<p>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.</p>
<p>Here, the best response of the receiver to any belief is a linear function (except for the trivial case that <span class="math-container">$G$</span> and <span class="math-container">$B$</span> 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.</p>
<p>However, the linearity also implies that every sender strategy gives the optimal payoff. Nothing the sender does matters.</p>
<p>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.</p>

CC BY-SA 4.0
2023-12-31T14:59:55.607Z
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.
Oliver Queen
CC BY-SA 4.0
2023-12-31T15:05:18.220Z
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 $\mathcal{I}$
Oliver Queen
CC BY-SA 4.0
2023-12-31T15:10:08.237Z
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.
Michael Greinecker
CC BY-SA 4.0
2024-01-02T15:51:57.597Z
Do you believe I am on the right track to solve the problem? I would appreciate it if you could give it a thought
Oliver Queen
CC BY-SA 4.0
2023-12-31T14:41:22.940Z
2023-12-31T14:41:22.940Z
Michael Greinecker
<p>As it is already pointed out, the problem of the sender reduces to the following</p>
<p><span class="math-container">$$\tau^*\in \text{argmax}\left(\mathbb{E}_{\tau}\left(\hat{V}(\mu_s)\right)\right) \tag{*}$$</span></p>
<p><span class="math-container">$$\text{such that $\sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta)$}\tag{**}$$</span></p>
<p>We know that the sender's utility is <span class="math-container">$u_s(\alpha, \theta) = \alpha$</span> and the receiver's utility is <span class="math-container">$u_r(\alpha, \theta) = -(\alpha-\theta)^2$</span>. As a consequence the receiver's optimal actions is</p>
<p><span class="math-container">$$\hat{\alpha}(\mu_s)=\mathbb{E}_{\mu_s}(\theta) \tag{r.o.a}$$</span></p>
<p>and hence</p>
<p><span class="math-container">$$u_s(\hat{a}(\mu_s),\theta) = \sum_{\theta\in \Theta}\mu_s(\theta)\hat{\alpha}(\mu_s)\tag{a}$$</span></p>
<p>By setting <span class="math-container">$\hat{V}(\mu_s) = \sum_{\theta\in \Theta}\mu_s(\theta)\hat{\alpha}(\mu_s)$</span> then</p>
<p><span class="math-container">$$\mathbb{E}_{\tau}(\hat{V}(\mu_s)) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s)\tag{b}$$</span></p>
<p>based on <span class="math-container">$(*)$</span>, <span class="math-container">$(**)$</span> and <span class="math-container">$(b)$</span> in order to find the optimal pair of <span class="math-container">$\tau^*$</span> since I assumed a binary state space, could I take the lagrangian</p>
<p><span class="math-container">$$\mathcal{L}(\tau(\mu(G)),\tau(\mu(B)), \lambda) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s) - \lambda\left(q - \tau(\mu(G))\mu_s(G) - \tau(\mu(B))\mu_s(B)\right)\tag{***}$$</span></p>
<p>where <span class="math-container">$\lambda \geq 0$</span> nad <span class="math-container">$q=\mu_0(G)$</span>.</p>
<p>Is <span class="math-container">$(***)$</span> equivalent to <span class="math-container">$(*)$</span>, <span class="math-container">$(**)$</span> and if yes ho do I proceed to find the posterior beliefs?</p>

CC BY-SA 4.0
2024-01-03T09:37:38.770Z
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.
Michael Greinecker
CC BY-SA 4.0
2024-01-03T10:48:45.377Z
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 

$$\mathbb{E}_{\tau}(\hat{V}(\mu_s)) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s) =  \sum_{\text{Supp($\tau$)}}\tau(\mu)\left(\mu_s(G)G+\mu_s(B)B \right)$$
Oliver Queen
CC BY-SA 4.0
2024-01-03T12:42:30.987Z
What does $\bar{\theta}$ mean?
Michael Greinecker
CC BY-SA 4.0
2024-01-03T12:49:27.697Z
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 $\hat{\alpha}(\mu_s)$?
Oliver Queen
CC BY-SA 4.0
2024-01-03T12:52:37.090Z
The expectation with respect to $\mu_s$ should presumably depend on $\mu_s$, which doesn't show up in the notation.
Michael Greinecker
CC BY-SA 4.0
2024-01-03T12:57:07.357Z
You mean the notation of the expected value should be written as $\bar{\theta}_{\mu_s}$ because the mean value comes from the posterior beliefs about $\theta$ that are reflected in the formula of the posteriors which is denoted by $\mu_s$, right?
Oliver Queen
CC BY-SA 4.0
2024-01-03T13:01:56.497Z
Something like that. But the problem is really not that complicated. W.l.o.g. let $G=1$ and $B=1$. Identify each belief with the probability of the high state. Then $\hat{V}(\mu)=\alpha(\mu)=\mu$. Since the expected payoff of the sender is then the expected posterior and the expected posterior must equal the prior, the sender's payoff is always equal to the prior. Nothing the sender can do can change that.
Michael Greinecker
CC BY-SA 4.0
2024-01-03T13:06:51.113Z
So in other words what kamenica and gentzkow try to do is to bring the value function to a formulation such that 

