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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?

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    $\begingroup$ What is the covariance of $X$ and $Y$? What is the covariance of $X$ and $Y-Z$? What is the correlation of $X$ and $Y-Z$? $\endgroup$
    – Henry
    Dec 12, 2021 at 4:34

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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 $(X-Y)'$

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  • $\begingroup$ The question doesn't say that (X,Y,Z) are jointly normal. Unless I missed something, we don't have that (X,Y) are bivariate normal, only that they're marginally normal with a specified correlation. $\endgroup$
    – Glen_b
    Dec 12, 2021 at 22:56
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    $\begingroup$ As @Glen points out, this approach requires an additional assumption. However, the idea is correct, because the question supplies full information about the variance-covariance matrix of $(X,Y,Z).$ Whether that has a multivariate Normal distribution is irrelevant. $\endgroup$
    – whuber
    Nov 19 at 17:10

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