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This is from the this paper in section $3$ about the two period example.

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 $(2)$?

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  • $\begingroup$ @game-theory can anyone confirm my calculations below? $\endgroup$ Feb 3, 2023 at 11:01

1 Answer 1

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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 $s_1$.

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