<p>Consider the following signalling game. Individuals choose an action <span class="math-container">$a \in \mathbb{R}$</span> to maximise
<span class="math-container">\begin{equation}
U(a, \theta) = \theta a - \frac{a^2}{2} + \mathbb{E}[\theta|a]
\end{equation}</span>
where <span class="math-container">$\theta$</span> is their type and <span class="math-container">$\mathbb{E}[\theta|a]$</span> represents the beliefs of observers about their type. For simplicity, assume that types are uniformly distributed over the <span class="math-container">$[0, 1]$</span> interval.</p>
<p>Q: does anyone know how to find (increasing) equilibria of this signalling game? I tried the following, but it gives me a strange answer.</p>
<p>My answer: let <span class="math-container">$f$</span> denote the equilibrium map from types to actions; and imagine pretending to be type <span class="math-container">$\hat{\theta}$</span> when one is really <span class="math-container">$\theta$</span>. One obtains utility
<span class="math-container">\begin{equation}
U(\hat{\theta}, \theta) = \theta f(\hat{\theta}) - \frac{f(\hat{\theta})^2}{2} + \hat{\theta}
\end{equation}</span>
Maximising with respect to <span class="math-container">$\hat{\theta}$</span> yields</p>
<p><span class="math-container">\begin{equation}
\theta f'(\hat{\theta}) - f'(\hat{\theta}) + 1 = 0
\end{equation}</span></p>
<p>If this is an equilibrium, the best choice is <span class="math-container">$\hat{\theta} = \theta$</span>, so</p>
<p><span class="math-container">\begin{equation}
\theta f'(\theta) - f'(\theta) + 1 = 0 \implies f'(\theta) = \frac{1}{1 - \theta}
\end{equation}</span></p>
<p>Integration then yields</p>
<p><span class="math-container">\begin{equation}
f(\theta) = -\ln(1 - \theta) + c
\end{equation}</span></p>
<p>By the boundary condition, <span class="math-container">$c = 0$</span> (individuals with type <span class="math-container">$\theta = 0$</span> should play <span class="math-container">$a = 0$</span>).</p>
<p>This all seems reasonable, except the equilibrium action then goes to infinity as <span class="math-container">$\theta \rightarrow 1$</span>! Moreover, one can check that playing <span class="math-container">$a = -\ln(1 - \theta)$</span> is <em>not</em> a best response given that all others play <span class="math-container">$a = -\ln(1 - \theta)$</span>. So what am I doing wrong?</p>
<p><strong>Update:</strong> as pointed out in the comments, the problem comes from an error I made when writing down the first order condition. If you do this correctly and solve the differential equation, you can indeed find an equilibrium action function.</p>

CC BY-SA 4.0
2022-11-17T14:50:57.970Z
Since this is really the reduced form of a game, you should think about how you model out-of-equilibrium-beliefs. For signaling games, this is essential. If you do this in the conventional way, there should be some simple pooling equilibrium supported by weird out-of-equilibrium beliefs.
Michael Greinecker
CC BY-SA 4.0
2022-11-17T16:02:30.340Z
@MichaelGreinecker of course, you are right that one should be formal. However, as a preliminary step, I am just trying to find a sensible looking equilibrium (which presumably can be supported through the right kind of off equilibrium path beliefs), and am not thinking too hard about uniqueness. I should add that, in this instance, there isn't a pooling equilibrium if beliefs are 'monotone': if all types in some interval $[\theta_{low}, \theta_{high}]$ pool on some action $a$, the type $ \theta_{high}$ would want to deviate to  $a + \epsilon$ for a sufficiently small $\epsilon > 0$.
afreelunch
CC BY-SA 4.0
2022-11-17T17:25:58.043Z
Sure, but there is a pooling equilibrium in which everyone plays $0$ and off equilibrium beliefs are such that everyone beliefs the type is $0$ for any other action.
Michael Greinecker
CC BY-SA 4.0
2022-11-17T19:50:19.253Z
@MichaelGreinecker I think that you are right that this is an equilibrium. My question, however, was more about increasing equilibria: I don't suppose you have a sense why my (rather heuristic) derivation didn't yield one? The steps seemed OK to me, so I'm a bit confused by why the answer is off. Perhaps there isn't an increasing equilibrium?
afreelunch
CC BY-SA 4.0
2022-11-17T22:31:22.833Z
I think it might be a simple calculation mistake when you calculate the first order condition. The derivative of $f(\hat{\theta})^2/2$ is not $f'(\hat{\theta})$. It is $f(\hat{\theta})f'(\hat{\theta})$.
Michael Greinecker
CC BY-SA 4.0
2022-11-17T22:43:00.287Z
That is embarrassing... thank you for spotting this, hopefully it will solve the problem!
afreelunch
CC BY-SA 4.0
2022-11-17T22:50:57.620Z
No problem; I didn't get it right away either.
Michael Greinecker
<game-theory><bayesian-game><signaling>
Signalling with continuous actions and continuous types
afreelunch
# Signalling with continuous actions and continuous types

