# 经济代写|博弈论代写Game Theory代考|ECON40010

## 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 $$\min {y \in \bar{Y}} \bar{U}2(y)=\min {w \in \mathbb{R}, y \in \bar{Y}}\left{w \mid w \geq \sum_{j=1}^n u_{i j} y_j \text { for all } j=1, \ldots, n\right} \text {. }$$

## 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 $$\max {x \in X} f(x) \quad \text { such that } g(x) \geq 0,$$
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
$\min {x \in \mathcal{F}} f(x)=-\max {x \in \mathcal{F}} \tilde{f}(x) \quad$ 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.

# 博弈论代考

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

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

$\backslash \max {x \backslash$ in $\backslash \operatorname{bar}{X}} U_{-} 1(x)=\backslash \max {z \backslash$ in $\backslash \operatorname{mathbb}{R}, x \backslash$ in $\backslash b a r{X}} \backslash$ left $\left{z \backslash \operatorname{mid} z \backslash l\right.$ eq $\backslash$ sum_ ${i=1}^{\wedge} m u_{-}{i j} x_{-} i \backslash t$ text ${f$

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

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

$$\max x \in X f(x) \quad \text { such that } g(x) \geq 0,$$

$\backslash$ mathcal ${F}=\backslash$ left $\left{x \backslash\right.$ in $X \backslash$ mid g_ $g_i(x) \backslash$ geq $0 \backslash$ text ${$ for all $}$ i $=1$, $\backslash$ ldots, $m \backslash$ right $}$ 。

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