out what the ecological risk assessors were actually saying. My first step was to translate their language of ecological risk assessment into that of ecology; that is, risk assessment at the population level. Below are some samples of my effort. In the study by Yap et al. [21], the marine alga, Isochrysis galbana, was exposed to zinc, and the results are shown in Fig. 2.4. The data in the figure provides good material for applying the logistic equation in Eq. 2.18. In the absence of zinc exposure (control), there should be no adverse effect from the zinc, and we use the standard logistic equation,

Fig. 2.4 Results of toxicity tests of zinc exposure for the marine alga, I. galbana. The numbers by the plots are zinc concentrations (mg/L). Dr. Yap kindly provided me with the original data, for which I am grateful

K . 1 + Ce−rt mg/L

20

Cell densities

N(t) =

.

0.1 15

0.16 0.26 0.7

10 5

2.0 1

2

3

4

Day

5

6

7

32

2 Population-level Assessment

Table 2.1 Estimated .α values at each treatment level

Concentration (mg/L) 0.1 (control) 0.16 0.26 0.7 2 a

.α

0a 0.053 0.11 0.17 0.49

Effect of zinc at control is assumed to be 0

Software-based model-fitting shows that .K = 20.48, .r = 0.5768 and .C = 4.461. Next, we modify the model to include the effects of toxicity: N(t) =

.

K(1 − α/r) . 1 + Ce−(r−α)t

At .t = 0, since N(0) =

.

K(1 − α/r) , 1+C

we have C = K(1 − α/r)/N(0) − 1

.

as the value of C. The value of .α is determined by data-fitting using the results in Fig. 2.4. The estimated .α values are shown in Table 2.1. If we perform a linear regression on this data, with the concentration as the explanatory variable and .α as the dependent variable, we can determine the relationship between the two variables: α = 0.2465c,

.

where c is zinc concentration. Since r = r − α

.

r  is 0 at zinc concentration of 2.340 mg/L, and we know that the population of I. galbana stops increasing at this concentration, termed the population threshold concentration (PTC).6

.

6 Here,

a multi-step method is used, first determining .C, K, and r, and then .α; however, a more systematic method is preferred (see, e. g., Houslay and Wilson [7]).

2.4 Case Studies Table 2.2 Effect of zinc on the number of eggs in one spawning. Data from [2]

33 Zinc Concentration (.μg/L) 2800 2800 1300 1300 660 680 320 340 180 180 32 (control) 28 (control)

Number of eggs/spawning 0 0 10 37 26 9 25 63 108 122 163 120

Zinc Toxicity on Fathead Minnow Another example is the study by Brungs [2], some of whose results are shown in Table 2.2. The data allows us to estimate the dose response of the zinc effect on the number of eggs. If the toxic mechanism behind the effect of zinc on this number is known, a function representing the mechanism should be used. However, in this case the mechanism was unknown, and in its absence, some simple functions, such as n = a + bx,

.

log n = a + bx, n = a + b log x, log n = a + b log x are often assumed, where x is the zinc concentration, n is the number of eggs, and a and b are constants.7 When the zinc concentration is 2800 .μg/L, the number of eggs is zero, which cannot be log-transformed; however, in such cases there are widely used techniques to either excise such data or replace zero with a small number such as 0.001. These are classical approaches, and there may be more sophisticated approaches available today.8 Here, for simplicity, we will take the classical approach and replace 0 eggs with 0.001. We perform a standard regression analysis for all

7 I did not noticed when I did the work, but the comparison of AIC in these equations is not good. We need to compute AIC using .n = exp(a+bx) instead of .log n = a+bx, and .n = exp(a+b log x) instead of .log n = a + b log x. 8 By using, for example, the “glm” function in the software R, we are free from these problems. We can use the “glm” without knowing the details, but I think we should try to understand what it is doing at least once.

34

2 Population-level Assessment

Fig. 2.5 A comparison of observed and predicted effects. The horizontal axis shows the zinc concentration in .μg/L, and the vertical axis the number of eggs (relative to the control)

Egg reduction rate

0

10

-1

10

-2

10

-3

10

-4

10

-5

10

500

1000

1500

2000

2500

Zinc concentration (µg/L)

models and use Akaike’s information criterion to determine the appropriate model. .

log n = 5.64 − 0.00409x, or n = e5.64−0.00409x

(2.40)

is used as a dose–response curve for the number of eggs. The number of eggs often depends on the environmental conditions. Laboratory fish, for example, are often maintained under ideal conditions (temperature, nutritional environment, etc.), and thus the number of eggs observed in [2] may be too high. Therefore, we focus on the rate of decrease in the number of eggs, and consider that the rate of decrease does not depend on the environmental conditions. By dividing Eq. 2.40 by the estimated number of eggs at the control zinc concentration (30 .μg/L as the mean of two replicates), we obtain the dose–response curve for the reduction in recruits (r): r(x) = 0.00399e5.64−0.00409x .

.

(2.41)

A comparison of observed data and predicted values is shown in Fig. 2.5. The effect of high zinc concentrations on survivability is not reported in [2]. However, as the 96-h half-lethal concentration (LC50) for juvenile fish (the most sensitive life stage) is 2.3 mg/L [1], it can be assumed that the fish do not die due to zinc exposure over the range of concentration with which we are concerned. The Life-History Parameters of the Fathead Minnow and Sensitivities The next task is to determine the life-history parameters of the fathead minnow. Miller and Ankley [15] reported the life-history matrix of the minnow: ⎞ 0.75 1.5 3 .M = ⎝ 0.39 0 0⎠. 0 0.39 0 ⎛

(2.42)

2.4 Case Studies

35

The matrix does not include the egg life stage but starts with age one. The leftmost value in the second column (0.39) should be read as the survivability (or hatchability) of the egg, multiplied by the survivability to the age of one. We first consider the basic properties of the matrix. As this is a 3.×3 matrix, there are three eigenvalues: .λ

=1.40 and .−0.325 ± 0.469i.

The maximum eigenvalue is 1.40. This implies that the population size increases by 1.4 times each year. The eigenvector corresponding to the eigenvalue can be obtained from Eq. 2.33 as ⎛

⎞ 1 .v = ⎝ 0.278 ⎠ . 0.0776 In practice, we use statistical software such as R to find the eigenvectors. The actual values of the elements may differ from those shown, but if the ratios of the elements remain the same, the eigenvector is correct, because only the direction of an eigenvector matters. Most software will return the value adjusted so that the length of the vector is 1. To obtain the adjusted eigenvector, we first prepare the sum of squares of each element (SS): SS = 12 + (0.278)2 + (0.0776)2

.

and then divide each element by the SS: ⎛ ⎞ ⎛ ⎞ 1 0.961 1 ⎝ .v = √ 0.278 ⎠ = ⎝ 0.268 ⎠ . SS 0.0776 0.0745 This is a right eigenvector, as computed by most software, which does not compute the left eigenvector, which we need in order to determine the sensitivity. To obtain the left eigenvector, we transpose the matrix .M as ⎛

⎞ 0.75 0.39 0 T .M = ⎝ 1.5 0 0.39 ⎠ , 3 0 0

(2.43)

and compute the eigenvectors of .MT using the standard built-in function for eigenvectors. In this case, the left eigenvector of .M is w = (0.346, 0.576, 0.740).

.

36

2 Population-level Assessment

Now we have the right and left eigenvectors. The sensitivity matrix (.S) (see Eq. 2.38) is ⎛ ⎞ 0.613 0.171 0.0476 .S = ⎝ 1.02 0.285 0.0793 ⎠ , (2.44) 1.31 0.366 0.102 and elasticity matrix (.e) is (see Eq. 2.39) ⎛ ⎞ 0.328 0.183 0.102 .e = ⎝ 0.285 0 0 ⎠. 0 0.102 0

(2.45)

The element in the second row and first column of .M (Eq. 2.42) is 0.39, and this represents the survivability from one to 2 years old. The corresponding element in the sensitivity matrix is 1.02 (Eq. 2.44), and this means that if the survivability is decreased by .δ, then .λ is decreased by .1.02δ. In both matrices, the elements corresponding to young individuals (.s11 , .s21 , .e11 and .e21 ) are high. This means that the impact on young individuals has a high impact on the population growth rate. In ecotoxicology, the effects on young individuals are often investigated, because chemicals tend to have stronger effects on younger than older individuals. The toxicity tests for young individuals are also meaningful for population growth in the case of this fish, because the impact on the survivability of one-year-old fish has a high impact on the population growth.

Quiz: Sum of All the Elements in Matrices What is the value of the sum of all the elements in .e?9 What is the value of the sum of all .aij × sij (.aij is the element in Eq. 2.42, .sij is the element in Eq. 2.44)?10

Combining the Toxic Effect and Life-History Matrix Now we know the reduction in recruits as a function of the zinc concentration (Eq. 2.41). The curve is then applied to the reproduction terms in Eq. 2.42: ⎛

⎞ 0.75r(x) 1.5r(x) 3r(x) .M(x) = ⎝ 0.39 0 0 ⎠. 0 0.39 0

9

1. 1.40. What is this value?

10

(2.46)

Fig. 2.6 The maximum eigenvalue as a function of the zinc concentration (horizontal axis). The horizontal dashed line shows a maximum eigenvalue of one

37

Max. eigenvalue

2.4 Case Studies 1.4 1.3 1.2

eigenvalue becomes 1

1.1 1.0 0.9 100

50

150

250

200

Zinc concentration (µg/L) Table 2.3 Zinc concentrations resulting in zero population growth [9]

Species type Algae Algae Microorganism Cladoceran Fish Fish

Species name Isochrysis galbana Chroococcus paris Colpoda cuculus Moina macrocopa Pimephales promelas Salvelinus fontinalis

PTC (mg/L) 2.35 4.74 6.57 0.94 0.172 0.295

The relationship between the zinc concentration (x) and eigenvalue is shown in Fig. 2.6. The eigenvalue is less than 1 between a zinc concentration of 172 and 173 .μg/L, and hence it can be concluded that the fathead minnow population cannot be sustained at zinc concentrations greater than 172 .μg/L.

2.4.2 Population-level Species Sensitivity Distribution Kamo and Naito [9] calculated the concentration of zinc at which the population growth rate of six species becomes zero. Although there are now extensive databases of life-history parameters (e.g., Add-my-Pets,11 COMPADRE, and COMADRE12 ), when we conducted this study, we obtained the life-history parameters through a literature survey, and only when both life-history parameters and toxicity data were available, to estimate the effects on each trait, were we able to estimate the concentration at which the population growth rate would be zero. Such good fortune occurred only in the case of six species (Table 2.3). The last task is to determine which concentration we will use for the risk assessment. The minimum concentration here, 0.175 mg/L for P. promelas, is merely the smallest known and not guaranteed to be the smallest among all species. Indeed,

11 https://www.bio.vu.nl/thb/deb/deblab/add_my_pet/index.html. 12 https://compadre-db.org/.

38

2 Population-level Assessment

Fraction affected

1.0 C. cuculus

0.8

C. paris

0.6

I. galbana M. macrocopa

0.4

P. promelas

0.2

S. fontinalis 0.1

1

10

100

Zinc concentration (mg/L) Fig. 2.7 Species sensitivity distribution of zinc concentrations leading to zero population growth. A log-normal distribution is often used for the SSD and is often expressed as a cumulative distribution. The concentration at 5% (i.e., the concentration which affects 5% of the species, or protects 95% of the species), which is often used as an index for risk assessment, is 0.107 .μg/L in the case of zinc

given the number of such species worldwide, it is almost certain that some are more susceptible. We thus need an extrapolation method to cover such more sensitive species. A common method of extrapolation utilizes statistical distributions, and the statistical distribution used for ecological risk assessment, termed the species sensitivity distribution (SSD) [17], will be the subject of Chap. 5. The estimated distribution for the six species above, assuming a log-normal distribution, is shown in Fig. 2.7. In the SSD approach, the concentration which is expected to protect 95% of the species (HC5) is used as the protection goal for risk assessment. In the case of zinc, it is 107 .μg/L. As often in statistics, such a result must be carefully qualified. Its greatest weakness lies in the fact that the population distribution is estimated based on data for only six species. Typically, we need a sample size of 100 or 1000 in order to accurately estimate the distribution. The difficulties here are discussed in detail in Chap. 5. Put simply, the SSD here provides only a rough idea of the effect of zinc on ecosystems. It is clearly unwise to determine a protection goal based on limited information; we need as much variety and abundance of information as possible about the adverse effects of the given chemicals. However, the result above can be used as one type of information for practical risk assessment.

2.4.3 Comparison of the NOEC and PTC We have concluded, then, that the fathead minnow population stops increasing at a zinc concentration of 172 .μg/L, and this concentration is termed the PTC. The difference between the PTC and NOEC is interesting. The toxicity tests on fathead minnows by Brungs [2] were conducted in hard water (hardness: 203 mg/L), and

2.4 Case Studies

PNEC>PTC 1000

NOEC (µg/L)

Fig. 2.8 Comparison of NOEC and population threshold concentration (PTC) values for the fathead minnow

39

Cr(VI)

Zn (hard)

500

Ni 100

Zn (soft)

Cd

50

PNEC d + α. We wish, then, to determine the optimal allocation rates of .τr and .τK and will do so using evolutionary dynamics.

2.5.3 Optimum Allocation Rates in a Crowded Population We first consider that the population size is saturated at the carrying capacity, and search for the optimum allocation rates under these conditions. As we are considering evolution in a crowded population, the maximum allocation rates are those for the K strategist. The population size at equilibrium is determined by Eq. 2.48. Now, suppose that mutants, with allocation rates slightly different from those of the resident type, invade the population. The population size of the mutants is described as y, and their allocation rates are .τr and .τK , respectively. Immediately after the mutants invade the population, the initial value of y is very small. We wish to know whether the tiny fraction of mutants will increase or decrease among the broader population (Fig. 2.9). If it decreases, the mutants cannot take over the population, but if it increases, they will succeed in this takeover. (In addition, it must be proven that the resident type cannot invade the mutant population, if the mutants are to exterminate the residents.)

42

2 Population-level Assessment invasion success

invasion fail

Fig. 2.9 Conceptual diagram of the mutant invasion. The mutants (black dots) invade a population composed of a single type of resident (white circles). If the mutant population does not increase in size, it fails to exterminate the population, which eventually reverts to residents only. If the mutant population increases, the mutants eventually take over the population

The mutant population dynamics can be obtained by changing x to y in Eq. 2.47, and changing the parameters to those related to y. We must ensure that the initial population size is saturated by the residents (.x ∗ ). The reproduction rate of the mutants (and residents) is almost zero when the mutants appear. The mutant population dynamics may then be expressed as .

  1 dy x∗ + y = τr ω 1 − y −  y. dt K τK ω

The density effects now include both the number of residents (.x ∗ ) and the number of mutants (y). As aforementioned, the number of mutants (y) is very small at the beginning of the invasion, and thus the second-order term of y (.y 2 ) is very small and may be approximated as zero. By doing this, we have .

    1 dy x∗ = τr ω 1 − −  y, dt K τK ω

(2.49)

and this is a further expression of the mutant dynamics. We have already determined the value of .x ∗ (Eq. 2.48). Substituting this value into Eq. 2.49, and arranging slightly, yields .

  dy = τr τK − τr τK ω. dt

(2.50)

As the condition under which the mutant population increases is .dy/dt > 0, it will increase if τr τK > τr τK .

.

2.5 Toxic Effects on r- vs. K-Selected Species

43

This means that evolution maximizes the value of .τr τK . Now, as .τK = 1 − τr , the term to be maximized is .τr (1 − τr ), and this is maximized when τr =

.

1 . 2

(2.51)

The conclusion thus far is that the respective optimum allocation rates for K strategists are .τr = τK = 1/2.

2.5.4 Optimum Allocation Rates in a Sparse Population In a sparse population, the respective population dynamics of residents and mutants are   1 dx x+y = τr ω 1 − x− x, . dt K τK ω   1 dy x+y = τr ω 1 − y −  y. dt K τK ω The sparse population means that both x and y are small, and hence the terms .x 2 , y 2 , and xy are approximated as zero, yielding   dx 1 = τr ω − x, . dt τK ω   dy 1 = τr ω −  y. dt τK ω These equations imply that the faster the rate of population growth the better, and thus evolution maximizes the term, τr ω −

.

1 , τK ω

(2.52)

and as .τK = 1 − τr , the term to be maximized is τr ω −

.

τr (1 − τr )ω2 − 1 1 = . (1 − τr )ω (1 − τr )ω

The maximum value can be obtained by solving .

d dτr



τr (1 − τr )ω2 − 1 (1 − τr )ω

 =0

44

2 Population-level Assessment

and the solutions are τr =

.

 ω−1 ω ω+1 ω

.

As .0 ≤ τr ≤ 1, .(ω + 1)/ω cannot be a solution, and hence the respective optimum allocation rates are τr =

.

ω−1 , ω

τK = 1 − τr =

1 . ω

(2.53)

Given these optimum allocation rates, the population growth rate (Eq. 2.52) is τr ω −

.

1 =ω−2 τK ω

and this must be positive for the population to increase, so .ω > 2 must be satisfied.

2.5.5 Toxic Effects: r vs. K We wish, then, to know how the r vs. K strategists behave in the presence of the toxicants. In conventional risk assessment, the effect on individual life-history traits is mainly investigated. To mimic such toxicity tests, we consider a situation where there is no reproduction (.r = 0), only mortality. The dynamic response of the K strategist is obtained by substituting Eq. 2.51 into Eq. 2.47, and as .d = 1/τK ω and .τK = 1/2 at optimum, we have .

2 dx = − x − αx dt ω

which can be solved as x(t) = x(0)e

.

 − 2t ω +α

,

where .x(0) is the initial value of x at time 0. The value of .α required for the number of individuals to be halved at time T is obtained by solving x(0)e

.

 − 2t ω +α

=

x(0) , 2

2.5 Toxic Effects on r- vs. K-Selected Species

45

and we have αK =

.

ω log(2) − 2T . ωT

Similarly, we can obtain the .α required for the number of individuals to be halved at time T for the r strategist, with .τK = 1/ω: αr =

.

log(2) − T . T

The difference between the respective .α values of the K and r strategist is αK − αr =

.

ω−2 , ω

(2.54)

and when .ω > 2, the difference is always positive and .αK is larger than .αr . This implies that more toxicant is needed for half-lethality in the case of the K strategist, who is, thereby, less sensitive to chemical toxicity than the r strategist. On the other hand, when both reproduction and mortality are taken into account in assessing the population-level risk, the equilibrium value for the K strategist (Eq. 2.48), given the optimum allocation rate, is ∗ xK =

.

K(ω2 − 2αω − 4) , ω2

and that of the r-strategist is xr∗ =

.

K(ω − α − 2) . ω−1

The value of .α required to reduce the population size by p (.0 ≤ p ≤ 1), in the case of the K strategist, is obtained by solving .

K(ω2 − 2αω − 4) K(ω2 − 4) =p 2 ω ω2

and is αK =

.

(1 − p)(ω2 − 4) 2ω

The similar value in the case of the r strategist is αr = (1 − p)(ω − 2).

.

46

2 Population-level Assessment

The difference between the respective .α values, under these conditions, is αK − αr = −

.

(1 − p)(ω − 2)2 , 2ω

(2.55)

and as .0 ≤ p ≤ 1, this is always negative, and the K strategist is thus identified as the more sensitive species. If, then, we consider the toxic effects on reproduction and mortality separately, as is often done in individual-level assessments, we conclude that K strategists are less sensitive. However, if we consider both effects in aggregate, in populationlevel risk assessment, we conclude that K strategists are more sensitive. Thus, the determination of r- vs. K-strategist sensitivity yields completely opposite results at the two different levels of assessment. By framing the problem explicitly in terms of the r- vs. K-strategies, we have answered a long-standing question about the respective chemical sensitivities of these strategies, and the answer is that the determination varies depending on the level of our assessment (individual- vs. population-level), the same conclusion as in Stark et al. [20]. In environmental management, models are often used as valuable predictors, but they can also be used to provide answers to questions (as we have done in this section), and this is exactly what theorists do. Throughout the chapter, I have described population models which are frequently used in population-level assessment. This is because I believe that risk assessment must always (at least) be done on the population level. The conventional assessment framework, which is often based on individual-level assessment in contrast to population-level assessment, is also important; but I consider these two assessments to be complementary, not contradictory, and disagree with those who think we only need one or the other. The very complexity of ecosystems argues for the use of more than a single evaluatory indicator, and indeed, for as many such indicators as possible.

References 1. Broderius S, Stmith LL (1079) Lethal and sublethal effects of binary mixtures of cyanide and hexavalent chromium, zinc, or ammonia to the fathead minnow (Pimephales promelas) and rainbow trout (Salmo gairdneri). J Fish Res Board Can 36:164–172 2. Brungs WA (1969) Chronic toxicity of zinc to the fathead minnow, Pimephales promelas rafinesque. Trans Am Fish Soc 98:272–279 3. Caswell H (1989) Matrix population models. Sinauer, Sunderland 4. DeAngelis DL, Bartell SM, Brenkert AL (1989) Effects of nutrient recycling and food-chain length on resilience. Am Nat 134(5):778–805 5. Grimm V, Johnston ASA, Thulke HH, Forbes VE, Thorbek P (2020) Three questions to ask before using model outputs for decision support. Nat Commun 11:4959. https://doi.org/10. 1038/s41467-020-17785-2 6. Gurney WSC, Nisbet RM (1998) Ecological dynamics. Oxford University Press, Oxford

References

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7. Houslay TM, Wilson AJ (2017) Avoiding the misuse of BLUP in behavioural ecology. Behav Ecol 28(4):948–952 8. Iwasaki Y, Hayashi TI, Kamo M (2010) Comparison of population-level effects of heavy metals on fathead minnow (Pimephales promelas). Ecotoxicol Environ Saf 73:465–471 9. Kamo M, Naito W (2008) A novel approach to determining a population-level threshold in ecological risk assessment: a case study of zinc. Hum Ecol Risk Assess 14:714–727 10. Kooijman SALM (2009) Dynamic energy budget theory for metabolic organisation, 3rd edn. Cambridge University Press, Cambridge 11. Kooijman SALM, Metz JAJ (1984) On the dynamics of chemically stressed populations: the deduction of population consequences from effects on individuals. Ecotoxicol Environ Saf 8(3):254–274 12. MacArthur RH, Wilson EO (2001) The theory of island biogeography. Princeton landmarks in biology. Princeton University Press, Princeton 13. Magliette R, Doherty F, McKinney D, Venkataramani E (1995) Need for environmental quality guidelines based on ambient freshwater quality criteria in natural waters - case study zinc. Bull Environ Contam Toxicol 54:626–632 14. May R (1972) Limit cycles in predator–prey communities. Science 171:900–902 15. Miller DH, Ankley GT (2004) Modeling impacts on populations: fathead minnow (Pimephales promelas) exposure to the endocrine disruptor 17 β-trenbolone as a case study. Ecotoxicol Environ Saf 59:1–9 16. Murray JD (2002) Mathematical biology: I. An introduction, 3rd edn. Interdisciplinary applied mathematics, vol 17. Springer, New York 17. Posthuma L, Suter II GW, Traas TP (eds) (2001) Species sensitivity distributions in ecotoxicology. Lewis Publishers, Boca Raton 18. Raimondo S, Schmolke A, Pollesch N, Accolla C, Galic N, Moore A, Vaugeois M, RuedaCediel P, Kanarek A, Awkerman J, Forbes V (2020) Pop-guide: population modeling guidance, use, interpretation, and development for ecological risk assessment. Integr Environ Assess Manag 17(4):767–784 19. Schaefer MB (1991) Some aspects of the dynamics of populations important to the management of the commercial marine fisheries. Bull Math Biol 53:253–279 20. Stark JD, Banks JE, Vargas R (2004) How risky is risk assessment: the role that life history strategies play in susceptibility of species to stress. Proc Natl Acad Sci U S A 101(3):732–736 21. Yap C, Ismail A, Omar H, Tan SG (2004) Toxicities and tolerances of Cd, Cu, Pb and Zn in a primary producer (Isochrysis galbana) and in a primary consumer (Perna viridis). Environ Int 29:1097–1104

Chapter 3

Population Models of Extinction

Abstract In the previous chapter we discussed the basics of population dynamics, whose models describe the average population size, which changes over time. In this chapter, we will consider population extinction by adding random movement in the population size. The study of extinction in conservation ecology and population genetics has a long history, and it is not the purpose of this chapter to summarize the many extant studies. Instead, the goal is to understand their essential aspects intuitively and thereby be able to perform some useful calculations on our own. To this end, the derivation of the relevant formulas has been considered in detail. If the reader is familiar with these derivations, they are encouraged to skip this chapter and move on to the next, which deals with case studies. First, we will present a model of the birth-death process, followed by a model of the stochastic process. Two diffusion process models play an important role in the stochastic process, and the ultimate goal of the chapter is to derive the mean time to extinction (MTE) using these diffusion process models. Keywords Population-level assessment · Population extinction · Birth-death process · Stochastic process · Mean time to extinction (MTE)

3.1 Introduction This chapter describes the most common and relatively simple model of population extinction. In Chap. 2, we discussed methods of ecological risk assessment using population models, determined the concentration of chemical substances at which the growth rate becomes zero (the population threshold concentration or PTC), and focused on a zero growth rate as the assessment endpoint. However, such methodology alone is not sufficient for considerations of extinction, because even if the environmental concentration of a given chemical reaches the PTC, the population will not necessarily go extinct. If the population size is K when the environmental concentration reaches the PTC, the population growth rate is zero, and in this case, the population size will remain at K forever unless affected by other, external factors. Some such factors, often the result of mere chance (for example, © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_3

49

50

3 Population Models of Extinction

that no individuals happen to give birth in a given year, or the mortality is unusually high due to lack of sunshine that year), must be considered in order to investigate the possibility of population extinction. In this context, an assessment endpoint is an event that we wish to avoid happening, and in population biology, the ultimate event to avoid is the extinction of the species. Even if the population is severely depleted, it can be increased if it is properly managed. However, once a species goes extinct, it is irreversible, and therefore ecology and conservation ecology have paid great attention to extinction. If we wish to avoid the extinction of biological populations, we must understand why and how extinctions occur, and for this reason, a great deal of theoretical work has been done in this regard. This chapter introduces several mathematical models related to population extinction, as outlined in the book, Ecosystems and Simulation [5]. As may be expected, given the discussion above, all the models incorporate an element of chance. The first model describes demographic stochasticity, the second describes environmental stochasticity, and the last incorporates both. The ultimate objective of the chapter is to derive the mean time to extinction of a given population, based on this last model. One of the general aims of this work is to detail the relevant derivations, which in some cases may be rather lengthy. Thus, readers already familiar with these models and their derivations, are welcome to skip this chapter and proceed to the next, in which some case studies of chemical substance assessment, using the composite model above, are discussed.

