Based on Kamenica and Gentzkow example about the prosecutor and the judge. The utilities of the sender and the receiver are $v(\alpha,\omega)=\alpha$ and $u(\alpha,\omega)=-(\alpha-\omega)^2$.

The judge is the receiver and the prosecutor is the sender, where $\omega =\{I, G\}$ and the $\mu_0(G) = 0.3$. The action set is $A = \{0,1\}$ where $0 : = \text{ acquit}$ and $1 : = \text{ convict}$. For any $\omega$, the prosecutor's utility is $v(1,\omega)= 1$ and $v(0,\omega)=0$ and the judge's utility is $u(0,I)=1=u(1,G)$ and $u(1,I)=0=u(0,G)$. The expected utility of the judge and the prosecutor are the following:

$$u(\alpha,\omega)=0.3\times 1 + 0.7\times 1 = 1\text{ and }v(\alpha,\omega)= 0.3\times 1 + 0.7\times 0 = 0.3$$

The prosecutor chooses a signal space $S=\{i,g\}$ and a distribution $\pi$ over $S$, that is $\pi(i|I)=4/7$, $\pi(i|G)=0$, $\pi(g|I)=3/7$, $\pi(g|G)=1$ and hence by the law of total probability

$$\tau(i) = \pi(i|G)\mu_0(G)+\pi(i|I)\mu_0(I)=0.4 \text{ and } \tau(g) = \pi(g|G)\mu_0(G)+\pi(g|I)\mu_0(I)=0.6$$

Hence, we have that

$$\mu_i(I) = \frac{\pi(i|I)\mu_0(I)}{\pi(i)}=1,\quad \mu_i(G) = \frac{\pi(i|G)\mu_0(G)}{\pi(i)}= 0$$


$$\mu_g(I) = \frac{\pi(g|I)\mu_0(I)}{\pi(g)}=\frac{1}{2},\quad \mu_g(G) = \frac{\pi(g|G)\mu_0(G)}{\pi(g)}= \frac{1}{2}$$

By $u$ and $v$ we have that the optimal response of the receiver is equal to $\alpha^{\star}(\mu_s) = \mathbb{E}_{\mu_s}(\omega)$ and thus sender's value function is

$$\hat{V}(\mu_s)=\mu_s(G)v(\alpha^{\star}(\mu_s(G)),G)+\mu_s(I)v(\alpha^{\star}(\mu_s(I)),I),\text{ for all $s\in\text{Supp}(\tau)$}$$

and therefore the sender's problem is simplified to

$$\text{max}_{\tau\in\Delta(\Delta(\Omega))} \{ \tau(i) \hat{V}(\mu_i) + \tau(g) \hat{V}(\mu_g) \}\text{ s.t. $\tau(i)\mu_i+\tau(g)\mu_g = \mu_0$ for all $\omega$}\tag{S.P.}$$

From the above numbers and if we substitute them for every signal $s=\{i,g\}$, the payoff of the sender is

$$\tau(i)\hat{V}(\mu_i)=\tau(i)\left( \mu_i(G)v(0,G)+\mu_i(I)v(0,I)\right)=0.4\times \left(0\times 0+ 1\times 0 \right)=0$$


$$\tau(g)\hat{V}(\mu_g)=\tau(g)\left( \mu_g(G)v(1,G)+\mu_g(I)v(1,I)\right)=0.6\times \left(\frac{1}{2}\times 1+ \frac{1}{2}\times 1 \right)=0.6$$

and $0.4 \times 1+ 0.6 \times(1/2) = 0.7$ which is the prior of innocence and $0.4 \times 0+ 0.6 \times(1/2) = 0.3$ which is the prior of guilty.

To wrap up the prosecutor will send the signal $i$ only if she is $100\%$ sure that the accused is guilty (based on the posterior belief of $I$) and the signal $g$ if the posterior belief of the accused to be either innocent or guilty is equally likely.

Although they do not present how they find the values for $\pi$ etc, i assume that the idea starts going backwordly and solving the problem having as constraints that $\pi(g|I)+\pi(i|I) = 1$ as well as $\pi(g|G)+\pi(i|G) = 1$.

My question is how do I solve the problem if the sender's role change and we have isntead of a prosecutor a defense attorney where her utility will be $v(\alpha, \omega)=1-\alpha$ and what becomes when both the defense attorney and the prosecutor comptete? (In the $2017$ paper where they talk about competition, I fail to understand their game when there exist two players, a defense attorney and a prosecutor, check in here.


1 Answer 1


Under the prior, the defendant is more likely to be innocent than guilty. So, without any additional information, the defense attorney in your example need not provide any information. A lot of signals are optimal here.

If the prosecutor and the defense attorney compete, one equilibrium is full revelation from both. No individual deviation from either sender changes anything here.

  • $\begingroup$ I understand what you say and I also come to the same conclusion. If we are in this case with the above numbers, then the judge - defense attorney problem seems to be satisfied in the exact same way with the judge - prosecutor one. The DA sends the signal $i$ with if the posterior belief of $I$ is equal to $1$ and the signal $g$ if the posterior belief of the accused to be either innocent or guilty is equally likely. The question that arises is, how did Kamenica and Getzkow solved the problem so as to reach to the optimal signal $\pi(i|I)=4/7$, $\pi(i|G)=0$, $\pi(g|I)=3/7$, $\pi(g|G)=1$? $\endgroup$ Commented Mar 16 at 18:57
  • $\begingroup$ Intuitively, we could say that $$\pi(i|I)+\pi(g|I)=1\tag{1}$$ and $$\pi(i|G)+\pi(g|G)=1\tag{2}$$ If the state is $G$, then form $(2)$ we could claim that always it is optimal for the prosecutor to send the signal $g$ and hence $\pi(g|G)=1$ and $\pi(i|G)=0$, but how do we colcude to $\pi(i|I)=4/7$ and $\pi(g|I)=3/7$? $\endgroup$ Commented Mar 16 at 19:00
  • $\begingroup$ Presumably, K&G solved the problem using concavification. $\endgroup$ Commented Mar 17 at 0:02
  • $\begingroup$ Sure this is what they do, but I do not understand....where does this property takes place in the solution? Do you understand it? $\endgroup$ Commented Mar 17 at 9:43
  • $\begingroup$ I'm not sure what "this property" refers to. $\endgroup$ Commented Mar 17 at 10:13

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