1
$\begingroup$

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?

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.