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Assume that X and Y are random variables which are normally distributed as N(0,s_X^2) and N(0,s_Y^2), respectively. Furthermore, assume that X and Y are independent. Could anyone please tell me if the variable given by the product XY and the variable X^2 are independent? Many thanks!

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They're not independent. Intuitively speaking, if $X^2=0$ , $XY$ must be $0$, so the two have a dependence.

More formally, There are plenty of other ways to do it, but I'll focus on a simple contradiction. Let $Z=XY,W=X^2$, then we are asking if $Z$ and $X$ are independent. If they are, we should have $\operatorname{var}(Z|W=w)=\operatorname{var}(Z)=\sigma_x^2\sigma_y^2$. However, let's say $w=0$, then $\operatorname{var}(Z|W=0)=0$ because $W=0\rightarrow X^2=0\rightarrow X=0\rightarrow XY=Z=0$, which makes $Z$ deterministic, and therefore have $0$ variance. So, we can conclude that, in general, $\operatorname{var}(Z|W=w)\neq\operatorname{var}(Z)$, and the two are not independent.

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  • $\begingroup$ You mean "$Z$ and $W$ are independent"? $\endgroup$
    – corey979
    Sep 27 '19 at 10:05
  • $\begingroup$ Yes, $Z$ and $W$. $\endgroup$
    – gunes
    Sep 27 '19 at 10:06
  • $\begingroup$ Thanks! But is it possible to choose the variables X and Y in such a way that XY and X^2 are independent? For instance, with reference to the proposed counterexample, if I assume that X is never 0? can I have independency? I am indeed wondering if XY and X^2 are never independent or are there any additional assumptions that can be made on X and Y so that the statement holds? If yes, which ones? In addition, is it possible to find real numbers a and b so that aX^2 and bXY are independent? Many thanks !!! $\endgroup$
    – Tiger07
    Sep 27 '19 at 13:56
  • $\begingroup$ @Tiger07, I suspect that if one variable was -1 (probability 50%) or +1 (probability 50%) that it's product with a 2nd RV would be independent of that RV. But it would probably not be independent of the square $\endgroup$
    – MikeP
    Sep 27 '19 at 14:24
  • $\begingroup$ If X,Y are I.I.D, are X^2, Y^2 i.i.d ? $\endgroup$
    – MSIS
    Oct 10 at 21:53

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