$$\mathbb{E}_{\tau}\hat{V}(\mu)= \sum_{\theta in \Theta}\mu\tau(\mu)$$ 

and this is done easily here becase of the fact that the value function $\hat{V}(\mu)=\mu$ after the calculations?
Oliver Queen
CC BY-SA 4.0
2024-01-03T13:21:05.087Z
The argument works here because the best reply function and the sender's utility are linear. K&G do not assume that. It follows from the concavification result that supplying information helps exactly when the concavification at the prior is larger than the value function. In particular, the value function needs to be nonconcave for this to work.
Michael Greinecker
CC BY-SA 4.0
2024-01-03T13:39:43.143Z
Well the sender's value function definetly needs to be either some step function or some kind of a continuous function that has maxima and minima that lead to non-concavity as you say...but it seems to be just tricky and quite restrictive as a model.
Oliver Queen
CC BY-SA 4.0
2024-01-02T15:50:45.973Z
2024-01-02T15:50:45.973Z
Oliver Queen
<p>Let's start with the basics.</p>
<p>Suppose that <span class="math-container">$\Theta$</span> is a finite set of states and <span class="math-container">$\theta$</span> is the element of the state set. To simplify the model, we assume that <span class="math-container">$\Theta = \{\theta_1 = G, \theta_2 = B \}$</span>. A sender's signal space is a partition <span class="math-container">$\pi=\{s_1, s_2\}$</span> of <span class="math-container">$\Theta\times [0,1]$</span> such that <span class="math-container">$(s_i)_{i=1}^2$</span> is a signal realization. Furtherly assume that <span class="math-container">$Y$</span> is a random variable that is indepedent of <span class="math-container">$\Theta$</span> an uniformly distributed over <span class="math-container">$[0,1]$</span> with <span class="math-container">$y$</span> being the realization of <span class="math-container">$Y$</span>. The signal <span class="math-container">$s\in \pi$</span> when <span class="math-container">$(\theta,y)\in s$</span> and let <span class="math-container">$\Lambda\{y|(\theta,y)\in s\} = \mathbb{P}(s|\theta)$</span>, where <span class="math-container">$\Lambda(\cdot)$</span> stands for the Lebesgue measure.</p>
<p>A distribution of posteriors is denoted by <span class="math-container">$\tau \in \Delta(\Delta(\Theta))$</span> and has finite support. Given a signal <span class="math-container">$\pi$</span>, any signal realization <span class="math-container">$s$</span> induces a posterior belief <span class="math-container">$\mu_s(\theta) \triangleq \mu(\theta|s)$</span>. Each signal <span class="math-container">$\pi$</span> leads to a distribution over posterior beliefs, namely each <span class="math-container">$\pi$</span> induces <span class="math-container">$\tau$</span> if <span class="math-container">$\text{Supp}(\tau)=\{\mu_s\}_{s\in \pi}$</span> and we write <span class="math-container">$\tau = &lt;\pi&gt;$</span>. Therefore, observing a signal realization <span class="math-container">$s$</span> with probability <span class="math-container">$\mathbb{P}(s)&gt;0$</span>  generates a unique posterior belief</p>
<p><span class="math-container">\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}</span></p>
<p>where <span class="math-container">$\mathbb{P}(s) = \sum_{\theta^{'}\in\Theta}\mathbb{P}(s|\theta^{'})\mu_0(\theta^{'}) $</span> is the marginal probability of <span class="math-container">$s$</span> and the distribution of posterior beliefs is</p>
<p><span class="math-container">\begin{equation}\tau(\mu)  \triangleq \sum_{\{s\in \pi : \mu_s = \mu\}}\mathbb{P}(s),\quad\text{for all $\mu$} \tag{2}\end{equation}</span></p>
<p>A distribution of posterior beliefs is <span class="math-container">$\textit{Bayes plausible}$</span> 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.</p>
