Suppose there is a signaling game with a finite message space $M$, finite action space $A$, and finite type space $T$. Even simpler, all sender types have identical preferences (the receiver just prefers different actions in response to different types). Can the receiver ever do strictly better by randomizing across responses? When an equilibrium exists where the receiver only takes pure actions?

Ubiquitous summarized my question nicely, "Is it ever the case that the equilibrium with the highest receiver payoffs necessarily involves mixed strategies?"

Let's go with sequential equilibrium. If you'd like some notation to start with.

$\sigma_{t}(m)$ is the probability that $t\in T$ sends $m\in M$.

$\sigma_R^m(a)$ is the probability that the receiver responds to $m$ with $a\in A.$ $\mu^m \in \Delta T$ gives the receiver's beliefs after observing $m$.

A sequential equilibrium requires $\sigma_t$ give optimal responses given $\sigma_R$, $\sigma_R$ is optimal given $\mu$ and $\mu$ is Bayesian given $\sigma$. This is really the definition of a weak sequential, but there is no distinction in a signaling game.

My intuition says no when there exists a equilibrium where the receiver only plays pure actions, but I've always been horrible with this kind of stuff. Maybe we also have to stipulate that it is not a zero-sum game, but I'm only saying that because I remember players being better off with the ability to randomize in those games. Perhaps this is a footnote in a paper somewhere?

Consider the game below where sender preferences are not identical. I apologize for the low quality. There are three sender types, each equally likely. We can create what I believe is the receiver (player 2) optimal equilibrium only if they randomize upon receiving message 1. Then types 1 and 3 will play $m_2$, creating a separating equilibrium. If the receiver uses a pure strategy in response to $m_1$, then a type 1 or 2 would deviate and make the receiver worse off.


enter image description here

  • $\begingroup$ Do the actions taken by the receiver as a function of the type have an impact on the message sent by the sender or are these independent? $\endgroup$ Commented Nov 26, 2014 at 3:08
  • $\begingroup$ I'm not exactly sure what you mean. There is one receiver type. Their strategy maps messages into a distribution over actions. They only have an impact on the message insofar as the senders are playing a best response. $\endgroup$
    – Pburg
    Commented Nov 26, 2014 at 3:10
  • 2
    $\begingroup$ Suppose that an equilibrium exists in which the receiver randomises over set of actions $\alpha$. This means, by definition, he must be indifferent between any two probability distributions over $\alpha$--including those in which all weight is put on a single action (pure strategies). So no, a mixed strategy can never be strictly better than the best pure strategy. Or did I mis-understand the question? $\endgroup$
    – Ubiquitous
    Commented Nov 26, 2014 at 8:20
  • $\begingroup$ @Ubiquitous That makes sense to me, but I was wondering if there might be some weird pathological cases. For example, I could only find a theorem, "For generic choices of payoffs in a finite extensive form game with perfect recall, payoffs are constant on each connected component of sequential equilibria." The generic caveat made me wonder. $\endgroup$
    – Pburg
    Commented Nov 26, 2014 at 12:47
  • 1
    $\begingroup$ @Pburg Yes, I see. It seems we had in mind different questions. I was thinking "is it ever the case that the receiver's unique best response to a given sender strategy is a mixed strategy?", whereas it seems your question is actually "is it ever the case that the equilibrium with the highest receiver payoffs necessarily involves mixed strategies?" $\endgroup$
    – Ubiquitous
    Commented Nov 26, 2014 at 19:53

2 Answers 2


Perhaps I have a counterexample!

Let there be three messages, $m_1, m_2,$ and $m_3$, and three sender types $t_1,t_2,t_3$ where $\Pr(t=t_3)=\frac{1}{2}-\epsilon$, $\Pr(t=t_2)=\frac{1}{4}$ and $\Pr(t=t_1)=\frac{1}{4}+\epsilon$. Sending $m_3$ results in a payoff $0$ for senders, we can think of it as exiting the game.

The set of receiver responses to a message $m=m_1,m_2$ is $\{a,r\}$

$u_t(a,m_1)=1 > u_t(a,m_2)=\beta>u_t(r,\cdot)=0$

$u_R(t_1,m_1,a)=u_R(t_2,m_2,a)=2$, $u_R(t_3,m_i,a)=1$,

$u_R(t_2,m_1,a)=u_R(t_2,m_1,a)=0$, $u_R(t_3,m_i,r)=2$,


Then in equilibrium, all senders must obtain the same utility, correct?. Otherwise, one will imitate the other's strategy.

So, the only pure strategy equilibrium is for all senders to choose $m_3$. In a pooling equilibrium on $m_1$ or $m_2$, the best response is to choose $r$. There is no pure strategy separating equilibrium except if $t_1$ and $t_2$ send $m_2$, and the receiver responds with $r$. Then $t_3$ is indifferent between all messages, because he will surely be met with payoff $0$. All of this gives the receiver payoff $\frac{3}{2}-\epsilon$

Then consider the case where $\sigma_R^{m_1}(a)=\beta$ and $\sigma_R^{m_2}(a)=1.$ Now, the senders are indifferent between sending those two messages. Then, let $\sigma_{t_3}(m_1)=\frac{\epsilon+1/4}{-\epsilon+1/2}=1-\sigma_{t_3}(m_1)$ and $\sigma_{t_i}(m_i)=1$ for $i=1,2$. Then the receiver strategy is rational.

