Let's start with the basics.

Suppose that $\Theta$ is a finite set of states and $\theta$ is the element of the state set. To simplify the model, we assume that $\Theta = \{\theta_1 = G, \theta_2 = B \}$. A sender's signal space is a partition $\pi=\{s_1, s_2\}$ of $\Theta\times [0,1]$ such that $(s_i)_{i=1}^2$ is a signal realization. Furtherly assume that $Y$ is a random variable that is indepedent of $\Theta$ an uniformly distributed over $[0,1]$ with $y$ being the realization of $Y$. The signal $s\in \pi$ when $(\theta,y)\in s$ and let $\Lambda\{y|(\theta,y)\in s\} = \mathbb{P}(s|\theta)$, where $\Lambda(\cdot)$ stands for the Lebesgue measure.

A distribution of posteriors is denoted by $\tau \in \Delta(\Delta(\Theta))$ and has finite support. Given a signal $\pi$, any signal realization $s$ induces a posterior belief $\mu_s(\theta) \triangleq \mu(\theta|s)$. Each signal $\pi$ leads to a distribution over posterior beliefs, namely each $\pi$ induces $\tau$ if $\text{Supp}(\tau)=\{\mu_s\}_{s\in \pi}$ and we write $\tau = <\pi>$. Therefore, observing a signal realization $s$ with probability $\mathbb{P}(s)>0$ generates a unique posterior belief

\begin{equation}\mu_s(\theta) = \frac{\mathbb{P}(s|\theta)\mu_0(\theta)}{\sum_{\theta^{'}\in\Theta}\mathbb{P}(s|\theta^{'})\mu_0(\theta^{'})},\quad\text{for all $s$ and $\theta$} \tag{1}\end{equation}

where $\mathbb{P}(s) = \sum_{\theta^{'}\in\Theta}\mathbb{P}(s|\theta^{'})\mu_0(\theta^{'}) $ is the marginal probability of $s$ and the distribution of posterior beliefs is

\begin{equation}\tau(\mu) \triangleq \sum_{\{s\in \pi : \mu_s = \mu\}}\mathbb{P}(s),\quad\text{for all $\mu$} \tag{2}\end{equation}

A distribution of posterior beliefs is $\textit{Bayes plausible}$ if the best projection about the posterior beliefs, given the prior distribution of beliefs, equals the prior beliefs, or in other words the beliefs satisfy the martingale property.

$$\mathbb{E}_{\tau}(\mu_s|\mu_0)=\sum_{\text{Supp}(\tau)}\mu_s\tau(\mu) =\mu_0 \tag{3}$$

The sender's utility is denoted by $v_1(\alpha, \theta)$ and the receiver's utility is denoted by $v_2(\alpha, \theta)$ where $\alpha$ denotes the action of the sender and $\theta$ the state of the world. Receiver forms the posterior belief $\mu_s$ using Bayes rule and then she takes an action that is $\alpha^*(\mu_s)= argmax_{\alpha\in A}\mathbb{E}_{\mu_s}v_2(\alpha,\theta)$.

They assume that they exist at least two actions and for every action $\alpha$, there exists a $\mu$ s.t. $\alpha\in \alpha^*(\mu)$ and the receiver's equilibrium outcome is denoted by $\hat{\alpha}(\mu)$. Any signal $s$ induces a posterior belief $\mu_s$, such that

$$\mathbb{E}_{\tau}(\mu_s(\theta)|\mu_0) = \sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta)= \sum_{s\in \pi: \mu=\mu_s}\pi(s) \frac{\pi(s|\theta)\mu_0(\theta)}{\sum_{\theta^{'}\in\Theta}\pi(s|\theta^{'})\mu_0(\theta^{'})} =\mu_0(\theta)\underbrace{\sum_{s\in \pi: \mu=\mu_s} \pi(s|\theta)}_{=1}=\mu_0(\theta)$$

And hence

$$\mathbb{E}_{\tau}(\mu_s(\theta)|\mu_0) = \sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta) \tag{4}$$

