3
$\begingroup$

Consider a game in static, complete information environment, and the following definition of a strategy never being a best response to a player:

A strategy $\sigma_i\in\Delta S_i$ is never a best response if there are no beliefs $\sigma_{-i}\in \Delta S_{-i}$ for player $i$ for which $\sigma_i\in BR_i(\sigma_{-i})$.

I want to understand the role of belief in this definition correctly. The definition of belief is given as:

A belief of player $i$ is a possible profile of his opponent's strategies, $\sigma_{-i}\in \Delta S_{-i}$.

My understanding is:

In situations where there is no strictly dominant strategy, a player asks herself "what do I think my opponents will do?". To resolve this uncertainty, the player assigns probability distribution over the set of pure strategies of opponent. One particular distribution produces a belief, one possible opponent's profile which is $\sigma_{-i}\in \Delta S_{-i}$.
Now, I look at my simplex and say, is my particular strategy $\sigma_{i}\in \Delta S_{i}$ a BR to that of opponent? I strike out my $\sigma_i$ if I cannot find any probability distribution over the opponent's pure strategies such that $\sigma_i$ would be my BR.
So, technically, I would have to compare my particular $\sigma_i$ against every possible probability distribution over the opponent's pure strategies to identify it as Not-BR, but in simple games, you often face with a situation where one of the pure strategies a player might have, say Column L of player 2, never gets underlined when you work out which ones for Player 2 is BR. This helps the process of eliminating all the strategies that are never a best response.

Is my understanding of belief and its role in finding the set of rationalizable strategies correct?

$\endgroup$
3
$\begingroup$

Your understanding is mostly correct, but the following sentence is not:

in simple games, you often face with a situation where one of the pure strategies a player might have, say Column L of player 2, never gets underlined when you work out which ones for Player 2 is BR.

Consider the following counterexample: \begin{array}{|c|c|c|c|} \hline &L&C&R\\\hline U&1,3&3,4&3,0\\\hline M&2,3&0,1&0,4\\\hline D&0,0&1,2&2,1\\\hline \end{array} In this game, $L$ is never a BR to any of player 1's pure strategies. However, if player 2 believes player 1 will play a mixed strategy that assigns $\frac12$ probability to $U$, $\frac12$ to $M$ and $0$ to $D$, then $L$ becomes a BR.

$\endgroup$
  • $\begingroup$ Yes, you are correct. I should have clarified what I meant by "simple". In introductory textbooks, you often see a situation where L is strictly dominated by the mixture between C and R. In your payoff matrix case, you have L that does not do that. But if you had payoffs for player to in L column such that each of them is strictly less than $.5u_2(C)+.5u_2(R)$, then you will have a case where $L$ is never underlined, plus it is strictly dominated by the mixture of $C$ and $R$. So, yeah, maybe the adverb "often" was being overzealous lol. $\endgroup$ – Frank Swanton Nov 13 '17 at 13:34
  • 1
    $\begingroup$ But I agree, generally having for player 2, $u_2((1,0,0),s_1)<u_2((0,1,0),s_1),u_2((0,0,1),s_1)$ where $s_1$ is the set of pure strategies for player 1 does not imply $L$ cannot be a BR to any element in player 1's simplex; if this is what you wanted to illustrate in your example.... $\endgroup$ – Frank Swanton Nov 13 '17 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.