I have a basic doubt on the definition of Bayesian Nash equilibrium.

Consider the following game:

1) $N$ players.

2) Each player $i$ has a type, assigned by nature and denoted by $\epsilon_i$.

$\epsilon_i$ can be thought of as a random variable, that is a map from the set of states of the world $\Omega$ to some real numbers. $Pr$ used below denotes the underlying "true" probability measure used by nature.

Let $E_i$ denote the support of $\epsilon_i$

Information structure assumption (minimal information): Assume that player $i$ only knows which is her own type. Assume also that the random variables $\epsilon_1,...,\epsilon_N$ are mutually independent. This implies that, by knowing her own type, a player does not get useful information about the other players' types.

3) Each player $i$ has to choose from a finite set of actions $A_i$.

4) $u_i: E_i\times (A_1\times ... \times A_N)\rightarrow \mathbb{R}$ is the utility of player $i$.

5) A pure strategy of player $i$ is a function $s_i: E_i\rightarrow A_i$.

What is the exact definition of Bayesian Nash equilibrium? Below is what I understood, with four questions.

Let $\sigma_{a_j}$ be the belief of player $i$ that player $j$ will play $a_j$ conditional on $i$'s information set (that is $\epsilon_i$). Note that $\sigma_{a_j}$ is not indexed by $i$: player $i$ and player $k$ have the same belief that player $j$ will play $a_j$, because types are mutually independent.

Let $\sigma$ be the vector collecting all such beliefs.

Given $\sigma$, let $(s^\sigma_1,...,s^\sigma_N)$ be the vector of pure strategies such that

$$ (1)\hspace{1cm}s^\sigma_i=\text{argmax}_{s_i} \mathbb{E}_{\sigma}\Big[u_i(s_i(E_i),A_{-i})\Big|E_i\Big] \hspace{1cm} \forall i $$ where $\mathbb{E}_{\sigma}$ is the expectation wrto to $A_{-i}$ computed using $\sigma$. Here I guess we are assuming that $(s^\sigma_1,...,s^\sigma_N)$ exists. It may not exist, in which case we can use the notion of mixed strategy.

Let $\sigma^*$ be a solution of the following system (assume it exists, again)

$$ (2)\hspace{1cm}\sigma_{a_i}=Pr(\text{$E_i$ takes a value such that $s_i^\sigma=a_i$}) \hspace{1cm} \forall a_i\in A_i,\forall i $$

Question A) We say that $\sigma^*$ is a Bayesian Nash equilibrium of the game. Is this correct? Or is it $(s^{\sigma^*}_1,...,s^{\sigma^*}_N)$?

Question B) Is $\sigma^*$ a "mixed strategy equilibrium" (sorry if this is incorrect or sloppy, I've put "..." on purpose)? I'm very confused regarding this point. On one hand, $\sigma^*$ is a probability distribution over actions; hence, one could think that each player chooses the actual action in a lottery based on $\sigma^*$. On the other hand, $(s^{\sigma^*}_1,...,s^{\sigma^*}_N)$ tells exactly what each player should actually play for a given realisation of $(\epsilon_1,...,\epsilon_N)$; hence, there is no lottery!

Question C) Sometimes, when describing a game with incomplete information, people also assume that each player $i$ has a prior $\mu_i$ regarding the probability distribution of $\epsilon_{-i}$. Is this somehow implicitly embedded into my characterisation of the belief vector $\sigma$?

