# MWG 8.B.7 - Any strictly dominant strategy must be a pure strategy

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?

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.