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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.

Community
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Joe St Amand
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2 Answers2

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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.

Ken Dunn
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For the general case that $X$ and $Y$ have a bivaraite normal distribution and have the same variances, it can be shown that $X+Y$ and $X-Y$ are independent (the means can be different and they can be dependent). It holds because of the following three facts:

1- $X+Y$ and $X-Y$ are uncorrelated, i.e., their covariance is zero.

2- $X+Y$ and $X-Y$ again jointly follow a bivariate normal distribution.

3- When $(W,U)$ follows a bivariate normal distribution, $W$ and $U$ are independent if and only if $W$ and $U$ are uncorrelated.

Amir
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