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Im given V and W are independent standard normal random variables where $x=\frac{(V+W)}{\sqrt(2)}$ and $y=\frac{(V-W)}{\sqrt(2)}$.

This is what I did: $M_{x,y}(s,t)=E(e^{sx+ty})=E(e^{s(\frac{V+W}{\sqrt(2)})+t(\frac{V-W}{\sqrt(2)})})=E(e^{v(\frac{s+t}{\sqrt(2)})+w(\frac{s-t}{\sqrt(2)})})=M_{V,W}(\frac{s+t}{\sqrt(2)},\frac{s-t}{\sqrt(2)})$

$M_{V,W}(\frac{s+t}{\sqrt(2)},\frac{s-t}{\sqrt(2)})=M_{V}(\frac{s+t}{\sqrt(2)})M_{W}(\frac{s-t}{\sqrt(2)})$

Since V and W are standard normal then their MGF's are in the form $e^{\frac{1}{2} t^2}$ and

$M_{V}(\frac{s+t}{\sqrt(2)})M_{W}(\frac{s-t}{\sqrt(2)})=e^{(\frac{s+t}{\sqrt(2)})^2}e^{(\frac{s-t}{\sqrt(2)})^2}$

Which when expanded and simplified I got

$e^{\frac{1}{2}(s^2+t^2)}$

So than the joint distribution of X and Y is (X,Y)~BVN(0,0,1,1,0). Is the method right because I feel like I should be getting something else?

USC
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1 Answers1

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Yes, this is correct. (There are some $\frac12$s missing in the middle, but the final result is right.)

Geometrically, the transformation from $(V,W)$ to $(X,Y)$ is a reflection about a line passing through the origin. However, the joint density of $(V,W)$ depends only on the distance from the origin, $\sqrt{V^2+W^2}$, so its distribution isn't changed by rotations around the origin or reflections about lines through the origin.

David Moews
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