# “Square root” of a normal RV?

Say $X_1,X_2$ are independently drawn from the same distribution (call it $X$) and that their product, $X_1X_2$ falls on a standard normal distribution.

Is it possible to get a pdf or cdf for $X$?

My progress: The $n$th moment of a standard normal is $0$ for odd $n$ and $n!!$ for even $n$. Then for even $n$:

$\mathbb{E}[(X_1 X_2)^n] = \mathbb{E}[X_1^n] \mathbb{E}[X_2^n] = \mathbb{E}[X^n]^2 = n!!$

Thus the $n$th moment of $X$ is $\mathbb{E}[X^n] = \sqrt{n!!}$ for even $n$ and zero otherwise. Therefore...

• Here is one possible approach, though I have no idea whether it is useful: (a) take $Y=\log_e |N|$ where $N$ has a standard normal distribution, (b) find the characteristic function of $Y$, (c) take its square root, and (d) determine whether this is the characteristic function of a probability distribution. If it is, say of $Y_1$, then take $X_1 = Z_1 \exp(Y_1)$ where $Z_1=+1 \text{ or }-1$ with equal probability, independently of $Y_1$; $X_2$ would have the same distribution but be independent of $X_1$. – Henry May 7 '13 at 22:28

From what you did, you can at least show that the moments indeed define the distribution uniquely. Note that the $n$-th moment of a standard normal distribution is actually $(n-1)!! = (n-1)(n-3)\ldots1$ for odd $n$, I think. To reinterate, you found that $$M_n = \begin{cases} \sqrt{(n-1)!!} &\text{if n odd} \\ 0 &\text{otherwise.} \end{cases}$$
The moment-generating function of $X_1$ (and $X_2$) is then $$M_X(t) := \sum_{k=0}^\infty M_k \frac{t^k}{k!}$$ and this series converges on a radius of $C^{-1}$ around zero, with \begin{align} C = \limsup \sqrt[n]{|M_n|} &= \limsup \sqrt[2n+1]{|M_{2n+1}|} = \limsup \left(\underbrace{(2n-1)(2n-3)\ldots1}_{n\text{ factors}}\right)^{\frac{1}{(2n+1)n}} \\ &\leq \limsup \sqrt[2n+1]{2n-1} \leq 1 \text{.} \end{align}
Since the moment-generating function does converge on some neighbourhood of zero, it uniquely defines the distribution of $X$. However, finding a closed formular for $M_x$ seems tricky - the factorials don't seems to cancel in any meaningful way.