# Conditional expectation on Gaussian random variables

If we suppose that the two independent random variables $X \sim \mathcal{N}(0,\sigma^2_x)$ and $N \sim \mathcal{N}(0,\sigma^2_n)$ and that $S = X + N$, how would I work out the conditional expectation $E[X\mid S=s]$?

#### Solutions Collecting From Web of "Conditional expectation on Gaussian random variables"

Since $X$ and $N$ are independent and normal, any combinations of $X$ and $N$ are normal. As any combination of $X+N$ and $\frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N$ is also a combination of $X$ and $N$, that is, any combinations of $X+N$ and $\frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N$ are normal, then $X+N$ and $\frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N$ are jointly normal. Moreover, note that
\begin{align*}
E\bigg((X+N)\Big( \frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N\Big) \bigg) &= 0.
\end{align*}
Then $X+N$ and $\frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N$ are independent. Consequently,
\begin{align*}
E\big( (X+N) \mid S\big) &= S,\\
E\bigg( \Big( \frac{1}{\sigma_x^2}X – \frac{1}{\sigma_n^2}N\Big) \mid S\bigg) &= 0.
\end{align*}
Solving this system, we obtain that
\begin{align}
E(X\mid S) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}S.
\end{align}
In other words,
\begin{align}
E(X\mid S=s) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}s.
\end{align}

Here’s another method:

https://stats.stackexchange.com/questions/30588/deriving-the-conditional-distributions-of-a-multivariate-normal-distribution

Not as “clever” as Gordon’s method above however.