<p>The <strong>short answer</strong> is that a mixed strategy <span class="math-container">$\sigma_i$</span> that uses with positive probability a dominated pure strategy <span class="math-container">$s_i$</span> can always be improved by <em>excluding</em> <span class="math-container">$s_i$</span> from its mixing support and redistributing its "probability weight" to one of its dominating strategies.</p>

<p><strong>More formally</strong>, suppose that strategy <span class="math-container">$s_i'$</span> dominates <span class="math-container">$s_i$</span> (you can think of <span class="math-container">$s_i'$</span> as a <em>pure</em> or <em>mixed</em> strategy itself, this does not make any difference for the argument; for simplicity I suppose <span class="math-container">$s_i'$</span> is a pure strategy), that is <span class="math-container">$\forall s_{-i}\in S_{-i}: \pi(s_i',s_{-i})&gt;\pi(s_i,s_{-i})$</span>.</p>

<p>Also, suppose that strategy <span class="math-container">$\sigma_i$</span> plays strategy <span class="math-container">$s_i$</span> with positive probability <span class="math-container">$p_{s_i}\in(0,1)$</span>. Build strategy <span class="math-container">$\sigma_i'$</span> as follows: identical to <span class="math-container">$\sigma_i$</span> but replace <span class="math-container">$s_i$</span> with <span class="math-container">$s_i'$</span> , that is play <span class="math-container">$s_i$</span> with probability 0 and play <span class="math-container">$s_i'$</span> with (additional) probability <span class="math-container">$p_{s_i}$</span>.</p>

<p>By linearity of the payoff function you have, <span class="math-container">$\forall s_{-i}\in S_{-i}$</span>:
<span class="math-container">$$
\pi(\sigma_i',s_{-i})=p_{s_i}\pi(s_i',s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) &gt; p_{s_i}\pi(s_i,s_{-i}) + (1-p_{s_i})\pi(\Lambda_i,s_{-i}) = \pi(\sigma_i,s_{-i})
$$</span>
where <span class="math-container">$\Lambda_i$</span>, loosely speaking, denotes the residual combination of strategies played in <span class="math-container">$\sigma_i$</span>, apart from <span class="math-container">$s_i$</span>, inherited by <span class="math-container">$\sigma_i'$</span> too.</p>

CC BY-SA 4.0
2019-05-09T06:15:30.063Z
2019-05-09T06:15:30.063Z
GabMac
<p>This problem is  of Fernando Vega Redondo(Economics and the theory of games)</p>
<p><strong>Exercise 2.1</strong>
Let G be a game in strategic form. Prove that, for every player
<span class="math-container">$i\in N$</span>, every mixted strategy <span class="math-container">$\sigma_{i}\in \Sigma_{i}$</span> that assigns positive weight to a pure strategy <span class="math-container">$s_{i}\in S_{i}$</span> that is dominated can be itself always be improved by another strategy <span class="math-container">$\sigma_{i}'$</span>.</p>
<p>This is if <span class="math-container">$s_{i}\in S_{i}$</span> is strongly dominated for some <span class="math-container">$\sigma_{i}$$\in \Sigma$</span>
with <span class="math-container">$\sigma_{i}(s_{i})&gt;0$</span>, <span class="math-container">$\exists$</span> <span class="math-container">$\sigma_{i}'\in \Sigma_{i}$</span> such that <span class="math-container">$\forall$$s_{-i}\in S_{-i}$$: \pi_{i}(\sigma_{i}',s_{-i})&gt;\pi_{i}(\sigma_{i},s_{-i})$</span>.</p>
<p>Q1: why they said that affirmation is obvious?</p>
<p>*I would like to know if anyone could tell me how to build a mixed strategy that dominates a strategy that assigns a strictly positive probability to a dominated strategy. It is not obvious to me.</p>
<p>Thanks!</p>

<game-theory>
Mixed strategy that assigns strictly positive weight to a pure strategy that is dominated
Franciscolli
# Mixed strategy that assigns strictly positive weight to a pure strategy that is dominated

(Asked by: username_0 on 2019-05-09)

This problem is of Fernando Vega Redondo(Economics and the theory of games)

**Exercise 2.1**
Let G be a game in strategic form. Prove that, for every player
$i\in N$, every mixted strategy $\sigma\_{i}\in \Sigma\_{i}$ that assigns positive weight to a pure strategy $s\_{i}\in S\_{i}$ that is dominated can be itself always be improved by another strategy $\sigma\_{i}'$.

This is if $s\_{i}\in S\_{i}$ is strongly dominated for some $\sigma\_{i}$$\in \Sigma$
with $\sigma\_{i}(s\_{i})>0$, $\exists$ $\sigma\_{i}'\in \Sigma\_{i}$ such that $\forall$$s\_{-i}\in S\_{-i}$$: \pi\_{i}(\sigma\_{i}',s\_{-i})>\pi\_{i}(\sigma\_{i},s\_{-i})$.

Q1: why they said that affirmation is obvious?

\*I would like to know if anyone could tell me how to build a mixed strategy that dominates a strategy that assigns a strictly positive probability to a dominated strategy. It is not obvious to me.

Thanks!

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2019-05-09. Score: 5

The **short answer** is that a mixed strategy $\sigma\_i$ that uses with positive probability a dominated pure strategy $s\_i$ can always be improved by *excluding* $s\_i$ from its mixing support and redistributing its "probability weight" to one of its dominating strategies.

**More formally**, suppose that strategy $s\_i'$ dominates $s\_i$ (you can think of $s\_i'$ as a *pure* or *mixed* strategy itself, this does not make any difference for the argument; for simplicity I suppose $s\_i'$ is a pure strategy), that is $\forall s\_{-i}\in S\_{-i}: \pi(s\_i',s\_{-i})>\pi(s\_i,s\_{-i})$.

Also, suppose that strategy $\sigma\_i$ plays strategy $s\_i$ with positive probability $p\_{s\_i}\in(0,1)$. Build strategy $\sigma\_i'$ as follows: identical to $\sigma\_i$ but replace $s\_i$ with $s\_i'$ , that is play $s\_i$ with probability 0 and play $s\_i'$ with (additional) probability $p\_{s\_i}$.

By linearity of the payoff function you have, $\forall s\_{-i}\in S\_{-i}$:
$$
\pi(\sigma\_i',s\_{-i})=p\_{s\_i}\pi(s\_i',s\_{-i}) + (1-p\_{s\_i})\pi(\Lambda\_i,s\_{-i}) > p\_{s\_i}\pi(s\_i,s\_{-i}) + (1-p\_{s\_i})\pi(\Lambda\_i,s\_{-i}) = \pi(\sigma\_i,s\_{-i})
$$
where $\Lambda\_i$, loosely speaking, denotes the residual combination of strategies played in $\sigma\_i$, apart from $s\_i$, inherited by $\sigma\_i'$ too.


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