# Probability distribution

I will appreciate any help on this (simple?) question:

Let $r$ be a real number with $0<r<1$. Set $m(n)=r(r+1)...(r+n-1)/n!$, $m(0)=1, m(1)=r, m(2)=r(r+1)/2$, e.t.c.

What is the probability distribution of the random variable X with

$$E(X^n)=m(n),\ \ \ n=0,1,... ?$$

In particular, does X admits a density? Is it discrete?

P.S. I know that the support of X is concentrated in the interval [0,1].

-

A famous distribution in $[0,1]$ is the beta distribution. That's where I started from. The density of a beta(a,b) is given by:

$$f(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1}(1-x)^{b-1}$$

Hence,

$$E[X^{n}] = \int_{0}^{1}{\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{n+a-1}(1-x)^{b-1}dx} =$$

$$=\frac{\Gamma(n+a)\Gamma(a+b)}{\Gamma(a)\Gamma(n+a+b)}\int_{0}^{1}{\frac{\Gamma(n+a+b)}{\Gamma(n+a)\Gamma(b)} x^{n+a-1}(1-x)^{b-1}dx} = \frac{\Gamma(n+a)\Gamma(a+b)}{\Gamma(a)\Gamma(n+a+b)}$$

The last equality holds because the integral is over the density of a beta(n+a,b). Now take $a=r$ and $b=1-r$ to get:

$$E[X^{n}] = \frac{\Gamma(n+r)\Gamma(1)}{\Gamma(r)\Gamma(n+1)} = \frac{r(r+1)\ldots(n+r-1)}{n!}$$

-
 Thanks very much! – zpuzouni Jun 27 '12 at 12:15