# Square root of normal distribution

Let $X$ and $Y$ be independent random variates with the same probability distribution, $P(x)$. Assuming that the product $Z=XY$ is a random variate with normal distribution, say $$f_Z(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$$ what is the form of $P(x)$? Does it exist in closed form?

EDIT: I just realized that a product distribution is necessarily divergent at zero, which means that the normal distribution cannot be a product distribution (at least when $P(x)$ is interpreted as a normal function; maybe there is a generalized function that solves the problem?). Do you agree?

• I guess we would just need $$\frac{1}{\sqrt{2\pi}}e^{-z^2/2}=\int_{-\infty}^\infty f_X(x)\,f_X(z/x)\,\frac1{|x|}\,dx$$ – Bjørn Kjos-Hanssen Mar 11 '15 at 19:51
• The existence of the square root follows from the infinite divisibility of the log-normal distribution, does it not? tandfonline.com/doi/abs/10.1080/03461238.1977.10405635 – Brendan McKay Mar 13 '15 at 1:00
• Brendan: I think log-normal distribution is in the opposite direction -- it is the distribution of $e^{tZ}$ for a nonzero real $t$, whereas here of relevance would be the infinite divisibility of $\ln|Z|$. – Iosif Pinelis Mar 10 '16 at 2:22

First: No helvio, the density of the product of two iid random variables (r.v.'s) is not necessarily divergent at 0. E.g., take the product of two iid r.v.'s each having a Gamma distribution with shape parameter 2.

Second: Let $U:=\ln|Z|$ and $W:=\ln|X|$. Then the characteristic function (c.f.) $f$ of $U$ is given by the formula $f(t)=2^{i t/2} \Gamma((1+i t)/2)/\sqrt{\pi}$ for real $t$. So, the c.f. of $W$ (say $g$) would be a square root of $f$, and then the density (say $q$) of $W$ would be given by the formula $q(x)=\frac1{2\pi}\,\int_{-\infty }^{\infty } g(t) e^{-i t x} \, dt$ for real $x$. I have tried to evaluate the latter integral numerically, and that seems to suggest that $q(3)<0$; however, I have not done a rigorous estimate of the approximation error.

On the other hand, numerical experiments using Bochner's theorem on the positive definite functions suggest that $g$ is a c.f.! If so, then the corresponding r.v. $W$ exists, and then it suffices to let $X$ have the distribution of $\varepsilon e^W$, where $\varepsilon$ is a Rademacher r.v. independent of $W$.

Addendum: The calculation of $q(3)$ mentioned previously was likely mistaken, as I forgot to make sure that $g(t)$ be taken to be the value of the square root of $z:=f(t)$ on the Riemann surface for $\sqrt z$, so as for $g$ to be a continuous function -- as it must be.

Another approach to the problem: Note that $\mu_j:=E|Z|^j=2^{j/2} \Gamma \left(\frac{j+1}{2}\right)/\sqrt{\pi }$ and then one would have $\nu_j:=E|X|^j=\sqrt{\mu_j}$ for $j>0$. To prove that such a desired r.v. $X$ exists, it is enough to show that for all nonnegative integers $n$ the determinants, say $d_n$ and $e_n$, of the matrices $(\nu_{i+j})_{i,j=0}^n$ and $(\nu_{i+j+1})_{i,j=0}^n$ are nonnegative (see e.g. Ch. V in Kreǐn, M. G. and Nudel'man, A. A., The Markov moment problem and extremal problems). These determinants are indeed positive for $n\le9$. One can hopefully find a recursion for $d_n$ and $e_n$ to show that they are indeed nonnegative for all $n$. Note here that $\mu_j$ is an integer or an integral multiple of $\sqrt{2/\pi}$.

• Thank you very much @iosif-pinelis for your detailed answer, I learned a great deal with it. So if I understood correctly the probability density $P(x)$ is given by the Fourier transform of the square root of the characteristic function of the normal probability density, $$P(x) = \int_{-\infty}^{+\infty} \left( \frac{2^{it/2}}{\sqrt\pi} \Gamma\Big(\frac{1+it}{2}\Big) \right)^\frac{1}{2} e^{-itx} dt$$ right? I can't expect a closed form solution, and even if it existed it's probably useless in practice. But the whole argument is sound and solid, so I'll accept it. – helvio Mar 13 '15 at 13:46
• helvio: $\frac{P(x)}{2\pi}$ would be the density of $\ln|X|$ -- provided that $P(x)$ is nonnegative for all real $x$. I think this nonnegativity is likely the case, but don't know how to prove that. The other approach that I suggested later, with the $\mu_j$'s and $\nu_j$'s, may be more promising. – Iosif Pinelis Mar 13 '15 at 16:58
• A complete answer to this question (and actually a more general one) is now available at arxiv.org/abs/1803.09838 – Iosif Pinelis Apr 2 '18 at 0:08