# Finding the moments of a distribution

1. Oct 19, 2012

### blalien

1. The problem statement, all variables and given/known data
The problem is to find the moments $E(X^k)$ of $f_x(x) = (\theta+1)(1-x)^\theta$, $0 < x < 1$, $\theta > -1$

2. Relevant equations
$E(X^k)=\int_0^1 x^k (\theta+1)(1-x)^\theta dx$
According to Mathematica, the solution is $\frac{\Gamma(1+k)\Gamma(2+\theta)}{\Gamma(2+k+ \theta )}$. I have no idea how to solve this integral by hand, however.

3. The attempt at a solution
If we let $W = -\log(1-x)$, then the distribution for $W$ is $f_w(w) = (\theta+1)e^{-(\theta+1)w}$, which is just the exponential distribution. The moments are $E(W^k) = \frac{\Gamma(k+1)}{(1+\theta)^k}$. The question is, if we know the relation between x and w and we know the moments for w, is it possible to find the moments for x? Or is there another way to solve this integral?

2. Oct 19, 2012

### Ray Vickson

You don't need to use Mathematica to get the value of E*W^k); it is expressible in terms of the Beta function, which is, in turn, expressible in terms of the Gamma function. The way you could do it manually is to "recognize" the Beta integral, then use standard theorems relating Beta to Gamma. Of course, if you are not familiar with the Beta integral, this will seem mysterious.

RGV

3. Oct 20, 2012

### blalien

That's it, thanks. I don't know how we were ever supposed to solve this without knowing the beta function.