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If $ X \sim N(\mu_X, \sigma^2_X) $ and $ Y \sim N(\mu_Y, \sigma^2_Y) $ ($X$ and $Y$ are independent), let $$ Z = \begin{pmatrix} 2X + Y \\ X - 3Y \end{pmatrix} $$

I can easily calculate $ \mu_{Z_{1,1}}, \mu_{Z_{2,1}}, \sigma^2_{Z_{1,1}}, \sigma^2_{Z_{2,1}} $, but I am having trouble figuring out how to calculate $ \text{Cov}(Z_{1,1}, Z_{2,1}) $ i.e. $ \text{Cov}(2X + Y, X - 3Y) $. I understand that this is equal to $$ E[(2X + Y)\times(X - 3Y)] - E[2X + Y]E[X - 3Y] $$ but am having trouble getting there, partially because I am worried that the two Gaussian random variables in $ Z $ are functions of the original 2 Gaussian RVs $ X, Y $.

Could somebody help me figure out how to find the covariance of this random vector?

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$\operatorname{Cov}(X,Y)$ is linear in both $X$ and $Y$. So by linearity, $$ \operatorname{Cov}(2X+Y,X-3Y) = 2\operatorname{Cov}(X,X)-5\operatorname{Cov}(X,Y) - 3\operatorname{Cov}(Y,Y) \\= 2\operatorname{Var}(X)-5\operatorname{Cov}(X,Y) - 3\operatorname{Var}(Y). $$ (By the way, computing the variances of $2X+Y$ and $X-3Y$ also requires using linearity.)

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  • $\begingroup$ And from this point, if X and Y are independent, the middle term can be canceled and I have my answer, yes? $\endgroup$ – Ian H Jan 14 '17 at 20:43
  • $\begingroup$ @IanH, if they're independent, yes. $\endgroup$ – zhoraster Jan 14 '17 at 20:45
  • $\begingroup$ Thanks zhoraster! Also, would this be considered a bivariate normal distribution or is there another name for when it is represented as a vector? $\endgroup$ – Ian H Jan 14 '17 at 20:59
  • $\begingroup$ @IanH, multivariate normal or multidimensional normal (with "multi" = 2). $\endgroup$ – zhoraster Jan 15 '17 at 15:28

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