<p>Since there are infinitely many numbers in <span class="math-container">$[0,1]$</span>, agent <span class="math-container">$i$</span> has infinitely many probabilities to randomize over actions in their action space <span class="math-container">$A_i$</span>. So far this has nothing to do with the game being Baysian.  In a Baysian game, a strategy <span class="math-container">$\sigma_i: \Theta_i \rightarrow \Pi(A_i)$</span> (or <span class="math-container">$s_i$</span> instead of <span class="math-container">$\sigma_i$</span>) maps a type into a probability distribution over (pure) actions. You can call the space of all such functions <span class="math-container">$\Sigma_i$</span> (or <span class="math-container">$S_i$</span>). There are many other notational conventions. So you always have to take care of careful definitions.</p>
<p>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 <span class="math-container">$x^2$</span> for <span class="math-container">$x\in[-1,1]$</span> although there are infinitely many numbers <span class="math-container">$x\in[-1,1]$</span>.</p>

CC BY-SA 4.0
2020-09-10T10:33:28.263Z
2020-09-10T10:33:28.263Z
Bayesian
<p>I am confused about the following statement from Wikipedia:</p>

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

<p>How do we check the Nash equilibrium inequality then?</p>

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

<p>In many texts, <span class="math-container">$S_i$</span> is defined as the function <span class="math-container">$S_i:\Theta_i\to\Pi(A_i)$</span>, where <span class="math-container">$\Theta_i$</span> is the type set and <span class="math-container">$\Pi(A_i)$</span> is the probability distribution over the actions.</p>

<p>How is it possible to have infinitely many mixed strategies? What is <span class="math-container">$S_i$</span> exactly, a set or a function?</p>

CC BY-SA 4.0
2019-02-07T00:06:06.923Z
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 $i$.
Herr K.
CC BY-SA 4.0
2019-02-07T00:09:46.503Z
@HerrK.You're right. I'll edit the question.
johnny09
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2019-02-07T00:11:40.897Z
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.
Herr K.
CC BY-SA 4.0
2019-02-07T00:15:25.760Z
@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.
johnny09
CC BY-SA 4.0
2019-02-07T01:48:10.240Z
I think you're confusing two notational conventions and this led to your confusion about the interpretation of the symbol $S_i$. If you take $S_i$ to be the set of pure strategies for player $i$, then it'd make sense to write $s_i\in S_i$ (i.e. the pure strategy $s_i$ in the set of pure strategies $S_i$). If you take $S_i$ to be a state-dependent mixed strategy, then it is a function of the states; however it makes no sense to write $s_i\in S_i$ under this interpretation.
Herr K.
CC BY-SA 4.0
2019-02-07T01:50:24.053Z
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.
Herr K.
CC BY-SA 4.0
2019-02-07T01:56:05.977Z
@HerrK.You're right on the second comment. I edited my question. I don't think I'm confused as I am just following this paper's formulation "Transition Models of Equilibrium Assessment in Bayesian Game," 2015 by Kiminao Kogiso and in particular Definition 2, Equation (3).
johnny09
CC BY-SA 4.0
2019-02-07T02:57:24.013Z
You should have included the information of the paper in your original question. On page 5997, the first paragraph of section II.A, the paper says "$S$ is a mixed strategy." That is simply wrong. It is clear from the context that $S$ is the set of mixed strategy profiles, not any particular mixed strategy. $S_i(\Theta_i)$ is the set of state-dependent mixed strategies for player $i$.
Herr K.
CC BY-SA 4.0
2019-02-07T03:09:52.487Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/89369/discussion-between-johnny09-and-herr-k).
johnny09
<game-theory><bayesian-game>
Mixed Strategies in Bayesian Games
johnny09
# Mixed Strategies in Bayesian Games

(Asked by: username_0 on 2019-02-06)

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?

**Tags:** game-theory, bayesian-game
> **Comments:**
> **username_1**: 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 $i$.  
> **username_0**: @HerrK.You're right. I'll edit the question.  
> **username_1**: 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.  
> **username_0**: @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.  
> **username_1**: I think you're confusing two notational conventions and this led to your confusion about the interpretation of the symbol $S_i$. If you take $S_i$ to be the set of pure strategies for player $i$, then it'd make sense to write $s_i\in S_i$ (i.e. the pure strategy $s_i$ in the set of pure strategies $S_i$). If you take $S_i$ to be a state-dependent mixed strategy, then it is a function of the states; however it makes no sense to write $s_i\in S_i$ under this interpretation.  
> **username_1**: 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.  
> **username_0**: @HerrK.You're right on the second comment. I edited my question. I don't think I'm confused as I am just following this paper's formulation "Transition Models of Equilibrium Assessment in Bayesian Game," 2015 by Kiminao Kogiso and in particular Definition 2, Equation (3).  
> **username_1**: You should have included the information of the paper in your original question. On page 5997, the first paragraph of section II.A, the paper says "$S$ is a mixed strategy." That is simply wrong. It is clear from the context that $S$ is the set of mixed strategy profiles, not any particular mixed strategy. $S_i(\Theta_i)$ is the set of state-dependent mixed strategies for player $i$.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/89369/discussion-between-johnny09-and-herr-k).  

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## 1 answers

### Answer by username_2 on 2020-09-10. Score: 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 $x\in[-1,1]$.


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