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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|LAGRANGE games

The analysis of zero-sum games is closely connected with a fundamental technique in mathematical optimization. A very general formulation of an optimization problem is
$$\max {x \in \mathcal{F}} f(x),$$ where $\mathcal{F}$ could be any set and $f: \mathcal{F} \rightarrow \mathbb{R}$ an arbitrary objective function. In our context, however, we will look at more concretely specified problems and understand by a mathematical optimization problem a problem of the form $$\max {x \in X} f(x) \quad \text { such that } \quad g(x) \geq 0$$
where $X$ is a nonempty subset of some coordinate space $\mathbb{R}^{n}$ with an objective function $f: X \rightarrow \mathbb{R}$.

The vector valued function $g: X \rightarrow \mathbb{R}^{m}$ is a restriction function and combines $m$ real-valued restriction functions $g_{i}: X \rightarrow \mathbb{R}$ as its components. The set of feasible solutions of $(22)$ is
$$\mathcal{F}=\left{x \in X \mid g_{i}(x) \geq 0 \text { for all } i=1, \ldots, m\right}$$
ReMARK 3.2. The model (22) formulates an optimization problem as a maximization problem. Minimization problems can, of course, also be formulated within this model because of
$$\min {x \in \mathcal{F}} f(x)=-\max {x \in \mathcal{F}} \tilde{f}(x) \quad \text { with the objective } \tilde{f}(x)=-f(x)$$
The optimization problem (22) defines a zero-sum game $\Lambda=$ $\left(X, \mathbb{R}{+}^{m}, L\right)$ with the so-called LAGRANGE function $$L(x, y)=f(x)+y^{T} g(x)=f(x)+\sum{i=1}^{m} y_{i} g_{i}(x)$$ as its utility. We refer to $\Lambda$ as a LAGRANGE game. ${ }^{6}$ The worst-case utility functions of the two players in $\Lambda$ are:
\begin{aligned} &L_{1}(x)=\min {y \geq 0} L(x, y)=f(x)+\min {y \geq 0} \sum_{i=1}^{m} y_{i} g_{i}(x) \ &L_{2}(y)=\max {x \in X} L(x, y)=\max {x \in X} f(x)+\sum_{i=1}^{m} y_{i} g_{i}(x) \end{aligned}

## 经济代写|博弈论代写Game Theory代考|Complementary slackness

The choice of an element $x \in X$ with at least one restriction violation $g_{i}(x)<0$ would allow the $y$-player in the LAGRANGE game $\Lambda=\left(X, \mathbb{R}{+}^{m}, L\right)$ to increase its utility value infinitely with the choice $y{i} \approx \infty$. So the risk avoiding $x$-player will always try to select a feasible $x$.

On the other hand, if all of the feasibility requirements $g_{i}(x) \geq 0$ are satisfied, the best the $y$-player can do is the selection of $y \in \mathbb{R}{+}^{m}$ such that the so-called complementary slackness condition $$\sum{i=1}^{m} y_{i} g_{i}(x)=y^{T} g(x)=0 \quad \text { and hence } \quad L(x, y)=f(x)$$
is met. Consequently, one finds:

The primal LAGRANGE problem is identical with the original problem:
$$\max {x \in X} \min {y \geq 0} L(x, y)=\max {x \in \mathcal{F}} L{1}(x)=\max {x \in \mathcal{F}} f(x) .$$ LEMMA 3.1. If $\left(x^{}, y^{}\right)$ is an equilibrium of the LAGRANGE game $\Lambda$, then $x^{}$ is an optimal solution of problem (22). Proof. For every feasible $x \in \mathcal{F}, y^{} \geq 0$ yields $\left(y^{}\right)^{T} g(x) \geq 0$ and, therefore, $$f\left(x^{}\right)=L{1}\left(x^{}\right)=L_{2}\left(y^{}\right) \geq f(x)+\left(y^{}\right)^{T} g(x) \geq f(x)$$ So $x^{}$ is optimal.

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

$$\max x \in \mathcal{F} f(x),$$

$\max x \in X f(x) \quad$ such that $\quad g(x) \geq 0$

$\backslash 1 \mathrm{eft}$ 的分隔符缺失或无法识别

$$\min x \in \mathcal{F} f(x)=-\max x \in \mathcal{F} \tilde{f}(x) \quad \text { with the objective } \tilde{f}(x)=-f(x)$$

$$L(x, y)=f(x)+y^{T} g(x)=f(x)+\sum i=1^{m} y_{i} g_{i}(x)$$

$$L_{1}(x)=\min y \geq 0 L(x, y)=f(x)+\min y \geq 0 \sum_{i=1}^{m} y_{i} g_{i}(x) \quad L_{2}(y)=\max x \in X L(x, y)=\max x \in X f(x)+\sum_{i=1}^{m} y_{i} g_{i}(x)$$

## 经济代写|博弈论代写Game Theory代考|Complementary slackness

$$\sum i=1^{m} y_{i} g_{i}(x)=y^{T} g(x)=0 \quad \text { and hence } \quad L(x, y)=f(x)$$

$$\max x \in X \min y \geq 0 L(x, y)=\max x \in \mathcal{F} L 1(x)=\max x \in \mathcal{F} f(x) .$$

$$f(x)=L 1(x)=L_{2}(y) \geq f(x)+(y)^{T} g(x) \geq f(x)$$

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