Question D) Now suppose that we are being agnostic about the information structure of the game: we could have complete information, or minimal information (as above), or hybrid cases between these two extremes. We know that each player $i$ receives a more or less informative random signal $\tau_i$ about $\epsilon_{-i}$ with support $T_i$. Can we still define a Bayesian equilibrium of such "relaxed" game? In a book I'm reading, a Bayesian equilibrium of the "relaxed" game is defined as a map $$ \sigma \equiv \Pi_{i=1}^n \sigma_i\text{, where $\sigma_i\equiv (\sigma^{a_i}_i \text{ }\forall a_i\in A_i) $ and $\underbrace{\sigma^{a_i}_i: E_i\times T_i \rightarrow [0,1]}_{\text{for each realisation of $\epsilon_i, \tau_i$ it gives the probability that player $i$ plays $a_i$}}$} $$ Does this sound correct to you? Does writing $\sigma \equiv \Pi_{i=1}^n \sigma_i$ underlie some implicit assumptions on independence or similar? How do we find $\sigma$ (i.e., what should it solve)?

Update following answer below: Thanks for the answer. To be sure I understood, let me update my example according to your explanations.

Assume there exists a pure strategy equilibrium.

First, given the belief of player $i$ regarding the other players' actions ($\mu_i$), I define player $i$'s pure strategy as the function $s_i: E_i \rightarrow A_i$. Also, $s\equiv (s_1,..., s_N)$.

Second, given the pure strategy adopted by the other players ($s_{-i}$),
I define the belief of player $i$ regarding player $j$'s action as the function $\mu^{j}_{i}: E_i \rightarrow \Delta(A_j)$. Let $\mu^{a_j}_i$ denote the $a_j$th function component of $\mu^{j}_{i}$.

I assume players form beliefs using the true probability distribution of types adopted by nature. Hence, $\mu^{a_j}_{i}(\epsilon_i)= Pr(s_j(\epsilon_j)=a_j| \epsilon_i)\overbrace{=}^{\epsilon_i\perp \epsilon_j} =Pr(s_j(\epsilon_j)=a_j)\equiv \sigma_{a_j}\in [0,1]$.

Notice that the function $\mu^{j}_{i}$ is flat on $E_i$ (because types are mutually independent) and $\mu^j_i=\mu^j_k$ (because all players form beliefs using the true probability distribution of types adopted by nature). Also, I assume that players do not coordinate, so that $\mu^{-i}_i=\Pi_{j\neq i} \mu^j_i$.

Now I define a Bayesian Nash equilibrium. Let $\sigma^*_i\in \Delta(A_i)$ (with $\sigma^*_{a_i}$ denoting the $a_i$th component) and $\sigma^*\equiv (\sigma^*_1,..., \sigma^*_N)$. $(\sigma^*,s^*)$ is a Bayesian Nash equilibrium if

$$\begin{cases} s^*_i(\epsilon_i)=argmax_{s_i(\epsilon_i)} \mathbb{E}_{\sigma^*}(u_i(s_i(\epsilon_i), s^*_{-i}(\epsilon_{-i})| \epsilon_i) & \forall i \\ \sigma^*_{a_i}= Pr(s^*_i(\epsilon_i)=a_i) & \forall a_i\in A_i, \forall i\end{cases}$$

(I) Is this recap correct?

Now, let's remove the assumption that types are mutually independent (but still we maintain that pure strategy equilibrium exists). We need to slightly modify the definition of beliefs.

Given the pure strategy adopted by the other players ($s_{-i}$), I define the belief of player $i$ regarding player $j$'s action as the function $\mu^{j}_{i}: E_i \rightarrow \Delta(A_i)$, where $\mu^{a_j}_{i}(\epsilon_i)= Pr(s_j(\epsilon_j)=a_j| \epsilon_i)$.

Now I define a Bayesian Nash equilibrium. Let $\sigma^{i,*}_j: E_i\rightarrow \Delta(A_j) $. Let $\sigma^{i,*}\equiv (\sigma^{i,*}_1,..., \sigma^{i,*}_N)$. Let $\sigma^*\equiv (\sigma^{1,*},..., \sigma^{N,*})$.

