Computing expectation of a random variable conditioned on the subtraction of two Let $(X,Y) $~$ N(\mu,\sum)$ where $\mu = (1,0)$ and $\sum = \begin{pmatrix}2&1\\1&2 \end{pmatrix}$. I want to compute the $E[X|X-Y]$
What I have done is that: I have $\mu_X = 1, \mu_Y=0$ and $\sigma_X^2 = 2$ and $\sigma_Y^2 = 2$ and so the correlation constant $\rho = 1/2$. I let Z=X-Y be ~$N(\mu_X-\mu_Y = 1, \sigma_X^2-\sigma^2_Y =0)$. Now, I just compute the E[X|Z]. Is this approach correct?
 A: If $(X,Y)$ is bivariate normal and $Z=X-Y$, then $(X,Z)$ is also bivariate normal.
You can check that $\frac{X-\mu_1}{\sigma_1}$ has the same distribution as
$$\frac{X-\mu}{\sigma_1} \overset{d}{=}
\text{Corr}(X,Z) \cdot \frac{Z - E[Z]}{\sqrt{\text{Var}(Z)}}
+ \sqrt{1 - \text{Corr}(X,Z)} \cdot U$$
where $U$ is standard normal independent of $Z$.
[This is the decomposition of bivariate normal distributions used to derive the conditional distribution.]
Then $E[(X-\mu)/\sigma_1 \mid Z] = \text{Corr}(X,Z) \cdot \frac{Z - E[Z]}{\sqrt{\text{Var}(Z)}}$. Some more manipulation yields the expression for $E[X \mid Z]$ that matches what is stated in the Wikipedia page linked above.

To actually compute the parameters of the bivariate distribution $(X,Z)$ above, note that

*

*$E[X]=\mu_1$,

*$E[Z] = E[X-Y] = \mu_1 - \mu_2$

*$\text{Var}(X) = \sigma_1^2$

*$\text{Var}(Z) = \text{Var}(X-Y) = \text{Var}(X) + \text{Var}(Y) - 2 \text{Cov}(X,Y) =  \sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2$.

*$\text{Cov}(X, Z) = \text{Cov}(X, X-Y) = \sigma_1^2 - \rho \sigma_1 \sigma_2$

*$\text{Corr}(X,Z) := \frac{\text{Cov}(X,Z)}{\sqrt{\text{Var}(X) \text{Var}(Z)}}$
