<p>I will try something like an answer though I miss some technicalities</p>
<p>So, if the receiver decides to play her myopic best responses then the payoff given by <span class="math-container">$(1)$</span> is</p>
<p><span class="math-container">$$\mathbb{E}\left[u_R(a_1^{'},a_2^{'},\theta)|s_1\right]=-\underbrace{\mathbb{E}\left(\left(\mathbb{E}(\theta_1|s_1)-\theta_1\right)^2|s_1\right)}_{v_1=\mathbb{V}(\theta_1|s_1)}-\underbrace{\mathbb{E}\left(\left(\mathbb{E}(\theta_2|s_1)-\theta_2\right)^2|s_1\right)}_{v_2=\mathbb{V}(\theta_2|s_1)}=\\=-v_1-\mathbb{V}(\theta_2|s_1)$$</span></p>
<p>and because <span class="math-container">$\theta_2=\rho\theta_1+\epsilon$</span> it holds that</p>
<p><span class="math-container">$\mathbb{V}(\theta_2|s_1)=\rho^2\underbrace{\mathbb{V}(\theta_1|s_1)}_{v_1}+\underbrace{\mathbb{V}(\epsilon|s_1)}_{\text{and because $\epsilon\sim N(0,\sigma^2)$ is independent of $\theta_1$ and hence of $s_1$ then $\mathbb{V}(\epsilon|s_1)=\mathbb{V}(\epsilon)=\sigma^2$ }}=\rho^2v_1+\sigma^2$</span></p>
<p>And as a consequence</p>
<p><span class="math-container">$$\mathbb{E}\left[u_R(a_1^{'},a_2^{'},\theta)|s_1\right]=-v_1-(\rho^2v_2+\sigma^2)$$</span></p>
<p>Does this seem to be ok? Do I need to clarify why <span class="math-container">$\mathbb{V}(\theta_1|s_1)=\mathbb{E}\left(\left(\mathbb{E}(\theta_1|s_1)-\theta_1\right)^2|s_1\right)$</span>? I would just say by definition but I do not know if I am missing anything.</p>
<p>I wonder why the expectation <span class="math-container">$\mathbb{E}[\theta_2|s_2]$</span> does not emerge it the payoff and the author only conditions with respect to <span class="math-container">$s_1$</span>.</p>