(Asked by: username_0 on 2022-11-17)

Consider the following signalling game. Individuals choose an action $a \in \mathbb{R}$ to maximise
\begin{equation}
U(a, \theta) = \theta a - \frac{a^2}{2} + \mathbb{E}[\theta|a]
\end{equation}
where $\theta$ is their type and $\mathbb{E}[\theta|a]$ represents the beliefs of observers about their type. For simplicity, assume that types are uniformly distributed over the $[0, 1]$ interval.

Q: does anyone know how to find (increasing) equilibria of this signalling game? I tried the following, but it gives me a strange answer.

My answer: let $f$ denote the equilibrium map from types to actions; and imagine pretending to be type $\hat{\theta}$ when one is really $\theta$. One obtains utility
\begin{equation}
U(\hat{\theta}, \theta) = \theta f(\hat{\theta}) - \frac{f(\hat{\theta})^2}{2} + \hat{\theta}
\end{equation}
Maximising with respect to $\hat{\theta}$ yields

\begin{equation}
\theta f'(\hat{\theta}) - f'(\hat{\theta}) + 1 = 0
\end{equation}

If this is an equilibrium, the best choice is $\hat{\theta} = \theta$, so

\begin{equation}
\theta f'(\theta) - f'(\theta) + 1 = 0 \implies f'(\theta) = \frac{1}{1 - \theta}
\end{equation}

Integration then yields

\begin{equation}
f(\theta) = -\ln(1 - \theta) + c
\end{equation}

By the boundary condition, $c = 0$ (individuals with type $\theta = 0$ should play $a = 0$).

This all seems reasonable, except the equilibrium action then goes to infinity as $\theta \rightarrow 1$! Moreover, one can check that playing $a = -\ln(1 - \theta)$ is *not* a best response given that all others play $a = -\ln(1 - \theta)$. So what am I doing wrong?

**Update:** as pointed out in the comments, the problem comes from an error I made when writing down the first order condition. If you do this correctly and solve the differential equation, you can indeed find an equilibrium action function.

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

> **Comments:**
> **username_1**: Since this is really the reduced form of a game, you should think about how you model out-of-equilibrium-beliefs. For signaling games, this is essential. If you do this in the conventional way, there should be some simple pooling equilibrium supported by weird out-of-equilibrium beliefs.  
> **username_0**: @username_1 of course, you are right that one should be formal. However, as a preliminary step, I am just trying to find a sensible looking equilibrium (which presumably can be supported through the right kind of off equilibrium path beliefs), and am not thinking too hard about uniqueness. I should add that, in this instance, there isn't a pooling equilibrium if beliefs are 'monotone': if all types in some interval $[\theta_{low}, \theta_{high}]$ pool on some action $a$, the type $ \theta_{high}$ would want to deviate to  $a + \epsilon$ for a sufficiently small $\epsilon > 0$.  
> **username_1**: Sure, but there is a pooling equilibrium in which everyone plays $0$ and off equilibrium beliefs are such that everyone beliefs the type is $0$ for any other action.  
> **username_0**: @username_1 I think that you are right that this is an equilibrium. My question, however, was more about increasing equilibria: I don't suppose you have a sense why my (rather heuristic) derivation didn't yield one? The steps seemed OK to me, so I'm a bit confused by why the answer is off. Perhaps there isn't an increasing equilibrium?  
> **username_1**: I think it might be a simple calculation mistake when you calculate the first order condition. The derivative of $f(\hat{\theta})^2/2$ is not $f'(\hat{\theta})$. It is $f(\hat{\theta})f'(\hat{\theta})$.  
> **username_0**: That is embarrassing... thank you for spotting this, hopefully it will solve the problem!  
> **username_1**: No problem; I didn't get it right away either.  

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## 0 answers