<p><span class="math-container">$$\mathbb{E}_{\tau}(\mu_s|\mu_0)=\sum_{\text{Supp}(\tau)}\mu_s\tau(\mu) =\mu_0  \tag{3}$$</span></p>
<p>The sender's utility is denoted by <span class="math-container">$v_1(\alpha, \theta)$</span> and the receiver's utility is denoted by <span class="math-container">$v_2(\alpha, \theta)$</span> where <span class="math-container">$\alpha$</span> denotes the action of the sender and <span class="math-container">$\theta$</span> the state of the world. Receiver forms the posterior belief <span class="math-container">$\mu_s$</span> using Bayes rule and then she takes an action that is <span class="math-container">$\alpha^*(\mu_s)= argmax_{\alpha\in A}\mathbb{E}_{\mu_s}v_2(\alpha,\theta)$</span>.</p>
<p>They assume that they exist at least two actions and for every action <span class="math-container">$\alpha$</span>, there exists a <span class="math-container">$\mu$</span> s.t. <span class="math-container">$\alpha\in \alpha^*(\mu)$</span> and the receiver's equilibrium outcome is denoted by <span class="math-container">$\hat{\alpha}(\mu)$</span>. Any signal <span class="math-container">$s$</span> induces a posterior belief <span class="math-container">$\mu_s$</span>, such that</p>
<p><span class="math-container">$$\mathbb{E}_{\tau}(\mu_s(\theta)|\mu_0) =  \sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta)= \sum_{s\in \pi: \mu=\mu_s}\pi(s) \frac{\pi(s|\theta)\mu_0(\theta)}{\sum_{\theta^{'}\in\Theta}\pi(s|\theta^{'})\mu_0(\theta^{'})} =\mu_0(\theta)\underbrace{\sum_{s\in \pi: \mu=\mu_s} \pi(s|\theta)}_{=1}=\mu_0(\theta)$$</span></p>
<p>And hence</p>
<p><span class="math-container">$$\mathbb{E}_{\tau}(\mu_s(\theta)|\mu_0) =  \sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta) \tag{4}$$</span></p>
<p>Now by going to <span class="math-container">$(1)$</span>, we have that</p>
<p><span class="math-container">$$v_s(\mu_s)=\mu_s(G) u_s(\hat{\alpha}(\mu_s(G)), G)+\underbrace{(1-\mu_s(G))}_{\mu_s(B)}u_s(\hat{\alpha}(\mu_s(B)), B)\tag{5}$$</span></p>
<p>Therefore we end up with <span class="math-container">$(5)$</span> where the utility of the sender depends only on the posterior beliefs <span class="math-container">$\mu_s=(\mu_s(G),\mu_s(B))$</span> and the state <span class="math-container">$\theta$</span>. However, by applying the expectations of the distribution of posterior beliefs on the latter equation, we do not longer have any concern about <span class="math-container">$\theta$</span> and thus</p>
<p><span class="math-container">$$\mathbb{E}_\tau v_s(\mu_s) =\mathbb{E}_{\tau}\left(\sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta)), \theta)\right) = \sum_{\text{Supp}(\tau)}\tau(\mu)\sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta),\theta)$$</span></p>
<p>By setting <span class="math-container">$\hat{V}(\mu_s)= \sum_{\theta\in\Theta}\mu_s(\theta) u_s(\hat{\alpha}(\mu_s(\theta),\theta)$</span> the problem of the sender reduces to the following</p>
<p><span class="math-container">$$\tau^*\in \text{argmax}\left(\mathbb{E}_{\tau}\left(\hat{V}(\mu_s)\right)\right) \tag{*}$$</span></p>
<p><span class="math-container">$$\text{such that $\sum_{\text{Supp}(\tau)}\tau(\mu)\mu_s(\theta) = \mu_0(\theta)$}\tag{**}$$</span></p>
<p>and the sender's problem reduces to <span class="math-container">$(*)$</span> and <span class="math-container">$(**)$</span></p>
<p>My questions is the following</p>
<p>How do i solve for the optimal <span class="math-container">$\tau=\tau^*$</span> in the case of the binary state space <span class="math-container">$\Theta={G,B}$</span>. Assume that teh prior belief about <span class="math-container">$G$</span> is <span class="math-container">$\mu_0(G)=q\in(0,1)$</span> and the utility of the sender is <span class="math-container">$u_s(\alpha, \theta) = \alpha$</span> and the recievers is <span class="math-container">$u_r(\alpha, \theta) = -(\alpha - \theta)^2$</span></p>