CC BY-SA 4.0
2023-02-01T16:58:36.347Z
2023-02-02T23:04:54.517Z
2023-02-02T23:04:54.517Z
Oliver Queen
<p>This is from the <a href="https://deliverypdf.ssrn.com/delivery.php?ID=858106007111022110110066098098014086018035030027084013029085099025007017094022078090028110123025107038033065127084108097026090041004008032002087088069081095095072007091016014125028121016066123096106093024007073109015127108119067115115064090110125027064&amp;EXT=pdf&amp;INDEX=TRUE" rel="nofollow noreferrer">this paper in section <span class="math-container">$3$</span> about the two period example</a>.</p>
<p>Suppose that we have the following two period, <span class="math-container">$t=1,2$</span>, sender(S) - receiver(R) model.</p>
<p>For an action path <span class="math-container">$a=(a_1,a_2)$</span> and a state path <span class="math-container">$\theta=(\theta_1,\theta_2)$</span>, the payoffs for the receiver and sender, respectively, are</p>
<p><span class="math-container">\begin{equation}u_R(a,\theta)=-\sum_{t=1}^{2}(a_t-\theta_t)^2,\quad u_S(a,\theta)=-\sum_{t=1}^{2}(a_t-\theta_t-\beta)^2,\quad\text{such that $\beta&gt;0$}\end{equation}</span></p>
<p>The initial state <span class="math-container">$\theta_1$</span> has a normal distribution <span class="math-container">$N(0, \sigma^2_1)$</span>. The second-period state is given by</p>
<p><span class="math-container">$$\theta_2=\rho\theta_1+\epsilon$$</span></p>
<p>where <span class="math-container">$\epsilon$</span> has a normal distribution <span class="math-container">$N(0, \sigma^2)$</span>, independent of <span class="math-container">$\theta_1$</span>.The persistence parameter <span class="math-container">$\rho$</span> is unrestricted. So, this setting is characterized by four parameters, that is <span class="math-container">$(\beta, \sigma^2_1, \rho, \sigma^2)$</span>.</p>
<p>The state distribution is common knowledge, but the state realizations are observed only by the sender. Before the first period, the sender commits to a dynamic information policy mapping each private history to a signal distribution. The game then proceeds as follows. In the first period, the sender observes the state <span class="math-container">$\theta_1$</span> and sends the receiver a signal <span class="math-container">$s_1$</span> drawn from the distribution prescribed by the information policy at the history <span class="math-container">$h_1 = \theta_1$</span>. The receiver observes <span class="math-container">$s_1$</span>, updates his belief about <span class="math-container">$\theta_1$</span>, and then chooses an action <span class="math-container">$a_1$</span>, which is observed by the sender. In the second period, the sender observes the state <span class="math-container">$\theta_2$</span> and sends a signal <span class="math-container">$s_2$</span> drawn from the distribution prescribed by the information policy at the history <span class="math-container">$h_2 = (\theta_1, s_1, a_1, \theta_2)$</span>. The receiver observes <span class="math-container">$s_2$</span>, updates his belief about <span class="math-container">$\theta_2$</span>, and chooses an action <span class="math-container">$a_2$</span>. Payoffs for both periods are then realized.</p>
<p>To find the sender's optimal policy we need to solve backwards. In the second period, the receiver just maximizes his flow payoff. So, after receiving signal <span class="math-container">$s_2$</span>, he matches his action with his updated expectation of <span class="math-container">$\theta_2$</span>. Let <span class="math-container">$v_2$</span> denote the posterior variance of <span class="math-container">$θ_2$</span> as assessed by the receiver after observing signal <span class="math-container">$s_2$</span>. By standard properties of the quadratic, the expected flow payoffs in the second period are <span class="math-container">$−v_2$</span> for the receiver and <span class="math-container">$−β^2 −v_2$</span> for the sender.</p>
<p>In the first period, the receiver’s action can affect both his flow payoff and the informativeness of the second-period signal. Upon receiving signal <span class="math-container">$s_1$</span>, the receiver may shift his action away from his bliss point <span class="math-container">$\mathbb{E}[\theta_1|s_1]$</span> if doing so is rewarded with a more precise signal in the second period. Let <span class="math-container">$v_1$</span> denote the conditional variance of <span class="math-container">$\theta_1$</span> given <span class="math-container">$s_1$</span>. Without any further information from the sender, the receiver can choose his myopic best responses <span class="math-container">$a^{′}_1 = \mathbb{E}[\theta_1|s_1]$</span> and <span class="math-container">$a^{′}_2 = \rho\mathbb{E}[\theta_1|s_1]$</span> to obtain the expected payoff</p>
<p><span class="math-container">$$\tag{1}E[u_R(a^{′}_1,a^{′}_2,\theta)|s_1] = −v_1 − (\rho^2v_1 + \sigma^2).$$</span></p>
<p>Therefore, a necessary condition for the receiver to choose action <span class="math-container">$a_1=\mathbb{E}[\theta_1|s_1]+b_1$</span> is that the resulting flow loss is no greater than the total loss from choosing <span class="math-container">$(a^{′}_1,a^{′}_2):$</span></p>
<p><span class="math-container">$$\tag{2}v_1+b_1^2\leq v_1 + (\rho^2 v_1+\sigma^2)$$</span></p>
<p><span class="math-container">$\textbf{Question:}$</span> How did <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> where calculated? Could anyone provide some help with the calculations that give <span class="math-container">$(1)$</span> and how did we come up with <span class="math-container">$(2)$</span>?</p>

CC BY-SA 4.0
2023-02-03T11:01:52.897Z
@game-theory can anyone confirm my calculations below?
Oliver Queen
<microeconomics><mathematical-economics><game-theory><bayesian-game>
Can anyone help with these calculations?
Oliver Queen
# Can anyone help with these calculations?

(Asked by: username_0 on 2023-01-30)

This is from the [this paper in section $3$ about the two period example](https://deliverypdf.ssrn.com/delivery.php?ID=858106007111022110110066098098014086018035030027084013029085099025007017094022078090028110123025107038033065127084108097026090041004008032002087088069081095095072007091016014125028121016066123096106093024007073109015127108119067115115064090110125027064&EXT=pdf&INDEX=TRUE).

Suppose that we have the following two period, $t=1,2$, sender(S) - receiver(R) model.

For an action path $a=(a\_1,a\_2)$ and a state path $\theta=(\theta\_1,\theta\_2)$, the payoffs for the receiver and sender, respectively, are

\begin{equation}u\_R(a,\theta)=-\sum\_{t=1}^{2}(a\_t-\theta\_t)^2,\quad u\_S(a,\theta)=-\sum\_{t=1}^{2}(a\_t-\theta\_t-\beta)^2,\quad\text{such that $\beta>0$}\end{equation}

The initial state $\theta\_1$ has a normal distribution $N(0, \sigma^2\_1)$. The second-period state is given by

$$\theta\_2=\rho\theta\_1+\epsilon$$

where $\epsilon$ has a normal distribution $N(0, \sigma^2)$, independent of $\theta\_1$.The persistence parameter $\rho$ is unrestricted. So, this setting is characterized by four parameters, that is $(\beta, \sigma^2\_1, \rho, \sigma^2)$.

