Let $(X,Y)$ be a Gaussian random vector with given means, variances and correlation, determine $E[X\mid X+Y]$ and $E[X\mid Y]$

Question

Let $(X,Y)$ be a Gaussian random vector, with $E[X]=E[Y]=0$, $var(X)=\sigma^2,\ var(Y)=\tau^2$, and $Cor(X,Y)=\rho$ (correlation coefficient). Find $E[X\mid X+Y]$ and $E[X\mid Y]$.

My attempt for $E[X\mid X+Y]$ is to first write $$E[X\mid X+Y]+E[Y\mid X+Y]=E[X+Y\mid X+Y]=X+Y$$ Since $(X,Y)$ is a Gaussian vector, $X+Y$ is also Gaussian. We can compute that $E[X+Y]=0$ and $var(X+Y)=\tau^2+\sigma^2+2\sigma\tau\rho$. Therefore, $X+Y\sim N(0,\tau^2+\sigma^2+2\sigma\tau\rho)$. But after this I don't know how to continue.

Another problem is that this does not seem to work for $E[X\mid Y]$.

Any help is appreciated.

Answer 1

Hints: Let $Z=cX+Y$ where $c$ is chosen such that $cov (X+Y,Z)=0$. (Compute $c$ by expanding the covariance). Then express $X$ as $t(X+Y)+sZ$. This means $X=(t+s)X+(t+sc)Y$ so we can take $s=\frac 1 {c-1}$ and $t= \frac 1 {1-c}$. Now note that $Z$ and $X+Y$ are independent (by joint normality) and $E(X|X+Y)=E(t(X+Y)+sZ|X+Y)=t(X+Y)+EZ=t(X+Y)$. [I will let you find out what you can do when $c=1$].

A similar argument can be used for $E(X|Y)$: Choose $d$ such that $X+dY$ is independent of $Y$. (You only have to make their covariance $0$). Then $E(X|Y)=E((X+dY)-dY|Y)==E(X+dY)-dY=-dY$.

Answer 2

$X|Y$ is also gaussian random variable.This is because Gaussian Conditional Property.This holds In general for any finite dimensional gaussian random vectors. Look up for derivations because they are a bit tedious and i am just stating $2D$ result here.You can find the general result here. MVN

$$\begin{pmatrix} X_1 \\ X_2 \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \mu_1 \\ \mu_2 \end{pmatrix} , \begin{pmatrix} \sigma^2_1 & \rho \sigma_1 \sigma_2 \\ \rho \sigma_1 \sigma_2 & \sigma^2_2 \end{pmatrix} \right) $$

The conditional expectation of X1 given X2 is:

$$\operatorname{E}(X_1 \mid X_2=x_2) = \mu_1 + \rho \frac{\sigma_1}{\sigma_2}(x_2 - \mu_2) =$$

In your case it reduces to following

$$\operatorname{E}[X|Y=y] = \rho \frac{\sigma}{\tau}.y$$