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I have three independent random variables X, Y and Z, uncorrelated between each other. Y and Z have zero mean and unit variance, X has zero mean and given variance. Do you know how to compute the correlation between the products XY and XZ? And whether it can be zero under any specific condition?

Roberto
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    Correlation is proportional to the covariance $$\operatorname{Cov}(XY,XZ)=E(XYXZ) - E(XY)E(XZ) = E(X^2)E(Y)E(Z) - E(X)E(Y)E(X)E(Z).$$ (Your independence assumptions justify the second equality.) Can you go on from there? – whuber Aug 28 '17 at 16:12
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    The question is more interesting without the zero mean assumptions – wolfies Aug 28 '17 at 16:31
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    Thank you whuber, I thus understand that the correlation is null. It was not a question from a course or textbook, I'm just developing a mixed-effect model for the data I'm working on and I am not good in statistics as you can see. I didn't know about [ self-study ] anyway, thank you for that. – Roberto Aug 29 '17 at 09:40

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Partially answered in comments:

Correlation is proportional to the covariance $$\operatorname{Cov}(XY,XZ)=E(XYXZ) - E(XY)E(XZ) = E(X^2)E(Y)E(Z) - E(X)E(Y)E(X)E(Z)$$ (Your independence assumptions justify the second equality.) Can you go on from there? – whuber

Edit Answer to question in comment:

does $X,Y,Z$ being independent implies that $X^2, Y, Z$ are too? If so, how?

Yes, functions of independent variables are independent, see Functions of Independent Random Variables, Mutual independence of functions of random variables

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    Sorry I know this is a really stupid question, but how do you go from $E(XY XZ)$ to $E(X^2)E(Y)E(Z)$? To do so, it seems that $E(XY XZ)=E(XX YZ)=E(X^2 YZ)=E(X^2)E(Y)E(Z)$. I know that $X,Y,Z$ are independent, but it seems we need to know that $X^2, Y, Z$ are independent in order to get $E(X^2 YZ)=E(X^2)E(Y)E(Z)$; does $X,Y,Z$ being independent implies that $X^2, Y, Z$ are too? If so, how? – Constantly confused Sep 15 '23 at 23:01
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    @Constantlyconfused: Have a look at the updated answer! – kjetil b halvorsen Sep 20 '23 at 23:08
  • Thank you very much! Really appreciate it – Constantly confused Sep 25 '23 at 21:38