3.2 A Model of the Birth and Death Process First, we will consider classic model of population extinction, explored by MacArthur and Wilson in “The Theory of Island Biogeography” [7], which considers the birth and death process. We consider, then, a population of size x at time t. Let r be the growth rate per individual per unit of time, and d be the death per unit of time (we consider a time interval of h). Now, assume that h is small enough that in h time, only one of the following will occur in this population: the population size will increase by 1, the population size will decrease by 1, or the population size will neither increase nor decrease. The population size increases in proportion to the growth rate per individual, the number of individuals in the population, and the time h, and hence increases at a rate of rxh. If the time interval, h, is small enough, the rate can be considered in terms of probability. By the same reasoning, the probability that the population size decreases is dxh, and the probability that the population size remains the same, after h, is .1 − rxh − dxh. Let .Tx−1 , .Tx and .Tx+1 , then, be the respective mean times to extinction (henceforth referred to as MTE) of populations of size .x − 1, x, and .x + 1. Assume that the current population size is x, and MTE is .Tx . After h, the respective population levels

3.2 A Model of the Birth and Death Process

51

will be .x − 1, .x + 1, and x, with the respective probabilities being dxh, rxh, and 1−rxh−dxh; hence MTE after h is .dxhTx−1 +rxhTx+1 +(1−dxh−rxh)Tx . Since this is MTE after time h has elapsed from the original time, the result of adding h to MTE should equal the original mean time, .Tx , and therefore we have

.

Tx = h + dxhTx−1 + rxhTx+1 + (1 − dxh − rxh)Tx ,

.

and an arrangement yields, 0 = 1 + (Tx+1 − Tx )rx − (Tx − Tx−1 )dx.

.

(3.1)

We now define that Dx = Tx − Tx−1 ,

.

and if we substitute this equation into Eq. 3.1, we have 0 = 1 + Dx+1 rx − Dx dx.

.

(3.2)

Assuming there is an upper limit (carrying capacity) to the population size, if the population size exceeds this limit, K, then it immediately reverts to K. This assumption implies that the respective MTE of populations of size .x = K + 1 and .x = K are equal (i.e., .TK = TK+1 ), and therefore we have DK+1 = TK+1 − TK = 0.

.

When .x = 0, the population has already become extinct and MTE is 0 (.T0 = 0); hence we have D 1 = T1 − T0 = T1 .

.

We can obtain MTE as follows. When .x = 2, from Eq. 3.1 we have the relationship, 0 = 1 + 2D2 r − 2D1 d, and arranging the equation yields

.

D2 =

.

d 1 D1 − . r 2r

Performing the same arrangement for .x = 3 yields d 1 D2 − = .D3 = r 3r

 2   d 1 1 d + , D1 − r r 2 r

52

3 Population Models of Extinction

and this equation is generalized for .x ≤ K as x−1  x−1  i−1  1  1 d d D1 − .Dx = . r r x−i r i=1

When .x = K + 1, since .DK+1 = 0, we have DK+1

.

K  K  i−1  1 d 1  d = D1 − = 0, r r K +1−i r i=1

and arrangement yields  −K   i−1  K 1 1 d d .D1 = , r r K +1−i r i=1



 i−1−K  K 1 d 1  = , r K +1−i r i=1

= T1 . Various MTE, with varying combinations of r and d when .x = 1, are illustrated in Fig. 3.1. MTE with varying population size (x) is depicted in Fig. 3.2. In this model, MTE becomes extremely large as the number of individuals increases, because the model assumes that the factors that cause variation in population size occur independently for each individual (i.e., each gives birth or dies independently of the rest). This type of variation is termed demographic stochasticity, and its magnitude is proportional Fig. 3.1 MTE for a population beginning with 1 individual (.T1 ). The numbers in the panel are varying ratios of .r/d (r is fixed at 2). The corresponding figure in “The Theory of Island Biogeography” [7] is reproduced as closely as possible (along with the most similar label font)

9

10

λ/μ=1000 10

2 1.5

1.01

7

10

5

10

λ=2

3

10 1

1 0.5

-1

10

1

10

100

1,000

10,000

3.3 Models of Stochastic Processes: Necessary Elements

λ/μ=1.08

Time to Extinction

Fig. 3.2 When the initial population size x is small, the mean time to extinction sharply increases, but the curves are flat for larger values of x

53

2000 1500 1000

λ/μ=1.05 λ/μ=1

500 0

0

20

40

x

60

80

100

to the square root of the population size x.1 When, for example, there are only 2 individuals in a given population, the probability that the 2 will die at the same time and the population go extinct is not negligible, but when there are 1000 individuals, the probability that they will all die at the same time is negligible. Another factor causing variation in the population size is environmental stochasticity, whose magnitude is proportional to the population size x. For example, if the survival rate for a given year is 80%, the average number of individuals that die in a year is 2 if the population is 10, and 20 if the population is 100. Even if the survival rate is the same, then, the number of individuals affected will vary depending on the population size x. In the coming sections, these stochastic models will be investigated, and the effect of environmental stochasticity is considered. Before investigating specific models, however, we should consider the elements necessary for understanding these models.

3.3 Models of Stochastic Processes: Necessary Elements In Chap. 2, we considered models based on continuous-time differential equations for population dynamics. Each described the average change in the population size and implicitly assumed that the number of individuals existed with a probability of one at the average (Fig. 3.3). In the case of a stochastic process, however, the number of individuals will not always be the average but rather distributed with a certain probability around the average. As, in statistical analysis, we often want to know the confidence interval of an estimated parameter, in ecosystem management,

1 Assume

two random variables, .X1 and .X2 , sampled independently from a normal distribution with a mean of 0 and variance of V . The variance in the sum of two random variables is generally .V ar(X1 +X2 ) = 2V +2 Cov(X1 , X2 ), but if the variables are independent, √ then .Cov(X1 , X2 ) = 0 and √ hence .V ar(X1 + X2) = 2V . The standard deviation in the sum is . 2V , and√proportional to . 2. If there are x variables, the standard deviation in their sum is proportional to . x.

54

3 Population Models of Extinction

1.0

Probability

Fig. 3.3 Conceptual diagram of the time evolution in population size in diffusion processes. The population dynamics in the previous chapter considered the change in population size in terms of vertical bars. In contrast, in diffusion processes, the time evolution of the distribution of population size (.p(x, t)) is considered

t=0

t=1

t=2

0.8 0.6 0.4 0.2

x

information about the range of population size provides us with information that we cannot obtain from the average alone. In this section, two diffusion equations play an important role. The first expresses the time evolution of the distribution of population size. Let .p(x, t), then, be the distribution of population size (x) at time t. The time evolution of this distribution is then .

∂p(x, t) ∂ 1 ∂2 {V (x)p(x, t)}. = − {M(x)p(x, t)} + ∂t ∂x 2 ∂x 2

(3.3)

This so-called forward equation is known as the Fokker–Planck equation. .M(x), which shows the average movement of x, is termed the drift. .V (x) shows the variance in the random movement of x. When .V (x) is not a function of x, but instead constant, V is termed the diffusion coefficient, often written as D. The second useful equation derives MTE of a population of initial size .x0 : .

− 1 = M(x0 )

dT (x0 ) V (x0 ) d 2 T (x0 ) + . dx0 2 dx02

(3.4)

This so-called backward equation expresses the probability of starting from an initial value (.x0 ) and reaching an objective value. Considering the time when the objective value is reached as the starting point, it appears as if the initial value is obtained by going back in time, from the future (when that point is reached) to the present; hence the term, backward equation. The initial value .x0 is the variable in the equation. We will discuss these diffusion equations in more detail below. First we must gain an intuitive understanding of stochastic processes in general. In Chap. 2, we considered the population growth model, .

dx = rx, dt

3.3 Models of Stochastic Processes: Necessary Elements

55

where x is the population size and r the intrinsic growth rate. The variable x and time t are continuous in the model. When we want to understand what the equation means, or find a numerical solution for the equation, we discretize these variables. Let .t, then, be the time interval for discrete time, and .xt be the population size at time t. In Chap. 2, we saw that the increment in x between time t and .t + t is xt = rxt t.

.

The value of x at time .t + 1 is then derivable as xt+t = xt + xt .

.

If we now assume that the time becomes T after repeating this process n times, the values of x at time T is xT = x0 +

n 

.

xt = x0 + r

n 

t=0

xt t,

t=0

when the starting point of the process is set as .t = 0 and the value of x at that point is .x0 . This is how we solve the differential equation numerically. In a stochastic process, on the other hand, we additionally consider random movement of x. Let .Wt be such random movement within the time .t. The value of x is then determined by the average movement we have thus far considered, plus the random movement .Wt : xt = rxt t + xt Wt .

.

(3.5)

By summing all the increments in x over time t as xT = x0 +

n 

.

t=0

rxt t +

n 

xt Wt ,

(3.6)

t=0

we obtain the value of x at time T . The second term on the right hand side describes the average movement, which we have previously considered. The third term represents the random movement over time, which remains for us to fully understand. As the simplest case, we first consider a model with .xt removed from Eq. 3.5: xt = rt + Wt ,

.

which describes a random walk, as discussed below.

(3.7)

56

3 Population Models of Extinction

Taylor Expansion Let .f (x) be a differentiable function. For a small value of .x, .f (x + x) can be rewritten as f (x + x) = f (x) + x

.

1 d 2 f (x) 1 d 3 f (x) df (x) + x 2 + x 3 + ··· . 2 dx 2 dx 6 dx 3

We often refer to this as expanding the function .f (x) into a Taylor series. When the function has two variables as .f (x, y), we have a similar Taylor series: f (x, y + y) = f (x, y) + y

.

+y 3

1 ∂ 2 f (x, y) ∂f (x) + y 2 ∂y 2 ∂y 2

1 ∂ 3 f (x, y) + ··· ; 6 ∂y 3

and this time, the symbol of differentiation is changed from d to .∂. When such functions have two or more variables, .∂ is used as the symbol of differentiation.

3.3.1 The Random Walk A simple example, often used in considering discrete random processes, is the onedimensional random walk, which describes the location of a particle that is moving to the right or left in a random way. We define the model as follows (Fig. 3.4). • Lattices of width .x are in one dimension. • Within a time step, .t, the particle moves by one lattice. If the particle’s current location is x and it moves to the right, its location at the next time will be .x + x. If it moves to the left, its location will be .x − x. • The probability that the particle moves to the right is r, and that it moves to the left l. If we exclude the case where the particle stays at the same lattice position over a given time step, we have .l + r = 1. l

-2Δx

-Δx

r

0

+Δx

+2Δx

Fig. 3.4 The concept of the random walk, where l is the probability that the particle moves to left, r that it moves to the right, and .r + l = 1

3.3 Models of Stochastic Processes: Necessary Elements

57

We now wish to know, for example, where the particle is when it has moved N times (i.e., the elapsed time is .t = Nt). If .r = 1, then it is obvious that the location of the particle is .Nx. If .0 < r < 1, the particle moves not only to the right but also to the left, and hence we can intuitively understand that the particle cannot arrive at .Nx but will be to the left of this location. In this example, we have been considering the location of a particle, but the problem (and solution) is the same when we ask how many dollars we have, on average, when we play a game in which we toss a coin and gain (loss) a dollar when the coin comes up heads (tails). Let us now assume that the particle moves to the right R times, to the left L times and is located at x, after N movements (note that .L + R = N), at time t. The probability, .p(x, t), that the particle is located at x is then expressible in terms of a binomial distribution: p(x, t) =

.

N! R L r l , R!L!

(3.8)

where N! = N × (N − 1) × (N − 1) × · · · × 3 × 2 × 1.

.

What we want to know is how the probability changes over time. For the particle to be at location x at time t, two possibilities are considered. The first is that, at time .t − t the particle was at location .x − x and then moved to the right. The other is that, at this time, the particle was at location .x + x and then moved to the left. The probability, then, that the particle is at .x − x at time .t − t is .p(x − x, t − t), and that it is at .x + x at time .t − t is .p(x + x, t − t), and recall that the probability of it moving to the left is l and to the right r. We can then intuitively understand that the relationship between the respective probabilities of the location of the particle at times t and .t − t is p(x, t) = rp(x − x, t − t) + lp(x + x, t − t).

.

(3.9)

Proof of Eq. 3.9 Since the particle is at x when it has moved R times to the right and L times to the left, for a total of N movements, the particle only needs to have moved .(R − L) times to the right and L times to the left to be at .x − x at .(n − 1) movements. The probability of this is p(x − x, t − t) =

.

(N − 1)! R−1 L r l , (R − 1)!L! (continued)

58

3 Population Models of Extinction

and similarly, the probability that a particle is at .x + x is p(x + x, t − t) =

.

(N − 1)! R L−1 r l . R!(L − 1)!

By adding these equations based on their weighted probabilities, we have rp(x − x, T − t) + lp(x + x, T − t) = r

.

(N − 1)! R−1 L r l (R − 1)!L!

+l

(N − 1)! R L−1 r l R!(L − 1)!

R N! R L L N! R L r l + r l N R!L! N R!L! R + L N! R L r l = N R!L! = p(x, T ). =

Note that .N = R + L.

3.3.2 Mean and Variance in Location Let us denote the location of the particle at time t by .xt , and the mean of the location by .E[xt ]. We consider that the particle moves to the right R times and to the left L times, and therefore the location is .xt = Rx − Lx. The probability that the particle moves to right is r, and therefore .E[R] = rN, and similarly, .E[L] = lN. The mean of .xt , then, is E[Rx − Lx] = E[R]x − E[L]x,

.

= rNx − lNx, = (r − l)Nx. Now, the time after N movements is .t = Nt, and thus we have .N = t/t, and the mean is E[xt ] = (r − l)(x/t)t,

.

= (2r − 1)(x/t)t; note that .l = 1 − r.

(3.10)

3.3 Models of Stochastic Processes: Necessary Elements

59

Let us denote the variance by .V [xt ]. This can be obtained as V [Rx − Lx] = V [Rx − (N − R)x],

.

= V [2Rx − Nx], = 4x 2 V [R]; note that .V [N x] = 0 because .Nx is a constant. As the variable R has a binomial distribution, we have V [R] = rlN = r(1 − r)N,

.

and eventually we have V [xt ] = 4r(1 − r)Nx 2 .

.

As .N = t/t, this equation becomes V [xt ] = 4r(1 − r)

.

x 2 t. t

(3.11)

The mean and variance imply that the location of a particle that commences random movement from time .t = 0 has a binomial distribution with mean .(2r −1)(x/t)t and variance .4r(1 − r)(x 2 /t)t. The important point is that if we know how many times the particle has moved (how much time has passed), we can describe the location of the particle. However, the precise location of the particle cannot be specified but is described in terms of a probability distribution, and the variance in the distribution is proportional to time t (the standard deviation is proportional to √ . t), making it more difficult to specify the location of the particle as time passes. In the case of a large N, the binomial distribution can be approximated by a normal distribution.2 Figure 3.5 shows examples of distributions of the location of the particle in a random walk, with two different numbers of movements obtained by Monte-Carlo simulations, and with a theoretically derived distribution (approximated by the normal distribution). The respective probabilities that the particle moves to the right and left are set as equal (.r = l = 1/2); hence the mean is 0 in both distributions, but the distribution with more movements is wider. Figure 3.6 shows a further example, with .r > l. When we apply this technique to the aforementioned coin-toss game, even if we use a loaded coin with a 2/3 chance of coming up heads, there is still a roughly 13.2% chance of losing money over ten tosses of the coin.3

2 The proof here (which utilizes Stirling’s formula), though not difficult, is cumbersome. It can easily be found online. 3 This probability is computed by integrating the distributions less than zero in Fig. 3.6.

60

3 Population Models of Extinction

0.12

t=10

Density

0.10 0.08 0.06 0.04

t=100

0.02 0.00 -40

-20

0

20

40

Location Fig. 3.5 Probability densities of .xt with .r = 1/2. The histograms are generated based on MonteCarlo simulations (.t = 10 for the light gray, and .t = 100 for the gray). The lines show normal distributions as approximations for binomial distributions. .x = t = 1

Location

10

5

2

4

6

8

10

-5

Time Fig. 3.6 Distributions of the location of a random-walk particle with .r = 2/3. The mean increases linearly with a slope of .(2r − 1)t. The distributions of the particle’s location are binomial but approximated by normal distributions in the figure. The distributions become wider as time passes, and the variance (which indicates the width of the distribution) is proportional to t. .x and .t are set at 1

Monte-Carlo Simulation of a Random Walk The simulation of a random walk is simple. We prepare a variable x and select a random number between 0 and 1. If the number is less than 0.5 we increase x by 1, and if larger than 0.5 we decrease x by 1 (when the probability of moving to the right and left is the same). After repeating this N times, we obtain the location of .xN . We continue this process, each time obtaining a different (continued)

3.3 Models of Stochastic Processes: Necessary Elements

61

location of .xN , but after many such iterations, we obtain the distribution of x. In computer language, this reads as .x = 0; if(.ξ < 0.5) .x = x + 1 else .x = x − 1 We repeat this code N times to obtain the location of .xN (where .ξ is a uniform random number, .0 < ξ < 1), and after many such iterations, we obtain the distribution of .xN (Fig. 3.5). This is known as the Monte-Carlo method.

We saw in this section that the precise location of a randomly moving particle cannot be specified (with probability 1) but can be expressed in terms of a probability distribution. The parameters of this distribution are determined by the time spent after the particle begins moving. Since a random walk is a discrete-time model, the time increment is a finite value. When the time interval between each step is made as close to zero as possible (i.e., .t → 0), we have the model of a diffusion process with continuous time. This continuous-time model describes the so-called Wiener process or Brownian motion. Such models may seem difficult to understand, but by discretizing them and observing what occurs between the time steps, we can gain an intuitive understanding of their nature.

3.3.3 The Forward (Fokker–Planck) Equation We have seen that there is a relationship between the current distribution and past distributions in a random walk (Eq. 3.9), namely, p(x, t + t) = rp(x − x, t) + lp(x + x, t).

.

(3.12)

When .x and .t are very small, we can expand the left hand side of Eq. 3.12 and obtain p(x, t + t) = p(x, t) + t

.

∂p(x, t) 1 2 ∂ 2 p(x, t) + t + ··· . ∂t 2 ∂t 2

Similarly, we can expand the first and the second terms on the right hand side as p(x − x, t) = p(x, t) − x

.

∂p(x, t) 1 2 ∂ 2 p(x, t) 1 3 ∂ 3 p(x, t) + x − x ··· , ∂x 2 6 ∂x 2 ∂t 3

and p(x + t, t) = p(x, t) + x

.

∂p(x, t) 1 2 ∂ 2 p(x, t) 1 3 ∂ 3 p(x, t) + x + x ··· . ∂x 2 6 ∂x 2 ∂x 3

62

3 Population Models of Extinction

Substituting these equations into Eq. 3.12 and, as they are very small, approximating (t)n for .n > 1 and .(x)n for .n > 2 by 0, we have

.

.

x ∂p(x, t) 1 x 2 ∂ 2 p(x, t) ∂p(x, t) = −(r − l) + . ∂t t ∂x 2 t ∂x 2

(3.13)

(r − l)(x/t) represents the average movement (or drift), and .x 2 /t represents the variance in the random movements. If we express these terms as M and V , respectively, then we have

.

.

∂p(x, t) V ∂ 2 p(x, t) ∂p(x, t) = −M + , ∂t ∂x 2 ∂x 2

(3.14)

and this is the forward or Fokker–Planck equation. When M and V are functions of x, this equation becomes Eq. 3.3, which we encountered at the beginning of Sect. 3.3.

3.3.4 Backward Equation Let us now define .T (x0 ) as MTE of a population of size .x0 at time 0. Under this assumption, the population size at the next time step is .x0 + x0 , and the mean time to extinction becomes .T (x0 + x). As we considered in the birth and death process model (Sect. 3.2), since MTE at time 0 becomes .T (x0 + x0 ) after a time step of .t, we have this relationship: T (x0 ) = t + T (x0 + x0 ).

.

(3.15)

Assuming, then, that .x0 is very small and we expand .T (x0 + x0 ) in a Taylor series, we have T (x0 + x0 ) = T (x0 ) + x0

.

dT (x0 ) x 2 d 2 T (x0 ) + , dx0 2 dx02

(3.16)

where .x n for .n > 2 is approximated by 0. We write the change in .x0 over a time step as M, and as the time step is .t, the average change in .x0 (.x0 ) is x0 = Mt.

.

(3.17)

Similarly, if we write the change in .x02 as V , we have x02 = V t.

.

(3.18)

3.3 Models of Stochastic Processes: Necessary Elements

63

Substituting Eqs. 3.16, 3.17, and 3.18 into Eq. 3.15, with some minor arrangement, yields .

−1=M

dT (x0 ) V d 2 T (x0 ) + , dx0 2 dx02

(3.19)

and this is the backward equation. When M and V are functions of .x0 , we obtain Eq. 3.4, as at the beginning Sect. 3.3.

Solving the Backward Equation: Obtaining MTE In this subsection, we solve Eq. 3.4, with an eye to obtaining MTE. We first consider a function that satisfies U (x0 ) =

.

dT (x0 ) , dx0

(3.20)

and try to find the solution for .U (x0 ). Eq. 3.4 may be rewritten, incorporating .U (x0 ), as V (x0 ) dU (x0 ) , 2 dx0

(3.21)

M(x0 ) dU (x0 ) 2 −2 U (x0 ). =− dx0 V (x0 ) V (x0 )

(3.22)

− 1 = M(x0 )U (x0 ) +

.

and arrangement yields .

Because .U (x0) is a function that yields two terms when differentiated by .x0 , we guess that .U (x0 ) has the form U (x0 ) = u1 (x0 )u2 (x0 ).

.

If we differentiate both sides of this equation by .x0 , we have .

dU (x0 ) du1 (x0 ) du2 (x0 ) = u2 (x0 ) + u1 (x0 ) . dx0 dx0 dx0

Equation 3.22 has .U (x0 ) on the right hand side, implying that .U (x0 ) is a function that generates itself when differentiated. A typical function that satisfies this condition is an exponential function. For example, if we differentiate the function f (x0 ) , we have .e .

def (x0 ) df (x0 ) f (x0 ) = e . dx0 dx0

64

3 Population Models of Extinction

If we now assume that u2 (x0 ) = ef (x0 ) ,

(3.23)

U (x0 ) = u1 (x0 )ef (x0 ) ,

(3.24)

.

then we have .

and if we differentiate both sides of this equation by .x0 , we have .

dU (x0 ) du1 (x0 ) f (x0 ) df (x0 ) = e + u1 (x0 )ef (x0 ) , dx0 dx0 dx0 du1 (x0 ) f (x0 ) df (x0 ) e + U (x0 ). dx0 dx0

=

(3.25)

By comparing Eqs. 3.22 and 3.25, we notice that the following relationships should hold: .

du1 (x0 ) f (x0 ) 2 e =− ,. dx0 V (x0 ) df (x0 ) M(x0 ) . = −2 dx0 V (x0 )

(3.26) (3.27)

By integrating both sides of Eq. 3.27, we have .f (x0 ) as

x0

f (x0 ) = −2

.

0

M(x) dx, V (x)

(3.28)

and substituting this equation into Eq. 3.23, we obtain u2 (x0 ) = e−2

x0

.

0

M(x) V (x) dx

.

(3.29)

Why We Have Eq. 3.28 Some may find it confusing that the upper limit of the interval of integration in Eq. 3.28 is the variable .x0 . To confirm that the equation is indeed correct, we must show that we obtain Eq. 3.27 if we differentiate both sides of Eq. 3.28 by .x0 . If we assume, then, that .g(x) = M(x)/V (x), then Eq. 3.28 is written as x0 .f (x0 ) = −2 g(x)dx. 0

(continued)

3.3 Models of Stochastic Processes: Necessary Elements

65

Let us assume that .g(x) can be integrated, and that the integrated form is .G(x) (i.e., .dG(x)/dx = g(x)). We then have f (x0 ) = −2(G(x0 ) − G(0)),

.

and by differentiating both sides, we have .

  df (x0 ) dG(x0 ) dG(0) . = −2 − dx0 dx0 dx0

Now, since .G(0) is not a function of .x0 , .dG(0)/dx0 = 0, and this equation is equal to Eq. 3.27.

Substituting Eq. 3.28 into Eq. 3.26 yields .

x0 du1 (x0 ) 2 e2 0 =− dx0 V (x0 )

M(x) V (x) dx

.

Integrating both sides yields u1 (x0 ) =

L

.

x0

2 2 0y e V (y)

M(x) V (x) dx

(3.30)

dy,

where .L > x0 (see the box above to confirm that this equation is correct). Summarizing our knowledge thus far, then: U (x0 ) = u1 (x0 )u2 (x0 ), L 2 2 0y e = x0 V (y) L 2 2 0y e = x0 V (y) L 2 2 0y e = x0 V (y)

.

M(x) V (x) dx

dy × e−2

M(x) V (x) dx

e−2

x0 0

x0 0

M(x) V (x) dx

x0 M(x) M(x) V (x) dx−2 0 V (x) dx

The exponents in the exponential function,

y

2

.

0

M(x) dx − 2 V (x)



x0 0

M(x) dx, V (x)

M(x) V (x) dx

,

dy,

dy.

(3.31)

66

3 Population Models of Extinction

are slightly complicated. As we did in the box above, then, we assume .g(x) =

x M(x)/V (x) and consider the function .G(x) = 0 g(z)dz. We can now arrange Eq. 3.31 as

y

2

.



x0

g(x)dx − 2

0

g(x)dx = 2(G(y) − G(0)) − 2(G(x0 ) − G(0)),

0

= 2 (G(y) − G(x0 )) , y =2 g(x)dx. x0

Equation 3.31 then becomes U (x0 ) =

L

.

x0

2 2 xy e 0 V (y)

μ(x) V (x) dx

dy,

(3.32)

and this is the solution for .U (x0 ). To obtain .T (x0 ), we simply integrate .U (x0 ) as T (x0 ) =

x0

U (z)dz

.

