Commit afa3d377 authored by Nick Ham's avatar Nick Ham

added changes

parent 5368e635
......@@ -38,20 +38,56 @@
Given a set $X$, we denote the set of probability distributions over $X$ as $\Delta(X)$, ie. $\Delta(X) = \{\sigma \in [0,1]^X: \Sigma_{x \in X}\sigma(x) = 1\}$. Need to clean this up.
Given a game $\Gamma = (N, A, u)$, for each player $i \in N$, the \textit{mixed strategies for player $i$}, which we denote by $\Delta(A_i)$, are the probability distributions over $A_i$, ie. $\Delta(A_i) = \{\sigma_i \in [0,1]^{A_i}: \Sigma_{s_i \in A_i} \sigma_i(s_i) = 1\}$. The \textit{mixed strategy profiles}, which we denote by $\Delta(A)$, are $\times_{i \in N}\Delta(A_i)$.
Given a game $\Gamma = (N, A, u)$, for each player $i \in N$, the \textit{mixed strategies for player $i$} are the probability distributions over $A_i$, ie. $\Delta(A_i) = \{\sigma_i \in [0,1]^{A_i}: \Sigma_{s_i \in A_i} \sigma_i(s_i) = 1\}$. The set of \textit{mixed strategy profiles} is the Cartesian product of the players' mixed strategies, ie. $\times_{i \in N}\Delta(A_i)$.
Need to argue $\Delta(A) = \{\sigma \in [0,1]^A: \Sigma_{s \in A}\sigma(s) = 1\} \cong \times_{i \in N}\Delta(A_i)$ equipped with for each $\sigma \in \times_{i \in N}\Delta(A_i)$, $\sigma(s) = \prod_{i \in N}\sigma_i(s_i)$ for all $s \in A$.
Need to do expected utility. For each player $i \in N$, the domain for their utility function can be extended linearly from $A$ to $\Delta(A)$ with $u_i(\sigma) = \Sigma_{s \in A}\sigma(s)u_i(s)$ for all $\sigma \in \Delta(A)$.
Define $f:\times_{i \in N}\Delta(A_i)\rightarrow\Delta(A)$ where for each $\sigma = (\sigma_1, \ldots, \sigma_n) \in \times_{i \in N}\Delta(A_i)$, we let $f(\sigma)(s) = f(\sigma_1, \ldots, \sigma_n)(s) = \prod_{i \in N}\sigma_i(s_i)$ for all $s \in A$.
\begin{proposition}
Given a game $(N, A, u)$, for each player $i \in N$, $\Delta(A)_{-i} \cong \Delta(A_{-i})$.
The function $f$ defined above satisfies:
\begin{enumerate}
\item For each $\sigma = (\sigma_1, \ldots, \sigma_n) \in \times_{i \in N}\Delta(A_i)$, $f(\sigma) = f(\sigma_1, \ldots, \sigma_n) \in \Delta(A)$ (ie. $f$ is well-defined);
\item For each $\sigma, \sigma' \in \times_{i \in N}\Delta(A_i)$, $f(\sigma) = f(\sigma')$ if and only if $\sigma = \sigma'$ (ie. $f$ is injective).
\end{enumerate}
\begin{proof}
\begin{enumerate}
\item \begin{align*}
\sigma_i(s_i) &\geq 0 \text{ for all } i \in N, s_i \in A_i \\
\Rightarrow f(\sigma)(s) = f(\sigma_1, \ldots, \sigma_n)(s) = \prod_{i \in N} \sigma_i(s_i) &\geq 0 \text{ for all } s \in A.
\end{align*}
Also:
\begin{align*}
\sum_{s \in A}f(\sigma)(s) = \sum_{s \in A}f(\sigma_1, \ldots, \sigma_n)(s) &= \sum_{s \in A}\prod_{i \in N}\sigma_i(s_i) \\
&= \sum_{s_1 \in A_1}\sigma_1(s_1)\left[\sum_{s_{-1} \in A_{-1}}\prod_{i \in N_{-1}}\sigma_i(s_i)\right] \\
&= \sum_{s_{-1} \in A_{-1}}\prod_{i \in N_{-1}}\sigma_i(s_i) \\
&= \sum_{s_2 \in A_2}\sigma_2(s_2)\left[\sum_{s_{-1-2} \in A_{-1-2}}\prod_{i \in N_{-1-2}}\sigma_i(s_i)\right] \text{ (need to fix this notation)} \\
&= \sum_{s_{-1-2} \in A_{-1-2}}\prod_{i \in N_{-1-2}}\sigma_i(s_i) \\
&\hspace{2mm} \vdots \\
&= \sum_{s_n \in A_n} \sigma_n(s_n) = 1.
\end{align*}
\item For each $\sigma, \sigma' \in \times_{i \in N}\Delta(A_i)$,
\begin{align*}
f(\sigma) &= f(\sigma') \\
\Rightarrow f(\sigma)(s) &= f(\sigma')(s) \text{ for all } s \in A \\
\Rightarrow \prod_{i \in N}\sigma_i(s_i) &= \prod_{i \in N}\sigma'_i(s_i)
\end{align*}
\textbf{RTS:} $\sigma = \sigma'$.
\end{enumerate}
\end{proof}
\end{proposition}
\begin{corollary}
$\times_{i \in N}\Delta(A_i) \subset \Delta(A)$.
\end{corollary}
%Given a game $\Gamma = (N, A, u)$, for each player $i \in N$, the \textit{mixed strategies for player $i$}, which we denote by $\Delta(A_i)$, are the probability distributions over $A_i$, ie. $\Delta(A_i) = \{\sigma_i \in [0,1]^{A_i}: \Sigma_{s_i \in A_i} \sigma_i(s_i) = 1\}$. The \textit{mixed strategy profiles}, which we denote by $\Delta(A)$, are $\times_{i \in N}\Delta(A_i)$.
%Need to argue $\Delta(A) = \{\sigma \in [0,1]^A: \Sigma_{s \in A}\sigma(s) = 1\} \cong \times_{i \in N}\Delta(A_i)$ equipped with for each $\sigma \in \times_{i \in N}\Delta(A_i)$, $\sigma(s) = \prod_{i \in N}\sigma_i(s_i)$ for all $s \in A$.
Need to do expected utility. For each player $i \in N$, the domain for their utility function can be extended linearly from $A$ to $\times_{i \in N}\Delta(A_i)$ with $u_i(\sigma) = \Sigma_{s \in A}f(\sigma)(s)u_i(s)$ for all $\sigma \in \times_{i \in N}\Delta(A_i)$. (it would be good to get rid of $f$ here).
What about $\times_{j \in N-\{i\}}\Delta(A_j)$? Do we have $\times_{j \in N-\{i\}}\Delta(A_j) \subset \Delta(A_{-i})$? Also given $i \in N$ and $\sigma \in \times_{i \in N}\Delta(A_i)$, we have $\sigma_{-i} = (\sigma_1, \ldots, \sigma_{i-1}, \sigma_{i+1}, \ldots, \sigma_n) \in \times_{j \in N-\{i\}}\Delta(A_j)$.
A \textit{pure strategy Nash equilibrium} is a strategy profile $s \in A$ where for each $i \in N$, $u_i(s_i, s_{-i}) \geq u_i(s_i', s_{-i})$ for all $s_i' \in A_i$. For example, in Example \ref{fullsymeg} the profile $(b,b,b)$ is a pure strategy Nash equilibrium.
For each player $i \in N$, the \textit{maximin value} for player $i$ is given by:
......
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