distribution of the product of two or three uniform r.v.'s on the interval [0,1].

[Porky Pig]

For the problem I'm working on, I had to compute the product of two or
three i.i.d. r.v.s, each is uniformly distributed on [0,1]. I've found
the article with the general formula for n such r.v.,s but, alas, I
couldn't make head nor tail of that. So, I ended up computing it from
the scratch and I would appreciate if somebody, probably familiar with
that type of problems, can take a quick look and tell me if my answers
are correct or not. For the product z of two i.i.d. r.v.s uniformly
distributed on [0,1] I got the formula for pdf

f(z) = ln(1/z)

where ln is the natural log. For the product z of three such r.v.s I
got the formula for pdf

f(z) = (1/2) ln^2 (1/z).

(Of course I made sure that those formulas are the valid pdf's by
integrating over the valid range of z and getting 1.)

TIA,

PPJr.

(Note: at least intuitively, those formulas do make sense to me. We
multiply some numbers between 0 and 1, and the result is smaller than
both, so having the result closer to 0 is more probably than closer to
1. That's one ln(1/z) says.)

[Einar Andreas Rødland]

Hi,

Your answer is correct.

If U=U1*...*Un where Ui are i.i.d, it is usually easier to switch to the
logarithm, e.g. X=-ln U and Xi=-ln Ui, where X=X1+...+Xn.

If Ui are uniform on [0,1], then Xi=-ln Ui have the exponential
distribution (Xi~exp(-x)). Adding n of these together gives you a
gamma-distribution X~x^{n-1}*exp(-x)/Gamma(n) where Gamma(n)=(n-1)!.
Convert back to U=exp(-X), and you get U~(-ln u)^{n-1}/Gamma(n).

Einar

[Porky Pig]

Thanks so much!

PPJr.