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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|Randomized matrix games

Recall from Ex. $5.2$ that a zero-sum game $\Gamma=(X, Y, U)$ with finite strategy sets $X$ and $Y$ does not necessarily admit an equilibrium.
Suppose the players randomize the choice of their respective strategies. That is to say, the $x$-player decides on a probability distribution $\bar{x}$ on $X$ and chooses an $i \in X$ with probability $\bar{x}{i}$. Similarly, the $y$-player chooses a probability distribution $\bar{y}$ on $Y$ and selects $j \in Y$ with probability $\bar{y}{j}$. Then the $x$-player’s expected gain is
$$\bar{U}(\bar{x}, \bar{y})=\sum_{i \in X} \sum_{j \in Y} u_{i j} \bar{x}{i} \bar{y}{j}$$
where the $u_{i j}$ are the coefficients of the utility matrix $U$.
So we arrive at a zero-sum game $\bar{\Gamma}=(\bar{X}, \bar{Y}, \bar{U})$, where $\bar{X}$ is the set of probability distributions on $X$ and $\bar{Y}$ the set of probability distributions on $Y . \bar{X}$ and $\bar{Y}$ are compact convex sets (cf. Ex. 3.4).
Moreover, the function $\bar{U}: \bar{X} \times \bar{Y} \rightarrow \mathbb{R}$ is linear, and thus continuous and both concave and convex in both components.

It follows that $\bar{\Gamma}$ is a convex game that satisfies the hypothesis of Theorem 3.1. Therefore, $\bar{\Gamma}$ admits an equilibrium. This proves the VON NEUMANN’s Theorem:

THEOREM $3.2$ (VON NEUMANN [32]). Let $U \in \mathbb{R}^{m \times n}$ be an arbitrary matrix with coefficients $u_{i j}$. Then there exist $x^{} \in X$ and $y^{} \in Y$ such that
$$\max {1 \leq i \leq m} \min {1 \leq y \leq m} \sum_{i, j} u_{i j} x_{i} y_{j}=\sum_{i, j} u_{i j} x_{i}^{} y_{j}^{}=\min {1 \leq j \leq n} \max {1 \leq i \leq m} \sum_{i, j} u_{i j} x_{i} y_{j} .$$
where $X$ is the set of all probability distributions on ${1, \ldots, m}$ and $Y$ the set of all probability distributions on ${1, \ldots, n}$.

## 经济代写|博弈论代写Game Theory代考|Computational aspects

While it is generally not easy to compute equilibria in zero-sum games, the task becomes tractable for randomized matrix games. Consider, for example, the two sets $X={1, \ldots, m}$ and $Y=$ ${1, \ldots, n}$ and the utility matrix
$$U=\left[\begin{array}{cccc} u_{11} & u_{21} & \ldots & u_{1 n} \ u_{21} & u_{22} & \ldots & u_{2 n} \ \vdots & \vdots & & \vdots \ u_{m 1} & u_{m 2} & \ldots & u_{m n} \end{array}\right] \in \mathbb{R}^{m \times n}$$
For the probability distributions $x \in \bar{X}$ and $y \in \bar{Y}$, the expected utility for the $x$-player is
$$\bar{U}(x, y)=\sum_{i=1}^{m} \sum_{j=1}^{n} u_{i j} x_{i} y_{j}=\sum_{j=1}^{n} y_{j}\left(\sum_{i=1}^{m} u_{i j} x_{i}\right) .$$
The worst case for the row player occurs when the column player selects a probability distribution that puts the full weight 1 on $k \in Y$ such that
$$\sum_{i=1}^{m} u_{i k} x_{i}=\min \left{\sum_{i=1}^{m} u_{i j} x_{i} \mid j=1, \ldots, n\right}=\bar{U}_{1}(x) .$$

Hence
$$\max {x \in \bar{X}} U{1}(x)=\max {z \in \mathbb{R}, x \in \bar{X}}\left{z \mid z \leq \sum{i=1}^{m} u_{i j} x_{i} \text { for all } i=1, \ldots, m\right} \text {. }$$
Similarly, the worst case for the column player it attained when the row player places the full probability weight 1 onto $\ell \in X$ such that
$$\sum_{j=1}^{n} u_{\ell j} y_{j}=\max \left{\sum_{j=1}^{n} u_{i j} y_{j} \mid i=1, \ldots, m\right}=\bar{U}{2}(y) .$$ This yields $$\min {y \in \bar{Y}} \bar{U}{2}(y)=\min {w \in \mathbb{R}, y \in \bar{Y}}\left{w \mid w \geq \sum_{j=1}^{n} u_{i j} y_{j} \text { for all } j=1, \ldots, n\right}$$

## 经济代写|博弈论代写Game Theory代考|Randomized matrix games

$$\bar{U}(\bar{x}, \bar{y})=\sum_{i \in X} \sum_{j \in Y} u_{i j} \bar{x} i \bar{y} j$$

$$\max 1 \leq i \leq m \min 1 \leq y \leq m \sum_{i, j} u_{i j} x_{i} y_{j}=\sum_{i, j} u_{i j} x_{i} y_{j}=\min 1 \leq j \leq n \max 1 \leq i \leq m \sum_{i, j} u_{i j} x_{i} y_{j} .$$

## 经济代写|博弈论代写Game Theory代考|Computational aspects

$$\bar{U}(x, y)=\sum_{i=1}^{m} \sum_{j=1}^{n} u_{i j} x_{i} y_{j}=\sum_{j=1}^{n} y_{j}\left(\sum_{i=1}^{m} u_{i j} x_{i}\right)$$

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