<p>Take the <a href="http://gametheory101.com/courses/game-theory-101/the-beer-quiche-game/" rel="nofollow noreferrer">Beer and Quiche Game</a> as an example.</p>
<p>Let's verify that the following is a weak PBE:</p>
<blockquote>
<ul>
<li>Both types (<span class="math-container">$S$</span> and <span class="math-container">$W$</span>) of player 1 choose beer (<span class="math-container">$B$</span>);</li>
<li>When player 2 sees a beer choice, he believes that player 1 is type <span class="math-container">$S$</span> with probability <span class="math-container">$0.9$</span>, i.e. <span class="math-container">$\mu_2(S\mid B)=0.9$</span>, and he chooses
<span class="math-container">$R$</span> in response;</li>
<li>In case that player 2 sees a quiche choice (<span class="math-container">$Q$</span>), he could have any belief that assigns a probability no more than <span class="math-container">$\frac12$</span> to player 1
being type <span class="math-container">$S$</span>, i.e. <span class="math-container">$\mu_2(S\mid Q)=p\in[0,\frac12]$</span>, and he would
choose <span class="math-container">$F$</span>.</li>
</ul>
</blockquote>
<p>The figure below depicts the above equilibrium in a game tree.</p>
<blockquote>
<p><a href="https://i.sstatic.net/J7v9I.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/J7v9I.png" alt="enter image description here" /></a></p>
</blockquote>
<p>Given player 2's strategy <span class="math-container">$(R,F)$</span>, i.e. <span class="math-container">$R$</span> after <span class="math-container">$B$</span> and <span class="math-container">$F$</span> after <span class="math-container">$Q$</span>, player 1 is better off choosing <span class="math-container">$B$</span>:
<span class="math-container">\begin{align}
u_1(B,(R,F)\mid S)=3&amp;&gt;0=u_1(Q,(R,F)\mid S)\\
u_1(B,(R,F)\mid W)=2&amp;&gt;1=u_1(Q,(R,F)\mid W).
\end{align}</span></p>
<p>Given player 1's strategy <span class="math-container">$\sigma_1(B\mid S)=\sigma_1(B\mid W)=1$</span>, i.e. choosing <span class="math-container">$B$</span> regardless of type, player 2's belief after <span class="math-container">$B$</span> can be derived from Bayes' rule:
<span class="math-container">\begin{align}
\mu_2(S\mid B)&amp;=\frac{0.9\cdot\sigma_1(B\mid S)}{0.9\cdot\sigma_1(B\mid S)+0.1\cdot\sigma_1(B\mid W)}=\frac{0.9\cdot 1}{0.9\cdot 1+0.1\cdot 1}=0.9
\end{align}</span>
Given this belief, it better for player 2 to choose <span class="math-container">$R$</span> after <span class="math-container">$B$</span>:
<span class="math-container">\begin{equation}
u_2(R,\sigma_1\mid B)=0.9(0)+0.1(0)=0&gt;-0.8=0.9(-1)+0.1(1)=u_2(F,\sigma_1\mid B).
\end{equation}</span>
Bayes' rule does not apply in the information set after <span class="math-container">$Q$</span>, since <span class="math-container">$Q$</span> is played with probability zero according to <span class="math-container">$\sigma_1$</span>. Hence any arbitrary belief would satisfy the requirement of weak PBE. However, to justify <span class="math-container">$F$</span> as a best response, the belief must be such that
<span class="math-container">\begin{align}
u_2(F,\sigma_1\mid Q)=p(-1)+(1-p)(1)&gt;p(0)+(1-p)(0)=u_2(R,\sigma_1\mid Q) \quad\Rightarrow\quad p\le\frac12.
\end{align}</span></p>
<p>Thus, we have shown that the proposed strategy profile and belief system is a weak PBE.</p>
<p>Next, we show that this weak PBE is also an SE. To this end, we need to show, in particular, that player 2's beliefs specified in the weak PBE can be justified by a sequence of player 1's completely mixed strategies that converge to her equilibrium strategy.</p>
<p>Let type <span class="math-container">$S$</span> adopt a mixed behavioral strategy that plays <span class="math-container">$B$</span> with probability <span class="math-container">$(1-\epsilon_S)$</span> and <span class="math-container">$Q$</span> with <span class="math-container">$\epsilon_S$</span>. Similarly, let type <span class="math-container">$W$</span> play <span class="math-container">$B$</span> with probability <span class="math-container">$(1-\epsilon_W)$</span> and <span class="math-container">$Q$</span> with <span class="math-container">$\epsilon_W$</span>. We want to show that there exist sequences of <span class="math-container">$\epsilon_S$</span> and <span class="math-container">$\epsilon_W$</span>, both converging to zero, such that player 2's equilibrium beliefs after <span class="math-container">$B$</span> and <span class="math-container">$Q$</span> are limits of the beliefs derived using Bayes' rule from these completely mixed strategies.</p>
<p>First consider player 2's belief after <span class="math-container">$B$</span>. Under player 1's completely mixed strategy, player 2's belief is
<span class="math-container">\begin{equation}
\widetilde \mu_2(S\mid B)=\frac{0.9(1-\epsilon_S)}{0.9(1-\epsilon_S)+0.1(1-\epsilon_W)},
\end{equation}</span>
which clearly converges to <span class="math-container">$\mu_2(S\mid B)=0.9$</span> as <span class="math-container">$\epsilon_S,\epsilon_W\to0$</span>.</p>
<p>Next, consider player 2's belief after <span class="math-container">$Q$</span>. Under player 1's completely mixed strategy, player 2's belief is
<span class="math-container">\begin{equation}
\widetilde \mu_2(S\mid Q)=\frac{0.9\epsilon_S}{0.9\epsilon_S+0.1\epsilon_W},
\end{equation}</span>
and we want this to equal <span class="math-container">$p\le\frac12$</span> as <span class="math-container">$\epsilon_S,\epsilon_W\to0$</span>.
Suppose <span class="math-container">$p&gt;0$</span>. Fix a small but positive <span class="math-container">$\epsilon_S$</span>, and set <span class="math-container">$\epsilon_W=\frac{9(1-p)}p\epsilon_S$</span>. Hence,
<span class="math-container">\begin{equation}
  \widetilde\mu_2(S\mid Q)=\frac{0.9\epsilon_S}{0.9\epsilon_S+0.1\cdot\frac{9(1-p)}p\epsilon_S}.
\end{equation}</span>
As <span class="math-container">$\epsilon_S\to0$</span>, which implies <span class="math-container">$\epsilon_W\to0$</span>, we have <span class="math-container">$\widetilde\mu_2(S\mid Q)\to\mu_2(S\mid Q)=p$</span>.
If <span class="math-container">$p=0$</span>, then we can set <span class="math-container">$\epsilon_W=\sqrt{\epsilon_S}$</span>, and this will yield the same convergence result.</p>
<p>Since player 1’s beliefs are trivially consistent, we conclude that the strategy profile and belief system of the weak PBE is an SE.</p>