0

and derive a general equation for MTE (.T (x0 )): T (x0 ) =

x0



L

.

0

z

2 2 zy e V (y)

M(x) V (x) dx

dydz.

(3.33)

Now we have all the tools required to properly consider, in the following sections, models of population dynamics.

3.4 A Model for Exponential Growth and Environmental Stochasticity Lande and Orzack [6] investigated the following model of population dynamics with exponential growth: .

dx = rx + x × “noise”, dt

where x is the population size, r is the growth rate, and “noise” represents random movement in the population size. Since this movement is proportional to the population size (x), noise is interpreted as a form of environmental stochasticity. In the discrete-time random-walk model, the random movement was described by

3.4 A Model for Exponential Growth and Environmental Stochasticity

67

W . In the stochastic model above, we describe the “noise” as .σe ξe (t), and .ξe (t) is white noise whose intensity is .σe . Dividing both sides of Eq. 3.33 by x, we have

.

.

1 dx = r + σe ξe (t), x dt

and since .d log x/dx = 1/x, this may be rewritten as .

d log x = r + σe ξe (t). dt

If we now take .z = log x and discretize the model, it equals Eq. 3.7, which we encountered in considering the random walk, and the random process is termed a geometric Wiener process or geometric Brownian motion. This model is widely studied in economics as a model for predicting stock prices.

White Noize Random noise is often termed “white noise,” which is noise that has no correlation with any other noises but itself. In the random walk example, we can think of it as a form of movement that has no tendency to go right after going right. “White” is named after the “color” of light that contains all the colors in the electromagnetic spectrum and thus is correlated to none of them, whereas we see red when the wavelength is long (low frequency) and blue or purple when the wavelength is short (high frequency). When all the frequencies are equally included in the light, the “color” is white, as in sun light. Similarly, there are noise “colors”; for example, a noise of lower frequency is called red noise, and one of higher frequency blue noise. There is also Brown noise, but this “Brown” is a person’s name (named after Dr. Robert Brown) and the noise is not brown.

The probability that the logarithm of the number of individuals (x) is .z = log x at time t is written as .p(z, t). Now, we consider that the population goes extinct when .z = 0(x = 1). In the model, the average movement of z is r, and the intensity of the random movement is .σe2 . If, then, we set .M = r and .V = σe2 in the Fokker–Planck equation (Eq. 3.14), we have .

∂p σe2 ∂ 2 p ∂p = −r + . ∂t ∂z 2 ∂z2

(3.34)

68

3 Population Models of Extinction t=20

r =1

0.08

p(z,t)

t=40

0.06

t=80

0.10

0.04

t=9

t=160 0.05

0.02

t=17 100

50

r =-1

t=5 0.15

z

150

200

t=13

2

250

4

6

8

z

10

12

14

Fig. 3.7 Examples of .p(z, t) changing over time. In the left panel, .r = 1, and in the right, .r = −1. In both panels, the initial value of z is 10. In the left panel, as the population growth rate is positive, the population size rapidly increases and the chance of extinction is thus very low, which results in .p(z, t) maintaining a Gaussian bell-type shape. The mean value of z corresponds to t (mean z values at times 20, 40, and 80 are 30, 50, and 90, respectively). In the right panel, on the other hand, since the growth rate is negative, ultimately leading to extinction, .p(z, t) does not have a bell shape, and the area under the curve becomes less as time passes

Although it is rare to find an analytical solution to the Fokker–Planck equation in general, the solution to the equation has been found in this case and is p(z, t) =

.





2π σe2 t e

−(z−z0 −rt)2 2σe2 t

−e

−

2rz0 (z+z0 −rt)2 − σe2 2σe2 t

 .

(3.35)

Figure 3.7 shows some examples of .p(x, t). Let us now write .Pz0 (t) for the probability that the population level is sustained (i.e., the probability of not going extinct) at time t, for a population that began with .z0 individuals. The probability that the population size is z at time t is .p(z, t), and hence .Pz0 (t) is a summation of .p(z, t) for .z > 0; that is, Pz0 (t) =

∞

p(z, t)dz

.

0

  



2rz − 20 1 − rt z −z0 − rt 0 = − e σe 1 − erf  , 1 − erf  2 2σe2 t 2σe2 t

(3.36)

where 2 erf(y) = √ π



y

.

e−k dk, and 2

0

erf(∞) = −erf(−∞) = 1. What we want to know is the “ultimate” probability of population extinction where ultimate means the after a very long time. By arranging the first erf function in

3.4 A Model for Exponential Growth and Environmental Stochasticity

69

Eq. 3.36 as

−z0 − rt  .erf 2σe2 t



√ √  −z0 / t − r t  = erf . 2σe2

√ √ If we now take the limit of .t → ∞, since .−z0 / t → 0 and .−r t → −∞, we have erf (−∞) = −1,

.

and this is the same for the second erf function. Substituting these values into Eq. 3.36 yields Pz0 (∞) = 1 − e

.

−

2rz0 σe2

(3.37)

,

and this is the probability that the population persists. The probability of extinction 2 2 is .1 − Pz0 (∞), which is .e−2rz0 /σe . When .r ≤ 0, .e−2rz0 /σe ≥ 1, implying that the probability of ultimate extinction is 1. Figure 3.8 shows examples of .Pz0 (t) for some values of .σe2 with .r = 0.1. When the elapsed time is short, .Pz0 (t) decreases rapidly (extinction occurs easily), the change in .Pz0 (t) becomes more gradual as time passes, and the value of .Pz0 (t) asymptotes to Eq. 3.37. In the initial phase of population growth, as the population size is small, the extinction probability is high; however, a population that survives this initial, high-risk phase may reach a considerable size, making the risk of extinction negligibly small. The growth rate of the model in this section does not change, regardless of how much the population size increases. In reality, the population growth rate is a function of the population size, and in many cases the growth rate also decreases as the size increases. We can interpret that the model for an initial phase of population growth, when the population size is small. We may think that the model lacks Fig. 3.8 Examples of .Pz0 (t). The numbers in the panel are 2 .σe values. The parameters are .r = 0.1, z = log(5) (the logarithm base is e)

1.0 0.1 0.8

Pz0(t)

0.6

0.5

0.4

1 2

0.2

3 0

20

40

60

Time

80

100

70

3 Population Models of Extinction

Fig. 3.9 .Pz0 (5) values as a function of .z0 = log(x0 ) and 2 2 .σe .(r = 0.1) . When .σe is 1, .Pz0 (5) = 0.9 when .z0 = 3.30. As .z0 = log(x0 ), the value of .x0 at .z0 is 26.83. You thus require roughly 30 individuals to make the above restoration project a success

7 0.4 0.5 0.6

5

σe

0.7 0.8

2

0.9

3

1 2

4

6

10

8

z0=log(x0)

applicability to realistic situations; however, it still offers some useful insights, for example, in the case of the problem discussed in the box below.

Required Initial Number of Individuals Imagine that you are involved in a project to restore a species that once lived here but is now extinct. How many individuals would you need, in order to reduce the risk of extinction within 5 years to less than 10%?4

Figure 3.9 shows the probability of the population persisting at .t = 5 in Eq. 3.36. Since we want the extinction probability to be less than 10%, we must find a combination of .z0 and .σe2 that results in .Pz0 (5) > 0.9. In the next section, we will look at the canonical model designed by Hakoyama and Iwasa [3], which is the main focus of this chapter and will require us to apply all the knowledge gained thus far, as it incorporates both demographic and environmental stochasticity, as well as carrying capacity (K).

3.5 Logistic Equation with Environmental and Demographic Stochasticity Hakoyama and Iwasa [3] investigated the following model: .

 √ dx x = rx 1 − + σe ξe (t) ◦ x + σd ξd (t) • x. dt K

(3.38)

4 You must find the combinations of .σ and .z that satisfy .P (5) > 0.9. The answer may be found e 0 z0 in Fig. 3.9.

3.5 Logistic Equation with Environmental and Demographic Stochasticity

71

Unlike in the population growth model discussed in Sect. 3.4, the population size is not allowed to infinitely increase, due to the carrying capacity (K). The model incorporates two types of stochasticity: environmental stochasticity, expressed in terms of .ξe (t) and .σe , which, respectively, correspond to the white noise and the intensity of the stochasticity, and demographic stochasticity, expressed in terms of the white noise .ξd (t) and intensity .σd . The environmental stochasticity is proportional to the population size (x), and the demographic stochasticity is √ proportional to the square root of the population size (. x), to have a greater impact when the population size is small and less as the population size increases. Equation 3.38 is called a stochastic differential equation. The second term on the right side has the symbol “.◦,” and the third term the symbol “.•.” These represent different formulations of stochastic differential equations, with the former known as the Stratonovich integral and the latter as the Itô integral. A comprehensive understanding of the difference between these two formulations is difficult and beyond the scope of this work; however, there are numerous works on stochastic differential equations (e.g., [4, 8]), and those who wish to gain this understanding are encouraged to consult them. In the literature review I conducted, the most comprehensive was the commentary by Dr. Lewis Smith5 [9]. A brief sketch of the essential difference may be found in the box below.

Itô vs. Stratonovich Consider the integral of a differential equation with white noise (.ξ(t)), which consists solely of discontinuous random movement: .

dx = a(x, t) + b(x, t)ξ(t). dt

Intuitively, the integral is a summation of many small changes in a given variable (see Eq. 3.6); thus, in this case, n 

xT = x0 +

.

n 

a(x, t)t +

t=0

b(x, t)Wt ,

t=0

where .Wt is the difference between .ξ(t) and .ξ(t + t). Continuous time refers to the limit where the time interval .t is zero, and in this case we write the summation as x(T ) = x(0) +

.

0

T

a(x, t)dt +

T

b(x, t)dW.

(3.39)

0

(continued)

5 https://oatml.cs.ox.ac.uk/blog/2022/03/22/ito-strat.html.

72

3 Population Models of Extinction

In the absence of random movement (.b(x, t)dW ), we can integrate a function by filling the function with rectangles and summing the areas of all the rectangles (see Fig. 3.1). However, in a process such as white noise, which is discontinuous throughout, the area of a rectangle cannot be determined in the same way as in a continuous process. Stratonovich and Itô formulated different integrals for such discontinuous random processes. Stratonovich considered movement which is random but has a correlation to time, which ceases to exist (i.e., becomes white noise motion) at the limit of .t → 0. Itô formulated the integral of a random process as the limit of .t → 0. Because the solutions of the integral differ in these two formulations, when the integral of the random process is written as

T

b(x, t)dW,

.

0

we must determine the proper interpretation of the integral in each case. In Stratonovich’s formulation, .b(x, t)dW is written as .b(x, t) ◦ dW , and in Itô’s as .b(x, t)•dW , with the following relationship between the two formulations: .

0

T

1 ∂b(x, t) b(x, t)dt + 2 ∂x



T

b(x, t) • dW =

0

T

b(x, t) ◦ dW.

(3.40)

0

Let us now consider a simple case where .a(x, t) = rx and .b(x, t) = σ x. If we interpret the process in Stratonovich terms, we have

T

x(T ) = x(0) + rxdt + 0 This equation is rewritten by Itô as



T

.

x(T ) = x(0) +

.

0

T



σ x ◦ dW.

0

 T 1 2 r + σ xdt + σ x • dW. 2 0

These equations imply that r in Stratonovich is equal to .r + σ 2 /2 in Itô. The results of the integrals differ in the two formulations, and confusion sometimes arises as to which interpretation we should adopt [1]. The problem, however, is not which interpretation to use, but rather that, once we decide which interpretation to use, we must determine the appropriate parameters, based on this decision [9].

Hakoyama and Iwasa [3] interpret environmental stochasticity as a timecorrelated process. However, the length of the time correlation can be considered to imply uncorrelated white noise if it is much longer than the generation time of

Population Size

3.5 Logistic Equation with Environmental and Demographic Stochasticity

73

500 400 300 200 100 100

200

300

400

500

Time Fig. 3.10 Example of time series data in the canonical model, where .r = 0.1, σe2 = 0.1, σd2 = 1, K = 100, and the initial value of x is K. The population goes extinct at time 559

the organisms in question. This interpretation is similar to that of the Stratonovich formulation, which is why the Stratonovich integral is used for environmental stochasticity. Let us now investigate the dynamic behavior of the canonical model. Because both the Stratonovich and Itô integrals are included in the model, we convert the Stratonovich integral to the Itô integral as .

 √ x 1 2 dx = rx 1 − + σe x + σe ξe (t) • x + σd ξd (t) • x. dt K 2

(3.41)

If we wish to solve the differential equation numerically, we discretize the system as     xt  1 2 + σe xt t + σe2 xt2 + σd xt ξ .xt+1 = xt + rxt 1 − (3.42) K 2 and solve the equation step by step, where .ξ is a random variable with 0 as √ the mean and . t as the standard deviation.6 An example of the trajectory of population dynamics is shown in Fig. 3.10, where the population goes extinct at time 559. The carrying capacity in the example is .K = 100, but the population size sometimes exceeds 500. Figure 3.10 represents one run in the simulation of Eq. 3.42. If we repeat the run independently many times, we obtain a distribution of the time that the population goes extinct (Fig. 3.11), and we can compute MTE using the frequency distribution. If we consider it enough to find MTE simply based on simulations, then we do not

 √ have .σe xt ξ + xξ = σe2 x 2 + xξ . This may require an explanation. Assume that there are two independent random variables, .X1 , X2 of which means are zero and the standard deviations, .σ1 and .σ2 , respectively. The sum of the two variables is also a random variable with 

6 We

zero mean and .σ12 + σ22 variance, and hence the standard deviation of this variable is . σ12 + σ22 .

74

3 Population Models of Extinction

Frequency

15,000 mean time: 563.81

10,000

5,000

max. time: 5895

0

2,000

4,000

6,000

Time at extinction Fig. 3.11 Histogram of the time to extinction based on numerical simulations of Eq. 3.42 with 100,000 iterations. The initial number of x is .K = 100. The maximum time is 5895.82, and the minimum time, 7.08; the 2.5% quantile is 36.30, and the 97.5%, 2016.25. The mean time to extinction is 563.8. The mean time to extinction by Eq. 3.32 is 563.244

need the analytical result for MTE. We, however, have already calculated MTE, and we will take a closer look at its analytical solution in the next section.

3.5.1 Mean Time to Extinction We have already derived a general solution for MTE (Eq. 3.33):

x0

T (x0 ) = 2



L

e2

.

0

z

y

M(x) z V (x) dx

V (y)

dydz.

In the case of the canonical model, the average movement is  x 1 2 + σe x, M(x) = rx 1 − K 2

.

(3.43)

and the variance in the random movement is V (x) = σe2 x 2 + σd2 x.

.

We first solve e2

.

y

M(x) z V (x) dx

.

(3.44)

3.5 Logistic Equation with Environmental and Demographic Stochasticity

75

If we substitute Eqs. 3.43 and 3.43 for the exponent on e, we have .

2

y

M(x) dx z V (x)   y rx 1 − Kx + 12 σe2 x dx, =2 σe2 x 2 + σd2 x z =

=

=

−2rσe2 (y − z) + (Kσe4 + 2r(σd2 + Kσe2 ))(log[σd2 + σe2 y] − log[σd2 + σe2 z]) , Kσe4 −2r(y − z) + log Kσe2

−2r (y − z) + log Kσe2

 Kσe +2r(σd +Kσe ) 4

σd2 + σe2 y σd2 + σe2 z

σd2 /σe2 + y σd2 /σe2 + z

2

2

Kσe4



 2r Kσe2

K+

, σd2 σe2

 +1

and if we define that .

R=

2r , Kσe2

D=

σd2 , σe2

then we have

  D + y R(K+D)+1 M(x) dx = −R(y − z) + log V (x) D+z

y

2

.

z

and substituting this back on e, we have e2

.

y

M(x) z V (x) dx

=e

 R(K+D)+1 −R(y−z)+log D+y D+z  log

D+y

,

R(K+D)+1

D+z = e−R(y−z) e , R(K+D)+1  D+y = e−R(y−z) . D+z

Now, the variance in the random movement is   2 2 2 = σe2 y(D + y). .V (y) = σe y y + σd /σe

(3.45)

(3.46)

76

3 Population Models of Extinction

Incorporating Eqs. 3.45 and 3.46 into the general equation for MTE (Eq. 3.33) yields

x0

T (x0 ) = 2



L

e−R(y−z)

.

0



D+y D+z

R(K+D)+1 (3.47)

dydz.

σe2 y(y + D)

z

Setting the upper bound (L) of the integral interval at infinity and replacing z by x yields Eq. 3 in Hakoyama and Iwasa [3].

The Numerical Solution for Eq. 3.47 Equation 3.47 can only be solved numerically. Since .T (x0 ) is a function of integrating .U (x0 ), the technique of first obtaining .U (x0 ) numerically and then integrating .U (x0 ) to obtain .T (x0 ) may be easier. As we saw in Eq. 3.31, .U (x0 ) is a product of two terms (.u1 (x0 ) and .u2 (x0 )). In the case of the canonical model, .u2 (x0 ) (Eq. 3.29) is u2 (x0 ) = e−2

x0 0

.

=e

M(x) V (x) dx

,

(1 + x0 /D)−(R(D+K)+1) .

Rx0

(3.48)

The integral of .u1 (x0 ) in Eq. 3.30 does not start at zero, hence we modify the equation so that the integral starts from zero:

L

u1 (x0 ) = 2

e2

y

M(x) 0 V (x) dx

.



L

=2

dy,

V (y)

x0

e2

y

M(x) 0 V (x) dx

V (y)

0



x0

dy − 2 0

e2

y

M(x) 0 V (x) dx

V (y)

dy.

(3.49)

Now, as the computer cannot work with an infinite L value, we must find a value of L large enough to satisfy e2

L

.

0

M(x) V (x) dx

V (L)

≈ 0.

In the case of the canonical model, the first term of Eq. 3.49 is

L

2

.

0

e2

y

M(x) 0 V (x) dx

V (y)



L

dy = 2 0

e−Ry (1 + y/D)(R(D+K)+1) dy. σe2 y(y + D)

(3.50)

3.5 Logistic Equation with Environmental and Demographic Stochasticity

U(x0)

Time

77

T(x0)

140

600

120

500

100

400

80

563.244

300

60 200

40

100

20 40

20

60

x0

80

100

120

20

40

60

x0

80

100

120

Fig. 3.12 .U (x0 ). (left panel) and .T (x0 ).(right panel). .U (x0 ). represents the slope of .T (x0 ). The mean time to extinction at .x0 = 100 is 563.2 (indicated by the dotted line). The mean time to extinction obtained by numerical simulation iterations (Fig. 3.11) is 563.8. The parameters are the same as in Fig. 3.10

If we, then, let .y be the width of the division of y, the integration is expressible in terms of the following summation:

L

2

.

0

e−Ry (1 + y/D)(R(D+K)+1) dy σe2 y(y + D)

≈2

N  e−R(iy) (1 + (iy)/D)(R(D+K)+1) i=1

σe2 (iy)((iy) + D)

y,

where .N = L/y. The second term in Eq. 3.49 is obtained during the integration of Eq. 3.50. Now two components, .u1 (x0 ) and .u2 (x0 ), are known. Multiplying these two components yields .U (x0 ), and by integrating (or summing) x from 0 to .x0 , we obtain the mean time to extinction .T (x0 ) in this case. Examples of .T (x0 ) and .U (x0 ) are shown in Fig. 3.12.

3.5.2 Time Evolution of the Probability Density and the Probability of Extinction If we want to know the time evolution of the probability density of the population size, we use the Fokker–Planck equation (Eq. 3.3). In the case of the canonical

78

3 Population Models of Extinction 1.0 t=1

0.010

Density

Prob. Persistence

0.012

0.008

t=10

0.006 0.004

t=100

0.002

t=1,000

50

100

150

200

250

0.8 0.6 0.4 0.2 0.0

300

500

x

1000

1500

2000

2500

3000

Time

Fig. 3.13 The left panel shows the probability density at various times, and the right panel shows the probability of population persistence as a function of time (i.e., the area under the curves in the left panel, see Eq. 3.52)

model, the equation is .

      ∂ x 1 ∂p 1 ∂ 2  2 2 2 =− rx 1 − + σe x p + σ x + σ x p , e d ∂t ∂x K 2 2 ∂x 2  1  4rx + K(σe2 − 2r) p = 2K   ∂p 1  2rx 2 + K 2σd2 − 2rx + 3σe2 x + 2K ∂x   2 ∂ p 1 . (3.51) + x σd2 + σe2 x 2 ∂x 2

Solving Eq. 3.51 numerically yields Fig. 3.13.7 While the elapsed time is short, the population distribution (.p(x, t)) is centralized around the initial value (.x0 = 100). As time passes, however, the peak of the distribution moves to the left, and the shape of the distribution flattens and widens (left panel in Fig. 3.13). The widening of the distribution over time reflects the case of the random walk, but in this latter case, the area under the distribution is not constant at one. This area indicates the probability of population persistence, and one minus this area indicates the probability of extinction. Let us, then, write the probability that the population is persistent at time t as .Px0 (t), whose value is computed by integrating .p(x, t) for all values of x greater than 0, or Px0 (t) =

∞

p(x, t)dx.

.

(3.52)

0

7 Partial differential equations are much more difficult to solve numerically than ordinary differential equations; however, there are several excellent guides, my favorite being Garcia [2].

References

79

This must be computed numerically (as there is no analytical solution), and the result is shown in the right panel in Fig. 3.13. Given a positive growth rate (r), the ultimate extinction probability in the exponential growth model (Fig. 3.7) did not increase over time. In contrast, in the logistic model, the probability of population persistence (.Px0 (t)) continues to decrease over time, even though the population growth rate is positive (.r > 0). This is because, in the exponential growth model, the size of the population is allowed to grow as large as possible, whereas in the logistic model, there is always a force driving the population size back to K, so the population size cannot grow so large that the risk of extinction is negligible. In other words, as long as there is a carrying capacity, no species can be sure to escape extinction. In the next chapter, we will consider some case studies of ecological risk assessment using the canonical model.

References 1. Braumann CA (2007) Itô versus Stratonovich calculus in random population growth. Math Biosci 206:81–107 2. Garcia AL (2015) Numerical methods for physics, 2nd edn. Addison-Wesley, Boston 3. Hakoyama H, Iwasa Y (2000) Extinction risk of a density-dependent population estimated from a time series of population size. J Theor Biol 204:337–359 4. Karlin S, Taylor HM (1981) A second course in stochastic processes. Academic Press, San Diego 5. Kusuda T, Iwasa Y (eds) (2002) Ecology and simulation. Asakura Shoten, Tokyo 6. Lande R, Orzack SH (1988) Extinction dynamics of age-structured populations in a fluctuating environment. Proc Natl Acad Sci U S A 85:7418–7421 7. MacArthur RH, Wilson EO (2001) The theory of island biogeography. Princeton landmarks in biology. Princeton University Press, Princeton 8. Öksendal B (2003) Stochastic differential equations. Springer, Berlin 9. Smith L (2018) Itô and Stratonovich; a guide for the perplexed. https://oatml.cs.ox.ac.uk/blog/ 2022/03/22/ito-strat.html

Chapter 4

Population-Level Assessment Using the Canonical Model

Abstract In the previous chapter, we derived the mean time to extinction (MTE). In this chapter, we will add chemical effects to this model in considering two case studies. First, we will discuss the parameter estimation method for this model, some of which will recall the discussion in Chap. 2. Then, using the model, we will investigate the effects of DDT among a herring gull population, introducing the concept of risk equivalence and exploring how this concept can be used to equivalently compare the impacts of different factors, such as chemical exposure and habitat destruction. The second case study investigates the effects of nonylphenol on Medaka populations, and here we will consider how chemical impact assessments are related to the impact assessments of conservation biology, such as those for the Red Database. This capability of relating chemical impacts to indicators developed elsewhere is the ultimate goal of this chapter. Keywords Canonical model · Mean extinction time (MTE) · Herring gull · Medaka · DDT · Nonylphenol

4.1 The Canonical Model and the Adverse Effect of Chemicals In this chapter, we consider examples of risk assessment using the canonical model. We begin by assuming that certain chemicals reduce the population size by rate .α, which means that the canonical model can be rewritten as .

 √ x dx = rx 1 − + σe ξe (t) ◦ x + σd ξd (t) • x − αx, dt K

(4.1)

where x is the population size, r and K are the intrinsic growth rate and carrying capacity, respectively, .ξ2 (t))and .ξd (t) are white noise factors representing environmental and demographic stochasticity, respectively, and .σe and .σd are their

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_4

81

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4 Population-Level Assessment Using the Canonical Model

respective intensities. When we want to solve Eq. (4.1) using simulations, we use the following equation:     xt  1 2 + σe − α t + σe2 xt2 + σd xt ξ, xt+1 = xt + xt r 1 − K 2

.

(4.2)

where .ξ is a random variable with mean zero and variance of .t. If, in the course of the simulations, .xt < 0, then t is the time that the population goes extinct. By repeating the simulation many times, we obtain the distribution of the extinction time and can thereby compute the mean time to extinction (MTE). We can rewrite Eq. (4.1) as   √ x dx = (r − α)x 1 − + σe ξe (t) ◦ x + σd ξd (t) • x, . dt K(1 − α/r) and if we define that r  = r − α,

.

K  = K(1 − α/r),

(4.3)

then the model returns to the original canonical model. The point is that if the population size decreases in proportion to .α, it has the effect of reducing both the growth rate and the carrying capacity. As we saw in the previous chapter, MTE is computed by

T (x0 ) =

.

2 σe2



x0

0



L

e

−R(y−z)



 D+y R(K+D)+1 D+z

y(y + D)

z

dydz,

(4.4)

where .

R=

2r  2r = , K  σe2 Kσe2

D=

σd2 . σe2

Among other things, this chapter will investigate the ecological effects of chemicals using MTE. First, however, we will discuss how we estimate the parameters of the model.

