Finding the moments of a distribution

[blalien]

Homework Statement

The problem is to find the moments $E(X^k)$ of $f_x(x) = (\theta+1)(1-x)^\theta$, $0 < x < 1$, $\theta > -1$

Homework 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.

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?

Thanks in advance!

 

[Ray Vickson]

Homework Statement

The problem is to find the moments $E(X^k)$ of $f_x(x) = (\theta+1)(1-x)^\theta$, $0 < x < 1$, $\theta > -1$

Homework 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.

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?

Thanks in advance!

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

 

[blalien]

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