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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?

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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.

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  • $\begingroup$ 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, 2017 at 13:34
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    $\begingroup$ 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, 2017 at 13:45

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