Optimal estimator of Poisson distribution As part of my homework, I was asked to solve the following:

a signal and noise are stochastically independent and it is known that $n\sim Poiss(\mu_n)$ and $x\sim Poiss(\mu_x)$ we define $y=x+n$

*

*Find the optimal estimator(MMSE) of $y$ given $x$.

*Find the linear optimal estimator(MMSE) of $x$ given $y$.


I'm not familiar with optimal estimators, and I couldn't find a good source to study it.
Any help will be appreciated.
 A: (1) The MMSE is given by $\widehat{y}=E\left[y\,\middle|~x\right]$ and the equation $y=x+n$ yields $$E\left[y \,\middle|~ x \right] = E\left[x \,\middle|~ x \right]+E\left[n  \,\middle|~ x \right] = x + E[n] = x+\mu_n~,$$
where the second identity holds in view of the independence of $x$ and $n$. 
In other words, $\widehat{y}=x+\mu_n$.
(2) The MMSE is given by $\widehat{x}=E\left[ x  \,\middle|~ y \right]$. The relation $y=x+n$ yields $E\left[x  \,\middle|~ y \right] = y-E\left[n  \,\middle|~  y  \right]$. As shown below, we have $E\left[n  \,\middle|~ y  \right] = y\left(\frac{\mu_n}{\mu_n+\mu_x}\right)$. 
Therefore, $\widehat{x}=y\left(\frac{\mu_x}{\mu_n+\mu_x}\right)$.

We now show that $E\left[n  \,\middle|~ y \right]=y\left(\frac{\mu_n}{\mu_n+\mu_x}\right)$. We have
$$E\left[n  \,\middle|~ y=y' \right]=\sum_{k\leq y'}k\mathbb{P}\left(n=k\left|y=y'\right.\right)~,$$
and therefore
\begin{align}
\mathbb{P}\left(n=k  \,\middle|~ 
 y=y' \right) & =\frac{\mathbb{P}\left(n=k,\,y=y'\right)}{\mathbb{P}\left(y=y'\right)} \\
&=\frac{\mathbb{P}\left(n=k,\,x=y'-k\right)}{\mathbb{P}\left(y=y'\right)} \\
&=\frac{\mathbb{P}\left(n=k\right)\mathbb{P}\left(x=y'-k\right)}{\mathbb{P}\left(y=y'\right)} \\
&= \frac{\left(e^{-\mu_n}\mu_n^k/k!\right)\left(e^{-\mu_x}\mu_x^{y'-k}/(y'-k)!\right)}{\left(e^{-\left(\mu_n+\mu_x\right)}\left(\mu_n+\mu_x\right)^{y'}/y'!\right)}=\left(\begin{array}{c} y' \\ k\end{array}\right) \left(\frac{\mu_n}{\mu_n+\mu_x}\right)^k\left(\frac{\mu_x}{\mu_n+\mu_x}\right)^{y'-k}~,
\end{align}
where the second to last equality above holds since $y\sim {\sf Poisson}(\mu_n+\mu_x)$, which follows from the independence of $x$ and $n$. In other words, for $y'$ fixed,  $\left(\mathbb{P}\left(n=k \,\middle|~y=y' \right)\right)_{k\leq y'}$ is ${\sf Binomial}(y',p)$ with $p= \mu_n \,/(\mu_n+\mu_x)$ and thus
$$E\left[n \,\middle| y=y' \right] = py' = y'\left(\frac{\mu_n}{\mu_n+\mu_x}\right)$$
That is, $E\left[n \,\middle| y \right]=y\left(\frac{\mu_n}{\mu_n+\mu_x}\right)$.