4.2 Parameter Estimation

83

4.2 Parameter Estimation We first consider the case where there are no adverse effects of the given chemical substances (i.e., .α = 0). In this context, we often encounter time series data similar to that in Fig. 4.1. Specifically, in chemical risk assessment, we often see a pattern similar to that in Fig. 4.1a as the toxicity data for algae, but rarely for other species. Although data in Fig. 4.1b can be found in field population surveys for specific species, the toxic effect of the chemicals on these species are often unknown. The data in Fig. 4.1a are valuable for estimating the population growth rate, but we are rarely lucky enough to possess data useful for parameter estimation, while at the same time knowing the toxicity of the given chemicals for the species in question. As a result, a large part of the work in this area involves the search for data that can be used in the assessment. Moreover, chemical effects are often investigated for a limited number of biological species, and there is a lack of data regarding the effects on diverse species. Overall, a major difficulty in conducting population-level assessment is that the relevant data must be obtained from a variety of sources and fitted together in a patchwork fashion.

4.2.1 Intrinsic Rate of Reproduction As we saw in Chap. 2, if we have a life-history matrix, the intrinsic population growth rate (r) is obtained by computing the maximum eigenvalue of the matrix. In addition, if we have time series data similar to that in Fig. 4.1a, we can estimate it by fitting the model, x(t) = x0 ert ,

(4.5)

.

Population Size

to the data. If the data is sparse, the accuracy of the fitting might not be as reliable. If there are only two data items, for example, with one the initial value and the other

(a)

(b)

Time

Time

Fig. 4.1 Typical examples of time series data for chemical risk assessment: (a) data for the initial phase of population increase (exponential growth phase) (b) data for a phase after having reached the carrying capacity (K), with fluctuation around this level

84

4 Population-Level Assessment Using the Canonical Model

twice this initial value (i.e., if the data is such that the time required to double the population size can be calculated), we may estimate the growth rate using the model, 2x0 = x0 erT ,

.

(4.6)

where T is the population-doubling time and .x0 is the initial population size.

4.2.2 Carrying Capacity and Environmental Stochasticity Intensity If we have data as in Fig. 4.1b, we can estimate K and .σe (and r) from the data. K is estimated as K = E[x],

.

where .E[x] is the mean of x, and x represents data items. If we describe the variance in the data as .V [x], the following relationship holds among the parameters: CV 2 = V [x]/E[x]2 = σe2 /(2r),

.

(4.7)

where CV is the coefficient of variation in the time series data. If we already know r, we can estimate .σe ; but even when r is unknown, we need not be disappointed because the data in Fig. 4.1b contain information not only about .σe but about r. We need, then, to compute the autocovariance (AC) of the data, as AC(τ ) = Cov[x(t), x(t + τ )],

.

= Ce−r|τ | , where .τ is a time lag and C is a constant .(σe2 K 2 + K)/(2r). By fitting the model to the observation data, C and r are estimated, and .σe is estimated from the value of C (for details, see Hakoyama and Iwasa [4]). While a separate discussion would be required to consider how much time series data is sufficient to make the estimated values sufficiently reliable, it is encouraging to note that, in principle, armed with the data in Fig. 4.1b, all the relevant parameters can be estimated. I say “all the relevant parameters,” though one might be concerned that the demographic stochasticity intensity has not been estimated. The reason is that this parameter need not, in fact, be estimated because .σd becomes one if the time unit in the model is the generation time of the species. When we estimate the parameters, the time unit is often a year, and we must convert the parameters in order to obtain the generation time as the time unit. Given, then, a life-history matrix with 1 year as the time unit, the generation time of a population that has reached a stable

4.3 The Effect of DDT Exposure on a Herring Gull Population

85

distribution (see Chap. 2) can be calculated as follows [8]: amax −(a+1)r f (a)l(a) a=0 ae , .Tg = a max −(a+1)r f (a)l(a) a=0 e

(4.8)

where a represents age, .f (a) is the fecundity at age a, and .l(a) is the survivability from age 0 to age a (the product of the annual survivability from age 0 to age a). Once we have obtained the generation time, all the parameters are converted simply by multiplying them by the generation time. For example, a growth rate with a time unit of 1 year (.ry ) is converted to one with a unit of generation time by r = r y Tg ,

.

(4.9)

and .σe2 = Tg (σe2 )y , where .(σe2 )y is the environmental stochasticity intensity with the time unit of 1 year. We will recall this discussion in our consideration of case studies in the coming sections.

4.3 The Effect of DDT Exposure on a Herring Gull Population As a first example, we consider the case studies by Nakamaru et al. [6], of DDT (p,p’-dichlorodiphenyltrichloroethane) exposure among herring gulls.

4.3.1 Parameters for Natural (DDT-Free) Population Nakamaru et al. first conducted a literature survey for the life-history parameters of a natural (i.e., DDT-free) population of herring gulls. The age-specific survivabilities were found in Kadlec and Drury [5]. A population-doubling time of 15 years was also reported in Kadlec and Drury, and the annual population growth rate was computed to be 0.0462 (Eq. (4.6)). Pimm et al. [7] reported that the CV for natural bird populations lies between 0.2 and 0.8, and the minimum CV value was adopted as the CV of the natural herring gull population. The age-specific fecundity was found in Chabrzyk and Coulson [3], and the generation time was computed to be 8.05, based on the survivabilities in Kadlec and Drury and the fecundities in Chabrzyk and Coulson. These parameters are summarized in Table 4.1. Only a few lines are required to summarize this information, but a great deal of time was required to obtain it. It is no exaggeration to say that 99% of the time in Nakamaru et al.’s study was spent in performing and analyzing the results of this literature survey.

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4 Population-Level Assessment Using the Canonical Model

Table 4.1 Parameters for the canonical model used in Nakamaru et al. [6]

Parameter .Tg .ry r CV 2 .σe

Value 8.05 0.0462 0.372 0.2 0.0298

Notes From Eq. (4.8) Growth rate per year Growth rate per generation Minimum CV by Pimm et al. [7] From Eq. (4.7) with .r = 0.372.

Table 4.2 The effects of DDT on bird survivorship. The environmental DDT concentrations were derived using a biomagnification factor of .2.38 × 105 . Survivability is defined 3 weeks after hatching, and the relative survivability is the survivability divided by the survivability at 0 ppm DDE (egg ppm) 0 46 144

DDT (environment ppm) 0 0.000193 0.000605

Relative fertility 1 0.605 0.24

Reproduction rate 0.372 −0.0236 −0.550

4.3.2 The Effects of DDT on a Herring Gull Population The adverse effects of DDT on a herring gull population were determined by a study [2] of the survivorship from embryonated eggs to 3 weeks post hatch. It reported that the survivorship with exposure to 0 ppm, 46 ppm, and 144 ppm of DDT was 38%, 23%, and 9%, respectively. The relative ratios of survivorship were 1 ( = 38%/38%), 0.61 ( = 23%/38%), and 0.24 (9%/38%), respectively, and the respective population growth rates at these concentrations were 0.372, .−0.0236, and .−0.550. These DDT concentrations in eggs were determined through biomagnification, and the biomagnification factor (BMF; here, .2.38 × 105 ) was estimated based on the relationship between the environmental DDT concentration of 0.00005 ppm and 11.9 ppm in the body. The concentrations are summarized in Table 4.2. Performing linear regression analysis on these results (with the fixed intercept at 0.372) yields r(DDT) = −1.57 × 103 DDT + 0.372.

.

(4.10)

Based on Eq. (4.3), the relationship between the DDT concentration and the adverse effects (.α) is α(DDT) = 1.57 × 103 DDT.

.

(4.11)

In Chap. 2, we derived the threshold concentration of chemicals leading to zero population growth rate (i.e., the population threshold concentration or PTC). In the case of the bird population, the threshold concentration of environmental DDT satisfies the following equation: r(DDT) = −1.57 × 103 DDT + 0.372 = 0,

.

4.3 The Effect of DDT Exposure on a Herring Gull Population

87

yielding a PTC of .0.237 × 10−3 ppm. Using this concentration as a target for chemical management is thus reasonable, and if the environmental concentrations can always be controlled below this value, the population will not become extinct. In reality, however, the environment is not always constant; even if the environmental concentrations are controlled below the PTC on average, and growth rates are kept positive on average, the stochastic nature of the environment could result in negative growth rates and the unlucky extinction of the population. The models presented in Chap. 2 cannot address this issue because they only consider average values. When there is environmental and demographic stochasticity, we need models that reflect the role of such stochasticity in possible species extinction. Using such models, we can better address such critical questions as, for example, how many years will it take for a population with a zero growth rate due to chemicals to go extinct? Knowing such times to extinction is important in environmental management because the longer the time, the more time we have to take preventive measures. Another issue here is that the PTC value derived in Chap. 2 only considers the effect on growth. It seems reasonable to think that extinction is unlikely to occur in a population where r is small but K is very large. Even assuming that the same level of excess mortality is caused by a given chemical, then, the effect on the population should differ for populations with large and small K values. Thus, the effect of carrying capacity on the probability of population extinction must also be considered, since the extinction of a population is related not only to the population growth rate but also to the magnitude of K. Furthermore, the carrying capacity and population growth rate do not affect the extinction risk independently, but rather in relation to each other, and hence we must simultaneously take into account the effect of both in our management. For example, we may encounter the following problems. Suppose we have an area contaminated by a chemical. The environmental concentration of the chemical is above the PTC, and hence, if no measures are taken, the population will go extinct. To reduce the environmental concentration of the chemical, we need a treatment plant. Assume that a larger treatment plant that uses more land can more effectively remove such chemicals, and thus that building such a plant would enable more efficient chemical removal. If, then, we simply want to remove the chemical, we can simply build a plant large enough to certainly reduce the concentration of the chemical. However, such a large plant would require more land and lead to habitat loss, and if the resulting habitat loss is so great that the population would go extinct, it is clear that such a plant should not be built. Thus, quantitative discussion is required, regarding what size of plant is appropriate to reduce the risk of extinction, and the canonical model can provide a solution to this problem.

Fig. 4.2 Mean time to extinction. The unit of time is generation time. The initial population size (.x0 ) equals K, and the vertical axis is on a logarithmic scale. The mean time increases almost exponentially as the carrying capacity (K) increases

4 Population-Level Assessment Using the Canonical Model

9

10

7

MTE

88

10

9

10

3

10

20

40

60

80

100

K (= x0)

4.3.3 Mean Time to Extinction (MTE) Using the parameters in Table 4.1, MTE with a carrying capacity of .K = 100 (implying 100 breeding females), the initial condition .x0 = K, and no DDT exposure is .2.32 × 109 generations, which corresponds to .1.87 × 1010 years, longer than the age of the Earth. Thus, the chance that a population of this bird species, with 100 breeding females, will go extinct is extremely low within the time frame we are aware of. Some sample mean time to extinction values with varying carrying capacity (and initial population size) are shown in Fig. 4.2. As K increases, the mean time to extinction increases (note that the vertical axis in Fig. 4.2 is on a logarithmic scale). It is worth noting that the term “mean time to extinction (MTE)” tends to suggest a 50% probability of extinction at the time, but this is not true. Figure 4.3 shows the probability distribution of the population size for MTE at .K = x0 = 20, obtained by solving the Fokker–Planck equation in Chap. 3. MTE is 1418 generations or 11,415 years. The sum of the probability distribution is 0.365, implying that the probability that the population has already gone extinct is 63.5% at MTE (the probability that the population goes extinct is somewhat greater than 50%). The lower 5th percentile of the distribution (the short vertical dotted line to the left) is .x = 7.6, and the 95th percentile is .x = 29.9 (the short vertical dotted line to the right), implying that in a population with .K = 20 and 20 initial breeding females, the population size is between 7.6 and 29.9 with a probability of 90%, if the population persists after 11,415 years. Figure 4.4 shows the time course of the probability of population persistence, as computed in two different ways: one (solid line in Fig. 4.4) is the sum of the distribution derived by the Fokker–Planck equation, as we saw in Fig. 4.3, and the other (dots in Fig. 4.4) derived by numerical simulation based on Eq. (4.2). In the latter approach, we stop the simulation when the population size becomes less than 0 and record the time at which the simulation was stopped. By repeating this procedure many times, we can derive the relationship between the time and the probability of population persistence. As the figure shows (and somewhat to my

4.3 The Effect of DDT Exposure on a Herring Gull Population average = 18.3

Probability

Fig. 4.3 Probability densities at 1418 generations in time. The average population size of the distribution is 18.3

89

0.0020 0.0015 7.2

29.2

0.0010 0.0005

Fig. 4.4 Probability that the population still persists at time t. The solid line is computed by solving the Fokker–Planck equation (Fig. 4.3) and the dots by numerical simulation based on Eq. (4.2)

Prob. of persistence

10

20

x

30

40

50

1.0 0.8 0.6 0.4 0.2 200

400

600

800 1000 1200 1400

Time surprise), the respective results are in almost perfect agreement, strongly suggesting that the results are correct.

4.3.4 Effect of DDT on MTE Figure 4.5 shows the relationship between MTE and the DDT concentration, when K = 50 and .x0 = K. The vertical axis is on a logarithmic scale, and MTE decreases almost exponentially as the DDT concentration increases. When we plot MTE as a function of the carrying capacity as reduced by DDT exposure (the numbers in Fig. 4.5 panel, here on the horizontal axis), we obtain the black dots shown in Fig. 4.6. The gray dots in the figure represent MTE at the same carrying capacity but in the absence of DDT exposure. The greater values of the gray dots indicate that, even with the same carrying capacity, a population affected by DDT will go extinct more quickly, and the reason is simple: DDT affects not only on the carrying capacity but also on the growth rate (Eq. (4.3)), implying that, even if we observe the same number of individuals in habitats with and without DDT (the magnitude of K being typically estimated based on a field survey), a population suffering from DDT exposure will have a shorter MTE.

.

Fig. 4.5 Environmental DDT concentrations (as parts per billion) and MTE (as generations). The carrying capacity (K) is 50 at no DDT in the environment. Small numbers in the panel show the carrying capacity computed by Eq. (4.3)

4 Population-Level Assessment Using the Canonical Model 6

10

50

5

10

45 40

MTE

90

4

10

35 30

3

10

25 20

2

10

0.00

0.05

0.10

15

10

0.15

0.20

40

50

DDT (ppb) 6

10

5

10

MTE

Fig. 4.6 MTE with stable habitat size (carrying capacity: K). With the same K value, the gray dots show MTE without DDT exposure and the black dots with DDT exposure. Since DDT exposure reduces both K and r (Eq. (4.3)), MTE is even less than if K simply decreased

4

10

3

10

2

10

10

20

30

Carrying capaity (K) The dotted horizontal line in Fig. 4.6, which marks MTE at 1000 generations, crosses the gray dot curve at roughly .K = 20 and the black dot curve at .K = 30. MTE result at .K = 30 is achieved when the environmental DDT concentration is 0.0948 ppm. Assume, then, that there is a population of the birds in a given habitat, with 50 breeding females (i.e., the carrying capacity is 50). As Fig. 4.6 shows, MTE of this population is about 106 generations, but with 0.0948 ppm DDT exposure, this drops to 1000 generations. As a 60% reduction in habitat (.K = 50 to .K = 20) also results in MTE of 1000 generations, 0.0948 ppm DDT exposure and 60% habitat loss have equal adverse effects on the bird population. Recalling, then, the question of whether to build a DDT treatment plant or accept adverse DDT effects (Sect. 4.3.2), this example shows that the respective risks presented by two qualitatively different effects (in this case, chemical exposure and habitat loss) can indeed be compared if MTE is utilized as the axis of assessment. This type of risk equivalence was devised by Nakamaru et al. [6]. In the next section, we will consider some examples.

4.4 Risk Equivalence

91

4.4 Risk Equivalence We will write MTE, then, as   T r, K, σe2 ,

.

where the growth rate is r, the environmental stochasticity intensity .σe2 , and the carrying capacity K, continue to use the DDT-affected bird population as an example (parameters in Table 4.1), and solve a few hypothetical problems, utilizing this notion of risk equivalence.

4.4.1 DDT Exposure vs. Habitat Loss Imagine, then, that a herring gull habitat is accidentally polluted by DDT, and the environmental DDT concentration becomes .0.118 × 10−3 ppm, which reduces, by precisely half, the original values of r and K; and we are required to compensate for the adverse effects of the DDT. What would be a reasonable form of compensation? MTE in the absence of any adverse effect is   T r = 0.372, K = 25, σe2 = 0.0298 = 4691.54.

.

(4.12)

Now, with DDT exposure which halves K and r (.α = 1/2r), MTE becomes T (r − α = 1/2r, K(1 − α/r) = 1/2K, σ 2 ) = 80.28.

.

We now wish to translate the adverse effect of the DDT into a habitat loss of x% and thus need to find the value of x that satisfies T (r, 0.01(100 − x)K, σ 2 ) = 80.28.

.

This value (.x = 65.3) implies that the DDT exposure is equivalent to the loss of 65.3% of the original habitat of .K = 25. Figure 4.7a illustrates the percentage of habitat loss for other concentrations of DDT. One answer to the question above, then, is “We pay what amounts to a punitive fee for the use of the (now negatively impacted) land (0.653 K) taken from the herring gulls.” This solution obviously does not contribute to the preservation of the gulls but has the advantage that the amount of the negative impact can be quantified by the amount of habitat loss. Nonetheless, some problems remain, such as who do we pay (the government?) and how do we determine the price of the land? The other way to compensate for the effect of chemical exposure is to expand the habitat to the original MTE value. In this case, we must determine the carrying

92

4 Population-Level Assessment Using the Canonical Model

% habitat loss

α=1/2r

α=1/2r

4000

80

3000

60

K’

40

2000

20

1000

0.05

0.10

0.15

0.20

0.05

0.10

0.15

0.20

DDT ppb Fig. 4.7 Risk equivalents between DDT concentrations and K. The maximum DDT concentration results in a 90% reduction in r (i.e., max .α = 0.9r). The left panel shows the percentage of habitat loss, with the vertical dotted line indicating the DDT concentration that results in half effect (.α = 1/2r). The corresponding percentage of habitat loss is about 60% (more than half the habitat destroyed). The right panel shows the habitat size (.K  ) that results in the same MTE without DDT exposure. Given the original habitat size of .K = 25, at the maximum DDT concentration, the compensatory habitat size must be more than 100 times greater (.K  > 2500) than the original for MTE to be the same

capacity (.K  ) that satisfies T (r − α, K  (1 − α/r), σ 2 ) = 4691.54,

.

which is .K  = 103.7, compared to the original carrying capacity of .K = 25. Thus, to ensure that the habitat expansion would result in the same MTE as the original (so that the DDT effect would be fully counteracted), more than four times the original habitat would have to be dedicated to this bird population. It is not a simple matter of doubling the habitat because the chemical has reduced the population by half. Figure 4.7b illustrates the carrying capacities required to compensate for varying DDT effects.

4.4.2 DDT Exposure vs. Environmental Stochasticity Global warming is thought to increase weather extremes. This implies greater intensity in environmental stochasticity (.σe2 ). As an example, we assume that the environmental stochasticity intensity doubles (i.e., .2σe2 ). With this intensity of environmental stochasticity, we have   T r, K, 2σe2 = 1990.62.

.

4.4 Risk Equivalence

93

We first compute the value of .α that results in the same MTE, which satisfies   T r − α, K(1 − α/r), σe2 = 1990.62.

.

This value is .α = 0.0283, and the DDT concentration resulting in this value can be determined (by Eq. (4.11)) to be 0.018 ppm. Thus, those who cannot accept the adverse effects of this DDT concentration should also not accept global warming that would double the intensity of environmental stochasticity. We can now, of course, also compute the corresponding habitat loss by finding the value of x that satisfies T (r, 0.01(100 − x)K, σ 2 ) = 1990.62.

.

This is .x = 14.4. Doubling the intensity of environmental stochasticity (from 0.0298 to 0.0596) is thus equivalent in impact to destroying 14.4% of the bird’s habitat.

4.4.3 How Long Do We Have for Habitat Restoration? A different but related topic to risk equivalency is the question of the time allowed for habitat restoration. In other words, what is the time allowed for cleaning up a contaminated area and returning it to its original state, before the population goes extinct. This is one of the most important pieces of information for environmental management. As we have seen, if the adverse effects of DDT (.α) equal .1/2r, then MTE is 80.28 generations, which corresponds to 646.25 years. When .α equals r, the population growth rate becomes 0, and MTE becomes 20.05 generations (about 160 years). In Chap. 2, the assessment endpoint was defined as the chemical concentration at which the population growth rate becomes zero. In the case of the bird population in this section, even after reaching the zero growth rate, it would take about 160 years on average for the population to go extinct. Thus, even if the growth rate becomes zero, it does not mean that extinction would occur immediately. If the restoration can be completed during this time, extinction of the population can be avoided, and 160 years is probably sufficient. However, it may well be unwise to consider the mean extinction time as the time allowed us for restoration because, as we saw in Fig. 4.4, the probability that a population has gone extinct by the time it reaches the mean extinction time is quite high. Thus, it may, for example, be more appropriate to think of the time allowed as the time until the extinction probability reaches 5%. This can be computed by the Fokker–Planck equation (Eq. (3.3)) or the numerical simulation based on Eq. (4.2), as we did in Fig. 4.4. In the case of the bird population here, the probability of population persistence is 0.96 at 8 generation time and 0.94 at 9 generation time.

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4 Population-Level Assessment Using the Canonical Model

In this case, then, to avoid population extinction, we must take successful measures within 64–72 years. In the next section, we will consider a further example: a case study of nonylphenol (NP) exposure in a Medaka fish population.

4.5 Effect of Nonylphenol Exposure on Medaka Fish Nonylphenol (NP) is a chemical produced mainly as material for the surfactant, nonylphenol ethoxylate, which degrades in the environment to NP. NP is highly toxic to aquatic organisms and also known to have endocrine disrupting effects. Watanabe et al. [10] developed the Medaka extended one-generation reproduction test (MEOGRT), using NP as the toxicant. Utilizing their results, we can consider a case study on the effects of NP on a Medaka population.

4.5.1 Medaka Life History The Medaka (Oryzias latipes), a Japanese killifish, is a model species for toxicity testing, and its life-history parameters in laboratory breeding systems have been well studied. The average life longevity is 12–16 months, and the fish start laying eggs 4 months after fertilization. Information about wild Medaka is somewhat sparse, but it is known that the fish does not have generation overlap (generation time .∼1 year), and the breeding period is limited to the hot season from June to September. Although the Medaka has high fertility, it is classified as a vulnerable species in Japan’s Ministry of the Environment Red List. Alonzo et al. [1] constructed a life-history matrix for the Medaka, based on 2month intervals, up to 20 months of age: ⎛

0 ⎜ 0.7 ⎜ ⎜ 0 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 .M = ⎜ ⎜ 0 ⎜ ⎜ 0 ⎜ ⎜ 0 ⎜ ⎝ 0 0

0 0 0.9 0 0 0 0 0 0 0

175 0 0 0.99 0 0 0 0 0 0

350 0 0 0 0.99 0 0 0 0 0

350 0 0 0 0 0.98 0 0 0 0

350 0 0 0 0 0 0.95 0 0 0

350 0 0 0 0 0 0 0.89 0 0

350 0 0 0 0 0 0 0 0.75 0

350 0 0 0 0 0 0 0 0 0.47

⎞ 350 0 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ ⎟ 0 ⎟ ⎟. 0 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ 0 ⎠ 0

Now, based on this matrix, the Medaka population growth rate is 5.42; that is, the population size becomes 5.42 times greater every 2 months. After 5 years, then, the

4.5 Effect of Nonylphenol Exposure on Medaka Fish

95

Medaka population is 1.05.×1022 , somewhat less than the Avogadro constant. After 6 years, the population is 2.05.×1026 , which is equal to the number of stars in the universe. The life-history parameters in Alonzo et al. [1] are likely based on the good conditions in the laboratory, where year-long reproduction is possible. Wild Medaka, however, does not reproduce in the cold season. In addition to the matrix above, then, we prepare another matrix (.N), whose elements for reproduction terms are 0. We assume that the starting point is April. As the Medaka does not reproduce in April, we use matrix .N without reproduction for a projection and thereby compute the June population to be x(t + 1) = Nx(t),

.

where .x is the population size of each age class, or .x = {x1 , x2 , ..., x20 }T . In June, the Medaka starts laying eggs, and the population size at the next time step (August) is x(t + 2) = Mx(t + 1) = MNx(t + 1).

.

Continuing this process until April of the following year, we obtain a matrix for the annual population dynamics: My = N3 M2 N,

.

(4.13)

and calculating the population growth rate based on .My yields an annual growth rate (.λ) of 420. This value is probably higher than the actual growth rate in nature because it is likely that the survivability of wild Medaka in winter is higher; however, as we do not know how much the survivability varies in reality, we must, for the present purposes, ignore this factor and adopt the population growth rate of 420. This growth rate is converted to an intrinsic growth rate as r = log(420) = 6.04.

.

(4.14)

As the carrying capacity (K) and environmental stochasticity intensity of the Medaka population are also unknown, the values will be specified in the respective assessment scenarios below.

4.5.2 NP Toxicity for the Medaka Fish Watanabe et al. [10] exposed Medaka to NP at 6 different concentrations: control (= 0), 1.27, 2.95, 9.81, 27.8, and 89.4 .μg/L. Their exposure tests included three generational groups: F0 (4 weeks exposure of adult fish), F1 (15 weeks exposure (in

4 Population-Level Assessment Using the Canonical Model

0.8 0.0

0.2

10

0.6

20

0.4

30

Survival

Number of fertilized eggs (eggs/day/pair)

1.0

96

0 1

2

5

10

20

50 100

(Concentration + 1) (µg/L)

1

2

5

10

20

50 100

(Concentration + 1) (µg/L)

Fig. 4.8 The effect of NP exposure on the number of fertilized eggs (right panel) and survivability (left panel). In both panels, the gray dots are values reported by Watanabe et al. [10], the solid line is the best-estimate curve, and the dashed lines show the 95% confidence interval. Estimated parameters: .b = 0.917, d = 27.2, e (number of fertilized eggs) = 35.9, .α = 3.20, and .β (survivability). = −6.567. The parameter estimation and figure drawing were performed by my colleague, Dr. Yuichi Iwasaki, for which I am very grateful

total) from egg to adult), and F2 (4 weeks exposure from egg to juvenile), and they measured the adverse effects on various traits. Of these traits, we will here utilize the results for the number of fertilized eggs and the survival rate of the F1 generation, to estimate the effect of NP exposure on population growth. To obtain a dose–response curve for the number of fertilized eggs (.f (C)), a 3parameter Weibull-1 model is used: f (C) = d exp{− exp[b log(C + 1) − e]},

.

where .d, b, and e are parameters and C is a concentration of NP. For a curve for the survival (.s(C)), a logistic curve is s(C) =

.

