This question from MWG 8.B.7

Any strictly dominant strategy must be a pure strategy.

How can I show this?

My explanation is as follows:

Suppose we have a strictly dominant strategy, $\sigma_i$ . Suppose further that $\sigma_i$ is not a degenerate pure strategy. Then $\sigma_i$ cannot strictly dominate any pure strategy for which $\sigma_i$ specifies playing with positive probability. Hence $\sigma_i$ cannot be strictly dominant. Thus, $\sigma_i$ must be a degenerate pure strategy if it is to be strictly dominant.

But I guess this is just an explanation what I thought.

How can I prove this sentences mathematically?


1 Answer 1


Fix any $\sigma_{-i}$. Assume $\sigma_i$ is strictly dominant but not a pure strategy. Let $X$ be the support of $\sigma_i$. Since $\sigma_i$ strictly dominates all pure strategies $s_i\in X$, we have $$\pi_i(\sigma_i,\sigma_{-i})>\pi_i(s_i,\sigma_{-i})$$ for all $s_i\in X$. This implies $$\sigma_i(s_i)\pi_i(\sigma_i,\sigma_{-i})>\sigma_i(s_i)\pi_i(s_i,\sigma_{-i})$$ for all $s_i\in X$. Summing up gives $$\sum_{s_i\in X}\sigma_i(s_i)\pi_i(\sigma_i,\sigma_{-i})>\sum_{s_i\in X}\sigma_i(s_i)\pi_i(s_i,\sigma_{-i})=\pi_i(\sigma_i,\sigma_{-i}),$$ implying $$\pi_i(\sigma_i,\sigma_{-i})\sum_{s_i\in X}\sigma_i(s_i)>\pi_i(\sigma_i,\sigma_{-i})$$ and hence $$\pi_i(\sigma_i,\sigma_{-i})>\pi_i(\sigma_i,\sigma_{-i}),$$ which is a contradiction.


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