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The mixed strategy $s=(s_1^*,...,s_n^*)$ is a Nash equilibrium if and only $\forall i \in \{1,...,n\}$ $ \forall s_i \in S_i$ $ u_i(s_1^*,...,s_{i-1}^*,s_i,s_{i+1}^*,...,s_n^*) \le u_i(s_1^*,...,s_{i-1}^*,s_i^*,s_{i+1}^*,...,s_n^*)$ where $s_{i,j}$ is the probability that the strategy $\sigma_{i,j}$ is chosen, $ \sigma \in \Sigma_1 \times...\times \Sigma_n$ the strategy set and $u_i$ is the payoff function.

Q: Let's restrict us on a 2x2 payoff matrix . If I determine the Nash equilibrium (algorithmically), I normally set $\mathbb E[u_i(S_{i_2})] = \mathbb E[u_i(S_{i_1})], i=1,2$. This delivers me a probability vector $(p, 1-p)$ for one player (let's say 1) so that if he choses it, no matter which probability $(q,1-q),$ $ q \in [0,1]$ player 2 takes, the expected payoff stays the same.

Generalizing this idea, I don't see why not $\forall i \in \{1,...,n\}$ $ \forall s_i \in S_i$ $ u_i(s_1^*,...,s_{i-1}^*,s_i,s_{i+1}^*,...,s_n^*) = u_i(s_1^*,...,s_{i-1}^*,s_i^*,s_{i+1}^*,...,s_n^*)$ holds?

In short, I am looking for a example where '$ <$'.

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Suppose a finite game for simplicity. The support of a mixed strategy $\sigma$ is the set of pure strategies that are assigned positive probability by $\sigma$. Equality is a necessary condition for a pure strategy to be in the support of $\sigma$. Inequality is a sufficient condition for a pure strategy for not being in the support.

For an example, consider the game $$\begin{array}{c|c|c} & L & R \\ \hline T & 1,0 & 0,1 \\ \hline M & 0,1 & 1,0\\ \hline B & -1,-1 & -1, -1\\ \hline \end{array}$$ The unique equilibrium is in the mixed strategies $\sigma_1^*$ (with probability $1/2$ for $T$ and $M$ and probability 0 for $B$) and $\sigma_2^*$ ((with probability $1/2$ for $L$ and $R$).

The support of $\sigma_1^*$ is $\{ T, M \}$. You can check that equality holds for $T$ and $M$: $u_1(T, \sigma_2^*) = u_1(M, \sigma_2^*) = 1/2$, but inequality holds for $B$ because $u_1(B, \sigma_2^*) = -1$. This implies that $B$ is played with zero probability in equilibrium.

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