1 , 1 + e(−[α+β log(C+1)])

where .α and .β are parameters to be estimated. The results are shown in Fig. 4.8. Incorporating these dose–response curves into the Medaka matrix (.My ), we obtain the population growth rates (.λ) and intrinsic growth rate (r) at each NP concentration (Table 4.3). Even at the highest dose, the population growth rate is still larger than 1, and hence the Medaka population can be maintained at the highest dose unless there is stochastic fluctuation. Since the carrying capacity and the intensity of the environmental stochasticity (.σe2 ) are unknown, we consider a scenario that assumes a site with an environmental carrying capacity of 1000 and MTE of 1000 years (which corresponds to 1000

4.5 Effect of Nonylphenol Exposure on Medaka Fish Table 4.3 Effect of NP exposure on the Medaka population growth rate (.λ) and intrinsic growth rate (.r = log λ)

97 Concentration (.μg/L) 0 (control) 1.27 2.95 9.81 27.8 89.4

Table 4.4 MTE with NP exposure

Concentration (.μg/L) 0 (control) 1.27 2.95 9.81 27.8 89.4

.λ

420 358 303 182 69.3 6.20

r 6.04 5.88 5.71 5.20 4.24 1.82

MTE (year) 1000 851 722 440 178 26.9

K 1000 854 696 400 146 17.5

Medaka generation because the average life time of wild Medaka is 1 year). The value of (.σe2 ) satisfying the scenario in the absence of NP can be obtained by solving   T 6.04, 1000, σe2 = 1000,

.

which yields .σe2 = 13.4. The calculated MTEs with NP exposure, using the growth rate in Table 4.3, are shown in Table 4.4. At the highest dose, then, MTE is about 30 years, only 3% of MTE without NP exposure. Many would consider such the great reduction in MTE is an unacceptable risk level; however, if we persist on the PTC concept discussed in Section 2, since the population growth rate is still greater than 1, it is concluded the risk level is acceptable. Thus, as the assessment method changes, the acceptable risk level also changes. Because we do not know which assessment method is the best, this result again suggests that it may well be unwise to assess the risk based on a single index alone; we must develop various indexes for ecological risk assessment. Currently, I have no clear sense of what level of risk we should accept in this example, but when one finds oneself at such a loss, the ideas developed by others may prove of use. One such idea is the notion of maximum sustainable yield, developed in conservation biology for fisheries. The theory here suggests that if the catch pressure (.α) is set at half the intrinsic growth rate of the fish population, the population is sustainable and the amount of catch maximized. If we apply the theory to the case of Medaka NP exposure, an acceptable level lies between 2.95 and 9.81 .μg/L.

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Another strategy might employ the criteria in the Red List of the International Union for Conservation of Nature and Natural Resources (IUCN). For example, if we want to avoid the Medaka being categorized as Critically Endangered, we consult Criteria A: 1. The rate of reduction is measured either over a 10 year span or across three different generations within that species. 2. The cause for this decline must also be known. 3. If the reasons for population reduction no longer occur and can be reversed, the population needs to have been reduced by at least 90% 4. If not, then the population needs to have been reduced by at least 80% As the population reduction at 27.8 NP .μg/L is about 85%, we can thus avoid the Medaka being categorized as Critically Endangered if the environmental NP concentration is less than 27.8 .μg/L In the USA, the acceptable environmental level of NP is 6.6 .μg/L in freshwater and 1.7 .μg/L in saltwater [9] and in Japan 0.7–1 .μg/L. Thus, the level of effect that we aim for in ecological management based on environmental criteria is far below that which would result in the species being listed as endangered, and once again we find very different targets, depending on the unit of assessment and are reminded of the importance of basing our discussion on a variety of units rather than relying on any one.

4.6 Conclusion This concludes our consideration of population-level ecological risk assessment, which has, I hope, provided the reader with a basic understanding of such assessment. The models we have dealt with here are relatively simple but may be extended and nuanced as the reader sees fit; I have merely attempted to provide essential knowledge of the model construction and analysis methods involved. I believe that the concept of risk equivalence outlined by Nakamaru et al. [6] is highly important, not only in ecological risk assessment but also in risk assessment in general. For all the fine work hitherto done in this field, we still have a very limited understanding of the absolute amount of risk, except for extreme risks like zero risk or a time to extinction of 1 day. For example, assume that MTE of a given species is 1000 years, or the probability of extinction at this moment is 1/1000. If we are asked whether this risk is acceptable or not, most of us would be unsure how to answer, and even those who felt confident to answer would most likely be more or less consciously comparing this risk to some other with which they are familiar (e.g., it is acceptable because it is sufficiently longer than an average human life). Integrating various effects into one indicator is of great benefit in risk assessment. Conservation ecologists, for example, are highly skilled at assessing the effects of species habitat loss but are often not as well equipped to assess the adverse effects of chemical substances and vice versa for ecotoxicologists. However, a single indicator

References

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that integrates these two effects can greatly facilitate mutual understanding, and the concept of risk equivalence has the potential to enhance communication between different disciplines. Nonetheless, in this chapter we focused solely on MTE; the development of other (especially such “hybrid”) indicators is an important task for risk assessment.

References 1. Alonzo F, Hertel-Aas T, Real A, Lance E, Garcia-Sanchez L, Bradshaw C, Battle JV, Oughton DH, Garnier-Laplace J (2016) Population modelling to compare chronic external radiotoxicity between individual and population endpoints in four taxonomic groups. J Environ Radioact 152:46–59 2. Beyer WN, Heinz GH, Redmon-Norwood AW (1996) Environmental contaminants in wildlife–interpreting tissue concentrations. SETAC Special Publication Series. SETAC, Lewis, Boca Raton 3. Chabrzyk G, Coulson JC (1976) Survival and recruitment in the herring gull Larus argentatus. J Anim Ecol 45:187–203 4. Hakoyama H, Iwasa Y (2000) Extinction risk of a density-dependent population estimated from a time series of population size. J Theor Biol 204:337–359 5. Kadlec JA, Drury WH (1968) Structure of the new England herring gull population. Ecology 49:644–676 6. Nakamaru M, Iwasa Y, Nakanishi J (2002) Extinction risk to herring gull populations from DDT exposure. Environ Toxicol Chem 21(1):195–202 7. Pimm SL, Jones HL, Diamond J (1988) On the risk of extinction. Am. Nat. 132:757–785 8. Staerk J, Conde DA, Ronget V, Lemaitre JF, Colchero F (2019) Performance of generation time approximations for extinction risk assessments. J Appl Ecol 56:1436–1446 9. United States Environmental Protection Agency Office of Water (2005) Aquatic life ambient water quality criteria—nonylphenol final. epa-822-r-05-005 10. Watanabe H, Horie Y, Takanobu H, Koshio M, Flynn K, Iguchi T, Tatarazako N (2017) Medaka extended one-generation reproduction test evaluating 4-nonylphenol. Environ Toxicol Chem 36:3254–3266

Part II

Models for Ecotoxicology

Chapter 5

Species Sensitivity Distribution in Ecological Risk Assessment

Abstract Species sensitivity distributions (SSDs) are statistical distributions of toxicity values for biological species. They are used in ecological risk assessment to derive predicted no effect concentrations (PNECs) or other indices of ecological risk. Risk assessment using SSDs is effective because it enables the use of statistical knowledge that has been developed over hundreds of years. However, there are a number of criticisms and concerns regarding their usage, and they have yet to be fully exploited in risk assessment practice. A particular concern is that we rarely have abundant toxicity data, and sensitivity distributions estimated from limited data have a low accuracy confidence level. Unfortunately, in many cases, this concern is poorly defined, and there has been little discussion of how low the confidence level actually is, in quantitative terms. However, even if the level is low, as long as we quantitatively evaluate how low it is, we can ensure reliability by using assessment factors (AFs), which are used to compensate for insufficiency in the quantities required to achieve a certain goal. In order to define the magnitude of a given AF, this insufficiency must be clearly defined, and thus far there has been a lack of discussion regarding such definition. In this chapter, we will discuss how to define the magnitude of AFs in ecological risk assessment using SSDs and how this magnitude varies with sample size and other factors. Keywords Species sensitivity distribution · Assessment factor · Uncertainty · Sample size · Small sample

5.1 Introduction Species sensitivity distributions (SSDs) have a long history. A seminal work on the subject was published in 2001 [11], and numerous previous studies had already been conducted. However, when I began my research on ecological risk assessment in the early 2000s, SSDs were still rarely used for risk assessments by public organizations and even now are not a primary method for such assessments. The toxicity of chemical substances differs among species. The statistical distributions of such hazards are called species sensitivity distributions. Statistics © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_5

103

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are ubiquitously employed in quantitative analysis employed worldwide, and most are expressed in terms of distributions when overall trends are sought. Thus, it is unusual to encounter a sample survey that does not incorporate some form of statistical distribution, making it all the more strange that SSDs are so little employed in environmental management. Most SSD studies investigate how such distributions should be practically applied in risk assessment. When we aim to use a new methodology, we typically seek to understand its advantages and disadvantages in comparison with conventional methodologies, and if the former outweigh the latter, we adopt the new method. Such comparative studies are important, but rare, in the case of SSDs. The purpose of this first half of the chapter is to illustrate how such comparisons can be made. In the latter half, we will discuss the extent to which we can trust the results of SSD-based assessments. In statistical approaches, parameters are estimated from observed data. Such estimation has intrinsic error, and this error is an indicator of the reliability, or trustworthiness, of the assessment results. Research regarding such error was conducted in the late 1980s–2000s, but little has been conducted since then, and one of our aims here is to understand how we can determine the accuracy of SSD-based assessments.

5.2 Basic SSD Elements SSDs are toxicity value distributions for selected organisms. Typically, in SSDbased risk assessment, the toxicity values are first log-transformed and then employed in the analysis. In this chapter, we mainly assume normal distributions for the SSDs, partly because such distributions are familiar and partly because they have been studied in detail, which offers us the benefit of exploiting the knowledge gained from past studies. When we use SSDs for risk assessment, it is enough to simply calculate the mean and variance, though the data to be included in the SSD analysis must be carefully selected. Those who have experience in conventional hazard assessment, where predicted no effect concentrations (PNECs) are predicted based on no observed effect concentrations (hereafter termed the minimum NOEC approach), know that we need a reliability assessment to determine whether a given toxicity test meets certain criteria. The question of what data should be used for SSD analysis is frequently raised, but I believe that any data that has previously been used in conventional assessments is worth employing. There is a wide variety of ecotoxicity data. Some tests, for example, use death as the endpoint, while others use growth, and each has different units for ecological impact; thus, it is unreasonable to fit the samples with different units in a single statistical distribution, and this is why we conducted population-level assessments that integrate the respective impacts of these different units as an intrinsic rate of increase in Chap. 2. However, when we consider a continuity from methods based on the minimum NOEC approach, toxicity values based on different endpoints must be considered to have the same

5.2 Basic SSD Elements Table 5.1 Example of toxicity values. The data are randomly generated from the standard normal distribution (.N (0, 1)), and the endpoints are arbitrarily selected. All are assumed to be chronic data. The logarithm base is 10

105 Species Algae 1 Algae 2 Algae 3 Crustacean 1 Crustacean 2 Crustacean 3 Crustacean 4 Fish 1 Fish 2

NOEC 0.24 0.21 0.16 0.44 2.30 0.88 0.079 0.051 0.050

log(NOEC) −0.62 −0.69 −0.80 −0.35 0.36 −0.05 −1.10 −1.29 −1.31

Endpoint Growth Growth Growth Immobilization Immobilization Death Death Death Hatchability

units because, in the minimum NOEC approach, we must choose the minimum NOEC to determine the PNEC, and we cannot determine the minimum value based on different units. Thus, if we decide that such data cannot be used in SSD analysis, due to such incompatible units, we simply cannot determine the minimum NOEC in the minimum NOEC approach. As aforementioned, there are a number of criticisms of SSDs, but some of these also serve as criticisms of the existing methods. We will discuss the criticisms in the next section; here, we will consider some examples of SSD-based assessment of ecotoxicity data (Table 5.1). The data were randomly determined and the endpoints arbitrarily selected. Because we are dealing with a log-normal distribution, the mean and standard deviation are calculated by log(NOEC) values. They are .

μ ˆ = −0.65 σˆ = 0.56.

We often imagine a bell-shaped function for normal distributions. The bell-shaped function is a probability density distribution (left panel in Fig. 5.1), and SSDs are

PNEC

log(NOEC)

log(NOEC)

Fig. 5.1 An example of an SSD. The left panel shows the SSD as a probability density function. SSDs are often expressed as in the right panel

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often described by a cumulative density distribution (right panel in Fig. 5.1). In the cumulative distribution, the x values are log(NOEC), and the y values are determined by .i/(n + 1), where n is a sample size and i is an ith NOEC sorted in ascending order. We often imagine a bell-shaped function for normal distributions. The bellshaped function is a probability density distribution (left panel in Fig. 5.1), and SSDs are often described in terms of a cumulative density distribution (right panel in Fig. 5.1). In such a distribution, the x values are log(NOEC), and the y values are determined by .i/(n + 1), where n is the sample size and i is the ith NOEC in ascending order [11]. In the minimum NOEC approach, the minimum log(NOEC) (e.g., .−1.31 for Fish 2 in Table 5.1) is divided by an assessment factor (AF) to derive the PNEC. As there are chronic data from three trophic levels in the example shown in Fig. 5.1, then, according to the Chemical Substances Control Law of Japan, the AF is 10 (see Chap. 1), the log(PNEC) is determined to be .

log(PNEC) = log(0.050/10) = −1.31 − 1 = −2.31,

and the PNEC is 0.0050 (dashed line in the left panel in Fig. 5.1). If the predicted environmental concentration (PEC) is higher than the PNEC, there is a risk, and if it is lower, there is no risk. This minimum NOEC approach, then, is a “yes or no” type of risk assessment. In the minimum NOEC approach, there is strictly no concept of “distribution,” but the approach implicitly assumes a step function such as that shown by the dashed lines in the right panel of Fig. 5.1. The yes or no type of risk assessment is also possible in the SSD method. The quickest way is to determine a cutoff point and compare this point with the PEC. The cutoff point is usually the concentration that affects 5% of the species or protects 95% of the species, and this concentration is often referred to as the 5% species hazard concentration (HC5). Depending on the amount and quality of the data, HC5 is further divided by an AF from 2 to 5 [6]. As will be discussed later, the SSD method’s use of an AF in this manner violates a quantitative risk assessment. SSDs can also be utilized in probabilistic risk assessment, especially when the respective PEC is expressed in the form of a distribution. If we denote the distribution of environmental concentration as ECD(x) and that of toxic values as SSD(x) (where x is a concentration), the probability that the species is affected by the chemical is  ∞ .risk = ECD(x)SSD(x)dx. −∞

This equation calculates the overlapping area of the two distributions; as in Fig. 5.2 for example, where the S-shaped functions is the SSD, the bell-shaped function is the PEC distribution, and the shaded area indicates the overall probability for the affected species. The respective means of the PEC distributions in the right and left panels are the same, but the standard deviations are different. The shaded area is

5.3 Criticisms of SSDs

log(Concentration)

107

log(Concentration)

Fig. 5.2 Examples of probabilistic risk assessment distributions. The bell-shaped functions are the environmental concentration distributions, and the S-shaped functions are the SSDs. The gray area indicates the overall magnitude of risk. A normal distribution was assumed for the environmental concentration. The environmental concentration distributions have the same mean, but different standard deviations. Greater variation (left panel) indicates greater risk, though the average behavior is the same

larger in the left panel than in the right panel, indicating that the standard deviation (an indicator of data variation) plays a role in determining the magnitude of risk.

5.3 Criticisms of SSDs As the examples in Figs. 5.1 and 5.2 suggest, we can use SSDs in various ways, and they offer a flexible tool for ecological risk assessment. However, the following are typical criticisms of the use of SSDs in this context. 1 A comparatively large amount of toxicity data is required. 2 Why HC5 (why not HC1 or HC10)? Specifically, what is the justification for the 5% rate of species affected? 3 Why use a normal distribution for the population? (“Population” here refers to the distribution from which the toxicity data are sampled.) 4 The samples (i.e., species used in toxicity tests) are biased. 5 Population-level effects and species interactions are ignored (i.e., there is a lack of ecological perspective). (“Population” here means a group of a biological species.) Whenever I see these criticisms, I despair at the immaturity of ecological risk assessment methods. First, it should be noted that the same criticisms that are raised against the SSD method are equally applicable to the minimum NOEC approach. I shall now take them one by one. Criticism 1 is often raised to highlight the main point of superiority of the minimum NOEC approach. In this approach, it is said, we can do a risk assessment when we have only one data item, whereas, in the SSD method, we need to estimate

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the mean and standard deviation and thus need at least two data items, and SSDs that are estimated based on such sparse data are inherently unreliable. I think, however, that the unreliability due to the sparse data is also true for the minimum NOEC approach, and hence we need a large assessment factor (AF) in such the cases. Moreover, in this approach, it is not possible to quantify what percentage of the ecological risk is improved by applying the AF, whereas, in the case of the SSD method, we can quantify the percentage of improvement by exploiting the statistical knowledge gained over the last several hundred years. We will later discuss how we quantify the reliability of SSDs. Criticism 2 is often presented as an interrelated set of concerns, with one focus being the following question: what if ecologically important species (such as keystone species) are among the 5% of species that are not protected. I agree that such the keystone species may be among the most sensitive 5% of species; however, if we are concerned about the existence of such sensitive keystone species, we should also be concerned about the possibility that the susceptibility of such species is even less than that of the organisms we have studied thus far in the minimum NOEC approach. In other words, if the SSD method is to be rejected because it ignores the existence of such the keystone species, then the minimum NOEC approach must also be rejected. Why 5% is a valid question; however, we cannot identify which approach is better unless we are certain of the number of species that are protected in the minimum NOEC approach. We will later discuss the quantitative nature of the minimum NOEC approach. Criticism 3 makes absolutely no sense. We have no way to identify the functional form of the population in general. There may be someone who think that we can determine the probability models using statistical tests such as the Kolmogorov– Smirnov test, but these tests can only show that the given data are not sampled from a certain distribution (e.g., a normal distribution), and they cannot confirm that the population has this distribution. There may also be someone who think we can determine the probability model by model selection based on, for example, the Akaike information criterion (AIC). In such selection, we are simply seeking the model that best explains the data we have now and will have in the future; we are not necessarily selecting the correct model (see the last chapter in this book for more details). Furthermore, it seems to me that selecting a different model for each substance is not a wise choice. For example, assume that we want to compare the respective risks presented by substances A and B. If we assume a normal distribution for the SSD of substance A, and a logistic distribution for B, then we cannot accurately assess the comparative risks involved because the substances and/or distribution shapes are different. Criticism 4 is an important point. I agree that the composition of the species used in toxicity testing is different from the species composition in nature. I have, however, similar concerns regarding the minimum NOEC approach, as it employs similarly biased data and thus only produces biased results. There is simply no way to know which method is more reliable, in this respect, without rigorous quantitative comparison.

5.3 Criticisms of SSDs

109

Criticism 5 is out of the question. If assessment methods that do not consider population-level effects and species interactions are not good, then the minimum NOEC approach is not good at all. I would myself like to add a further criticism. 6. As we saw in the sensitivity analysis in Chap. 2 of the matrix model, the contribution of unit change to the population growth rate is different for each life history parameter. The respective NOECs used to define the reproduction rate and survivability have different meanings, and thus comparing these NOECs in terms of absolute values is not meaningful. This criticism focuses on a problem in current risk assessment, first noted by Stark et al. [14]. The principle that “this NOEC is less than that NOEC, so we’ll use this value” seemed strange to me when I first began doing risk assessment. Some may think that acute toxicity values, such as LC50 and EC50, are comparable between species because the exposure time and endpoints are the same, but these toxicity values are in fact not comparable because the contribution of the respective life-history traits differ among species. To address this problem, I proposed an SSD based on the population threshold concentrations (PTCs) that result in a zero population growth rate (Chap. 2). Such PTCs share a common factor in the sense of species extinction and are thus comparable, meaning that we can interpret the related SSD as a type of species diversity curve. In that study, I took the PTC as the endpoint, but other endpoints, such as 50% reduction in the population growth rate (based on the MSY theory), mean time to extinction, and extinction risk within 100 years, are also possible. I do not, however, think that the current risk assessment using NOECs is completely meaningless and should be rejected. Unfortunately, we have yet to develop a single, all-purpose, and fully accurate assessment method. However, the minimum NOEC and SSD methods are not mutually exclusive. The important thing, I think, is not to be restricted to any one approach, to the exclusion of the others, but instead to create a framework that will make the best use of the information provided by all such approaches. In order to do so, we must be able to make decisions such as “this is better in this case” or “this is better in other cases.” In the next section, then, we will focus on assessment accuracy and investigate whether the minimum NOEC or SSD approach is more robust in ecological risk assessment.

Catastrophic Shift and Goal of Protection I noted above that, at the population level, the SSD forms a kind of biodiversity curve. The distribution is a function of monotonic increase (as it should be, since we assume a normal distribution). If, however, we consider species interaction, the relationship between ecological stress and ecosystem state is not monotonic, but a discontinuous function with bistability under some (continued)

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conditions [12] (Fig. B1). For example, such bistability has been found [4] in CASM [3] studies, and it can even be observed in much simpler ecosystems with two-species interaction [10].

Reduction in population size (%)

100

LB

80 60 HB

40 20

Low

Stress

High

Fig. B1 Example of catastrophic shift: a model used in May [10], for vegetation biomass that is stressed by herbivore grazing pressure. The solid curves represent stable equilibrium and the dashed curve unstable equilibrium. When the stress is greater than at point HB, the biomass that remains at the roughly 50% reduction rate undergoes a sharp decline to a state of almost 100% reduction. This is termed a catastrophic (or regime) shift. In order to enhance the reduced population size, the stress must be reduced to less than LB

In considering SSDs, which assume a monotonically increasing function, the debate arises as to what percentage of the control target should be specified (in most cases, HC5), but if the function has a discontinuity point such as in Fig. B1, it may be rather easy to set the management goal. There are two easy targets to come up with: the upper end (point LB) and lower end (point HB) of the bistable region. The idea of aiming for the upper end is to avoid the point where a catastrophic shift will occur. The idea of aiming at the lower end is to avoid the possibility of catastrophic shift itself. In the mesocosm test, a toxicity test using multiple species such a catastrophic shift may be observed, and if so, how do we determine the PNEC?

5.4 The Probability that We Achieve the Protection Goal Let us define that the goal is to protect x% of the species. If we aim to protect 95% of the species, then x is 95, and we aim to set PNEC lower than HC5. However, the HC5 here means the HC5 of the population (the distribution from which the data are sampled), not the HC5 estimated by the data, and as the population HC5 is usually unknown, we sometimes fail to set PNEC lower than HC5.

5.4 The Probability that We Achieve the Protection Goal

111

We first investigate, then, whether the conventional minimum NOEC approach or the SSD method has the greater probability of achieving the desired level of protection.

5.4.1 The Minimum NOEC Approach The minimum NOEC approach relies on the minimum NOEC for the assessment, but this value is simply the minimum thus far determined; there is no guarantee that it will continue to be the minimum in the future. The minimum NOEC method typically overlooks more sensitive species that may be found in the future. What we want to know is how risky it is. We consider the following assumptions: • There is a population for the species sensitivity to a given chemical. • There is no way to know the sensitivity of a species of any member of this population in advance. Given these assumptions, a toxicity test is equivalent to a random sample of the toxicity values among the population. These assumptions may not actually hold, but they are required in order to make the computation of failure probability possible. However, even if there is a sample bias, we can consider some bias correction method, as long as this bias is explicitly described. We assume, then, that the log-transformed toxicity values have a normal distribution. The protection level for the species is 95%, and hence HC5 is a protection level for concentrations. Even if the goal is protection of 99% or 90% of the species, we can conduct a similar computation to determine the failure probability. In the minimum NOEC approach, the minimum NOEC is divided by an assessment factor (AF) to derive the PNEC, and the magnitude of the AF depends on the amount of data but may also differ from country to country. In Japan, for example, the AF is 10 if the chronic toxicity values for all three trophic groups are available. We consider, then, the case where toxicity values for three trophic groups are known and set the AF at 10. We then define that we fail the management when the PNEC is set higher than the population HC5 (i.e., less than 95% of the species are protected). We compute the predicted no effect concentration (PNEC) as .

log PNEC = log NOECmin − 1

with AF = 10 (logAF = 1), where NOEC.min represents the minimum NOEC among the dataset. Now, we want to make the PNEC less than HC5. The condition is .

log PNEC < log HC5,

(5.1)

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which can be arranged as .

log NOECmin − 1 < log HC5, log NOECmin < log HC5 + 1.

(5.2)

We assume that the population mean and standard deviation, respectively, are 0 and σ . We write the cumulative density function of the distribution as .f (x; 0, σ 2 ), where x is a log-transformed concentration. The upper limit of the NOEC.min is given by Eq. (5.2) and the cumulative probability (p) at the upper limit is

.

p = f (log HC5 + 1; 0, σ 2 ).

.

The p is the probability that one randomly sampled NOEC will be less than logHC5+1 (i.e., the probability of successful). In the case of the standard normal distribution (i.e., .σ = 1), logHC5 .= −1.64 and the value of p is p = 0.26.

.

The probability of failure is .(1 − p). If there are n samples (toxicity data for n species), the probability that we fail for n data items is pn = (1 − p)n .

.

Fig. 5.3 Failure probability in the minimum NOEC approach. When the standard deviation of population distribution is low, there is no possibility of failure, but as it increases, the probability of failure increases. The n values on the right are the sample size

Failure probability

The probabilities for various sample sizes and standard deviations are shown in Fig. 5.3. When the sample size is larger (i.e., if we carry out the toxicity tests for a greater number of species), there is a lesser probability that we fail to set the PNEC lower than the protection level (HC5). The standard deviation of the population distribution also affects the probability of failure. We will discuss this at greater length later on. An important implication of this result is that the probability that we do not set the PNEC lower than HC5 is not negligible. For example, when the number of data

0.6

n=3

0.5

n=4

0.4

n=5 n=6

0.3

n=7 n=8 n=9 n=10

0.2 0.1 0.0

0.2

0.4

0.6

0.8

1.0

Standard deviation

1.2

1.4

5.4 The Probability that We Achieve the Protection Goal

113

items is 3 and the population standard deviation is 1.5, the failure probability is about 0.6. Thus, the situation where 5% of the species are not protected should be a frequent occurrence in the minimum NOEC approach.

