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

<p><strong>Proof:</strong> SE <span class="math-container">$\implies$</span> WPBE is trivial since SEs are PBEs by construction, and thus are also WPBEs.</p>

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

<p><span class="math-container">$$
\begin{align}
\mu^n(t|m) = \frac{\pi(t)\beta^n_s(m|t)}{\sum_{t' \in T}\pi(t')\beta^n_s(m|t')} &amp;&amp;\text{whenever} \sum_{t' \in T} \pi(t')\beta^n_{s}(m|t') &gt; 0 
\end{align}
$$</span> </p>

<p>since this is a two-player game between players <span class="math-container">$r$</span> and <span class="math-container">$s$</span>. The condition on <span class="math-container">$\mu^n$</span> above is known as <em>weak (Bayesian) consistency</em>.</p>

<p>To construct such a totally mixed strategy profile, <span class="math-container">$\beta^n$</span>, consider first the following message-sending strategy profile for player <span class="math-container">$s$</span> as a function of <span class="math-container">$n$</span> conditional on the state of the world <span class="math-container">$t$</span>:</p>

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

<p>where, borrowing notation from <a href="https://www.sciencedirect.com/science/article/pii/002205319190155W" rel="nofollow noreferrer">Fudenberg and Tirole (1991)</a>, </p>

<p><span class="math-container">$$\mathcal{N_s}(t) \equiv \#\{m \in M\ |\ \beta^*_s(m|t) = 0 \text{ and } \sum_{t' \in T} \pi(t')\beta^*_{s}(m|t') &gt; 0 \}.$$</span> </p>

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

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

<p><span class="math-container">$$
\beta^n_r(a|m) = 
\begin{cases}
\left(1 - \frac{1}{n}\right)\beta^*_r(a|m) &amp;&amp; \text{if } \beta^*_r(a|m) &gt; 0 \\
\frac{1}{n\mathcal{N_r}(m)} &amp;&amp; \text{if } \beta^*_r(a|m) = 0,
\end{cases}
$$</span>
where <span class="math-container">$\mathcal{N_r}(m) \equiv \#\{a \in A\ |\ \beta^*_r(a|m) = 0\}$</span>. Clearly <span class="math-container">$\beta^n_r \to \beta^*_r$</span> as <span class="math-container">$n \to \infty$</span>.</p>

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

<p>and thus consistency is demonstrated. Hence the WPBE is also a SE for this game.<span class="math-container">$\ \blacksquare$</span></p>

CC BY-SA 4.0
2020-06-02T09:52:27.900Z
Thanks so much! It seems really involved.
lalala8797
CC BY-SA 4.0
2020-05-24T06:48:54.160Z
@WalrasianAuctioneer I fixed the typos you mentioned. Very helpful, thanks. You may want to delete the comments so as to not confuse future readers.
Kenneth Rios
CC BY-SA 4.0
2020-05-23T18:44:08.077Z
2020-05-24T06:45:20.213Z
2020-05-24T06:45:20.213Z
Kenneth Rios
<p>The definition of a signaling game:</p>

<p>A signaling game is an extensive game with the following structure:</p>

<p>(i) A move of chance first determines a state of the world/type <span class="math-container">$t ∈ T$</span>, which is drawn from T according to a probability distribution <span class="math-container">$π ∈ ∆T$</span>,</p>

<p>(ii) The first player, referred to as the sender (player s), observes the true state/his type <span class="math-container">$t ∈ T$</span>, and chooses a signal/message <span class="math-container">$m ∈ M$</span> to send to the second player, referred to as the receiver (player r),</p>

<p>(iii) The receiver observes the sender’s message m (but not the sender’s type t), and chooses an action <span class="math-container">$a ∈ A$</span>,</p>

<p>(iv) The players’ payoffs are given by functions <span class="math-container">$u_i: T × M × A → R$</span>, where <span class="math-container">$i ∈ {s, r}$</span>.</p>

<p>I need to prove:</p>

<blockquote>
  <p>If the sets <span class="math-container">$T$</span>, <span class="math-container">$M$</span>, and <span class="math-container">$A$</span> are finite, then an assessment <span class="math-container">$&lt;\beta_s, \beta_r, \mu&gt;$</span> is a WPBE if and only if it is an SE.</p>
</blockquote>

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

<p>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.</p>

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

CC BY-SA 4.0
2020-05-21T23:30:28.700Z
Could I ask where you got this question from? I'm genuinely surprised such a result is true.
Walrasian Auctioneer
CC BY-SA 4.0
2020-05-21T23:57:50.237Z
This is my homework. Australian National University. ECON8011 microeconomic theory
lalala8797
CC BY-SA 4.0
2020-05-22T03:46:22.080Z
Thanks so much! Looking forward to your full proof.
lalala8797
CC BY-SA 4.0
2020-05-22T04:10:21.550Z
[Fudenberg and Tirole (1991)](https://www.sciencedirect.com/science/article/pii/002205319190155W) go through the difficult part of the biconditional for the general case on pages 244-246.
Kenneth Rios
CC BY-SA 4.0
2020-05-23T19:05:30.207Z
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.
Kenneth Rios
<game-theory><bayesian-game><signaling>
How to prove the WPBE is SE in signaling game?
lalala8797
# How to prove the WPBE is SE in signaling game?

(Asked by: username_0 on 2020-05-21)

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?

**Tags:** game-theory, bayesian-game, signaling

> **Comments:**
> **username_1**: Could I ask where you got this question from? I'm genuinely surprised such a result is true.  
> **username_0**: This is my homework. Australian National University. ECON8011 microeconomic theory  
> **username_0**: Thanks so much! Looking forward to your full proof.  
> **username_2**: [Fudenberg and Tirole (1991)](https://www.sciencedirect.com/science/article/pii/002205319190155W) go through the difficult part of the biconditional for the general case on pages 244-246.  
> **username_2**: 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.  

---

## 1 answers

### Answer by username_2 on 2020-05-23. Score: 4 | Accepted answer

**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)](https://www.sciencedirect.com/science/article/pii/002205319190155W),

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

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.$\ \blacksquare$

> **Comments:**
> **username_2**: @username_1 I fixed the typos you mentioned. Very helpful, thanks. You may want to delete the comments so as to not confuse future readers.  
> **username_0**: Thanks so much! It seems really involved.  

---

