When should a receiver randomize across actions in a signaling game? 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.

$\sigma_R^{m_1}(a)=.5=\sigma_R^{m_1}(r)=.5$
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 $\sigma^m_R(a') &gt; 0$ and $\sigma^m_R(a'') &gt; 0$ 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

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

Under some continuity assumption, there must also exist

$$ a '''' &lt; a''' $$ 

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.