CC BY-SA 4.0
2022-01-07T23:05:31.653Z
Thank you, very clear explanation!
Econometric Novice
CC BY-SA 4.0
2022-01-07T18:40:35.070Z
2022-01-07T18:40:35.070Z
Herr K.
<p>My question is quite simple. Could someone given an example of how to determine a Sequential Equilibrium given a set of Perfect Bayesian Equilibria?</p>
<p>The definition of sequential equilibrium where consistent off-equilibrium beliefs are limits to sequences of completely mixed strategies is a bit abstract, so I am unsure how to use it in practice.</p>
<p>Any help would be appreciated.</p>

CC BY-SA 4.0
2022-01-07T01:54:16.063Z
Would https://economics.stackexchange.com/a/9843/ answer your question?
Herr K.
CC BY-SA 4.0
2022-01-07T10:35:06.437Z
I have seen the question before, and though the intuition is there, I was hoping for something a little more formal.
Econometric Novice
<game-theory><nash-equilibrium><bayesian-game>
Sequential and Perfect Bayesian Equilibrium: an example?
Econometric Novice
# Sequential and Perfect Bayesian Equilibrium: an example?

(Asked by: username_0 on 2022-01-06)

My question is quite simple. Could someone given an example of how to determine a Sequential Equilibrium given a set of Perfect Bayesian Equilibria?

The definition of sequential equilibrium where consistent off-equilibrium beliefs are limits to sequences of completely mixed strategies is a bit abstract, so I am unsure how to use it in practice.

Any help would be appreciated.

**Tags:** game-theory, nash-equilibrium, bayesian-game

> **Comments:**
> **username_1**: Would https://economics.stackexchange.com/a/9843/ answer your question?  
> **username_0**: I have seen the question before, and though the intuition is there, I was hoping for something a little more formal.  

---

## 1 answers

### Answer by username_1 on 2022-01-07. Score: 4 | Accepted answer

Take the [Beer and Quiche Game](http://gametheory101.com/courses/game-theory-101/the-beer-quiche-game/) as an example.

Let's verify that the following is a weak PBE:

> * Both types ($S$ and $W$) of player 1 choose beer ($B$);
> * When player 2 sees a beer choice, he believes that player 1 is type $S$ with probability $0.9$, i.e. $\mu\_2(S\mid B)=0.9$, and he chooses
>   $R$ in response;
> * In case that player 2 sees a quiche choice ($Q$), he could have any belief that assigns a probability no more than $\frac12$ to player 1
>   being type $S$, i.e. $\mu\_2(S\mid Q)=p\in[0,\frac12]$, and he would
>   choose $F$.