5.4.2 The SSD Approach In the SSD approach, we first estimate mean and standard deviation from dataset, and then we derive the estimated HC5. As the first case, we adopt the estimated ˆ without AF as PNEC. If the PNEC is lower than HC5 of population HC5 (.HC5) distribution, then it is considered that we fail the management. Let us examine the failure probability here, through a computer experiment (Monte-Carlo simulation), whose procedure is as follows: • Generate n random numbers from the distribution for a population of which the mean is .μ and the standard deviation .σ . n is the number of tested species. • Calculate the mean (.μ) ˆ and standard deviation (.σˆ ) from the generated data. ˆ and compare this value to the population HC5. • Estimate .HC5 By repeating this process 100,000 times and counting the number of failures, we obtain the failure probability. The results, shown in Fig. 5.4, reveal that, in this case, the failure probability in the SSD method is never less than 50%, even when the sample size is large. The reason for this is simple. Consider Fig. 5.5, which shows estimated HC5 ˆ is distributed around the population HC5, no histograms, with no AF. Since .HC5 matter how much the sample size increases, it will never exceed 50%. This implies that we must apply some AF if the PNEC is to have a greater probability of being lower than the HC5. 100

Failure probability (%)

Fig. 5.4 Failure probability in the SSD approach, with no AF. Even if the sample size (n) is very large, the failure probability will never be less than 50%, and the results are roughly similar for all values of .μ and .σ

80 60 40

50%

20 4

6

8

10

100

Sample size (n)

1000

114

5 Species Sensitivity Distribution in Ecological Risk Assessment HC5 2.5 2.0 1.5 1.0 0.5 -6

-4

-2

0

2

log(Concentration) Fig. 5.5 Estimated HC5 histograms, with no AF (as PDF). The gray region represents the ˆ when the sample size is 3 and the black when the sample size is 1000. A standard observed .HC5 normal distribution is assumed for the population distribution. A vertical bar represents the HC5 of the population distribution. A larger sample size results in closer estimation of HC5, but roughly half the estimated values are less than the HC5 of population distribution

5.4.3 The t-Distribution Aldenberg and Jaworska [1] is one of the most notable papers on the magnitude of the assessment factor (AF) in the SSD method. In order to properly understand this study, however, we must first understand the t-distribution, which is the subject of this subsection. Suppose, then, that there are n samples (.X1 , X2 , . . . , Xn ) from a normal distribution whose mean is .μ and standard deviation .σ , .N(μ, σ 2 ). The estimated mean and standard deviations are 1 Xi , n n

.

μ ˆ =

i=1

1  (Xi − μ) ˆ 2. n−1 n

σˆ 2 =

(5.3)

i=1

If we express each data item as Z=

.

X−μ , σ

(5.4)

then Z is a random variable having the standard normal distribution, .N (0, 1). We write the sum of .Z 2 as Y =

.

 n   Xi − μ 2 σ i=1

(5.5)

5.4 The Probability that We Achieve the Protection Goal

115

and Y , then, has a .χ 2 distribution with n degrees of freedom. We arrange Y as

.

 n n   1  Xi − μ 2 = 2 (Xi − μ ˆ +μ ˆ − μ)2 , σ σ i=1

i=1

n n 2  1  2 (X − μ) ˆ + (Xi − μ)( ˆ μ ˆ − μ) i σ2 σ2

=

i=1

+ =

i=1

n 

1 σ2

(μ ˆ − μ)2 ,

i=1

n   i=1

ˆ Xi − μ σ

2



μ ˆ −μ +n σ

2 .

(5.6)

The left hand side (LHS) of Eq. (5.6) is Y and has .χ 2 distribution, and hence the right hand side (RHS) also has .χ 2 with n degrees of freedom. The first term in the RHS is a random variable having .χ 2 distribution with .(n − 1) degrees of freedom, and the second term is a .χ 2 distribution with 1 degree of freedom. We write the first term of the RHS in Eq. (5.6) as W =

.

 n   ˆ 2 Xi − μ . σ

(5.7)

i=1

If we divide both sides of Eq. (5.3) by .σ 2 , we have

.

 n  σˆ 2 ˆ 2 1  Xi − μ 1 = = W. n−1 σ n−1 σ2

(5.8)

i=1

This equation indicates that .σˆ 2 /σ 2 equals W divided by the degrees of freedom of W (i.e., .n − 1). Since W is a random variable with a .χ 2 distribution and with .(n − 1) degrees of freedom, .σˆ 2 /σ 2 also has a .χ 2 distribution and with .(n − 1) degrees of freedom. Suppose, then, that we repeat the estimation of the mean of random variables sampled n times from an arbitrary distribution of which the mean is .μ and the standard deviation .σ . The estimated mean (.μ) ˆ will differ with each estimation. The distribution of .μ ˆ is a normal distribution of which the mean is .μ and the standard √ deviation .σ/ n; that is, .

μ ˆ ∼ N(μ, σ 2 /n).

(5.9)

116

5 Species Sensitivity Distribution in Ecological Risk Assessment

When n is infinite, the normal distribution becomes .N(μ, 0), implying that, with very large n, the estimated mean approaches the population mean. This is the intuitive explanation of the central limit theory. The intuitive understanding is easy, but the proof is difficult (see, for example, Feller [7]). When we standardize .μ ˆ using Eq. (5.4), we have μ ˆ −μ Z= , σ 2 /n

.

(5.10)

where Z is a random variable with the standard normal distribution. From Eq. (5.8), we have σ 2 = (n − 1)σˆ 2 /W,

.

and by combining this equation with Eq. (5.10), we have .

μ ˆ −μ Z  =√ , 2 W/(n − 1) σˆ /n

(5.11)

√ where .Z/ W/(n − 1) is a variable with a t-distribution and .(n − 1) degrees of freedom. The normality of Z is derived by the noncentral limit theorem, and the assumption of a normal distribution for the population is not necessary; however, for W to have a .χ 2 distribution, the population must be normal, and hence we need to assume that the population is normal when we want to use t-distributions in some statistical analysis. There is great benefit in using a t-distribution in this regard; for example, it can be used to determine whether the estimated mean is significantly different from the population mean (i.e., a t-test). In SSD analysis, if we assume a normal distribution for the population, we can use the t-distribution to calculate various useful statistics. Thus, though we cannot determine the probability model of the population, I consider it useful to assume a normal distribution, in order to gain the benefits of the t-distribution.

A t-Test for Potato Chips Let us consider, then, an application of the t-distribution. You manage a potato chip company which sells potato chips in 100-g bags. You are responsible for making sure that the chip weight per bag is 100 g, but the weight per bag naturally varies somewhat, making it difficult to achieve the goal of producing (continued)

5.4 The Probability that We Achieve the Protection Goal Fig. 5.6 The potato company t-distribution, with its lower limit (.tl ) and higher limit (.th ), and with the shaded area representing 95% of the t-distribution. If .tpotato falls outside the shaded area, the average value is significantly different from 100; but in this case, as .tpotato = −1.21, the value is not significant

117

0.4 0.3

t(x)

0.2 0.1 0.0 -4

tpotato

tl -2

0

x

th 2

4

exactly 100-g bags. Thus, you seek to manage the chip quantity per bag so that the bag weights average 100 g. You have purchased and weighed 10 bags of your potato chips from 10 supermarkets near you, and the results are weight = .{89, 90, 105, 86, 107, 97, 100, 102, 91, 104} g,

with a mean of 97.1 g and a variance of 57.43. Is this mean value of 97.1 g, then, significantly different from 100 g?1

We can determine the significance using a t-test. By substituting the mean and standard variation into Eq. (5.11), we have 97.1 − 100 tpotate = √ = −1.21 57.43/10

.

and we compare this value to that obtained from the t-distribution. When the significance level is 5%, the lower 2.5%ile (.tl ) and upper 97.5%ile (.th ) of the tvalues of which the degree of freedom is .9(= 10 − 1) are2 .

− tl = th = 2.262.

The value of .tpotate is between .tl and .th (Fig. 5.6), implying that the average weight per bag of potato chips produced by your company is not significantly different from 100 g, and your company is true to its customers.

1

Not significant. See the main text for details. you are using R, type the command qt(0.025, df .= 9, ncp .= 0).

2 If

118

5 Species Sensitivity Distribution in Ecological Risk Assessment

Null Hypotheses of t-Test Null hypothesis is required for all hypothesis testing. In the case of the potato chips bag, the null hypotheses is .H0

: the average weight per bag of potato chips is 100 g.

In the example, the average value was not significantly different from 100 g. Which is the correct interpretation of the result?3 1. The average is 100 g. 2. The average is near, but we cannot know how near, to 100 g. Most people make the correct interpretation when the null hypothesis is rejected; however, many make the incorrect interpretation when the null hypothesis is not rejected (i.e., in the case of species toxicity, many think that the NOEC means that the concentration has no effect). Whenever we do hypothesis tests, in other words, we must keep in mind what the null hypothesis and understand the proper meaning of “not significant.”

In this way, the t-distribution is of great benefit because it allows us to estimate confidence intervals and do statistical tests, even if we do not know the standard deviation of the population distribution. Thus far we have focused on how much, if at all, the sample mean deviates from the target value. In the SSD approach, the target value is not the mean value, but the 5 %ile point of the distribution (HC5), and thus we need to know how much the estimated HC5 deviates from the population HC5. Aldenberg and Jaworska [1] summarized well our approach to such deviations, and we will discuss this approach in the context of that study.

Is Population Normal? In the above example, we assumed that the weight of potato chips per bag had a normal distribution. What, then, should we do when facing the following problem? “You report the non-significant average deviation result to the authorities, to show that you are true to your customers. However, the authorities say that there is no guarantee that the weight of potato chips has a normal distribution, and thus the results cannot be accepted until such a distribution has been proven.” (continued)

3

2.

5.4 The Probability that We Achieve the Protection Goal

119

If, then, we establish the rule that t-tests can only be performed when the population is known to be normally distributed, there will be few situations in which such tests can be utilized. The authorities’ point is simply another way of saying that the SSD population does not necessarily have a normal distribution.

5.4.4 The Noncentral t-Distribution and the Extrapolation Factor The 5 %ile (HC5) point of the normal distribution of which the mean is .μ and the standard deviation .σ is calculated by .

log HC5 = μ + σ K0.05 ,

(5.12)

where .K0.05 is the 5%ile point of the standard normal distribution. This equation is ˆ is given by also true for the estimated values, and the estimated HC5 (.HC5) .

ˆ =μ log HC5 ˆ + σˆ K0.05 .

(5.13)

Now, we set the protection level to be 95% of the species (i.e., our management is ˆ as the PNEC, successful when the PNEC is set lower than the HC5). If we set .HC5 without an assessment factor, then, as we saw in Fig. 5.5, the failure probability is about 50%. Thus, we introduce an assessment factor (AF): .

ˆ − log AF = μ log PNEC = log HC5 ˆ + σˆ K0.05 − log AF.

(5.14)

Instead of an AF, Aldenberg and Jaworska [1] utilized the extrapolation factor, .ks : .

log PNEC = μ ˆ − σˆ ks

(5.15)

and sought the magnitude of .ks that satisfied the desired probability for achieving the target protection level. Comparing Eqs. (5.14) and (5.15), we obtain the following relationship between the extrapolation factor and an AF: ks = −K0.05 + log AF/σˆ .

.

(5.16)

Now, we want to set the PNEC lower than HC5. From Eqs. (5.12) and (5.15), we find that the condition for this is .

μ ˆ − σˆ ks < μ + σ K0.05 ,

(5.17)

120

5 Species Sensitivity Distribution in Ecological Risk Assessment

and we seek the .ks that satisfies this condition. The equation is arranged in this manner: .

μ ˆ − μ − σ K0.05 < σˆ ks ,

μ ˆ −μ σ − K0.05 < ks , σˆ σˆ (μ ˆ − μ)/σ − K0.05 < ks , σˆ /σ √ √ √ (μ ˆ − μ)/(σ/ n) − nK0.05  < ks n, 2 2 σˆ /σ √ Z+δ < ks n, √ W/(n − 1)

(5.18)

√ where .δ = − nK0.05 and is constant, Z is a random variable with the standard normal distribution, and W is a random variable with a .χ 2 distribution. If we compare the LHS of this equation to the RHS in Eq. (5.11), we find that the only difference is that this equation adds .δ to the numerator. A variable with such a constant has a noncentral t-distribution with .(n − 1) degrees of freedom. Suppose, then, that we want to set the PNEC lower than the population HC5 with 95% probability. Hereafter, we will consider the protection of 95% of the species with 95% probability to be the protection goal. This condition is mathematically expressed as 

 √ Z+δ .Pr √ < nks = 0.95, W/(n − 1)

(5.19)

and we need to find the √ .ks that satisfies this condition. The shape of the distribution for the variable .(Z+δ)/ W/(n − 1) is shown in Fig. 5.7, where the 95%ile value of the distribution is denoted by .th (the shaded area indicates 95%). The figure shows Fig. 5.7 Example of a noncentral t-distribution, where .th represents the 95%ile value of the distribution

0.25 0.20

t(x)

0.15 0.10

th

0.05 2

4

6

8

x

10

12

14

5.4 The Probability that We Achieve the Protection Goal Fig. 5.8 Results of the Monte-Carlo simulations using the extrapolation factors in Aldenberg and Jaworska [1] (cf. Fig. 5.4). See Eq. (5.14) for the PNEC derivation. The probability that PNEC > HC5 is 5%

121

Failure probability (%)

20 15 10 5

3

4

5

6

7

8

9

10

Sample size ( n) Fig. 5.9 .ks values as a function of sample size. The value approaches .−K0.05 = 1.64 as .n → ∞

25 20 15

ks

10 5 0 2

4

6

8

10

12

14

Sample Size (n)

√ √ that if we set the value of . nks to be .tl , then the variable is lower than . nks with 95% probability; that is, the value of .ks that satisfies Eq. (5.19) is √ ks = th / n.

.

(5.20)

Now we repeat the Monte-Carlo simulations whose results were summarized in Figs. 5.4 and 5.5, but here the PNEC is determined by Eq. (5.15). The results are shown in Fig. 5.8, where we can see that, in this case, the failure probability is exactly 5% for any sample size. If we wish to ensure that the failure probability is 5%, we must utilize the extrapolation factors in Aldenberg and Jaworska [1]. However, a problem immediately arises because these factors are too large for practical risk assessment. Figure 5.9 shows .ks up to 20 in sample size. When the sample size is just 2, .ks = 26.26. When the estimated mean (.μ) ˆ and standard deviation (.σˆ ), respectively, are 0 and 1, the actual value of the PNEC =.10−26.7 , almost zero. When the sample size is 10, 2.91 .≈ 1000, a value still too large for practical risk assessment .ks = 2.91 and .10 (compared to AF .= 10 in the conventional minimum NOEC approach when data from all three trophic groups are available). As aforementioned, however, these

122

5 Species Sensitivity Distribution in Ecological Risk Assessment

extrapolation factors are necessary to achieve the goal of protecting 95% of the species with 95% probability. In the next subsection, we will discuss the possibility of relaxing this requirement.

ks for a logistic distribution and the corresponding confidence interval When the population is a logistic distribution, we do not have a distribution that is analytically evident like the t-distribution, and therefore, we need to generate a distribution corresponding to the t-distribution by Monte Carlo simulations. The .ks value we need to know is the one satisfying (see Eq. (5.18))   (μ ˆ − μ) σ Pr X = − K0.05 < ks = 0.95, σˆ σˆ

.

where .μ and .σ , respectively, are the mean and the standard deviation of the population, .μ ˆ and .σˆ , respectively, are the mean and the standard deviation estimated from the sample, and .K0.05 is the 5%ile point of the standard logistic distribution (.μ = 0 and .σ = 1). We need to • Sample n values from a logistic distribution with m as the scale parameter and s as the location parameter. • Estimate the mean and standard deviation from the sample. • Calculate the value of X in the equation above. By iterating the processes many times, we have a distribution of X, and the 95%ile point of the distribution of X is the value of .ks . The .ks values derived by the Monte-Carlo simulations vary with each derivation of .ks . For example, when .n = 5, the derived .ks values for HC5 with 1,000,000 Monte-Carlo iterations were between 4.46511 and 4.47874 for 10 independent derivations. As noted by Aldenberg and Slob [2], every second decimal in such iterations seems not to be guaranteed; however, one might conjecture that this level of variation has a negligible impact on the derivation of PNEC values less than the population HC5. The respective confidence intervals for the logistic distribution and normal distribution for .μ ˆ = 0 and .σˆ are shown in Fig. B2. As we know, the logistic distribution has a wider tail, and the confidence intervals are also wider in this distribution than in the normal distribution. The difference is, however, not great. If the exposure assessment can be accurate enough to make this difference non-negligible, then the difference is worth considering, but unless this is the case, we likely need not be concerned about whether the probability model for a given population is logistic or normal. (continued)

Fraction affected

5.4 The Probability that We Achieve the Protection Goal

1.0

123

0.20

0.8

0.15

0.6 0.10

logistic

0.4 0.05

0.2

normal 0.00 -4

-2

2

0

4

-5

-4

-3

-2

-1

0

log(Concentration) Fig. B2 Comparison of the respective confidence intervals for two population distributions (logistic and normal). The gray lines represent the normal distribution and the black lines represent the logistic distribution (mean is 0 and standard deviation is 1 for both distributions). Solid lines are the median, and dashed lines are 90% confidence limits when the sample size (n) is 5. The horizontal dashed line is drawn where the fraction affected is 5%. The right panel shows a close-up of the low concentration area in the left panel. The left panel shows that the curves of the two distribution types are very similar

5.4.5 The Distribution of the Estimated HC5 The shape of the distribution of the estimated HC5 is our focus in this subsection. ˆ is computed by The estimated HC5 (.HC5) .

ˆ =μ log HC5 ˆ + σˆ K0.05 ,

which we can arrange as .

√

ˆ = σ/ n log HC5

 √ σˆ μ ˆ K0.05 . √ √ + σ/ n − 1 σ/ n σ/ n − 1

(5.21)

As we saw in Eqs. (5.10) and (5.11), when .μ = 0, .

μ ˆ √ ∼ N(0, 1), σ/ n

and .

σˆ ∼ χ (n − 1). √ σ/ n − 1

(5.22)

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5 Species Sensitivity Distribution in Ecological Risk Assessment

ˆ is a variable which equals the sum of two These equations imply that .log HC5 √ random variables, one with a normal distribution (multiplied by .σ/ n)) and the√other with a .χ distribution with .(n − 1) degrees of freedom (multiplied by .σ/ n − 1). Let us, then, denote the probability distribution function of the standard normal distribution by .byN (x; 0, 1) and that of the .χ distribution with .(n − 1) degrees of freedom by .χ (x; n − 1). We then define two functions: f (x) = N(x/C1 ; 0, 1)

.

g(x) = χ (x/C2 ; n − 1), √ √ where .C1 = σ/ n and .C2 = σ K0.05 / n − 1. The probability density function of ˆ is then .log HC5  H (z) = h(z)/

∞

.

−∞

(5.23)

h(x)dx,

where  h(z) =

∞

.

−∞

f (x)g(z − x)dx.

ˆ The idea is to define the distance Figure 5.10 shows the distribution of .log HC5. between .K0.05 and the 95%ile value (i.e., d) as the AF. When the population standard deviation (.σ ) is 1, logHC5 is .K0.05 , and if we consider that the 95%ile ˆ the maximum point of the distribution in Fig. 5.10 is the maximum value of .log HC5, ˆ is d. If we use d, difference between the population logHC5 (.K0.05 ) and .log HC5 then, as the AF, and subtract it from .K0.05 to derive the logPNEC, we can expect ˆ the logPNEC will not exceed that, even if we utilize the maximum value of .log HC5, Fig. 5.10 The distribution of with .σ = 1. The black line is the distribution computed by Eq. (5.23), and the histogram is the result of the Monte-Carlo simulations using Eq. (5.21). The results reflect a standard deviation of 1 in the population distribution. The distance, d, between .K0.05 and the 95%ile of the distribution is the assessment factor (AF)

K0.05 Probability density

ˆ .log HC5

95%ile

d

0.5 0.4 0.3 0.2 0.1 -4

-3

-2

-1

log(Concentration)

0

1

5.4 The Probability that We Achieve the Protection Goal

125

Failure probability

0.25

n=3

0.20

n=4 n=5

0.15

n=10

0.10 0.05

n=3

0.2

0.4

n=5

0.6

n=7 n=9

0.8

n=11 n=13

1.0

1.2

n=15

1.4

Standard deviation Fig. 5.11 PNEC values derived by Eq. (5.24). The failure probabilities are the mean value of 10,000 Monte-Carlo iterations. The standard deviations on the horizontal axis are those for the population distribution. The gray lines represent the failure probabilities derived by the conventional minimum NOEC. The sample sizes are superimposed on the lines

K0.05 . In this approach, the logPNEC is determined by

.

.

log PNEC = μ ˆ + σˆ (K0.05 − d).

(5.24)

The probability that the PNEC is higher than HC5 (failure probability), calculated using Eq. (5.24), is shown in Fig. 5.11. I expected that the failure probability would be 95%, but the probabilities were higher than 0.95. However, the failure probabilities of these PNEC values are less than the PNEC determined by the conventional minimum NOEC approach (gray lines in Fig. 5.11), when the sample size is small and the population standard deviation is large. When I first came upon the idea of PNEC determination based on the difference between the population logHC5 and the upper limit of the estimated logHC5, I thought the idea was new. However, I later found that some theoretical studies had taken the same approach in the late 1980s and early 1990s [2, 9, 15, 16]. For example, Van Straalen and Denneman [15] tried to determine the PNEC in much the same manner as I have described here; however, all these studies assumed a logistic population distribution, which is why I did not explore them in designing my approach, as it assumed a normal distribution. Wagner and Lokke [16] showed that we could use a noncentral t-distribution when the population distribution was normal. These studies are summarized in [1]. It should be noted that the original paper written by myself and a colleague [13] had a number of errors; thus, if this paper should be consulted, Belanger and Carr [5], which pointed out our errors, should also be consulted, along with our reply in [8]. Of note in Fig. 5.11 is that, in the minimum NOEC approach, the failure probability varies with the standard deviation (.σ ) of the population distribution, increasing as .σ increases, whereas, in the method proposed here, this probability remains constant with respect to variation in .σ . In the next section, we will discuss why.

126

5 Species Sensitivity Distribution in Ecological Risk Assessment

5.5 The Relationship Between σˆ , Sample Size, and the AF We have already seen a relationship between the extrapolation factor and the AF, in Eq. (5.20). This equation may be rewritten as .

log AF = σˆ (K0.05 + ks ),

(5.25)

which indicates that logAF is a linear function of the estimated standard deviation (.σˆ ), implying that if we want to achieve the protection goal (here, protection of 95% of the species with 95% probability), the magnitude of AF should be varied depending on .σˆ . In other words, if AF is kept constant for any .σˆ , the magnitude AF is relatively small for chemicals whose toxicity values vary widely from species to species (i.e., large .σˆ ), and therefore the level of protection is relatively poor, while for chemicals with small .σˆ , the magnitude of AF is relatively large, resulting in over-protection. Thus, to achieve a specific level of protection, the magnitude of AF must be changed depending on .σˆ ; however, unfortunately, the notion of altering the magnitude of the AF has yet to take hold in current ecological risk assessment. Variation in sensitivity (represented as the standard deviation) is a type of uncertainty, and therefore large AF is required when .σ is large. Two SSDs with different .σ values are illustrated in Fig. 5.12. If we conduct a random sampling (which corresponds to a new toxicity test) from the SSD in the left panel in this figure, the sampled value (NOEC) will be different every time. This implies that the toxicity for a specific species is difficult to predict, and we need a large number of NOEC values, for different species, in order to make an accurate estimation of the population mean and standard deviation. On the other hand, when different species have roughly the same NOEC value and thus the standard deviation is small (right panel in Fig. 5.12), NOEC prediction for a newly tested species is easy, and we do not need to conduct toxicity testing as many times. Thus, .σ is a measure of uncertainty and the larger .σ is, the more uncertain is the NOEC result. We have seen (Figs. 5.3 and 5.11) that, in the minimum NOEC approach, the failure probability varies with the population standard deviation but is kept constant

Fraction Affected

1.0

σ=1

σ=0.001

0.8 0.6 0.4 0.2 0.0 -4

-2

0

2

4

-4

Concentration Fig. 5.12 Two SSDs with different .σ values

-2

0

2

4

5.5 The Relationship Between σˆ , Sample Size, and the AF

127

in the SSD approach. The reason for this is that the former approach makes no allowance for uncertainty due to variation in sensitivity among species. As a large variation in sensitivity means that the uncertainty is large, a large AF is needed in order to compensate for such uncertainty. The minimum NOEC approach, however, focuses solely on the smallest NOEC value, discarding the rest of the data. It is impossible, in principle, to know the variation in species sensitivity, and as a result, when the variation is large, the fraction of organisms protected is small, and conversely, when the variation is small, the percentage of organisms protected is large. This is a critical limitation of the minimum NOEC approach. In the SSD method, on the other hand, the magnitude of the extrapolation factor is weighted by .σˆ (Eq. (5.20)), ensuring constant failure probabilities. In this approach, the AF may also be determined independently of species sensitivity variation, when deriving the PNEC (Eq. (5.14)). This, however, leads to the same problem as in the minimum NOEC approach. Accepting this problem as inevitable is equivalent to thinking that a smaller fraction of protected species may be acceptable in the case of chemicals with a large variability in species sensitivity. If we want to adopt such a management approach, we can do so; however, this adoption should be explicitly justified. Another typical source of uncertainty involves the sample size (the number of NOEC values). This uncertainty is widely recognized, and there is much debate about the sample size (n) required to accurately estimate SSDs. However, since the overall uncertainty is determined by both n and .σˆ , the required n cannot be determined without consideration of .σˆ ; though these two uncertainties seem different, they are mutually convertible. In seeking to achieve any goal, we need an adequate amount of information, and this is no less true when seeking to achieve protection goals in ecological risk assessment, such as our present goal of protection of 95% of the species with 95% probability. Now, n and .σˆ each have their own information (or uncertainty) requirements, and the total amount of required information is determined by these two factors. When the sample size is small and .σˆ is large, the goal of protection cannot be achieved due to lack of information, and the uncertainty coefficient is used to compensate for this deficiency. Hence, the magnitude of AF is, in a sense, a metric of the amount of required information that is lacking (i.e., uncertainty), and the amount of lacking is given by Eq. (5.25). Figure 5.13 shows the relationship between n, .σˆ , and the logAF required to achieve the protection goal. At point A, n is 6 and .σˆ is about 0.5, and at point B, n is 11 and .σˆ is about 1.0. But as the magnitude of logAF is the same in each case (logAF = 1, AF = 10), each pair of n and .σˆ has the same amount of required information. This means that the amount of information gained by increasing the number of toxicity data items from 6 to 11, for example, is equivalent to the amount of information lost by increasing .σˆ from 0.5 to 1.0, and in this way, the amount of information related to the sample size may be converted to the amount related to the standard deviation, in absolute value terms.

