$X,Y$ are random independent variables with normal distribution. Can we conclude that $XY$ has also normal distribution ?

I know definition of normal distributio, however I have a problem with it. Can you explain it ?

  • You accepted the answer below 40 minutes after it was posted. Sure you checked it thoroughly before doing so? (Three (baffling) upvotes are no reason to proceed this way, if you ask me.) – Did Aug 30 '16 at 17:46
  • After all, answer is correct. –  Aug 30 '16 at 17:47

1 Answers1


Note that $XY =\frac{1}{4}\Big( (X+Y)^2 + (X-Y)^2 \Big)$. If $X$ and $Y$ have same mean and variance, say mean $0$ and variance $1$, then $(X+Y)^2$ and $(X-Y)^2$ are Chi-Square with one degrees of freedom. So $XY$ is a sum of Chi-Squares, which is not Gaussian.

P.S: I have deleted my previous answer because i assumed they were dependent and picked a counter example. Here is the case when $X$ and $Y$ are independent

Ahmad Bazzi
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  • So answer is Yes or No ? What does it mean *Chi-Square* ? –  Aug 30 '16 at 17:02
  • The answer is no. $XY$ is not Gaussian. – Ahmad Bazzi Aug 30 '16 at 17:03
  • First, there is a sign mistake in the RHS of the identity. Second, are you sure that the difference of two chi-squares is not gaussian, whatever their dependency? (You are probably referring to the result that a **sum** of **independent** chi-squares is a chi-square but the two reasons I mentioned ruin the approach.) This seems at least as difficult as proving the desired result itself. – Did Aug 30 '16 at 17:43