The figure below depicts the above equilibrium in a game tree.

> [![enter image description here](https://i.sstatic.net/J7v9I.png)](https://i.sstatic.net/J7v9I.png)

Given player 2's strategy $(R,F)$, i.e. $R$ after $B$ and $F$ after $Q$, player 1 is better off choosing $B$:
\begin{align}
u\_1(B,(R,F)\mid S)=3&>0=u\_1(Q,(R,F)\mid S)\\
u\_1(B,(R,F)\mid W)=2&>1=u\_1(Q,(R,F)\mid W).
\end{align}

Given player 1's strategy $\sigma\_1(B\mid S)=\sigma\_1(B\mid W)=1$, i.e. choosing $B$ regardless of type, player 2's belief after $B$ can be derived from Bayes' rule:
\begin{align}
\mu\_2(S\mid B)&=\frac{0.9\cdot\sigma\_1(B\mid S)}{0.9\cdot\sigma\_1(B\mid S)+0.1\cdot\sigma\_1(B\mid W)}=\frac{0.9\cdot 1}{0.9\cdot 1+0.1\cdot 1}=0.9
\end{align}
Given this belief, it better for player 2 to choose $R$ after $B$:
\begin{equation}
u\_2(R,\sigma\_1\mid B)=0.9(0)+0.1(0)=0>-0.8=0.9(-1)+0.1(1)=u\_2(F,\sigma\_1\mid B).
\end{equation}
Bayes' rule does not apply in the information set after $Q$, since $Q$ is played with probability zero according to $\sigma\_1$. Hence any arbitrary belief would satisfy the requirement of weak PBE. However, to justify $F$ as a best response, the belief must be such that
\begin{align}
u\_2(F,\sigma\_1\mid Q)=p(-1)+(1-p)(1)>p(0)+(1-p)(0)=u\_2(R,\sigma\_1\mid Q) \quad\Rightarrow\quad p\le\frac12.
\end{align}

Thus, we have shown that the proposed strategy profile and belief system is a weak PBE.

Next, we show that this weak PBE is also an SE. To this end, we need to show, in particular, that player 2's beliefs specified in the weak PBE can be justified by a sequence of player 1's completely mixed strategies that converge to her equilibrium strategy.

Let type $S$ adopt a mixed behavioral strategy that plays $B$ with probability $(1-\epsilon\_S)$ and $Q$ with $\epsilon\_S$. Similarly, let type $W$ play $B$ with probability $(1-\epsilon\_W)$ and $Q$ with $\epsilon\_W$. We want to show that there exist sequences of $\epsilon\_S$ and $\epsilon\_W$, both converging to zero, such that player 2's equilibrium beliefs after $B$ and $Q$ are limits of the beliefs derived using Bayes' rule from these completely mixed strategies.

First consider player 2's belief after $B$. Under player 1's completely mixed strategy, player 2's belief is
\begin{equation}
\widetilde \mu\_2(S\mid B)=\frac{0.9(1-\epsilon\_S)}{0.9(1-\epsilon\_S)+0.1(1-\epsilon\_W)},
\end{equation}
which clearly converges to $\mu\_2(S\mid B)=0.9$ as $\epsilon\_S,\epsilon\_W\to0$.

Next, consider player 2's belief after $Q$. Under player 1's completely mixed strategy, player 2's belief is
\begin{equation}
\widetilde \mu\_2(S\mid Q)=\frac{0.9\epsilon\_S}{0.9\epsilon\_S+0.1\epsilon\_W},
\end{equation}
and we want this to equal $p\le\frac12$ as $\epsilon\_S,\epsilon\_W\to0$.
Suppose $p>0$. Fix a small but positive $\epsilon\_S$, and set $\epsilon\_W=\frac{9(1-p)}p\epsilon\_S$. Hence,
\begin{equation}
\widetilde\mu\_2(S\mid Q)=\frac{0.9\epsilon\_S}{0.9\epsilon\_S+0.1\cdot\frac{9(1-p)}p\epsilon\_S}.
\end{equation}
As $\epsilon\_S\to0$, which implies $\epsilon\_W\to0$, we have $\widetilde\mu\_2(S\mid Q)\to\mu\_2(S\mid Q)=p$.
If $p=0$, then we can set $\epsilon\_W=\sqrt{\epsilon\_S}$, and this will yield the same convergence result.

Since player 1’s beliefs are trivially consistent, we conclude that the strategy profile and belief system of the weak PBE is an SE.

> **Comments:**
> **username_0**: Thank you, very clear explanation!  

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