$(\sigma^*,s^*)$ is a Bayesian Nash equilibrium if

$$ \begin{cases} s^*_i=argmax_{s_i} \mathbb{E}_{\sigma^{i,*}}(u_i(s_i(\epsilon_i), s^*_{-i}(\epsilon_{-i})| \epsilon_i) & \forall i \\ \sigma^{i,*}_{a_j}(\epsilon_i)= Pr(s^*_j(\epsilon_j)=a_j|\epsilon_i) & \forall i, \forall a_j\in A_j,\forall j\neq i \end{cases} $$

(II) Is this second scenario analysed correctly?

(III) Following your definition of Bayesian Nash equilibrium, I think that $s^*$ is what you call "SET OF STRATEGIES" and $\sigma^*$ is what you call "SET OF BELIEFS", correct?

(IV) I have a last doubt: take a game where we remain agnostic about the information structure, i.e., we could have complete information, or minimal information (as above), or hybrid cases between these two extremes. We know that, before taking a decision, each player $i$ receives a realisation of more or less informative random signal $\tau_i$ about $\epsilon_{-i}$ with support $T_i$. Upon receiving such a realisation, each player $i$ updates her prior beliefs. Are we still assuming that, under any information structure, each player knows fully $Pr$ so that he can update "correctly" her prior?

  • $\begingroup$ Good practice on this site is to accept an answer when your question has been answered, and post a new question if you have further questions. It’s not fair to answerers if you keep asking more questions that they won’t get credit for answering. $\endgroup$ – dismalscience Apr 16 '19 at 14:52
  • $\begingroup$ I'm sorry but the answer is not clear to me, I cannot accept it as it is. That is why I have added some questions, just to help clarifying. $\endgroup$ – user3285148 Apr 16 '19 at 15:21
  • $\begingroup$ Your recaps are not correct, and missed the main point of the answer provided below. Beliefs $\mu_i$ are a distribution over the space of types $\epsilon_{-I}$ (since this is what is unknown to the players when she is making her decisions). There is no reason to overcomplicate a simple definition - look up Tirole's book. It provides a wonderful introduction to Bayesian Games. $\endgroup$ – Tomcat Aug 10 at 18:07

Bayesian Nash equilibrium is a set of strategies $\{\sigma_i\}$ one for each player and some beliefs $\{\mu_i\}$ also one for each player such that $\sigma_i$ is a best response for player $i$ given his belief, $\mu_i$, and the beliefs are Bayesian for all players, given their information.

Each strategy $\sigma_i$ is a function from the set of types to a (possibly mixed) strategy: $\sigma_{a_i}: E_i\rightarrow \Delta(A_i)$ I'm using $\Delta$ to denote that it is a probability distribution over actions. So that for any element $\epsilon_i$ the function $\sigma_{a_i}(\epsilon_i)$ gives you the probability with which each action is played by player $i$ (let $\sigma$ denote the vector of all such functions). Note that for players to find their best responses they need to form beliefs about what other players are doing. In equilibrium those beliefs are correct, so they know that other players are using the strategies $\{\sigma_j\}$ Hence the expectation in the definition you gave (which only work for pure strategies, but can easily be extended to mixed strategies).

The expectation is taken because even though you know $\sigma$, the other players might be using mixed strategies. Furthermore, even if all players are choosing pure strategies, you only know they are using $\sigma$ which tells you their best response given their type. But you don't know their type, so you need to take expectations over that. That is where $\mu$ comes in because you need to form beliefs about the other player's types. Here, since types are independent, learning your type provides no information about other players' types, so the prior is equal to the posterior. Thus, beliefs are Bayesian if you compute the expected value (in your first equation) according to the "true" probability with which nature chooses types, and the mixed strategies chosen by each player, according to $\sigma$.

Addressing your questions directly:

A) $\sigma$ is a Bayesian equilibrium. Only if you know that it is in pure strategies, then you can say that $(s_1^\sigma, \dots, s_N^\sigma)$ is a Bayesian equilibrium. Note that both of them are functions from $E_i$ to the set of mixed strategies and the set of pure strategiers, respectively.

