If X and Y are independent random variables both with the same mean (0) and variance, how about $X^2$ and $Y^2$? I tried calculating E($X^2Y^2$)-E($X^2$)E($Y^2$) but haven't been able to get anywhere.
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$\begingroup$ They must be independent. How could you calculate $E(X^2Y^2)$? $\endgroup$– BerciCommented Oct 24, 2012 at 21:37
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8$\begingroup$ If $X$ and $Y$ are independent, then so are $g(X)$ and $h(Y)$ for (measurable) functions $g(\cdot)$ and $h(\cdot)$. Means and variances don't come into the picture and your attempted calculation of $\text{cov}(X^2,Y^2)$ will not prove independence even though the covariance will turn out to be $0$. $\endgroup$– Dilip SarwateCommented Oct 24, 2012 at 21:41
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$\begingroup$ Thank you! That's very helpful to know. $\endgroup$– JarrisCommented Oct 24, 2012 at 21:56
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$\begingroup$ @Dilip: You could post that as an answer so the question doesn't remain unanswered. $\endgroup$– jorikiCommented Oct 24, 2012 at 22:42
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$\begingroup$ @DilipSarwate what's measurable in elem probability? math.stackexchange.com/questions/3944284/… $\endgroup$– BCLCCommented Dec 11, 2020 at 12:34
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As per joriki's suggestion, my comment (with additional information) is posted as an answer.
If $X$ and $Y$ are independent, then so are $g(X)$ and $h(Y)$ independent random variables for (measurable) functions $g(⋅)$ and $h(⋅)$. In particular, $X^2$ and $Y^2$ are independent random variables if $X$ and $Y$ are independent random variables. Means and variances don't come into the picture at all, and your attempted calculation of $\text{cov}(X^2,Y^2)$ will not prove independence even though the covariance will turn out to be $0$.
answered Oct 25, 2012 at 1:36
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1$\begingroup$ "In particular, $X^2$ and $Y^2$ are independent random variables if $X^2$ and $Y^2$ are independent random variables." I am 99% sure that this should read "In particular, $X^2$ and $Y^2$ are independent random variables if $X$ and $Y$ are independent random variables"? $\endgroup$ Commented Nov 6, 2013 at 0:20
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$\begingroup$ @Silverfish You are absolutely correct. Thanks for proof-reading carefully. I have corrected the typos. $\endgroup$ Commented Nov 6, 2013 at 2:13
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$\begingroup$ What's 'measurable' in elementary probability? see here: math.stackexchange.com/questions/3944284/… $\endgroup$– BCLCCommented Dec 11, 2020 at 12:28