CC BY-SA 4.0
2023-12-31T09:55:14.620Z
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.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T10:38:49.353Z
If this is the only problem in my definition I will make a correction right now
Oliver Queen
CC BY-SA 4.0
2023-12-31T10:39:57.623Z
@MichaelGreinecker I believe that we are ok now about the deifinition of the model...right?
Oliver Queen
CC BY-SA 4.0
2023-12-31T10:46:57.373Z
Partitions of the state space are still insufficient as available signals.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T10:48:55.520Z
This might be useful: https://youtu.be/KfdZ4lh84kE?si=Iq22YCwHJ1uo08PX
Michael Greinecker
CC BY-SA 4.0
2023-12-31T10:56:37.260Z
@MichaelGreinecker I've seen this video before...does not answer my questions. The model is written above in the most simplified way and this is how they define it if you check his papers of $2011$ and $2017$ (two papers in 2017 were publised with the topic of competintion in persuasion). Above is their model as I have understand it to its last point. But I can go no deeper....
Oliver Queen
CC BY-SA 4.0
2023-12-31T11:24:21.093Z
It seems to be the formulation in 2017 but not in the famous 2011 paper, in which a signals is given by the conditional distributions over a realization space. Anyways, the solution will depend on the preferences of the receiver if you want an explicit solution.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T11:41:11.670Z
Well, the representation seems to be equivalent to me....i can not understand the difference that you mention between $2011$ and $2017$. It seems like the same setup but maybe I am wrong.  As per the preferences of the receiver, Kamenica nd Gentzkow in all their series of Bayesian persuasion papers, they assume that the preferences are fully alligned aong senders and receivers in the biggest part of their discussion. Anyway the problem of the one sender is reduced to $(*)$, $(**)$ I believe. I think, I have understood their model until this point
Oliver Queen
CC BY-SA 4.0
2023-12-31T11:46:10.157Z
The models are equivalent, though not quite the same. The partition version doesn't translate to a infinite state space though. If the preferences of senders and receivers are perfectly alignd, full revelation is always optimal. This does not hold in the classic prosecutor/judge example.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T11:53:59.170Z
When the example expands to the prosecutor/judge/ =defense attorney, where the prosecutor and the defense attorney oppose one another, at the end of the day each one of them will provide as much information as she can to change the beliefs of the judge. Hence full revelation is the only equilibrium if the prosecutor thinks that leaving evidence out of her opposition and the defense attorney will find out then she will use it. Contrarily, if the defense attorney does not want to provide information to mislead the judge, then she knows that the prosecutor will unleash this info if she finds out
Oliver Queen
CC BY-SA 4.0
2023-12-31T11:54:59.523Z
In other words, competition will reveal the truth because none will try to leave any ecidence out of her case...but in fact this is not always true in every economic setting
Oliver Queen
CC BY-SA 4.0
2023-12-31T11:55:22.237Z
I mean the example at the beginning of the 2011 paper. There, partial revelation is optimal.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T11:57:40.970Z
Yes, partial revelation is optimal in the first one. Sorry for the passionate response...but I've read all three papers in the past few months and I am in the crazy mode :P
Oliver Queen
CC BY-SA 4.0
2023-12-31T12:53:28.883Z
Changed a little bit the answer and I defined the same utilities that Kamenica and Gentzkow used in $2011$ paper where the sender's payoff depends only on the expected state.
Oliver Queen
<microeconomics><mathematical-economics><game-theory><bayesian-game><information>
Unwrapping Kamenica and Gentzkow model
Oliver Queen
# Unwrapping Kamenica and Gentzkow model

(Asked by: username_0 on 2023-12-30)

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 $u\_r(\alpha, \theta) = -(\alpha - \theta)^2$

