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...

  • $\begingroup$ 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 $X_1$. $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.