The receiver's expected utility from $m_1$ given $a$ or $r$ is 1.5. The expected utility from $m_2$ is slightly above 1.5, given $a$. So the ex ante expected payoff is above $\frac{3}{2}-\epsilon$, better than the pure equilibrium described above. Furthermore, this separation is only maintained by mixing. Any other pure strategy taken by the receiver will induce sender pooling, meaning the only a pure strategy equilibrium is when the receiver chooses $r$.

I should have $\beta$s in the picture below for the lefthand side sender payoffs to $a$. I think the $\beta<1$ is the key ingredient.

enter image description here


I think this cannot happen with risk averse senders, risk neutral receiver, and $A$ rich enough.

For example, and to stick to the canonical signaling model, suppose that $A$ is the positive real line and senders' utility $u$ is increasing in $a$ while receiver's have linear utility decreasing in $a$.

(Admittedly, this is only a partial answer as the framework is much less general that the one in your question, so it might not be satisfactory to you. I still provide an argument in case you were ok with these assumptions)

To derive a contradiction, suppose that at an equilibrium $\sigma^m_R(a') > 0$ and $\sigma^m_R(a'') > 0$ for some $a' \neq a'' \in A$. Let

$$a''' \equiv \frac{\sigma^m_R(a')}{\sigma^m_R(a') + \sigma^m_R(a'') } a' + \frac{\sigma^m_R(a'')}{\sigma^m_R(a') + \sigma^m_R(a'') } a''.$$

By risk aversion

$$ u[ a''' ] > \frac{\sigma^m_R(a')}{\sigma^m_R(a') + \sigma^m_R(a'') } u(a') + \frac{\sigma^m_R(a'')}{\sigma^m_R(a') + \sigma^m_R(a'') } u(a'').$$ $$ [\sigma^m_R(a') + \sigma^m_R(a'')] u( a''' ) > \sigma^m_R(a') u(a') + \sigma^m_R(a'') u(a'').$$

Under some continuity assumption, there must also exist

$$ a '''' < a''' $$

such that

$$ [\sigma^m_R(a') + \sigma^m_R(a'')] u( a'''' ) = \sigma^m_R(a') u(a') + \sigma^m_R(a'') u(a'').$$

So consider $\sigma^m_R{'}$ constructed in the following way

  • $\sigma^m_R{'}(a') = \sigma^m_R{'}(a'') = 0$,
  • $\sigma^m_R{'}(a'''') = \sigma^m_R(a'''') + [\sigma^m_R(a') + \sigma^m_R(a'')]$
  • For all other $\tilde{a}$, $\sigma^m_R{'}(\tilde{a}) = \sigma^m_R(\tilde{a})$

Receivers would prefer $\sigma^m_R{'}$ over $\sigma^m_R$ if it did not alter the signals sent by the senders, because it involves lower expected compensations. But by construction, senders are indifferent between $\sigma^m_R{'}$ and $\sigma^m_R$, so they should send the same signals as in $\sigma^m_R$. Thus $\sigma^m_R$ cannot be an equilibrium which shows that we cannot have two different actions played with positive probability at an equilibrium.

  • $\begingroup$ In this model, wouldn't the receiver always just choose $a=0$? $\endgroup$
    – Pburg
    Commented Nov 26, 2014 at 4:56
  • $\begingroup$ I don't this is necessarily the case. If the receiver always choses $a$ no matter the signal, she does not incentivize "high" types to reveal their type trough a "higher" signal. This may be optimal in a pooling equilibrium, but not in a separating equilibrium. See for instance section 13.C of Mas-Colell, Whinston and Green, although the setup is again a little different from yours (e.g. there are two firms competing for the workers of different types) $\endgroup$ Commented Nov 26, 2014 at 14:10
  • $\begingroup$ What does "receiver's have linear utility decreasing in a" mean then? $\endgroup$
    – Pburg
    Commented Nov 26, 2014 at 14:12
  • $\begingroup$ Sorry that was not very clear. In the Spence signaling model I have in mind, the action the receiver take consists in paying a wage w to the sender. The receiver utility depends on the type of the sender t, minus the wage paid t−w. Basically, the receiver is risk neutral : she cares only about the expected wage she will have to pay, and the expected type she will employ. $\endgroup$ Commented Nov 26, 2014 at 14:23
  • $\begingroup$ Okay, I suppose I've seen this as quadratic loss, $-(t-w)^2.$ Thanks for the suggestion, though I'm looking for something a little more general but with discrete actions. $\endgroup$
    – Pburg
    Commented Nov 26, 2014 at 14:57

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