# Variance for a product-normal distribution

I have two normally distributed random variables (zero mean), and I am interested in the distribution of their product; a normal product distribution.

It's a strange distribution involving a delta function.

What is the variance of this distribution - and is it finite?

I know that

$Var(XY)=Var(X)Var(Y)+Var(X)E(Y)^2+Var(Y)E(X)^2$

However I'm running a few simulations and noticing that the sample average of variables following this distribution is not converging to normality - making me guess that its variance is not actually finite.

-
The characteristic function of this distribution is $f(z)=\dfrac{1}{\sqrt{1+t^2}}$, you can find the mean and variance from this. (If they exist) –  Inquest Feb 11 '13 at 0:16
How do I do that? Sorry - I am not that familiar with characteristic functions. –  s4027340 Feb 11 '13 at 0:43

Hint: We need to know something about the joint distribution. The simplest assumption is that $X$ and $Y$ are independent. Let $W=XY$. We want $E(W^2)-(E(W))^2$. To calculate $E((XY)^2)$, use independence.

-
I already know that Var(XY)=Var(X)Var(Y)+Var(X)E(Y)^2+Var(Y)E(X)^2. Because the means are 0 this means that Var(XY)=Var(X)Var(Y). But is this the variance of a product-normal like in the mathworld link? –  s4027340 Feb 11 '13 at 1:16
Neither you nor MathWorld specify independence. But MathWorld in giving the joint density assumes independence. Without independence, or some other specific assumption about the joint distribution, one cannot find the variance of the product. That's why I put independence into the answer. Perhaps I will make that clearer. –  André Nicolas Feb 11 '13 at 1:28
Okay, so assuming independence, the variance of that quite weird product-normal distribution is actually simply the product of the two variances of the original two normal distributions. –  s4027340 Feb 11 '13 at 4:59
Yes. And without replacing the assumption of independence by some other explicit assumption, we cannot compute the variance of $XY$. –  André Nicolas Feb 11 '13 at 5:02
I see, thank you –  s4027340 Feb 11 '13 at 6:49