<p>Yes. Generally, the set of best replies is always a mixture over the set of pure best replies.  In particular, there is never a strict incentive to play a mixed strategy, a player playing a mixed best response is indifferent between all pure strategies in the support of the mixed strategy. All this follows from the linearity of expected utility.</p>

CC BY-SA 3.0
2017-11-15T07:58:37.707Z
2017-11-15T07:58:37.707Z
Michael Greinecker
<p>Consider a static, complete information game with 2 players.
<br/>Strategy sets are $S_1=\{U,D\},S_2=\{l,m,r\}$.</p>

<p>Payoffs are irrelevant to this question as I am trying to get the concept of rationalizability correct.</p>

<p>Suppose I want to verify whether $m$ is a rationalizable strategy for player 2.</p>

<p>Then, I want to ask the following question:</p>

<p>$\exists \sigma_1=(q^*,1-q^*)\in\Delta(S_1)$ such that for $\sigma^*_2=(0,1,0),$ $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,\sigma_2)$ for all $\sigma_2\in\Delta(S_2)?$</p>

<p>Now, suppose I have payoff matrix such that I could find $(q^*,1-q^*)$ such that it satisfies both:</p>

<p>(1) $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,(1,0,0))$
<br/>(2) $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,(0,0,1))$.</p>

<p>This means, I could find a valid range of $q^*$ such that for player 2, choosing $m$ provides a weakly better payoff for her compared to the degenerate (i.e. pure) strategies of $l$ or $r$.</p>

<p>My question is:</p>

<p>If I could find such $q^*$ that satisfies both (1),(2), then I do not have to check for any other strategy profiles in $\Delta(S_2)$, that is any convex combo of $(1,0,0)$ and $(0,0,1)$? My intuition is that for any $\alpha\in[0,1]$, I could simple have:</p>

<p>(1)' $\alpha u_2(\sigma_1,\sigma^*_2)\geq \alpha u_2(\sigma_1,(1,0,0))$
<br/>(2)' $(1-\alpha)u_2(\sigma_1,\sigma^*_2)\geq (1-\alpha)u_2(\sigma_1,(0,0,1))$.</p>

<p>and show that (1)'+(2)'implies the degenerate strategy $m$ for player 2 is a best response to some belief, $\sigma_1\in\Delta(S_1)$. Hence, the bottom line is (1),(2) is sufficient, and I do not have to check the convex combo of the two other pure strategies.</p>

CC BY-SA 3.0
2017-11-15T03:53:09.567Z
I think your understanding is correct. In fact, the argument is similar to strategic dominance with mixed strategies. See the discussion on p.240-242 of MWG.
Herr K.
CC BY-SA 3.0
2017-11-15T13:00:46.520Z
@HerrK. Yes. Thanks for the reference. I read it, and it is saying something to that effect. It makes sense that if you are trying to see whether a mixed strategy for player $I$ is strictly dominated by another in his simplex, you only need to check against the degenerate (pure) strategies of the opponent as their convex combo will preserve the ordering. Convenience we can take advantage of since we don't have to check against all probability distributions in the opponent's simplex.
Frank Swanton
<game-theory>
Verifying rationalizability of pure strategy
Frank Swanton
# Verifying rationalizability of pure strategy

(Asked by: username_0 on 2017-11-15)

Consider a static, complete information game with 2 players.
  
Strategy sets are $S\_1=\{U,D\},S\_2=\{l,m,r\}$.

Payoffs are irrelevant to this question as I am trying to get the concept of rationalizability correct.

Suppose I want to verify whether $m$ is a rationalizable strategy for player 2.

Then, I want to ask the following question:

$\exists \sigma\_1=(q^\*,1-q^\*)\in\Delta(S\_1)$ such that for $\sigma^\*\_2=(0,1,0),$ $u\_2(\sigma\_1,\sigma^\*\_2)\geq u\_2(\sigma\_1,\sigma\_2)$ for all $\sigma\_2\in\Delta(S\_2)?$

Now, suppose I have payoff matrix such that I could find $(q^\*,1-q^\*)$ such that it satisfies both:

(1) $u\_2(\sigma\_1,\sigma^\*\_2)\geq u\_2(\sigma\_1,(1,0,0))$
  
(2) $u\_2(\sigma\_1,\sigma^\*\_2)\geq u\_2(\sigma\_1,(0,0,1))$.

This means, I could find a valid range of $q^\*$ such that for player 2, choosing $m$ provides a weakly better payoff for her compared to the degenerate (i.e. pure) strategies of $l$ or $r$.

My question is:

If I could find such $q^\*$ that satisfies both (1),(2), then I do not have to check for any other strategy profiles in $\Delta(S\_2)$, that is any convex combo of $(1,0,0)$ and $(0,0,1)$? My intuition is that for any $\alpha\in[0,1]$, I could simple have:

(1)' $\alpha u\_2(\sigma\_1,\sigma^\*\_2)\geq \alpha u\_2(\sigma\_1,(1,0,0))$
  
(2)' $(1-\alpha)u\_2(\sigma\_1,\sigma^\*\_2)\geq (1-\alpha)u\_2(\sigma\_1,(0,0,1))$.

and show that (1)'+(2)'implies the degenerate strategy $m$ for player 2 is a best response to some belief, $\sigma\_1\in\Delta(S\_1)$. Hence, the bottom line is (1),(2) is sufficient, and I do not have to check the convex combo of the two other pure strategies.

**Tags:** game-theory

> **Comments:**
> **username_1**: I think your understanding is correct. In fact, the argument is similar to strategic dominance with mixed strategies. See the discussion on p.240-242 of MWG.  
> **username_0**: @HerrK. Yes. Thanks for the reference. I read it, and it is saying something to that effect. It makes sense that if you are trying to see whether a mixed strategy for player $I$ is strictly dominated by another in his simplex, you only need to check against the degenerate (pure) strategies of the opponent as their convex combo will preserve the ordering. Convenience we can take advantage of since we don't have to check against all probability distributions in the opponent's simplex.  

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## 1 answers

### Answer by username_2 on 2017-11-15. Score: 3 | Accepted answer

Yes. Generally, the set of best replies is always a mixture over the set of pure best replies. In particular, there is never a strict incentive to play a mixed strategy, a player playing a mixed best response is indifferent between all pure strategies in the support of the mixed strategy. All this follows from the linearity of expected utility.


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