128

5 Species Sensitivity Distribution in Ecological Risk Assessment

Fig. 5.13 A contour plot of the required magnitude of logAF (numbers in the panel). Point A marks the pair of .n = 6 and .σ ˆ ≈ 0.5 and point B the pair of .n = 13 and .σ ˆ ≈ 1.0, with dashed lines marking the axes values

1.5

4.0 3.0

2.0

1.5

1.3

1.0

1.1 B

0.8

0.9

^ σ

0.6

0.7 A

0.5

0.4

0.3

0.2

0.1 3

7

12

17

Sample size (n)

5.5.1 Reduction of the Protection Goal In this subsection, we will explore the relationship between the protection goal (here, protection of 95% of the species), the probability of achieving this goal (thus far, 95%), and the corresponding magnitude of AF. When we have no AF (AF .= 1, logAF .= 0), Eq. (5.20) becomes ks = −K0.05 ,

.

(5.26)

implying that the extrapolation factor is reduced from .−K0.05 +log AF/σˆ to .−K0.05 . As was shown in Fig. 5.4, the probability that we can protect 95% of the species is about 50% in this case. When .n = 10 and .σˆ = 1, the probability is precisely computed √ to be 45.6%, using a noncentral t-distribution with 9 degrees of freedom and .− 10K0.05 as the noncentral parameter. This may be expressed mathematically (see Eq. (5.19)) as 

 √ Z+δ .Pr √ < − nK0.05 = 0.456, W/(n − 1) √ where .δ is the noncentral parameter and equals .− nK0.05 for protection of 95% of the species. The reduction in the protection probability is due to the amount of information lost by removing the AF. The elimination of AF, then, degrades the accuracy of the assessment, due to the loss of information. In this example, the amount of lost information is converted into a reduction in the probability of protecting 95% of the species. It is also possible to convert the amount of lost information into a reduction in the fraction of protected

5.5 The Relationship Between σˆ , Sample Size, and the AF

129

species. In other words, with the present goal of protecting 95% of the species with 95% probability, eliminating the AF changes the goal to protecting 95% of the species with 45.6% probability. Similarly, we can convert the amount of lost information into the protection of y% of the species with 95% probability. How we determine the value of y is the focus of this subsection. The noncentral parameter, .δ, represents the fraction of unprotected species. Since, in the case of the present √ goal, this fraction was 5% (=protection of 95% of the species), and .δ was .− nK0.05 . Now we change the noncentral parameter to √  .δ = − nK0.01y , where y is a percentage of unprotected species, and will search for the value of .δ  (or y) that satisfies   √ Z + δ Pr √ < − nK0.05 = 0.95. W/(n − 1)

.

However, as the shape of the noncentral t-distribution varies with the noncentral parameter, rather complex numerical exploration is necessary to find the noncentral parameter .δ  . The flow to find .δ  is illustrated in Fig. 5.14. √ The values of .δ’ (or y) are determined as follows. We set .δ  = − nK0.10 with .y = 10 for example (with 10 meaning protection of 90% of the species) and then compute√ the 95%ile value of the noncentral t-distribution with √ noncentral parameter  .δ = − nK0.10 . If the 95%ile value is greater than .− nK0.05 , we increase y; if not, we decrease y; and we repeat this √ procedure until the 95%ile value of the noncentral t-distribution becomes .− nK0.05 . The results for various sample sizes (n) √ are shown in Fig. 5.16. When .n = 10, for example, the value of y in  .δ = − nKy is 21.2, implying that the choice of no AF makes the protection goal √

n k's

√

=

√

− n K0.05

0.4

t(x)

n ks

0.3 0.2 0.1 0

2

4

x

6

8

10

√ Fig. 5.14 A flow to find values of .δ  = − nK0.01y . The noncentral t-distribution for the original protection goal (protection of 95% of the species with 95% probability) is shown by the gray curve. √ The extrapolation factor (.ks ) to achieve this goal is determined by Eq. (5.20) (i.e., .ks = t0.95 / n) Here, then, we must find a new distribution, where .t0.95 is the 95%ile value of the distribution). √ for which the cumulative probability at .− nK √0.05 is 95% (shown by the black curve), which is equivalent to √ finding a distribution having .− nK0.05 as the 95%ile value. The 95%ile value is denoted by . nks

130

5 Species Sensitivity Distribution in Ecological Risk Assessment

Protection goal (%)

Fig. 5.15 Percentage of species that can be protected with 95% probability, in the absence of an AF. The horizontal dashed line shows the original protection goal of protection of 95% of the species with 95% probability

100

95%

90 80 70 60 50 5

10

15

20

Sample size ( n) “protection of 78.8% species with 95% probability.” When .n = 20, the protection goal is “protection of 85.6% species with 95% probability” (Fig. 5.15). I have employed the goal of the protection of 95% species with 95% probability merely as an example and do not wish to suggest that it is the ultimate protection goal, which should be determined by social consensus. Once we set a protection goal, however, the goal should be the same for all substances. It should not be the case that some substances are heavily protected against while others are not, simply due to the small sample size or large variation in species sensitivity in the latter cases. Instead, I believe that the quantitative discussion presented here should form an important part of the risk assessment process.

5.5.2 How Far from HC5 Thus far, we have investigated how to determine the extrapolation and assessment factors that result in a lower PNEC than the population HC5. Our discussion has been based on this aim. Sometimes, however, we want to know if the PNEC is “close enough to,” instead of simply lower than, the population HC5. Consider, for example, the HC5 histograms in Fig. 5.5. The light gray histogram has some ˆ values distant from the population HC5, but in the dark gray histogram, the .HC5 ˆ values are concentrated around the population HC5. We can consider that the .HC5 ˆ values are case depicted by the gray histogram is not acceptable because some .HC5 distant from the population HC5, but the case depicted by the dark gray histogram is ˆ values are “close enough” to the population HC5, even acceptable because the .HC5 though some are greater than the population HC5. In this last subsection, then, we discuss how to quantify the deviation between the population and estimated HC5. ˆ and logHC5, as We define the relationship between the distance (d) and log.HC5 ˆ − log HC5. d = log HC5

.

(5.27)

5.5 The Relationship Between σˆ , Sample Size, and the AF

131

Using the relationships in Eqs. (5.12) and (5.13), we can arrange Eq. (5.27) as follows: d=μ ˆ + σˆ K0.05 − (μ + σ K0.05 ),

.

d (μ ˆ − μ) σˆ = + K0.05 − K0.05 , σ σ σ σˆ (μ ˆ − μ) − σ K0.05 d − K0.05 = , σ σ σ d (μ ˆ − μ)/σ − K0.05 − K0.05 = , σˆ σˆ /σ √ √   √ d (μ ˆ − μ)/(σ/ n) − nK0.05 n − K0.05 = . σˆ σˆ /σ

(5.28)

As the RHS of this equation is a random variable with a noncentral t-distribution (see Eq. (5.18)), the √ LHS also has a noncentral t-distribution, with .(n − 1) degrees of freedom and .− nK0.05 as the noncentral parameter. We write the 5%ile and 95%ile values of the distribution as .t0.05 and .t0.95 , respectively, and define that √ nk0.05 = t0.05 , √ nk0.95 = t0.95 .

.

The maximum value of the random variable, under one-sided 5% significance level, is   √ √ d − K0.05 = t0.95 = nk0.95 , . n σˆ and the maximum value of d is d = σˆ (K0.05 + k0.95 ).

.

(5.29)

This equation is the same as Eq. (5.25), implying that the maximum d is equal to logAF (i.e., AF is determined so as to offset the maximum value of the deviation between the estimated and population HC5). Similarly, the minimum value of d is d = σˆ (K0.05 + k0.05 ).

.

(5.30)

We often want to know if the estimated value and true value are within a factor of x or not. The condition that the maximum d is lower than .log x is σˆ (K0.05 + k0.95 ) < log x.

.

132

5 Species Sensitivity Distribution in Ecological Risk Assessment

To achieve the condition, .σˆ must satisfy .

σˆ
f50 [BL]T ,

.

the mixture effect is synergistic, and if [MBL] < f50 [BL]T ,

.

178

7 Mathematical Models for Chemical Mixtures

Fig. 7.13 The mixture effects of exposure to two metals whose concentrations are determined by Eq. (7.45). The white area indicates an antagonistic mixture effect, and the black area a synergistic effect. Additivity only holds in the boundary between the black and white areas (i.e., practically non-existent). The parameters are: .[L]T = 10−4 , [BL]T = 10−6 , K1L = 104 , K1BL = 106 , f50 = 0.5

logK BL 2

antagonistic

synergistic

logK L2 the mixture effect is antagonistic. We then conduct numerical simulations using various values of .KiL and .KiBL . Figure 7.13 shows the results for different values of .K2L and .K2BL (see caption for the parameter values). The black area indicates that the mixture effect is synergistic (.[MBL] > f50 [BL]T ), and the white area indicates that the mixture effect is antagonistic (.[MBL] < f50 [BL]T ). Now, recall that we are assuming that the mixture effect of metals is additive; yet we can see, from the above, that the mixture effect can only be non-additive. This implies the seemingly self-contradictory result that the metal mixture effect is non-additive, though the toxic mechanisms are the same (hence the effect should be additive). How is this possible?

7.6.4 The Reason for Non-additivity with Identical Toxic Mechanisms We now discuss why non-additivity occurs, even though the respective toxicity mechanisms are assumed to be the same. We can understand why by obtaining the total metal concentrations required to cause a half effect, under the additive assumption. We have hitherto assumed that half effect occurs when [MBL] = f50 [BL]T

.

is satisfied, where .[MBL] is the sum of .[M1 BL] and .[M2 BL].

(7.46)

7.6 Metal Mixture

179

Now, let us assume that the .[M1 BL] fraction of [MBL] that satisfies Eq. (7.46) is p, and thus the [M.2 BL] fraction is .(1 − p), then [M1 BL] = f50 p[BL]T ,

.

[M2 BL] = f50 (1 − p)[BL]T .

(7.47)

The concentration of free biotic ligand ([BL]) is [BL] = (1 − f50 )[BL]T .

(7.48)

.

By substituting these into Eq. (7.38) for Metal 1, we have K1BL (1 − f50 )[BL]T [M1 ] = pf50 [BL]T ,

.

and by arranging this, the free-ion concentration of Metal 1 is [M1 ] =

f50 p , (1 − f50 )K1BL

(7.49)

[M2 ] =

(1 − p)f50 . (1 − f50 )K2BL

(7.50)

[M1 L] = K1L [L][M1 ], .

(7.51)

[M2 L] =

(7.52)

.

and that of Metal 2 is .

From Eq. (7.37), we have .

K2L [L][M2 ].

The total concentration of ligand (.[L]T ) must satisfy [L]T = [L] + [M1 L] + [M2 L].

(7.53)

.

The .[Mi L] (.i = 1, 2) values are derived from Eqs. (7.51) and (7.52), and, as we already know .[M1 ] and .[M2 ] (from Eqs. (7.49) and (7.50)), we can express the concentration of ligand ([L]) as [L] =

.

(1 − f50 )K1BL K2BL [L]T (1 − f50 )K1BL K2BL + f50 (1 − p)K2L K1BL + f50 pK1L K2BL

.

(7.54)

180

7 Mathematical Models for Chemical Mixtures

We now substitute .[L] and .[Mi ] (.i = 1, 2) into Eqs. (7.51) and (7.52) and obtain [M1 L] =

.

[M2 L] =

f50 pK2BL K1L LT K1BL [(1 − f50 )K2BL + f50 (1 − p)K2L ] + f50 pK2BL K1L f50 (1 − p)K1BL K2L LT K1BL [(1 − f50 )K2BL + f50 (1 − p)K2L ] + f50 pK2BL K1L

,

.

(7.55)

Since the total metal concentrations must satisfy [Mi ]T = [Mi ] + [Mi L] + [Mi BL]

.

the respective concentrations resulting in additivity are  [M1 ] = f50 p

.

T

+

1 (1 − f50 )K1BL K2BL K1L [L]T

K1BL [(1 − f50 )K2BL + f50 (1 − p)K2L ] + f50 pK2BL K1L

.



[M2 ]T = f50 (1 − p) +

 + [BL]

T

,

(7.56) 1 (1 − f50 )K2BL K1BL K2L [L]T

K1BL [(1 − f50 )K2BL + f50 (1 − p)K2L ] + f50 pK2BL K1L

 + [BL]T

.

(7.57) A standard concentration setting for metal mixture toxicity is 1/2 the ECx (Eq. (7.45)) as the total concentration of each metal. The total concentration of Metal 1, then, is   K1L [L]T f50 1 T T .[M1 ] = + + [BL] . (7.58) 2 (1 − f50 )K1BL (1 − f50 )K1BL + f50 K1L

7.6 Metal Mixture

181

If we input .p = 1/2 into Eq. (7.56), the total concentration of Metal 1 resulting in additivity is f50 × .[M1 ] = 2 T

+



1 (1 − f50 )K1BL 2K1L K2BL [L]T

  + [BL] K1BL 2(1 − f50 )K2BL + f50 K2L + K1L K2BL

 T

. (7.59)

Equations (7.58) and (7.59) are similar, but the second term in the parenthesis is different; this term, which represents the concentration of metal binding to the ligand, is more complex in Eq. (7.59) that in Eq. (7.59). Recall that the total concentration in Eq. (7.59) is half the .EC501 value for Metal 1, and .EC501 is derived from the toxicity value in the absence of Metal 2. This is why the second term in Eq. (7.59) does not reflect the Metal 2 parameters, and when we specify the mixture test concentrations based on Eq. (7.59), we implicitly assume that the metal speciation occurs independently of the other metals. Such speciation, however, is competitive between metals, and the total concentration should therefore be designated with this in mind. Since the total concentration in Eq. (7.59) is derived under the condition that two metals are present, competitive binding to the ligand is here taken into account, and thus the second term in Eq. (7.59) also reflects the parameters for the other metal. The specification of mixture test concentrations based on Eq. (7.58) is, therefore, inappropriate because it implicitly assumes that the metal speciation occurs independently. When such tests are conducted based on Eq. (7.58), the mixture effect never becomes a half effect, even if the toxic mechanisms of the metals are the same and hence the mixture effect is additive when the concentration is specified by Eq. (7.59). Unfortunately, the test concentrations in most of the mixture tests conducted thus far have been based on Eq. (7.58), meaning that we have little information to properly determine if the mixture effects of metals are additive, and as a result, lack the knowledge necessary to properly discuss the toxicity mechanisms of metals. When we observe non-additive toxic effects of chemical mixtures, we often think there is some kind of interaction. However, as we have seen in this chapter, we do not necessarily need such interaction to have non-additive effects. In the enzyme-substrate reaction model, for example, the mere fact that the chemicals have different sites of action in a chain reaction can result in non-additive mixture effects. Moreover, in the case of BLM analysis of metal mixtures, even ignoring the role of metal speciation, there is a possibility that the metal mixture effects will be identified as non-additive, due to improper concentration specifications in the mixture tests. In any models in this chapter, interactions that would cause a direct reaction between chemicals are not considered. Of course, there could be cases where such direct interaction results in non-additivity, but this does not mean that if the mixture effect

182

7 Mathematical Models for Chemical Mixtures

is non-additive there is necessarily interaction. Understanding why and how nonadditive effects occur forms a critical part of the broader understanding of the toxic effects of chemicals, and mathematical models are powerful tools in this respect. It will soon be 100 years since Loewe’s seminal paper was published. I would hope that our research on the toxic effects of chemical mixtures will continue to advance, with an eye to a centennial celebration of Loewe’s additivity.

References 1. Allison J, Brown D, Novo-Gradac K (1991) MINTEQA/PRODEFA2, A Geo- chemical assessment model for environmental systems: version 3. 0. EPA/600/3091/021. U.S. Environmental Protection Agency, Athens 2. Bliss CI (1939) The toxicity of poisons applied jointly. Ann Appl Biol 26:585–615 3. Cedergreen N (2014) Quantifying synergy: A systematic review of mixture toxicity studies within environmental toxicology. PLoS One 9:e96580 4. den Berg MV, Birnbaum L, Bosveld AT, Brunström B, Cook P, Feeley M, Giesy JP, Hanberg A, Hasegawa R, Kennedy SW, Kubiak T, Larsen JC, van Leeuwen FX, Liem AK, Nolt C, Peterson RE, Poellinger L, Safe S, Schrenk D, Tillitt D, Tysklind M, Younes M, Waern F, Zacharewski T (2006) The 2005 world health organization reevaluation of human and mammalian toxic equivalency factors for dioxins and dioxin-like compounds. Toxicol Sci 93:223–241 5. Draper NR, Smith H (1998) Applied regression analysis. Wiley Series in Probability and Statistics, 3rd edn. Wiley-Interscience, New York 6. Kodell RL, Krewski D, Zielinski JM (1991) Additive and multiplicative relative risk in the two-stage clonal expansion model of carcinogenesis. Risk Anal 11:483–490 7. Kortenkamp A, Backhaus T, Faust M (2009) State of the art report on mixture toxicity. Report, European Commission, Brussels 8. Loewe S, Muischnek H (1926) Effect of combinations: mathematical basis of the problem. Arch Exp Path Pharm 114:313–326 9. Norwood WP, Borgmann U, Dixon DG, Wallace A (2003) Effects of metal mixtures on aquatic biota: a review of observations and methods. Hum Ecol Risk Assess 9(4):795–811 10. Takeshita J-I, Seki M, Kamo M (2016) Criteria for deviation from predictions by the concentration addition model. Environ Toxicol Chem 35:1806–1814 11. Tipping E (1994) WHAMC—a chemical equilibrium model and computer code for waters, sediments, and soils incorporating a discrete site/electrostatic model of ion-binding by humic substances. Comput Geosci 20:973–1023 12. Warne MSJ, Hawker DW (1995) The number of components in a mixture determined whether synergistic and antagonistic or additive toxicity predominate: the funnel hypothesis. Ecotoxicol Environ Saf 31:23–28 13. Yin N, Ma W, Pei J, Ouyang Q, Tang C, Lai L (2014) Synergistic and antagonistic drug combinations depend on network topology. PLoS One 9(4):e93960

Chapter 8

Statistics and Related Topics

Abstract In natural science research, statistical approaches such as hypothesis testing, regression analysis, and model selection are frequently used, and the results of statistical analysis, such as “statistically significant” (or .p < 0.05), can be encouraging or the opposite. We often frame a given study with a hypothesis, conduct experiments to obtain the relevant data, and then perform statistical analysis based on this data. When our results have a p-value of less than 0.05, we consider our hypothesis to be correct, and thus we often consider that .p < 0.05 indicates acceptance of the hypothesis; however, in fact, .p < 0.05 only when the hypothesis is rejected. This sentence may seem self-contradictory, but those who grasp its deeper meaning will not need to read this chapter. The logical structure of statistical methods is somewhat complicated, but if we do not understand this structure correctly, we may misinterpret the results we have obtained. Many harmful effects of such misinterpretations have, in fact, been noted in recent years. Using simple examples, this chapter explains how we perform regression analysis and obtain confidence intervals, and how we select models and highlight features that are often misinterpreted in statistical analysis. Keywords Hypothesis testing · Null hypothesis · Regression analysis · Confidence interval · p-value · Model selection

8.1 Introduction Statistics are powerful and, therefore, can be very harmful if misused. When reading papers on environmental science, we sometimes come across studies that strike us as statistically mistaken. In this chapter, then, we discuss some basic elements of statistics, and how they are misused. In scientific work, we often frame a hypothesis and then test it against the evidence, frequently employing statistical analysis to make true–false judgments, and using the p-value as a critical factor in these judgments. However, the p-value is not the only factor in our determination of the truth [10]. We ourselves must ultimately decide what is true and what is not, testing our hypothesis against the © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Kamo, Theories in Ecological Risk Assessment, Theoretical Biology, https://doi.org/10.1007/978-981-99-0309-2_8

183

184

8 Statistics and Related Topics

results of our statistical analysis. It would thus be a mistake to shirk or abandon this effort and outsource our decision making to statistical results, effectively making ourselves the slaves of statistics. In my opinion, the development of statistical software in recent years has made this situation worse. When conducting a test, it is important to understand the concept behind it, especially when a relevant null hypothesis is required to ensure the veracity of the test. When using statistical software, however, we are encouraged to simply enter the data, hit the Enter key, and immediately obtain the .p < 0.05 result. This means that the null hypothesis has been rejected but is this the wisest course of action? Consider the case of linear regression analysis, where we often wish to determine whether the p value of a given regression coefficient is less than 0.05 or not. What does a value of .p < 0.05 for the coefficient actually mean? And conversely, what does .p > 0.05 mean? If we cannot answer these questions in a straightforward manner, then we are the slaves of our own statistics. Let us first, then, consider some fundamental statistical concepts.

8.2 Hypothesis Testing In environmental science, statistics are widely used for the following two purposes. When organisms are exposed to a given chemical and the effects are observed, we may wish to know whether these effects differ from the effects observed in the case of control organisms not exposed to the chemical. Organisms are exposed to a given chemical at several levels of concentration. How do we quantify the relevant relation between concentration and effect? In the former case, we will typically perform analysis of variance (ANOVA), which is a form of hypothesis testing often used for discrete observations. In the latter case, regression analysis is typically performed. Such observations involve a collection of discrete points, and we need to connect the points. Using regression models, we can interpolate and extrapolate to estimate the relevant quantitative relationships. First, then, we will briefly explain the concept of hypothesis testing. It has recently been argued that hypothesis testing, and the p-value used as a criterion of confidence, is so misunderstood [10] that is actually harmful [1]. Further, in environmental science, the meaning of no observed effect concentrations (NOECs), which are determined by hypothesis testing, has not been properly understood, and there has long been debate over the possible banning of their use [6, 8, 9]. However, NOECs remain a major assessment factor in ecotoxicology and have a significant impact on the determination of environmental criteria. Personally, I think that banning NOECs is extreme; they have a long history of use in characterizing environmental hazards, and banning their use would cause a serious breach in continuity from the past. If, however, we wish to use them well in such characterization, we need to fully understand their limitations. Put simply, we should not use them without first understanding clearly what they mean.

8.2 Hypothesis Testing

185

Hypotheses are essential to statistical testing. There may be some who think that the hypothesis in such testing formally amounts to the assertion that “there is a difference between the treatment and control effects,” because we wish to know whether there is such a difference; however, this is completely mistaken. The proper hypothesis in an ANOVA test is the assertion that “there is no difference.” This is referred to as the null hypothesis (.H0 ). Suppose, for example, that we conduct a toxicity test and count the number of organism deaths. We assume that the deaths are stochastic events and occur according to a specific statistical distribution such as a binomial or normal distribution. The assumption that there is no difference in the effects means that the control and treatment data are both assumed to come from the same distribution with the same parameters (e.g., mean and variance). Then, we calculate the probability (p-value) that the number of observed deaths has the given distribution, with the understanding that the lower the probability, the more unlikely it is that the data are sampled from the same distribution (the ASA statement [10]). We usually specify a threshold, .α, as the so-called significance level, and when the p-value falls below this threshold, we reject the null hypothesis and conclude that the death rate is significantly different from the control. In ecotoxicities, the minimum concentration at which the null hypothesis is rejected is the LOEC, and the maximum concentration at which it is not rejected is the NOEC. In ecological risk assessment, NOEC values are typically used to infer “safe” chemical concentrations. We need to determine the value of .α in advance of the tests. In most cases, the value is 5%, which implies that an event that happens only once in 20 times is rare enough that we may safely reject the null hypothesis. Even if there is no effect in a given test, if the test is repeated 20 times, the 5% p-value implies that we should have at least one significant result. However, as this value is merely a probability, it is not certain that we will have one such result in 20 trials, but it is certain that if we try to obtain a significant result in such a cheating (“p-hacking” [10]) manner, our results will be discredited. Few misunderstand what it means for the null hypothesis to be rejected, but many fail to understand what it means when the null hypothesis is not rejected. In other words, when the exposure death rate in our example is not significantly different from the control rate, what does this mean? Non-experts, and even experts, often think that not significant means no effect [9]. The mistake in understanding not significant to mean no effect can be understood by looking at the logic table in Table 1.1. When a proposition is true, the contraposition is also true, but the inversion and conversion may or may not be true. Now, the null hypothesis in deriving NOEC is, “There is no effect from exposure to the chemical.” Let the proposition, then, be “If the null hypothesis is rejected (p), there is an effect (q).” Since, in the case of the NOEC, we are talking about the maximum concentration at which the null hypothesis is not rejected, we are dealing with the conversion of the proposition (“If the null hypothesis is not rejected, then there is no effect.”), which may or may not be true (Table 8.1). Believing that not significant means no effect is harmful in several ways. In Chap. 2, we saw that the accumulation of the statistically invisible effect below the

186 Table 8.1 True/False table. The proposition .p ⇒ q is assumed to be true

8 Statistics and Related Topics Proposition Inversion Conversion Contraposition

⇒q ⇒p not .p ⇒ not q not .q ⇒ not p

.p .q

True True or False True or False True

NOEC can cause the population to decline; yet if we believe there is no effect below the NOEC, then this possibility is overlooked. Further, this belief may also result in the dismissal of meaningful research. There are two possible responses to the rejection of the null hypothesis. One is to give up the research, believing that if our working hypothesis is untenable, it is better to give up as soon as possible, because no matter what we do, the null hypothesis will never be rejected (except for the one-in-twenty coincidence). The other is to continue, after reconfiguring the research plan and experimental design. If we believe that the hypothesis is tenable, we should try again. I believe that having a strong (but not careless) belief in the value of one’s research is a necessary attribute of a successful researcher. It seems a needless misfortune to abandon research due to the misconception that the working hypothesis is untenable simply because the null hypothesis is not rejected. Of course, knowing when to quit is also a valuable attribute in a researcher. There are many misconceptions about statistics, but I think most stem from improper understanding of the nature of the null hypothesis, or of what it means for the null hypothesis not to be rejected. A total ban on the use of the NOEC is extreme, but for those who do not understand the meaning of non-rejection of the null hypothesis, it might be better to ban the NOEC’s use. For example, among the chemical mixture toxicity studies, there are some that have detected mixture effects of exposure to sub-NOEC chemical levels. For those who believe that NOEC is an indicator of no effect, this should be a surprising result because zero is zero no matter how many times we do the calculation, and there is no way to observe any effects. Put simply, if one is surprised by those results, they should not be using NOECs.

