Unwrapping Kamenica and Gentzkow model

Question

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 $u_r(\alpha, \theta) = -(\alpha - \theta)^2$

Answer 1

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.

Answer 2

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?