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I'm working on the following problem: Given $X,Y,Z$ three independent Gaussian RV with distribution $N(0,1)$, prove that $A = (X-Y)^2 + (X-Z)^2 + (Y-Z)^2$ is independent from $B = X+Y+Z$.

I am a bit stuck with how to approach this sort of problem. I am thinking of showing that $\operatorname{Cov}(A,B) = 0$ as both $A$ and $B$ are also normally distributed [Edit: turns out this may not be true so I'm on the wrong track!]. However, I can't move past a certain stage in the simplification, specifically computing $\operatorname{Cov}(X,XY)$ and $\operatorname{Cov}(X,Y^2)$ and terms like that. How should I approach this? Any help would be appreciated.

StubbornAtom
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space_gladiator
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1 Answers1

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Show that $X-Y$ is independent of $X+Y+Z$ (easy, since they are jointly Gaussian). Same for $Y-Z$ and for $X-Z$. This implies that the vector random variable $V= (X-Y, Y-Z, X-Z)$ is (a multivariate Gaussian) independent from (the scalar Gaussian) $X+Y+Z$ (this can be seen, for example, by observing that they are jointly 4D Gausssian and covariance matrix is block-diagonal, so the density splits as a product). Then any function of $V$ is also independent of $X+Y+Z$. QED.

Note that this is equivalent to showing that sample mean and sample variance are independent when sampling from a normal distribution, e.g. Proof of the independence of the sample mean and sample variance.

Max
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  • Hi again @Max , I was trying to use the theorem "If X and Y are independent RV then U = f(X) and V = g(Y) are also independent" but couldn't find a multivariate version so wasn't sure if my proof held. After proving the 3 components as independent from X+Y+Z, can I say f(A,B,C) = A^2 + B^2 + C^2 is independent from g(D) = D? where A = X-Y, B = X-Z, C = Y-Z, D = X+Y+Z? or does this not hold as per the theorem? – space_gladiator Sep 23 '20 at 20:20
  • It’s very generally true: https://stats.stackexchange.com/questions/94872/functions-of-independent-random-variables – Max Sep 23 '20 at 22:29
  • Actually my wording was a bit sloppy: you need independence from the joint. I have edited the answer to clarify. – Max Sep 24 '20 at 07:55
  • The warning here is that without independence of joint, there can be 3 r.v.s $X$, $Y$, $Z$ such that they are pairwise independent but $Z$ is not independent of $X+Y$. – Max Sep 24 '20 at 08:19