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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 经济代写|博弈论代写Game Theory代考|Evolutionarily Stable Strategies

Consider a large population in which individuals are paired at random for a twoplayer game. An individual following strategy $x^{\prime}$ in this game gets payoff $W\left(x^{\prime}, x\right)$ when the partner follows strategy $x$. Maynard Smith (1982) defines a strategy $x^{}$ to be an ESS if, for each strategy $x^{\prime} \neq x^{}$, one of the two conditions (ES1) and (ES2) holds: (ES1) $W\left(x^{\prime}, x^{}\right)}, x^{}\right)$ (i) $W\left(x^{\prime}, x^{}\right)=W\left(x^{}, x^{}\right)$ and (ii) $W\left(x^{\prime}, x^{\prime}\right)<\left(W\left(x^{}, x^{\prime}\right)\right.$ The motivation behind this definition is as follows. Suppose the resident strategy is $x^{}$ and $x^{\prime}$ is a mutant strategy that is different from $x^{}$. Initially the mutant will be very rare. It will thus almost always partner residents (as do residents). If condition (ES1) holds it does worse than residents, and so will be selected against and disappear. Suppose that condition (ES1) fails to hold, but condition (ES2)(i) holds. Then the mutant will have equal fitness to residents when very rare. If mutant numbers increase by random drift, both mutants and residents will occasionally partner other mutants. Condition (ES2)(ii) says that when this happens mutants do worse than residents in such contests. This will tend to reduce mutant numbers so that the mutant again becomes very rare. Overall, the mutant cannot invade, in the sense that the proportion of mutants will always remain very small. These ideas are formalized in Box $4.1$ where we derive an equation for the rate of change in the proportion of mutants over time. Note that if $x^{}$ is an ESS it is also a Nash equilibrium strategy since the Nash condition in eq (2.4) is that all mutants must either satisfy (ES1) or (ES2)(i). The converse is not true. As we will illustrate, there are many examples of Nash equilibrium strategies $x^{}$ that are not ESSs. However, if $x^{}$ is the unique best response to itself, then (ES1) holds for all mutants, so that $x^{*}$ is an ESS. At least one Nash equilibrium exists for (almost) any game, but because the ESS criterion is stronger than the Nash equilibrium criterion, there are many games for which there is no ESS.

## 经济代写|博弈论代写Game Theory代考|Rate of change of the proportion of mutants

Suppose that a proportion $\epsilon$ of a large population uses strategy $x$ and a proportion $1-\epsilon$ uses strategy $x^{}$ in a two-player game with payoff function $W$. Then the expected payoff to each strategy can be written as \begin{aligned} w(\epsilon) &=(1-\epsilon) W\left(x, x^{}\right)+\epsilon W(x, x) \ w^{}(\epsilon) &=(1-\epsilon) W\left(x^{}, x^{}\right)+\epsilon W\left(x^{}, x\right) . \end{aligned}
The difference $\Delta=w(\epsilon)-w^{}(\epsilon)$ in payoff is $$\Delta=(1-\epsilon)\left[W\left(x, x^{}\right)-W\left(x^{}, x^{}\right)\right]+\epsilon\left[W(x, x)-W\left(x^{*}, x\right)\right] .$$
If condition (ES1) holds, the first square bracket on the right hand side of eq (4.1) is negative, so that $\Delta<0$ for sufficiently small $\epsilon$. If condition (ES2) holds, the first square bracket is zero and the second is negative, so that $\Delta<0$ for all $\epsilon$. Thus the Maynard Smith conditions ensure that $\Delta<0$ if the proportion of mutants is sufficiently small; in fact these conditions are equivalent. However, the conditions say nothing about what will happen if the proportion of mutants is allowed to become large.

We now consider the rate of change in $\epsilon$ over time. Let $n(t)$ and $n^{}(t)$ denote the numbers of individuals following strategies $x$ and $x^{}$, respectively at time $t$, so that $\epsilon(t)=n(t) /(n(t)+$ $\left.n^{}(t)\right)$. Taking the time derivative, we get $d \epsilon / d t=\epsilon(1-\epsilon)\left[\frac{1}{n} d n / d t-\frac{1}{n^{}} d n^{} / d t\right]$. Let the per-capita rate of increase of numbers following a strategy be equal to the payoff to the strategy in the game plus (or minus) a quantity that is the same for each strategy. That is $d n / d t=\left[w(\epsilon(t))+w_{0}(t)\right] n(t)$ and $d n^{} / d t=\left[w^{}(\epsilon(t))+w_{0}(t)\right] n^{}(t)$. Here $w_{0}(t)$ might depend on $n(t)$ and $n^{}(t)$, and so incorporate density-dependent effects. Using our definition of $\Delta$ we get that $$\frac{d \epsilon}{d t}=\epsilon(1-\epsilon) \Delta .$$ Thus the condition for a strategy $x^{}$ to be an ESS is equivalent to the condition that in a large population comprising a mixture of $x$ and $x^{*}$ strategists, the proportion of $x$ strategists will tend to zero if the initial proportion is sufficiently small. Equation (4.2) is a special case of the continuous-time replicator equation, when there are two types (Section 4.4). Derivation of the discrete time analogue is set as an exercise (Exercise 4.2).

## 经济代写|博弈论代写Game Theory代考|Rate of change of the proportion of mutants

$$w(\epsilon)=(1-\epsilon) W(x, x)+\epsilon W(x, x) w(\epsilon) \quad=(1-\epsilon) W(x, x)+\epsilon W(x, x) .$$

$$\Delta=(1-\epsilon)[W(x, x)-W(x, x)]+\epsilon\left[W(x, x)-W\left(x^{*}, x\right)\right]$$

㧴们现在考虑变化率 $\epsilon$ 随着时间的推移。让 $n(t)$ 和 $n(t)$ 表示遵㑑策略的个人数量 $x$ 和 $x$ ，分别在时间 $t$ ，以便 $\epsilon(t)=n(t) /(n(t)+n(t))$. 取时间导数，我们得到 $d \epsilon / d t=\epsilon(1-\epsilon)\left[\frac{1}{n} d n / d t-\frac{1}{n} d n / d t\right]$. 假设遵循策略的人均数字增长率等于游戏中策略的收益加上 (或减去) 每个策略相同的数量。那是
$d n / d t=\left[w(\epsilon(t))+w_{0}(t)\right] n(t)$ 和 $d n / d t=\left[w(\epsilon(t))+w_{0}(t)\right] n(t)$. 这里 $w_{0}(t)$ 可能取决于 $n(t)$ 和 $n(t)$ ，因此结合了密度相关效应。使用我们的定义 $\Delta$ 我们明白了
$$\frac{d \epsilon}{d t}=\epsilon(1-\epsilon) \Delta .$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师