Mean Square Error Optimal Estimator of X given Y $X$, $A$, and $Z$ are scalar independent random variables.
$Y = AX + Z$.
$$ A = \begin{cases}1 & \text{with probability } p \\ 0 & \text{with probability } 1-p\end{cases} $$
$X$ has mean $\mu$ and variance $\sigma_x^2$.  $Z$ has mean $0$ and variance $\sigma_z^2$.
Find the optimal estimator of $X$ given that $Y$ is observed.
I know the optimal estimator to be $E\left[X\mid Y=y\right]$ but I don't know how to determine it.
I'm especially confused because if $A=0$ then $Y$ does not provide any information about $X$.
As always, thank you for your help.
 A: $\newcommand{\E}{\mathbb E}$
$\newcommand{\var}{\operatorname{var}}$
$\newcommand{\cov}{\operatorname{cov}}$
\begin{align}
\E(X\mid Y=y) & = \E( \E(X\mid Y=y, A)) \\[8pt]
& = \E( X\mid Y=y, A=0)\Pr(A=0)+\E(X\mid Y=y,A=1)\Pr(A=1) \\[8pt]
& = \E(X)(1-p)+ \E(X\mid X+Z=y)p.\tag{1}
\end{align}
So we want the conditional expectation of $X$ given $X+Z$.  We're not given their joint distribution, but we have at least some information about it.  The correlation between $X$ and $X+Z$ is
$$
\rho = \frac{\cov(X,X+Z)}{\sigma_X\sqrt{\sigma_X^2+\sigma_Z^2}} = \frac{\sigma_X^2}{\sigma_X\sqrt{\sigma_X^2+\sigma_Z^2}} = \frac{\sigma_X}{\sqrt{\sigma_X^2+\sigma_Z^2}}.
$$
We can do a linear regression to find a predicted value $\widehat X$ of $X$ based on $X+Z$:
$$
\widehat{X} = \mu_X +\rho\sigma_X\left( \frac{(X+Z)-(\mu_{X+Z})}{\sigma_{X+Z}} \right).
$$
This would be the conditional expected value of $X$ given $X+Z$ if we knew that $X$ and $Z$ were normally distributed (given that they're independent; without that, we'd wonder whether joint normality is given).  I think without further information about the distributions, we probably can't find this conditional expected value.
But what about "optimality" of estimation?  I think one could show this has the smallest mean squared error of any estimator that is an affine function of the observed value of $X+Z$.
Then after this, we'd want the weighted average of this estimator and $\E(X)$, as mentioned in $(1)$ above.
