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I have a a bivariate normally distributed random vector $X = (X,Y)$ and with Expected Value $(X)= (\mu(x),\mu(y))$, and Covariance Matrix $2\times 2$. (not independent)

Now I want to show which distribution $(X-Y, X+Y)$ has.

I first took the bivariate normal density function and simply replaced $X$ by $X-Y$ and $Y$ by $X+Y$ but I don't quite know how to continue .

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    $\begingroup$ The notation $X=(X,Y)$ seems troublesome. I recommend editing your question and using a different symbol (there are two $X$s). $\endgroup$ – parsiad Jun 6 '16 at 20:13
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Jun 6 '16 at 20:14
  • $\begingroup$ That is actually what I wanted to do, it is the same X and Y. However, I do agree I should change variables while calculating $\endgroup$ – Heike Lehner Jun 6 '16 at 20:17
  • $\begingroup$ The expected value(s) should be easy enough $\endgroup$ – Henry Jun 6 '16 at 20:20
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The mean vector is easy: $E(X\pm Y) = E(X) \pm E(Y)$ so $$ \left< (X-Y,X+Y) \right> = (\mu(x)-\mu(y),\mu(x)+\mu(y)) = (<X>-<Y>, <X>+<Y>) $$

As to the covariance matrix:

$$<(X-Y)^2> - (<X-Y>)^2 = \\<X^2> + <Y^2> - 2<XY> - (<X>)^2 - (<Y>)^2 + 2<X><Y> =\\ [ <X^2> - (<X>)^2] + [ <Y^2> - (<Y>)^2] -2[<XY>-<X><Y>] $$ and each expression in square brackets is recognized as an entry in the covariance matrix for $X,Y$. Similarly for each of the other three elements in the covariance matrix of $X-Y, X+Y$.
$$<(X+Y)^2> - (<X+Y>)^2 =\\ <X^2> + <Y^2> + 2<XY> - (<X>)^2 - (<Y>)^2 - 2<X><Y> =\\ \sigma_x^2 + \sigma_y^2 -2\sigma_{xy} $$ and $$<(X-Y) (X+Y)> - <X-Y><X+Y> =\\ <X^2> - <Y^2> - (<X> - <Y>) (<X> + <Y>) \\= [<X^2>-(<X>)^2 ] - [<Y^2>-(<Y>)^2 ] = \sigma_x^2 - \sigma_y^2 $$

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