Mixed Strategies in Bayesian Games

Question

I am confused about the following statement from Wikipedia:

"A mixed strategy is an assignment of a probability to each pure strategy. This allows for a player to randomly select a pure strategy. Since probabilities are continuous, there are infinitely many mixed strategies available to a player."

How do we check the Nash equilibrium inequality then?

For example, in Bayesian games with player set $N=\{1,2,\ldots,n\}$, the equilibrium is defined as the strategy profile $s=(s_i,s_{-i})$ such that, we have $$EU_i(s_i,s_{-i})\geq EU_i(s_i',s_{-i}),$$ for all $s_i'\in S_i$, for all $i\in N$ and $s_i'\neq s_i$.

In many texts, $S_i$ is defined as the function $S_i:\Theta_i\to\Pi(A_i)$, where $\Theta_i$ is the type set and $\Pi(A_i)$ is the probability distribution over the actions.

How is it possible to have infinitely many mixed strategies? What is $S_i$ exactly, a set or a function?

Answer 1

Since there are infinitely many numbers in $[0,1]$, agent $i$ has infinitely many probabilities to randomize over actions in their action space $A_i$. So far this has nothing to do with the game being Baysian. In a Baysian game, a strategy $\sigma_i: \Theta_i \rightarrow \Pi(A_i)$ (or $s_i$ instead of $\sigma_i$) maps a type into a probability distribution over (pure) actions. You can call the space of all such functions $\Sigma_i$ (or $S_i$). There are many other notational conventions. So you always have to take care of careful definitions.

Just because there are infinitely many ways to randomize over the actions (pure strategies) does not mean that you have to compare all of them one-by-one in an inequality like yours. You can invoke monotonicity arguments or simply maximize over the probabilities. In a similar vein, you can find an extremum of the function $x^2$ for $x\in[-1,1]$ although there are infinitely many numbers $x\in[-1,1]$.