Correlation between normal random variables Suppose $X$, $Y$, and $Z$ are normal random variables with means and variances of $\mu_X, \mu_Y, \mu_Z$ and $\sigma^2_X, \sigma^2_Y, \sigma^2_Z$ respectively. $X$ and $Y$ have correlation $\rho$. $X$ and $Z$ are independent, and $Y$ and $Z$ are independent. Then what is the correlation of $X$ and $Y-Z$? I know that $Y-Z$ has distribution $\mathcal{N}(\mu_Y - \mu_Z, \sigma^2_Y + \sigma^2_Z)$.
$corr(X, Y-Z) = \frac{cov(X,Y-Z)}{\sigma_X\sigma_{Y-Z}}$. $cov(X,Y-Z) = cov(X,Y) - cov(X,Z)$.
$corr(X,Y) = \rho = \frac{cov(X,Y)}{\sigma_X\sigma_Y}$. So $cov(X,Y) = \rho \sigma_X \sigma_Y$. $cov(X,Z) = 0$ since they are independent.
So $cov(X,Y-Z) = cov(X,Y)$.
So $corr(X, Y-Z) = \frac{cov(X,Y)}{\sigma_X\sqrt{\sigma^2_Y + \sigma^2_Z}} = \frac{\rho \sigma_X \sigma_Y}{\sigma_X\sqrt{\sigma^2_Y + \sigma^2_Z}} = \frac{\rho \sigma_Y}{\sqrt{\sigma^2_Y + \sigma^2_Z}}$.
Is this right?

You have a joint (normal) distribution of $(X,Y,Z)'$, can you think of a transformation to get the joint distribution of $(X,Y-Z)$?
Hint: If you take A = $[1,-1,0]$ you get the distribution of $(X-Y)'$