8.3 Regression Analysis Regression models, including linear models, are often used in statistical analysis. However, since many real-world processes are non-linear, linear models often do not provide sufficient estimation accuracy. Nonetheless, such models are intuitively easy to understand, and it is attractive that statistical information such as confidence intervals can be explicitly calculated relatively easily by a lot of efforts by predecessors. In recent years, computer advances have made it easier to perform non-linear regression analysis using complex models, but the fundamental concept is the same as in linear regression analysis; thus, an understanding of linear regression is the first step in understanding even the most sophisticated regression analysis.

8.3 Regression Analysis

187

error

y (x, y)

x Fig. 8.1 An image of regression analysis, where .x¯ and .y¯ are the respective means of the observed x and y values, and the solid circles represent the observed data points. We first draw a line through the point (.x, ¯ y), ¯ and then rotate the line around the point to find the slope that minimizes the total distance (error) between the line and the observed y values (i.e., here, the solid line)

Regression analysis is performed to determine the quantitative relationship between an explanatory or independent variable, x, and an objective or dependent variable, y. Since the sample is usually a collection of data points, a model is needed to fill in the gaps between these points. In linear regression, we first compute the means of the observed x and y and then rotate a line about that point (mean x, mean y) to find the slope which minimizes the differences between the observed y and the y estimated by the model (Fig. 8.1) [5]. Let us, then, write y = a + bx

.

as an expression of the regression model. Suppose, now, that we have n data items and number them as .xi , .yi for the i-th observed data item. The total error is defined as the sum of the respective squared deviations between the observed and predicted values of y; that is, se =

n 

.

(yi − (a + bxi ))2 ,

(8.1)

i=0

where se is the sum of the squared errors (deviations). We then seek the a and b parameters that will minimize se. The important point is that in ordinary linear

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8 Statistics and Related Topics

regression, we are concerned about the error in y, but not the error in x. In other words, there is no concept of error in x. If we write the sample variance of x as .Sxx , and the covariance of x and y as .Sxy , the parameters that minimize the error are a = y¯ − bx, ¯

.

b=

Sxy , Sxx

and we can find the proof in any textbooks.1 Thus, in linear regression analysis, parameters and other information can often be obtained analytically, which is one of the reasons why linear regression is preferred. The solution is given explicitly, and our confidence in the solution is often felt to be an order of magnitude greater than in the case of computer-generated results based on complicated calculations difficult to understand. Especially in the days of more limited computer power, finding an analytical solution was a very important task. Concentration-response relationships for the toxic effect of chemicals are often, however, non-linear, with S-shaped functions (or to be precise, we assume they have S-shaped functions). In such cases, efforts have been made to linearize them by arranging the variables; for example, if a normal cumulative distribution is assumed for the S-shaped function, linearization is possible using probit conversion. If a logit cumulative distribution is assumed for the S-shaped function, logit conversion is used for the linearization. The logit model is p=

.

1 , (1 + exp[−a − bx])

(8.2)

where p is the rate of a certain effect such as death, and a and b are parameters to be estimated. We convert the p as  p , . log 1−p 

1 There have been so many textbooks on statistics published that it is difficult to choose which book to read. I like Grafen and Hails [5] and Chatterjee et al. [2], which covers the basics and applications. What we read totally depends on our tastes and interests. It is important to find one’s own favorite books.

8.3 Regression Analysis

189

and in so doing obtain the following linear function:  .

log

p 1−p

 = log(p) − log(1 − p),   1 1 − log 1 − , = log 1 + exp[−a − bx] 1 + exp[−a − bx] = log

1 exp[−a − bx] − log , 1 + exp[−a − bx] 1 + exp[−a − bx]

= − log(exp[−a − bx]), = a + bx.

(8.3)

The weakness in such logit conversion is that it cannot be handle .p = 0 or 1 (logit(p) = .−.∞ or .∞) values, which are often found in toxicity tests (i.e., all die or all survive, respectively). The classic way to deal with this is to replace .p = 0 with .p = 0.0000001 and .p = 1 with .p = 0.999999.

8.3.1 Regression Analysis Based on the Maximum Likelihood Given the current speed of computers, logistic regression analysis using the maximum likelihood method is often performed. If the endpoint of the toxicity test is death, for example, the data consists of binary values: dead or alive. Let us assume, then, that the death rate at a specific chemical concentration is 0.4, and the total number of fish is 10. The probability that 3 fish die can be obtained using a binomial distribution: p=

.

10! 0.43 0.610−3 . (10 − 3)!(3!)

To obtain the likelihood, we repeat this calculation. Let us, then, perform logistic regression analysis on the results of the nonylphenol toxicity test, shown in Chap. 5 (Table 8.2). We assume a log-logistic function for the concentration-response relationship, with the caveat that, though this function is widely used for this relationship, it has yet to be proven that the relationship is, in fact, log-logistic; use of the function is simply based on our experience. First, then, we log-transform all the toxic values, and then employ the logistic function for the transformed values. In the logit model, the probability of death at a given concentration (c) is given as, p(c) =

.

1 . 1 + exp[−a − bc]

(8.4)

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8 Statistics and Related Topics

Table 8.2 Dead counts for 10 Daphnia magna exposed to nonylphenol Data item 1 2 3 4 5

Concentration (mg/L) 0.046 0.059 0.085 0.11 0.17

Log10(concentration) .−1.34 .−1.23 .−1.07 .−0.96 .−0.77

Dead counts 0 2 3 8 10

We first select arbitrary values of a and b: let both a and b be 10 as an initial choice. If the concentration of nonylphenol is 0.046 (mg/L), the log-transformed value (with base 10) is .−1.34. Substituting this value into Eq. (8.4) with .a = 10 and .b = 10 yields .p(−1.34) = 0.033, implying that the expected death rate is 3.3%. As the number of observed deaths is 0, the probability (.p1 ) that we have correctly estimated the number of deaths can be determined, using binomial distribution, to be p1 =

.

10! 0.03310 0.96710−10 = 0.71. (10 − 0)!0!

Here, the probability has a subscript 1, because this is the probability for Data Item 1. We add similar subscripts for the other data items, and by repeating the same computation for the other items, we obtain the probabilities (.p1 , p2 , . . . , p5 ) of the respective death rates. Under the assumption that the death rate at each concentration is independent of the others, the total probability that we have correctly estimated the observed rate is the joint probability of these probability values, which is given by multiplying the values: l = p1 × · · · × p 5 ,

.

where l is the likelihood of such correct estimation. This l value is often very small and decreases as the number of observed data items increases. We log-transform the value of the likelihood to obtain the log-likelihood (ll): ll = log(p1 ) + log(p2 ) + · · · + log(p5 ),

.

which, in this example, the value of it is .

log(ll) = −6.47.

We repeat this process for slightly different .a, b values, and if the new ll is larger than the previous one, we adopt the new values. This is continued until we can find no .a, b values that can increase the value of ll. To obtain such .a, b values, we need to perform a numerical search. I personally prefer to write the code for such searches myself, but this is often not necessary, because most statistical software (such as R)

8.4 The Confidence Interval 10 5

logit (p)

Fig. 8.2 Conventional regression analysis (dashed line) compared with maximum likelihood estimation (dark Line). For the two extreme data values, .p = 0 is replaced by .p = 0.0001, and .p = 1 by .p = 1 − 0.0001. The conventional method seems to put on more weight on the two extreme points

191

0 -5 -10 -1.4

-1.3

-1.2

-1.1

-1.0

-0.9

-0.8

- 0.7

Log(concentration) can do it for us (using the glm function, for example). Nonetheless, I consider the exercise to be worthwhile, at least once. A comparison of the results of classical regression analysis and maximum likelihood estimation is shown in Fig. 8.2. In ecotoxicology, there is debate about whether the focus of the evaluation should be EC10 or NOEC. In a sense, this is a battle between the ANOVA group and the regression group; however, the choice of which to use in ecological risk assessment should, I believe, be case dependent. From a practical point of view, EC10 and NOEC are not opposing concepts. It would certainly be simpler to rely exclusively on one or the other (two indicators often confuse the risk assessors), but from the point of view of ecosystem conservation, it is better to have multiple indicators, and therefore excluding either one seems unwise. At the same time, I am concerned about the widespread misconception that NOECs are an indicator of no effect. Any management practices designed on the basis of false knowledge are not merely incorrect but potentially dangerous; thus my earlier suggestion, to restrict NOEC use to those with a proper understanding of what it means.

8.4 The Confidence Interval Since the regression line is also an estimate, there is uncertainty, and this uncertainty is expressed in terms of the confidence interval. In the previous section, we saw that the regression coefficients are estimated so as to minimize the y-axis error. The breadth of the uncertainty, then, is expressed in terms of vertical distance. The regression analysis gives no information on horizontal uncertainty. Since a regression line is a model that predicts the value of y for given x, the concept of uncertainty in the x-axis (horizontal) direction does not exist. Examples of the confidence interval are shown in Fig. 8.3A and B.

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8 Statistics and Related Topics

B

Death rate

A

Concentration

Concentration

Fig. 8.3 Examples of regression analysis confidence intervals. The intervals are expressed in terms of y-axis values (e.g., the two triangles in Panel A mark the upper and lower limits of the interval). By computing the confidence interval for all the concentrations, we obtain the two dotted lines showing the range of the upper and lower confidence limits (Panel B)

The Wald-based confidence interval is calculated as follows: p(c) ˆ =

.

1 ˆ 1 + exp[−aˆ − bc]

(8.5)

,

where .zˆ represents the estimated value of z (where z is a, b, and .P (c)). Let the sample size be n, and the toxicity test concentrations be .c1 , c2 , . . . , cn , in order from ˆ 1 ), p(c ˆ 2 ), . . . , p(c ˆ n ). lowest to highest. The respective predicted effect rates are .p(c We prepare two matrices: ⎛

1 ⎜1 ⎜ .c = ⎜ . ⎝ ..

⎞ c1 c2 ⎟ ⎟ .. ⎟ , . ⎠

(8.6)

1 cn and ⎛

ˆ 1 )) 0 ... np(c ˆ 1 )(1 − p(c ⎜ )(1 − p(c ˆ )) . .. 0 n p(c ˆ 2 2 ˆ =⎜ .I(a, ˆ b) ⎜ .. .. . .. ⎝ . . 0

0

0 0 .. .

⎞ ⎟ ⎟ ⎟ ⎠

ˆ n )). . . . np(c ˆ n )(1 − p(c (8.7)

ˆ i )(1 − p(c ˆ i ) term in the second matrix is a variance of the binomial The .np(c distribution when the probability is .pˆ i . Using these two matrices, we obtain the

8.4 The Confidence Interval

193

Fig. 8.4 Determination of the concentration confidence interval (see also Fig. 7.4 in Chap. 7). .C L and .C U represent the lower and upper limits of the interval. Since the vertical axis is logit(p), the S-shaped function in Fig. 8.3 is here expressed as a linear function

10

logit (p)

5 0

cL

EC50

cU

-5 -10 -3.5

-3.0

-2.5

-2.0

-1.5

Concentration variance-covariance matrix for the parameters: ˆ = .c I(a, ˆ b)c



T

σˆa σˆ ab σˆ ab σˆb

 .

(8.8)

When we write the logistic model as p=

.

1 , 1 + exp[−f (c)]

we determine the confidence interval using  f (c) = a + bc ± K σˆa + 2σˆ ab c + σˆb c2 ,

.

(8.9)

where K is the 95%ile value of the standard normal distribution. The confidence intervals in Fig. 8.4 are derived from this equation. In environmental science, we often wish to know the confidence interval of the concentration that causes a certain effect (i.e., the confidence interval in a horizontal direction), rather than the confidence interval of the effect at a certain concentration. As noted above, in regression analysis, the regression line is determined based on minimization of the y-axis error, and the confidence intervals are determined based on the distribution of this error. However, since such analysis yields no knowledge of the x-axis error, it would seem to be impossible in principle to obtain the horizontal confidence interval, and the means of calculating it has been a longstanding question for me. We know that this interval is determined by the intersections of a horizontal line at a certain y-value (effect) and the vertical confidence intervals (Fig. 8.4). I personally learned this from the valuable textbook by Draper and Smith [3]. However, the book does not explain whether the confidence interval so obtained meets the general requirements for a confidence interval. A 95% confidence interval, for example, means that there is a 5% probability that the value of the variable in question is out of the range of the interval. Thus, recalling the case discussed in Chap. 5 (Sect. 5.4.3), when the weight per bag of potato chips is stated as 100 .± 5 g

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8 Statistics and Related Topics

(the level of significance is usually stated explicitly only in scientific studies, but in most cases is 5%), it means that the lucky probability of buying a bag with more than 105 g of chips is 2.5%, and the unlucky probability of buying a bag with less than 95 g of chips is 2.5%. Despite my longstanding interest in this question (and the latter’s practical implications to ecological management), I have never come across a means of determining whether the horizontal confidence interval meets this requirement; perhaps it does, but I doubt it. As it stands, I can only conclude that the horizontal confidence interval is simply a definition (without theoretical justification). One assumes there is some extant work in which it was first defined (and its use, one would hope, justified), but I have not been able to find it. There is, for me, a further mystery surrounding the confidence interval. The values that I calculate for the intersection points between the horizontal EC50 line and the dashed lines in Fig. 8.4 differ slightly different from those computed by the standard software functions. The reason for such difference is somewhat involved. We first need to understand how the confidence interval is computed by the software functions. Some compute this confidence interval in the following way. The y-axis confidence interval is calculated using the variance-covariance matrix in Eqs. (8.8) and (8.9). By arranging Eq. (8.9) as  f (c) = a + b(c ± K σˆa + 2σˆab c + σˆb c2 /b),

.

the formula inside the parenthesis in the second term is  c ± K σˆa + 2σˆab c + σˆb c2 /b.

.

(8.10)

This looks as if it represents the concentration confidence interval, and indeed, the confidence interval for the concentration at a given effect is computed, by some software functions, using this formula. The resulting confidence interval is, however, slightly different from that derived by the method shown in Fig. 8.4. Figure 8.5 illustrates the reason for this difference. Let O be the intersection of the horizontal 50%-effect line and the vertical line at c (i.e., EC50). Point A is the upper confidence interval limit of the effect at this concentration, and the distance between O and A, as computed by Eq. (8.10), is OA = logit(pU ) − logit(p)  = a + bc + K σˆa + 2σˆ ab c + σˆb c2 − (a + bc)  = K σˆa + 2σˆ ab c + σˆb c2 .

.

(8.11)

Point B is the lower confidence interval limit of the concentration, and the distance between O and B, as computed by Eq. (8.10), is  OB = K σˆa + 2σˆ ab c + σˆb c2 /b.

.

(8.12)

8.4 The Confidence Interval

1.0

logit ( p)

Fig. 8.5 Comparison of the concentration confidence interval ranges determined by a method by Draper and Smith [3] and that by standard software function

195

A

cL

0.5 0.0

50% line

B

-0.5

s

Be

O

-1.0 -2.7

cU

t t fi

c -2.6

-2.5

-2.4

-2.3

-2.3

Concentration Hence, the line that passes through A and B has the slope  OA K σˆa + 2σˆ ab c + σˆb c2 = b. . =  OB K σˆa + 2σˆ ab c + σˆb c2 /b

(8.13)

This line is parallel to the best-estimate line. The confidence interval limits represented by the dashed lines are not straight, but broad U-shaped curves. The lines demarcating the concentration confidence interval computed by Eq. (8.10) are approximations of these curves using a linear model; thus, the lower confidence limit derived from Eq. (8.10) will always be higher than .C L , which marks the intersection of the horizontal 50%-effect line and the dashed line that defines the upper confidence interval limit. Similarly, the upper concentration confidence interval limit will always be lower than .C U . As aforementioned, several software functions determine the concentration confidence interval using Eq. (8.10), and the approximation approach illustrated above is the standard method for this determination, though so far as I know, its validity has never been confirmed. However, I am not sure that the alternate method shown in Fig. 8.4 is wrong and thus deserving of rejection. The confidence interval is a highly important element in risk assessment, and it is at least problematic that there are several ways for determining it, each of which generates different values. Despite the fact that confidence intervals for concentrations (or explanatory variables in general) are frequently used in various fields of empirical research, including environmental science, there appears to be a lack of attention paid to the methods and implications of their derivation. Once we learn that there are two different and seemingly legitimate ways to derive the confidence interval, we may be inclined to be rather critical when reviewing papers that employ them. When we find a confidence interval derived by Eq. (8.10), for example, we might say “this interval was derived by approximation, but precise values should be derived.” Only a few people other than the readers of this book will understand this comment, and the rest may be excused for wondering what it even means.

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8 Statistics and Related Topics

8.5 AIC and Model Selection The AIC, which originally meant “an information criterion” but is now considered to be an abbreviation of “Akaike’s information criterion” is a widely used criterion for model selection. It is often considered that a model selected based on the AIC reflects the truth, but this is a misconception; the AIC is not a criterion for selecting the correct model [7]. The easiest way to maximize the predictive power of the data that we now have is to increase the number of parameters and variables. Using machine learning, which has become popular in recent years, it is possible to perform regression analysis that takes into account all the points of the data in hand. However, when the number of parameters and variables increases, such analysis explains well the variation in the data in hand (i.e., the higher .R 2 value) but loses its predictive power for newly observed data. For example, suppose we have the data shown in Fig. 8.6. We consider the following models for fitting: lm1 : y = a + b1 x,

.

lm2 : y = a + b1 x + b2 x 2 , .. . lm10 : y = a +

10 

bi x i .

Observed y

i=1

Observed x Fig. 8.6 An example of data in hand

(8.14)

8.5 AIC and Model Selection

197

Observed y

lm1

10

lm10 Observed x Fig. 8.7 Comparison of the results for lm1 (dashed line) and lm10 (solid line). lm10 succeeds in passing through almost all the points

We perform regression analysis on each model to estimate the parameters. The predictions by lm1 and lm10, with estimated parameters, are shown in Fig. 8.7. Model lm10 successfully draws a line through almost all the points. If we use the criterion that a model that passes through as many points as possible is a good model, then lm10 is the best, and the .R 2 value for lm10 is higher than that for the simplest model (lm1). However, lm10 is never selected for model selection using AIC. Although lm10 seems to be more predictive at a glance, the AIC of lm1 is lower than that of lm10, and hence we must select lm1 as a better model. Suppose, now, that we conduct additional research and obtain the new data shown in Fig. 8.8. Figure 8.9 incorporates the new data, as well as the regression lines of lm1 and lm10, from Fig. 8.8. lm10, which had the highest .R 2 value for the previous dataset shown in Fig. 8.6, is making mostly poor predictions for the new data. The reason for this is that lm10 overfits the previous dataset with excessive parameters. Model selection based on the AIC can balance the two conflicting pressures, to reduce the number of parameters and avoid such overfitting, on the one hand, and to increase the number of parameters to increase the predictive power, on the other. For the data in Fig. 8.6, the AIC of lm2 (among lm1 to lm10) is the lowest, and therefore from the AIC’s point of view, we should select lm2. Figure 8.10 compares the predictions of lm1 and lm2. Compared to lm1, lm2 has a lower AIC and higher .R 2 value. This, however, does not mean that lm2 is the more correct model. In fact, the data in Fig. 8.6 are generated by a linear model (lm1) with noise: y = 10 + 2x + ξ ,

.

(8.15)

8 Statistics and Related Topics

New y

198

New x

New y

Fig. 8.8 Newly obtained data

New x Fig. 8.9 Comparison of the results for lm1 (dashed line) and lm10 (solid line) for new data. lm10 may no longer be a good model

199

Observed y

8.5 AIC and Model Selection

Observed x Fig. 8.10 Comparison of lm1 (dashed line) and lm2 (solid line). The true model is a linear model in Eq. (8.15), but lm2 is selected by AIC

and ξ ∼ N (0, 42 ).

.

Not only does AIC not select the correct model but the very concept of what constitutes the correct model is not factored into model selection based on the AIC. To see how well the AIC selects the correct model, we will repeat the above process 10,000 times, generating data from Eq. (8.15) and using the AIC to select the model. The results are shown in Fig. 8.11. The correct model in this case is a linear model corresponding to lm1 in Eq. (8.15). When the sample size is 20, the correct model will be selected about 60% of the time, regardless of the standard deviation in the error term (Eq. (8.15)). As the sample size increases to 200, the frequency of correct models being selected increases, but is still around 70%. These results show that considering the model selected by the AIC to be correct has a high probability of being wrong. It may seem rather surprising that the AIC selects the correct model with a relatively high probability of 60% or 70%, even though it has no concept of what the correct model is. Higher-order models are also selected frequently, with lm10 being chosen in about 5% of the cases. When such a high-order model happens to be selected, we may debate why (e.g., discuss the importance of the specific variables included in the model), but such debate may be quite pointless. In the case of Fig. 8.10, for example, such discussion is clearly useless because correct model is a linear model.

200

8 Statistics and Related Topics 0.8

0.6

n=20

n=200

σ=4

σ=4

n=20

n=200

σ=0.01

σ=0.01

Frequency

0.4

0.2

0.0 0.8

0.6

0.4

0.2

0.0 2

4

6

8

10

2

4

6

8

10

Model number Fig. 8.11 Frequency histogram of the selected model. The horizontal axis represents the model number (2 represents lm2 in Eq. (8.14)). The value of the explanatory variable (x) was randomly selected from a uniform random distribution between 0 and 10. The n values in the panels represent the sample size generated from the model in Eq. (8.15), and the .σ values represent the standard deviations for the error term in Eq. (8.15)

We sometimes encounter the following question when performing model selection using the AIC: Some of the regression coefficients in the model selected by the AIC are “not significant.” When making predictions using this model, should this parameter be added or not?

The answer is: It should be added, if the aim is better prediction.

Some may think that non-significant regression coefficients should not be incorporated into a model and some hesitate to incorporate such the non-significant coefficients into a model. However, model selection is not about selecting the correct model, but about maximizing predictive power, so if incorporating the regression coefficient increase predictive power, there is no reason to hesitate to do so. Although this is only my opinion, I think the reason why we hesitate to incorporate the explanatory variables with non-significant coefficients in models is that we incorrectly think that non-significant regression coefficients mean the magnitude of the coefficients are zero. Again, insignificance does not guarantee that the null hypothesis is true. Since model selection using AIC does not ask whether the regression coefficients reflect a causal effect, minimizing the AIC is the single most important criterion. But in return, there is no point in asking what the regression coefficients mean because we can never be sure if it reflects the truth correctly.

8.6 SSDs, Hypothesis Testing, and Model Selection

201

8.6 SSDs, Hypothesis Testing, and Model Selection In the species sensitivity distributions (SSDs) introduced in Chap. 5, hypothesis testing and model selection are also used to determine the probability model [4]. In assessments using SSDs, the concentration that affects 5% (HC5) of the species is often used as a management target. The concern, in this respect, is that each probability model is to some extent skewed, and the HC5 value, which lies near the lower end of such distributions, differs with each model. For example, compared to the normal distribution, the logistic distribution has a longer tail, and thus the logistic distribution’s HC5 value is lower than that of the normal distribution. Thus, determining a probability model of the population from which the SSD study data are sampled is an important task. In hypothesis testing, for example, Anderson-Darling’s goodness-of-fit test is used. The null hypothesis for this test is .H0 :

the observed data are sampled from this (e.g., normal) distribution.

If this hypothesis is rejected, the data can be judged not to be a sample from that distribution; however, if it is not rejected, we have actually obtained no new and useful result by means of the statistical test. The hypothesis testing, then, identifies distributions we should not use but does not identify distributions we should use. When selecting a probability model through model selection, even if a normal distribution, for example, is selected, this simply means that this distribution type is the most predictable for a given dataset; it does not tell us if this type is true. Truth, in this respect, is still “in the bush.” As far as I know, there is no way to identify what the probability model of the population is. Indeed, it is not hard to see how meaningless it is to ask what the population is. If we wish to determine functional shapes of the population, statistical analysis alone is not enough; we need a mechanistic approach. For example, the Weibull distribution is theoretically derived by analyzing a mechanism-based model that describes the process for determining the time required for a machine to break down. Similarly, if the proportion of species affected by a given chemical could be described mechanistically, it would be possible to choose a probability model that likely reflects the population. Without this, however, it is impossible to identify the true population distribution. In SSD-based risk assessment, an assessment factor (AF) is sometimes used to determine the predicted no effect concentration (PNEC), which may be used as a management goal. In Chap. 5, we saw how the inadvertent use of this factor makes it impossible to know exactly how well we are protecting a given ecosystem. In Chap. 5, the discussion was conducted from a frequentist standpoint, and from this standpoint, the magnitude of the assessment factor must be determined so that it nullifies the difference between the estimated HC5 and the population HC5. From the standpoint of model selection, then, how should we determine the magnitude of the assessment factor? My answer is simply, “Don’t use it.” In model selection, we do not ask if the selected model reflects the truth; the data, and the data alone, does

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8 Statistics and Related Topics

this. If we think the data is dubious, then the selected model is also dubious. If we think we need an AF, it means that the data does not provide sufficient information for setting management goals. In this case, we cannot perform model selection, because any model selected based on such dubious data is likely suboptimal. There are basically two approaches to SSD-based ecological risk assessment, the frequentist approach and the model selection approach, and each has different underlying concepts. Moreover, the magnitude of the AF can vary with the respective approach. Thus far, however, there has been a lack of discussion regarding the proper use of the AF in this regard.

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