Let $X,Y$ be both normally distributed and independent of each other such that $X\sim N(\mu_x,\sigma_x^2), Y\sim N(\mu_y,\sigma_y^2)$. Consider $V=X+Y$, then compute $\mathbb{E}(X\mid V)$.
I'm not sure how to tackle this really, any help appreciated for differing means $\mu_x,\mu_y$! Is this right?
\begin{align}
\mathbb{E}(X\mid X+Y) &= \mathbb{E}(\sigma_x Z + \mu_x \mid \sigma_y Z + \mu_y )
\\&= \sigma_x\mathbb{E}(Z \mid Z \le t) + \mu_x \mathbb{E}(1 \mid Z \le t)+\sigma_y\mathbb{E}(Z \mid Z \le t) + \mu_y \mathbb{E}(1 \mid Z \le t) \\
\end{align}
Best Answer
The final result stated is not true as seen by taking $Y=2X$, for example. However it is true if $X$ and $Y$ have the same mean and variance (so that they are i.i.d.). In this case $E(X|X+Y)=E(Y|X+Y)$ since $X$ and $Y$ are i.i.d.. This gives $E(X|X+Y)=\frac 1 2 (E(X|X+Y)+E(Y|X+Y))=\frac 1 2 E(V|V)=\frac 1 2 V$.