<p>I assume that implicit in the lemma is the condition &quot;holding other players' strategies fixed&quot;. I'll also assume that the strategy space is finite.</p>
<p>Given a profile of the other players' strategies <span class="math-container">$\sigma_{-i}$</span>, player <span class="math-container">$i$</span>'s utility from playing a mixed strategy <span class="math-container">$\sigma_i$</span> is given by:
<span class="math-container">\begin{equation}
u_i(\sigma_i,\sigma_{-i})=\sum_{s_i\in S_i}\sigma_i(s_i)u_i(s_i,\sigma_{-i})
\end{equation}</span>
where <span class="math-container">$\sigma_i(s_i)$</span> is the probability of playing pure strategy <span class="math-container">$s_i$</span> according to <span class="math-container">$\sigma_i$</span>.
Now let <span class="math-container">$$s_i^*\in\mathop{\arg\max}_{s_i\in S_i}\;u_i(s_i,\sigma_{-i}).$$</span> Such a maximum must exist because <span class="math-container">$S_i$</span> is finite.
It follows that, for all <span class="math-container">$\sigma_i$</span>,
<span class="math-container">$$u_i(s_i^*,\sigma_{-i})\ge\sum_{s_i\in S_i}\sigma_i(s_i)u_i(s_i,\sigma_{-i}) \tag*{$\blacksquare$}$$</span></p>

CC BY-SA 4.0
2022-07-28T00:26:32.353Z
2022-07-28T01:31:12.953Z
2022-07-28T01:31:12.953Z
Herr K.
<p>I saw the lemma:</p>
<p>&quot;In extensive form games with complete and perfect information, any mixed strategy
for player i will result in a lower or equal utility for player i compared to some pure strategy available
to player i.&quot;</p>
<p>I understand it intuitively, that a mixed strategy decision, would be sub-par to the best-response I can take, but I don't know how to prove it.</p>

<game-theory><utility><expected-utility>
Mixed strategy in extensive form games with complete and perfect information
Jorge González
# Mixed strategy in extensive form games with complete and perfect information

(Asked by: username_0 on 2022-07-27)

I saw the lemma:

"In extensive form games with complete and perfect information, any mixed strategy
for player i will result in a lower or equal utility for player i compared to some pure strategy available
to player i."

I understand it intuitively, that a mixed strategy decision, would be sub-par to the best-response I can take, but I don't know how to prove it.

**Tags:** game-theory, utility, expected-utility


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## 1 answers

### Answer by username_1 on 2022-07-28. Score: 4 | Accepted answer

I assume that implicit in the lemma is the condition "holding other players' strategies fixed". I'll also assume that the strategy space is finite.

Given a profile of the other players' strategies $\sigma\_{-i}$, player $i$'s utility from playing a mixed strategy $\sigma\_i$ is given by:
\begin{equation}
u\_i(\sigma\_i,\sigma\_{-i})=\sum\_{s\_i\in S\_i}\sigma\_i(s\_i)u\_i(s\_i,\sigma\_{-i})
\end{equation}
where $\sigma\_i(s\_i)$ is the probability of playing pure strategy $s\_i$ according to $\sigma\_i$.
Now let $$s\_i^\*\in\mathop{\arg\max}\_{s\_i\in S\_i}\;u\_i(s\_i,\sigma\_{-i}).$$ Such a maximum must exist because $S\_i$ is finite.
It follows that, for all $\sigma\_i$,
$$u\_i(s\_i^\*,\sigma\_{-i})\ge\sum\_{s\_i\in S\_i}\sigma\_i(s\_i)u\_i(s\_i,\sigma\_{-i}) \tag\*{$\blacksquare$}$$


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