Now by going to $(1)$, we have that

$$v_s(\mu_s)=\mu_s(G) u_s(\hat{\alpha}(\mu_s(G)), G)+\underbrace{(1-\mu_s(G))}_{\mu_s(B)}u_s(\hat{\alpha}(\mu_s(B)), B)\tag{5}$$

Therefore we end up with $(5)$ where the utility of the sender depends only on the posterior beliefs $\mu_s=(\mu_s(G),\mu_s(B))$ and the state $\theta$. However, by applying the expectations of the distribution of posterior beliefs on the latter equation, we do not longer have any concern about $\theta$ and thus

$$\mathbb{E}_\tau v_s(\mu_s) =\mathbb{E}_{\tau}\left(\sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta)), \theta)\right) = \sum_{\text{Supp}(\tau)}\tau(\mu)\sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta),\theta)$$

By setting $\hat{V}(\mu_s)= \sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta),\theta)$ the problem of the sender reduces to the following

$$\tau^*\in \text{argmax}\left(\mathbb{E}_{\tau}\left(\hat{V}(\mu_s)\right)\right) \tag{*}$$

$$\text{such that $\sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta)$}\tag{**}$$

and the sender's problem reduces to $(*)$ and $(**)$

My questions is the following

How do i solve for the optimal $\tau=\tau^*$ in the case of the binary state space $\Theta={G,B}$. Assume that teh prior belief about $G$ is $\mu_0(G)=q\in(0,1)$ and the utility of the sender is $u_s(\alpha, \theta) = \alpha$ and the recievers is $u_r(\alpha, \theta) = -(\alpha - \theta)^2$

  • $\begingroup$ There are only two possible partitions of a binary state space, one completely informative and one completely uninformative. Information must be modeled differently here. And I don't get what is measurable here. With two states, concavification usually works quite well. $\endgroup$ Commented Dec 31, 2023 at 9:55
  • $\begingroup$ If this is the only problem in my definition I will make a correction right now $\endgroup$ Commented Dec 31, 2023 at 10:38
  • $\begingroup$ @MichaelGreinecker I believe that we are ok now about the deifinition of the model...right? $\endgroup$ Commented Dec 31, 2023 at 10:39
  • 1
    $\begingroup$ This might be useful: youtu.be/KfdZ4lh84kE?si=Iq22YCwHJ1uo08PX $\endgroup$ Commented Dec 31, 2023 at 10:48
  • 1
    $\begingroup$ I mean the example at the beginning of the 2011 paper. There, partial revelation is optimal. $\endgroup$ Commented Dec 31, 2023 at 11:55

2 Answers 2


It is optimal for the sender not to reveal any information at all. This follows from the result that the highest payoff the sender can achieve is given by the concavification of the value function evaluated at the prior.

Here, the best response of the receiver to any belief is a linear function (except for the trivial case that $G$ and $B$ are the same number), and the sender's corresponding utility is linear, too. So, the sender's value function is linear and, therefore, equal to its own concavification.

However, the linearity also implies that every sender strategy gives the optimal payoff. Nothing the sender does matters.

Another way to think about the problem is to note that only the expected posterior is relevant in this setup, and the expected posterior has to equal the prior.

  • $\begingroup$ The problem seems to be very restrictive by definition and there are not too much things to say in this setup. I am trying to figure out a way to solve a problem on strategic information transmission in a financial market, but this sepcific setup does not help too much to model both the strategic interaction of the senders and the strategic interaction with the receivers that might have quadratic preferences and be risk averse. $\endgroup$ Commented Dec 31, 2023 at 14:59
  • $\begingroup$ Maybe I should re-state my problem on how could I translate the expectations from expectations under the posterior beliefs, namely $\mathbb{E}_{\mu_s}(u_s(\alpha, \theta)$ to conditional expectations under an information set, namely $\mathbb{E}[u_s(\alpha, \theta)| \mathcal{I}]$. In this case the beliefs are reflected in the set $\mathcal{I}$ $\endgroup$ Commented Dec 31, 2023 at 15:05
  • $\begingroup$ One of the central points in this literature is that it is without loss of generality for the sender simply tells the receiver what to believe (their posterior) and for the receiver to update, given this information, to exactly this posterior. Changin the modeling of information does not change that. $\endgroup$ Commented Dec 31, 2023 at 15:10
  • $\begingroup$ Do you believe I am on the right track to solve the problem? I would appreciate it if you could give it a thought $\endgroup$ Commented Jan 2 at 15:51

