<p>This is homework.</p>
<p>Question: Suppose, <span class="math-container">$\sigma_i^*$</span> is a max-min strategy of player <span class="math-container">$i$</span>. Prove that if <span class="math-container">$s_i \in \operatorname{supp}(\sigma_i^*)$</span>, then <span class="math-container">$s_i$</span> cannot be weakly dominated by any pure/mixed strategy in <span class="math-container">$S_i$</span>.</p>
<p>My attempt: I am trying to prove by contraction.</p>
<p>Let <span class="math-container">$v$</span> be the max-min value for player <span class="math-container">$i$</span>.</p>
<p>By definition:
<span class="math-container">$$v = \min_{\sigma_{-i}} u_i(\sigma_i^*, \sigma_{-i})$$</span></p>
<p>From indifference lemma, I can show that:</p>
<p><span class="math-container">$$v = \min_{\sigma_{-i}} u_i(s_i, \sigma_{-i}) = \min_{\sigma_{-i}}u_i(\sigma_i^*, \sigma_{-i})$$</span></p>
<p>Thus, <span class="math-container">$$\min_{\sigma_{-i}} u_i(s_i, \sigma_{-i}) \geq \min_{\sigma_{-i}} u_i(s_i', \sigma_{-i})\,\,\,\,\,\,\,\, \forall s_i' \in \Delta S_i$$</span></p>
<p>Suppose <span class="math-container">$\exists \,\, \bar{\sigma_i}$</span> which weakly dominates <span class="math-container">$s_i$</span>. Thus,</p>
<p><span class="math-container">$$u_i(\bar{\sigma_i}, \sigma_{-i}) \geq u_i(s_i, \sigma_{-i})\,\,\,\,\,\,\,\, \forall \sigma_{-i}$$</span></p>
<p>I am able to show that <span class="math-container">$$\min_{\sigma_{-i}}u_i(\bar{\sigma_i}, \sigma_{-i}) =\min_{\sigma_{-i}} u_i(s_i, \sigma_{-i})=v$$</span></p>
<p>I am stuck here. Intuitively, I need to show that if <span class="math-container">$s_i$</span> is weakly dominated, it cannot be the max-min strategy as there would exist another that will give greater utility than <span class="math-container">$\min_{\sigma_{-i}} u_i(s_i, \sigma_{-i})$</span> - which will be a contradiction.</p>
<p>Please feel free to use a completely different approach if you think that the above is not efficient.</p>

CC BY-SA 4.0
2022-09-14T12:10:40.667Z
Hi! Do you know the exact definition of weak dominance? Seems like you should use that. Also, a lot of the stuff, e.g., about $s_i'$, seems not to lead to anything, so you can omit that.
Giskard
CC BY-SA 4.0
2022-09-15T06:27:48.653Z
@Giskard: so I realised that the statement is actually wrong (it is easy to construct a counter example). When I went back to the question it had the word 'each'. That is, $s_i$ is in *each* of its max-min strategies. Then it was easy to prove. So the question that I am asking is effectively wrong.
Dayne
<game-theory><dominant-strategy>
Maxmin Strategy cannot be Weakly Dominated
Dayne
# Maxmin Strategy cannot be Weakly Dominated

(Asked by: username_0 on 2022-09-14)

This is homework.

Question: Suppose, $\sigma\_i^\*$ is a max-min strategy of player $i$. Prove that if $s\_i \in \operatorname{supp}(\sigma\_i^\*)$, then $s\_i$ cannot be weakly dominated by any pure/mixed strategy in $S\_i$.

My attempt: I am trying to prove by contraction.

Let $v$ be the max-min value for player $i$.

By definition:
$$v = \min\_{\sigma\_{-i}} u\_i(\sigma\_i^\*, \sigma\_{-i})$$

From indifference lemma, I can show that:

$$v = \min\_{\sigma\_{-i}} u\_i(s\_i, \sigma\_{-i}) = \min\_{\sigma\_{-i}}u\_i(\sigma\_i^\*, \sigma\_{-i})$$

Thus, $$\min\_{\sigma\_{-i}} u\_i(s\_i, \sigma\_{-i}) \geq \min\_{\sigma\_{-i}} u\_i(s\_i', \sigma\_{-i})\,\,\,\,\,\,\,\, \forall s\_i' \in \Delta S\_i$$

Suppose $\exists \,\, \bar{\sigma\_i}$ which weakly dominates $s\_i$. Thus,

$$u\_i(\bar{\sigma\_i}, \sigma\_{-i}) \geq u\_i(s\_i, \sigma\_{-i})\,\,\,\,\,\,\,\, \forall \sigma\_{-i}$$

I am able to show that $$\min\_{\sigma\_{-i}}u\_i(\bar{\sigma\_i}, \sigma\_{-i}) =\min\_{\sigma\_{-i}} u\_i(s\_i, \sigma\_{-i})=v$$

I am stuck here. Intuitively, I need to show that if $s\_i$ is weakly dominated, it cannot be the max-min strategy as there would exist another that will give greater utility than $\min\_{\sigma\_{-i}} u\_i(s\_i, \sigma\_{-i})$ - which will be a contradiction.

Please feel free to use a completely different approach if you think that the above is not efficient.

**Tags:** game-theory, dominant-strategy
> **Comments:**
> **username_1**: Hi! Do you know the exact definition of weak dominance? Seems like you should use that. Also, a lot of the stuff, e.g., about $s_i'$, seems not to lead to anything, so you can omit that.  
> **username_0**: @Giskard: so I realised that the statement is actually wrong (it is easy to construct a counter example). When I went back to the question it had the word 'each'. That is, $s_i$ is in *each* of its max-min strategies. Then it was easy to prove. So the question that I am asking is effectively wrong.  

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## 0 answers

