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

CC BY-SA 4.0
2022-08-22T12:25:58.850Z
2022-08-25T08:51:41.573Z
2022-08-25T08:51:41.573Z
VARulle
<p>This question from MWG 8.B.7</p>
<blockquote>
<p>Any strictly dominant strategy must be a pure strategy.</p>
</blockquote>
<p>How can I show this?</p>
<p>My explanation is as follows:</p>
<p>Suppose we have a strictly dominant strategy, <span class="math-container">$\sigma_i$</span> . Suppose further that <span class="math-container">$\sigma_i$</span> is not a degenerate pure strategy. Then <span class="math-container">$\sigma_i$</span> cannot strictly dominate any pure strategy for which <span class="math-container">$\sigma_i$</span> specifies playing with positive probability. Hence <span class="math-container">$\sigma_i$</span> cannot be strictly dominant. Thus, <span class="math-container">$\sigma_i$</span> must be a degenerate pure strategy if it is to be strictly dominant.</p>
<p>But I guess this is just an explanation what I thought.</p>
<p>How can I prove this sentences mathematically?</p>

<microeconomics><game-theory><self-study><nash-equilibrium>
MWG 8.B.7 - Any strictly dominant strategy must be a pure strategy
studentp
# MWG 8.B.7 - Any strictly dominant strategy must be a pure strategy

(Asked by: username_0 on 2022-08-19)

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?

**Tags:** microeconomics, game-theory, self-study, nash-equilibrium


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## 1 answers

### Answer by username_1 on 2022-08-22. Score: 7 | Accepted answer

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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