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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 $E[X\mid S=s]$?

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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:

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

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