B) Hope my discussion above has clarified this. In general you want to think of $\sigma$ as the equilibrium strategy (pure or mixed) and here it always entails uncertainty, because you don't know other player's types, so you might not know the pure strategy they are using.

C) The thing about the prior is implicitly assumed in how you compute $E_{\sigma}(u_i(s_i(E_i), \sigma_{-i}(E_{-i})|E_i)$ because $A_{-i}$ depends on the other players' types which you don't know, but your belief about their type might depend on your own type.

D)The notion of Bayesian Nash equilibrium can be used for general information structures, The notation $\sigma=\prod_{i=1}^n\sigma_i$ is implying that each player chooses their strategy $\sigma_i$ independently. (players do not collude or coordinate) this is a relatively mild assumption made most of the time. Sometimes it is useful to think that players correlate their actions (this is called a Bayes Correlated Equilibrium). Other than than the notation simply implies that $\sigma$ is the vector of equilibrium strategies for each player, and each of them is a best response of other player's strategies.

Hope this helps

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  • $\begingroup$ Thanks: I have added un update to my question, which tries to reformulate the problem using your advice. Could you check whether I have understood correctly your answer? $\endgroup$ – user3285148 Apr 15 '19 at 13:10
  • $\begingroup$ No, it is not correct, $\mu_i$ is your belief about other player's types, not their actions. Your beliefs about their actions is given by $\sigma_i$ since in equilibrium these beliefs are correct, that why we don't need to index them by the players, so we can simply write $\sigma$. So let me write it formally. An equilibrium is a profile of (mixed) strategies and beliefs $(\sigma,\mu)$ where $\sigma=(\sigma_1, \dots, \sigma_N)$ and $\mu=(\mu_1, \dots, \mu_N)$ and each $\sigma_i: E_i\rightarrow \Delta(A_i)$ is a strategy for each player. $\endgroup$ – Regio Apr 17 '19 at 15:32
  • $\begingroup$ I.e. function from each player's types, to their action (pure or mixed, if for some reason you want to restrict actions to only be pure, you can replace the notation of $\sigma$ for $(s_1, \dots, s_N)$, but for simplicity let's not do it.). And each belief $\mu_i\in\Delta(\prod_{j\neq i} E_j)$ gives the probability that each of the other players are of each of their types. Now, the restriction imposed by equilibrium is that each action in the support of $\sigma^*_i(\epsilon_i)$, let's denote it $s^*_i(\epsilon)$ $\endgroup$ – Regio Apr 17 '19 at 15:34
  • $\begingroup$ (i.e. every action that is played by player $i$ when his type is $\epsilon_i$ with positive probability) must satisfy $s_i^*(\epsilon)\in argmax \int_{\epsilon_{-i}\in E_{-i}}\int_{\sigma^*_{-i}(\epsilon_{-i})} u(s_i^*(\epsilon_i), s_{-i}(\epsilon_{-i})|\epsilon_i) \sigma_{-i}(\epsilon_{-i})\mu_i^*(\epsilon_{-i}|\epsilon_i)ds^*_{-i}(\epsilon_{-i})d\epsilon_{-i}$. Notice that we intergrate over the mixed strategies given each profile of types of other players, and then we integrate over all possible profiles of other players' types. $\endgroup$ – Regio Apr 17 '19 at 15:40
  • $\begingroup$ The beliefs $\mu_i:E_i\rightarrow \Delta(\prod_{j\neq i}E_j)$ are your beliefs about other player's types given that you are of type $\epsilon_i$, and it must be obtained by Bayes updating from the prior (you call it $Pr$ I'm going to call it $\mu_{i0}\in \Delta(\prod_{j\neq i}E_j)$. In your particular case since information is minimal, you simply update this belief based on your type if you had more information, you will update your belief based on that information too. Further, since you assumed that types are independent, there is nothing to learn, so $\mu_i$ equals the prior $\mu_{i0}$. $\endgroup$ – Regio Apr 17 '19 at 15:46

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