calculate the conditional expectation $E[D |s] =?$

Question

I know that

Given the joint normality $(D, \epsilon)$, it is defined that $s = D + \epsilon$

where $\epsilon \sim N(0,\sigma_{\epsilon}^2)$ and $D \sim N(\bar{D}, \sigma^2)$

$D$ and $\epsilon$ are independent.

How can I calculate the conditional expectation $E[D |s] =?$

Answer 1

We have $s \sim N(\bar{D}, \sigma^2 + \sigma^2_\epsilon)$. Further, $s$ and $D$ are jointly normal (since any linear combination of $s$ and $D$ can be written as a linear combination of $D$ and $\epsilon$ which are independent).

The covariance of $s$ and $D$ is $\text{Cov}(s, D) = \text{Cov}(D+\epsilon, D) = \text{Var}(D) = \sigma^2$, so the correlation is $\frac{\sigma^2}{\sigma\sqrt{\sigma^2 + \sigma_\epsilon^2}} = \frac{\sigma}{\sqrt{\sigma^2 + \sigma_\epsilon^2}} =: \rho$.

Thus, we can express $D$ in terms of $s$ via $$\frac{D-\bar{D}}{\sigma} \overset{d}{=} \rho \frac{s-\bar{D}}{\sqrt{\sigma^2 + \sigma_\epsilon^2}} + \sqrt{1-\rho} \cdot Z$$ where $Z \sim N(0,1)$ is independent of everything else.

From here, we have $E[D \mid s] = \bar{D} + \frac{\sigma \rho}{\sqrt{\sigma^2 + \sigma_\epsilon^2}} (s-\bar{D}) = \bar{D} + \frac{\sigma^2}{\sigma^2 + \sigma_\epsilon^2}(s-\bar{D})$.

[See here for a reference for the conditional expectation in a bivariate normal distribution.]