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.

  • 1
    $\begingroup$ Would economics.stackexchange.com/a/9843 answer your question? $\endgroup$
    – Herr K.
    Jan 7 at 1:54
  • $\begingroup$ I have seen the question before, and though the intuition is there, I was hoping for something a little more formal. $\endgroup$ Jan 7 at 10:35

1 Answer 1


Take the Beer and 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

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.

  • $\begingroup$ Thank you, very clear explanation! $\endgroup$ Jan 7 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.