As it is already pointed out, the problem of the sender reduces to the following

$$\tau^*\in \text{argmax}\left(\mathbb{E}_{\tau}\left(\hat{V}(\mu_s)\right)\right) \tag{*}$$

$$\text{such that $\sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta)$}\tag{**}$$

We know that the sender's utility is $u_s(\alpha, \theta) = \alpha$ and the receiver's utility is $u_r(\alpha, \theta) = -(\alpha-\theta)^2$. As a consequence the receiver's optimal actions is

$$\hat{\alpha}(\mu_s)=\mathbb{E}_{\mu_s}(\theta) \tag{r.o.a}$$

and hence

$$u_s(\hat{a}(\mu_s),\theta) = \sum_{\theta\in \Theta}\mu_s(\theta)\hat{\alpha}(\mu_s)\tag{a}$$

By setting $\hat{V}(\mu_s) = \sum_{\theta\in \Theta}\mu_s(\theta)\hat{\alpha}(\mu_s)$ then

$$\mathbb{E}_{\tau}(\hat{V}(\mu_s)) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s)\tag{b}$$

based on $(*)$, $(**)$ and $(b)$ in order to find the optimal pair of $\tau^*$ since I assumed a binary state space, could I take the lagrangian

$$\mathcal{L}(\tau(\mu(G)),\tau(\mu(B)), \lambda) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s) - \lambda\left(q - \tau(\mu(G))\mu_s(G) - \tau(\mu(B))\mu_s(B)\right)\tag{***}$$

where $\lambda \geq 0$ nad $q=\mu_0(G)$.

Is $(***)$ equivalent to $(*)$, $(**)$ and if yes ho do I proceed to find the posterior beliefs?

  • $\begingroup$ The Lagrangian is a function and not an optimality condition. You will have to write out $\hat{V}$ to solve it. Then you will see that it doesn't matter what the sender does. $\endgroup$ Commented Jan 3 at 9:37
  • $\begingroup$ In case of the binary state $\{B, G\}$ the value function becomes $$\hat{V}(\mu_s)=\sum_{\theta\in \{G,B\}}\mu_s(\theta)\hat{\alpha}(\mu_s)=\mu_s(G)\bar{\theta}+\mu_s(B)\bar{\theta}$$ because from the receiver's optimal response is equalt to the epxctation of the state, that is $$\hat{\alpha}(\mu_s)=\mathbb{E}_{\mu_s}(\theta)=\sum_{\theta \in \{B,G\}}\mu_s(\theta)\theta = \mu_s(G)G+\mu_s(B)B=\bar{\theta}$$ and thus $$\mathbb{E}_{\tau}(\hat{V}(\mu_s)) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\left(\mu_s(G)G+\mu_s(B)B \right)$$ $\endgroup$ Commented Jan 3 at 10:48
  • $\begingroup$ What does $\bar{\theta}$ mean? $\endgroup$ Commented Jan 3 at 12:42
  • $\begingroup$ If you check the optimal action of the receiver, its value is equal to the mean value of the state variable and i denote this with $\bar{\theta}$. Isn't that clear from the formula of $\hat{\alpha}(\mu_s)$? $\endgroup$ Commented Jan 3 at 12:49
  • $\begingroup$ The expectation with respect to $\mu_s$ should presumably depend on $\mu_s$, which doesn't show up in the notation. $\endgroup$ Commented Jan 3 at 12:52

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