Finding $\mathbb{E}(X\mid X+Y)$ for independent normal variables $X,Y$

Question

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}

Answer 1

let $d=\frac{\sigma^2_x}{\sigma^2_y}$ so $$cou(X-dY,X+Y)=\sigma^2_x-d\sigma^2_y=\sigma^2_x-\frac{\sigma^2_x}{\sigma^2_y}\sigma^2_y=0$$

since $$(X-dY,X+Y)$$ is bi-variate normal so they are independent.

so $$E(X-dY|X+Y)=E(X-dY)=\mu_x -d \mu_y $$

so $$ E(X|X+Y)-d E(Y|X+Y)=\mu_x -d \mu_y \hspace{.5cm} (1)$$

and obviously $$E(X+Y|X+Y)=X+Y $$ so $$E(X|X+Y)+ E(Y|X+Y)=X+Y \hspace{.5cm} (2)$$

combine (1) and (2) to get answer.

I think $$E(X|X+Y)=\frac{d}{d+1} (X+Y)+\frac{\mu_x -d \mu_y}{d+1}$$

Answer 2

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 $E(X|X+Y)=\frac 1 2 (E(X|X+Y)+E(Y|X+Y))=\frac 1 2 E(V|V)=\frac 1 2 V$.