**Tags:** microeconomics, mathematical-economics, game-theory, bayesian-game, information
> **Comments:**
> **username_1**: 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.  
> **username_0**: If this is the only problem in my definition I will make a correction right now  
> **username_0**: @MichaelGreinecker I believe that we are ok now about the deifinition of the model...right?  
> **username_1**: Partitions of the state space are still insufficient as available signals.  
> **username_1**: This might be useful: https://youtu.be/KfdZ4lh84kE?si=Iq22YCwHJ1uo08PX  
> **username_0**: @MichaelGreinecker I've seen this video before...does not answer my questions. The model is written above in the most simplified way and this is how they define it if you check his papers of $2011$ and $2017$ (two papers in 2017 were publised with the topic of competintion in persuasion). Above is their model as I have understand it to its last point. But I can go no deeper....  
> **username_1**: It seems to be the formulation in 2017 but not in the famous 2011 paper, in which a signals is given by the conditional distributions over a realization space. Anyways, the solution will depend on the preferences of the receiver if you want an explicit solution.  
> **username_0**: Well, the representation seems to be equivalent to me....i can not understand the difference that you mention between $2011$ and $2017$. It seems like the same setup but maybe I am wrong.  As per the preferences of the receiver, Kamenica nd Gentzkow in all their series of Bayesian persuasion papers, they assume that the preferences are fully alligned aong senders and receivers in the biggest part of their discussion. Anyway the problem of the one sender is reduced to $(*)$, $(**)$ I believe. I think, I have understood their model until this point  
> **username_1**: The models are equivalent, though not quite the same. The partition version doesn't translate to a infinite state space though. If the preferences of senders and receivers are perfectly alignd, full revelation is always optimal. This does not hold in the classic prosecutor/judge example.  
> **username_0**: When the example expands to the prosecutor/judge/ =defense attorney, where the prosecutor and the defense attorney oppose one another, at the end of the day each one of them will provide as much information as she can to change the beliefs of the judge. Hence full revelation is the only equilibrium if the prosecutor thinks that leaving evidence out of her opposition and the defense attorney will find out then she will use it. Contrarily, if the defense attorney does not want to provide information to mislead the judge, then she knows that the prosecutor will unleash this info if she finds out  
> **username_0**: In other words, competition will reveal the truth because none will try to leave any ecidence out of her case...but in fact this is not always true in every economic setting  
> **username_1**: I mean the example at the beginning of the 2011 paper. There, partial revelation is optimal.  
> **username_0**: Yes, partial revelation is optimal in the first one. Sorry for the passionate response...but I've read all three papers in the past few months and I am in the crazy mode :P  
> **username_0**: Changed a little bit the answer and I defined the same utilities that Kamenica and Gentzkow used in $2011$ paper where the sender's payoff depends only on the expected state.  

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## 2 answers

### Answer by username_1 on 2023-12-31. Score: 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.

> **Comments:**
> **username_0**: 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.  
> **username_0**: 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 $\mathcal{I}$  
> **username_1**: 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.  
> **username_0**: Do you believe I am on the right track to solve the problem? I would appreciate it if you could give it a thought  

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### Answer by username_0 on 2024-01-02. Score: 0

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?

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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   $$\mathbb{E}_{\tau}(\hat{V}(\mu_s)) = \sum_{\text{Supp($\tau$)}}\tau(\mu)\hat{V}(\mu_s) =  \sum_{\text{Supp($\tau$)}}\tau(\mu)\left(\mu_s(G)G+\mu_s(B)B \right)$$  
> **username_1**: What does $\bar{\theta}$ mean?  
> **username_0**: 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 $\hat{\alpha}(\mu_s)$?  
> **username_1**: The expectation with respect to $\mu_s$ should presumably depend on $\mu_s$, which doesn't show up in the notation.  
> **username_0**: You mean the notation of the expected value should be written as $\bar{\theta}_{\mu_s}$ because the mean value comes from the posterior beliefs about $\theta$ that are reflected in the formula of the posteriors which is denoted by $\mu_s$, right?  
> **username_1**: Something like that. But the problem is really not that complicated. W.l.o.g. let $G=1$ and $B=1$. Identify each belief with the probability of the high state. Then $\hat{V}(\mu)=\alpha(\mu)=\mu$. Since the expected payoff of the sender is then the expected posterior and the expected posterior must equal the prior, the sender's payoff is always equal to the prior. Nothing the sender can do can change that.  
> **username_0**: So in other words what kamenica and gentzkow try to do is to bring the value function to a formulation such that   $$\mathbb{E}_{\tau}\hat{V}(\mu)= \sum_{\theta in \Theta}\mu\tau(\mu)$$   and this is done easily here becase of the fact that the value function $\hat{V}(\mu)=\mu$ after the calculations?  
> **username_1**: The argument works here because the best reply function and the sender's utility are linear. K&G do not assume that. It follows from the concavification result that supplying information helps exactly when the concavification at the prior is larger than the value function. In particular, the value function needs to be nonconcave for this to work.  
> **username_0**: Well the sender's value function definetly needs to be either some step function or some kind of a continuous function that has maxima and minima that lead to non-concavity as you say...but it seems to be just tricky and quite restrictive as a model.  

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