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?

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    $\begingroup$ The quantifiers at the end of your criterion for equilibrium in Bayesian games seem wrong. It should have been for all $s_i'\in S_i$ and for all $i$. $\endgroup$ – Herr K. Feb 7 '19 at 0:06
  • $\begingroup$ @HerrK.You're right. I'll edit the question. $\endgroup$ – johnny09 Feb 7 '19 at 0:09
  • $\begingroup$ Also, in the second-to-last paragraph, did you mean $s_i:\Theta_i\to\Pi(A_i)$, i.e. with lower case $s_i$ as opposed to upper case? It matters. $\endgroup$ – Herr K. Feb 7 '19 at 0:11
  • $\begingroup$ @HerrK. The reference I am using uses an upper case for the function of mixed strategies. For example, $S$ is a mixed strategy and the strategy of a player $i$, $s_i(\theta_i)\in S_i(\Theta_i)$ for all $\theta_i\in\Theta_i$ is a probability distribution. $\endgroup$ – johnny09 Feb 7 '19 at 0:15
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    $\begingroup$ Btw, $s_{-i}\ne s_i$ should be removed from the definition of equilibrium in Bayesian games. It makes no sense. $s_{-i}$ is a vector of $n-1$ elements whereas $s_i$ is a scalar. $\endgroup$ – Herr K. Feb 7 '19 at 1:50

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