The state distribution is common knowledge, but the state realizations are observed only by the sender. Before the first period, the sender commits to a dynamic information policy mapping each private history to a signal distribution. The game then proceeds as follows. In the first period, the sender observes the state $\theta\_1$ and sends the receiver a signal $s\_1$ drawn from the distribution prescribed by the information policy at the history $h\_1 = \theta\_1$. The receiver observes $s\_1$, updates his belief about $\theta\_1$, and then chooses an action $a\_1$, which is observed by the sender. In the second period, the sender observes the state $\theta\_2$ and sends a signal $s\_2$ drawn from the distribution prescribed by the information policy at the history $h\_2 = (\theta\_1, s\_1, a\_1, \theta\_2)$. The receiver observes $s\_2$, updates his belief about $\theta\_2$, and chooses an action $a\_2$. Payoffs for both periods are then realized.

To find the sender's optimal policy we need to solve backwards. In the second period, the receiver just maximizes his flow payoff. So, after receiving signal $s\_2$, he matches his action with his updated expectation of $\theta\_2$. Let $v\_2$ denote the posterior variance of $θ\_2$ as assessed by the receiver after observing signal $s\_2$. By standard properties of the quadratic, the expected flow payoffs in the second period are $−v\_2$ for the receiver and $−β^2 −v\_2$ for the sender.

In the first period, the receiver’s action can affect both his flow payoff and the informativeness of the second-period signal. Upon receiving signal $s\_1$, the receiver may shift his action away from his bliss point $\mathbb{E}[\theta\_1|s\_1]$ if doing so is rewarded with a more precise signal in the second period. Let $v\_1$ denote the conditional variance of $\theta\_1$ given $s\_1$. Without any further information from the sender, the receiver can choose his myopic best responses $a^{′}\_1 = \mathbb{E}[\theta\_1|s\_1]$ and $a^{′}\_2 = \rho\mathbb{E}[\theta\_1|s\_1]$ to obtain the expected payoff

$$\tag{1}E[u\_R(a^{′}\_1,a^{′}\_2,\theta)|s\_1] = −v\_1 − (\rho^2v\_1 + \sigma^2).$$

Therefore, a necessary condition for the receiver to choose action $a\_1=\mathbb{E}[\theta\_1|s\_1]+b\_1$ is that the resulting flow loss is no greater than the total loss from choosing $(a^{′}\_1,a^{′}\_2):$

$$\tag{2}v\_1+b\_1^2\leq v\_1 + (\rho^2 v\_1+\sigma^2)$$

$\textbf{Question:}$ How did $(1)$ and $(2)$ where calculated? Could anyone provide some help with the calculations that give $(1)$ and how did we come up with $(2)$?

**Tags:** microeconomics, mathematical-economics, game-theory, bayesian-game

> **Comments:**
> **username_0**: @game-theory can anyone confirm my calculations below?  

---

## 1 answers

### Answer by username_0 on 2023-02-01. Score: 0

I will try something like an answer though I miss some technicalities

So, if the receiver decides to play her myopic best responses then the payoff given by $(1)$ is

$$\mathbb{E}\left[u\_R(a\_1^{'},a\_2^{'},\theta)|s\_1\right]=-\underbrace{\mathbb{E}\left(\left(\mathbb{E}(\theta\_1|s\_1)-\theta\_1\right)^2|s\_1\right)}\_{v\_1=\mathbb{V}(\theta\_1|s\_1)}-\underbrace{\mathbb{E}\left(\left(\mathbb{E}(\theta\_2|s\_1)-\theta\_2\right)^2|s\_1\right)}\_{v\_2=\mathbb{V}(\theta\_2|s\_1)}=\\=-v\_1-\mathbb{V}(\theta\_2|s\_1)$$

and because $\theta\_2=\rho\theta\_1+\epsilon$ it holds that

$\mathbb{V}(\theta\_2|s\_1)=\rho^2\underbrace{\mathbb{V}(\theta\_1|s\_1)}\_{v\_1}+\underbrace{\mathbb{V}(\epsilon|s\_1)}\_{\text{and because $\epsilon\sim N(0,\sigma^2)$ is independent of $\theta\_1$ and hence of $s\_1$ then $\mathbb{V}(\epsilon|s\_1)=\mathbb{V}(\epsilon)=\sigma^2$ }}=\rho^2v\_1+\sigma^2$

And as a consequence

$$\mathbb{E}\left[u\_R(a\_1^{'},a\_2^{'},\theta)|s\_1\right]=-v\_1-(\rho^2v\_2+\sigma^2)$$

Does this seem to be ok? Do I need to clarify why $\mathbb{V}(\theta\_1|s\_1)=\mathbb{E}\left(\left(\mathbb{E}(\theta\_1|s\_1)-\theta\_1\right)^2|s\_1\right)$? I would just say by definition but I do not know if I am missing anything.

I wonder why the expectation $\mathbb{E}[\theta\_2|s\_2]$ does not emerge it the payoff and the author only conditions with respect to $s\_1$.


---

