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Prove that if X and Y are Normal and independent random variables, X+Y and X−Y are independent. Note that X and Y also have the same mean and standard deviation.

Note that this is a duplicate of Prove that if $X$ and $Y$ are Normal and independent random variables, $X+Y$ and $X-Y$ are independent, however, there isn't a complete solution to the answer given and I do not understand exactly what the hints are suggesting.

My attempt was to check if $f_{x+y,x-y}(u,v) = f_{x+y}(u)f_{x-y}(v)$, however, this does not seem to be working out too nicely.

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You left out an assumption. $X$ and $Y$ need to have the same variance for this to be true. – Robert Israel Oct 24 '12 at 1:04
Yes you are correct, in my specific case X and Y actually have the same mean and standard deviation. – user38784 Oct 24 '12 at 1:12
WLOG you can assume the mean is $0$. Then write out the joint pdf of $X$ and $Y$ and note that it has radial symmetry. $(X,Y) \to (X+Y,X-Y)/\sqrt{2}$ is a rotation. – Robert Israel Oct 24 '12 at 1:39
up vote 1 down vote accepted

Define $U = X + Y, V = X - Y$. Then, $X = (U + V)/2, Y = (U - V)/2$. Find the Jacobian $J$ for the transformation.

Then, $f_{U,V}(u,v)=f_{X}(x=(u+v)/2)f_{Y}(y=(u-v)/2)|J|$.

You will find that $f_{U,V}(u,v)$ factors into a function of $u$ alone and a function of $v$ alone. Thus, by the Factorization thm, $U$ and $V$ are independent.

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I came back to update, I was able to work this out with the hint you gave. Thanks! – user38784 Oct 29 '12 at 2:33
You're most welcome – Ken Dunn Oct 30 '12 at 5:38

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