The definition of a signaling game:

A signaling game is an extensive game with the following structure:

(i) A move of chance first determines a state of the world/type $t ∈ T$, which is drawn from T according to a probability distribution $π ∈ ∆T$,

(ii) The first player, referred to as the sender (player s), observes the true state/his type $t ∈ T$, and chooses a signal/message $m ∈ M$ to send to the second player, referred to as the receiver (player r),

(iii) The receiver observes the sender’s message m (but not the sender’s type t), and chooses an action $a ∈ A$,

(iv) The players’ payoffs are given by functions $u_i: T × M × A → R$, where $i ∈ {s, r}$.

I need to prove:

If the sets $T$, $M$, and $A$ are finite, then an assessment $<\beta_s, \beta_r, \mu>$ is a WPBE if and only if it is an SE.

The SE must be a WPBE, this is trivial. But I'm struggling with proving WPBE is SE in this game.

I wanna define a strictly mixed strategy sequence and the corresponding belief system to prove that when the receiver's information set cannot be reached, the belief is still consistent. But I'm stuck here.

Can anyone please give me some hints about how to prove this? And how to define such a sequence?

  • $\begingroup$ Could I ask where you got this question from? I'm genuinely surprised such a result is true. $\endgroup$ Commented May 21, 2020 at 23:30
  • $\begingroup$ This is my homework. Australian National University. ECON8011 microeconomic theory $\endgroup$
    – lalala8797
    Commented May 21, 2020 at 23:57
  • $\begingroup$ Thanks so much! Looking forward to your full proof. $\endgroup$
    – lalala8797
    Commented May 22, 2020 at 3:46
  • $\begingroup$ Fudenberg and Tirole (1991) go through the difficult part of the biconditional for the general case on pages 244-246. $\endgroup$ Commented May 22, 2020 at 4:10
  • $\begingroup$ In my answer below, I've essentially mapped the general case discussed in Fudenberg and Tirole (1991) to the particular game posited in the question. $\endgroup$ Commented May 23, 2020 at 19:05

1 Answer 1


Claim: If choice sets $T, M,$ and $A$ are finite, then an assessment $\{\beta^*_{r}, \beta^*_{s}, \mu^*\}$ is a WPBE (weak perfect Bayesian equilibrium) of the two-stage signalling game between receiver $r$ and sender $s$ if and only if it is a SE (sequential equilibrium).

Proof: SE $\implies$ WPBE is trivial since SEs are PBEs by construction, and thus are also WPBEs.

To prove that a WPBE is also a SE for this signalling game over finite choice sets $T, M,$ and $A$, we must demonstrate that there exists a tuple $(\beta^n_r, \beta^n_s, \mu^n)$ such that $\{\beta^*_{r}, \beta^*_{s}, \mu^*\} = \lim\limits_{n \to \infty} (\beta^n_r, \beta^n_s, \mu^n)$ for some totally mixed strategy $\beta^n$ such that, for all $n$,

$$ \begin{align} \mu^n(t|m) = \frac{\pi(t)\beta^n_s(m|t)}{\sum_{t' \in T}\pi(t')\beta^n_s(m|t')} &&\text{whenever} \sum_{t' \in T} \pi(t')\beta^n_{s}(m|t') > 0 \end{align} $$

since this is a two-player game between players $r$ and $s$. The condition on $\mu^n$ above is known as weak (Bayesian) consistency.

To construct such a totally mixed strategy profile, $\beta^n$, consider first the following message-sending strategy profile for player $s$ as a function of $n$ conditional on the state of the world $t$:

$$ \beta^n_{s}(m|t) = \small{ \begin{cases} \frac{n-1}{n} \left(\frac{n-1}{n}\right) \beta^*_s(m|t) && \text{if } \beta^*_s(m|t) > 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') > 0 \\ \frac{n-1}{n} \left(\frac{1}{n \mathcal{N_s}(t)}\right) && \text{if } \beta^*_s(m|t) = 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') > 0 \\ \frac{n-1}{n} \left(\frac{\pi(m) \mu^*(t|m)}{n\pi(t)} \right) && \text{if } \mu^*(t|m) > 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') = 0 \\ \frac{1}{n} \left[1 - \sum\limits_{m' \in M} \left(\frac{n-1}{n} \frac{\pi(m') \mu^*(t|m')}{\pi(t)}\right) \right]&& \text{otherwise} \end{cases} } $$

where, borrowing notation from Fudenberg and Tirole (1991),

$$\mathcal{N_s}(t) \equiv \#\{m \in M\ |\ \beta^*_s(m|t) = 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') > 0 \}.$$

The finiteness of $M$ is invoked here so as to provide a well-defined domain over $\#\{\cdot\}$. These four cases span the parameter-space of $\beta_s^*$ and $\mu^*$ in this game. Thus $\sum_{m \in M} \beta^n_{s}(m|t) = 1$ and $\beta^n_{s}(m|t) > 0$ for all $n$ and for all $m \in M$, establishing that $\beta^n_s$ is a totally mixed strategy profile for player $s$. By construction of the asymptotic behavior of $\beta^n_{s}(m|t)$, it is easy to verify that $\beta^n_{s} \to \beta^*_s$ as $n \to \infty$.

The residual strategy profile for player $r$, $\beta_r^n(a|m)$, can be simply constructed by assigning probabilities only across the parameter-space of $\beta^*_r$ since the belief $\mu^*$ is just-identified by $\beta^*_s$:

$$ \beta^n_r(a|m) = \begin{cases} \left(1 - \frac{1}{n}\right)\beta^*_r(a|m) && \text{if } \beta^*_r(a|m) > 0 \\ \frac{1}{n\mathcal{N_r}(m)} && \text{if } \beta^*_r(a|m) = 0, \end{cases} $$ where $\mathcal{N_r}(m) \equiv \#\{a \in A\ |\ \beta^*_r(a|m) = 0\}$. Clearly $\beta^n_r \to \beta^*_r$ as $n \to \infty$.

Lastly, by construction of $\beta^n_s(m|t)$, it is straightforward to verify that the sequence $$ \begin{gather} \mu^n(t|m) = \frac{\pi(t)\beta^n_s(m|t)}{\sum_{t' \in T}\pi(t')\beta^n_s(m|t')} \to \mu^*(t|m) \text{ as } n \to \infty \end{gather} $$

and thus consistency is demonstrated. Hence the WPBE is also a SE for this game.$\ \blacksquare$

  • $\begingroup$ @WalrasianAuctioneer I fixed the typos you mentioned. Very helpful, thanks. You may want to delete the comments so as to not confuse future readers. $\endgroup$ Commented May 24, 2020 at 6:48
  • 1
    $\begingroup$ Thanks so much! It seems really involved. $\endgroup$
    – lalala8797
    Commented Jun 2, 2020 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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