How is the product of two independent samples, X and Y, from U(0, 1) distributed?
0 ≤ XY ≤ 1.
The point (X, Y) is uniformly distributed in the unit square. The curve xy=c is a hyperbola dividing the square into two parts. We seek the area of the lower part next to the origin.
The lower part is composed of a c by 1 rectangle and the area under the hyperbola from c to 1.
For c ∊ [0, 1] (the probability that XY < c) = CDF(c) = c + ∫[x=c, x=1] (c/x) dx
= c + c(∫[x=c, x=1] (1/x) dx) = c + c(log(1) − log(c)) = c − c(log(c)).
PDF(c) = (d/dc)(c − c(log(c))) = 1 − ((c(1/c)) + log(c)) = −log(c).
Note that this PDF is unbounded near zero.

### Contagion

Symmetry considerations indicate that if the distributions for X and Y are U(−1, 1) and U(0, 1) then the PDF will be 0 outside [−1, 1] and −log(|x|)/2 inside. We get the same PDF when both distributions are U(−1, 1).

Suppose that PDFX is the PDF for X and that for some δ>0, (−1 ≤ x ≤ 1) → PDFX(x) > δ
and that Y is a sample from U(−1, 1)
then the PDF for XY will also have the singularity at 0, perhaps diminished by factor δ. If PDFY has an ε like the δ for X then the distribution for XY will have a singularity at 0 diminished by a factor of δε.

Note that N(m, d) which is the normal distribution with mean m and standard deviation d is bounded away from 0 by some positive δ in the interval [−1, 1]. We reluctantly conclude that the PDF of the product of samples from any two normal distributions is unbounded near zero!

Curiously it seems not to matter much; δε is usually very very small.