"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...
 A: 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.
