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I have the following question given to me about the beta distribution. I have been stuck on this one for hours now.

Let $X$ have a beta distribution with parameters $\alpha$ and $\beta$, let $r$ and $s$ be given positive numbers. Find the general expression for $E(X^{r}(1-X)^{s})$.

Beta distribution form I have been using is:

$$\frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}x^{\alpha -1}(1-x)^{\beta -1}\qquad\alpha >0, \beta >0$$

$x$ is between $0$ and $1$.

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A ratio of two beta functions (times 1). – Did Sep 18 '12 at 12:57
Hint (with correction of typo not noticed within 5 minutes): You will see that $E[X^r(1-)^s] = \int_0^1 x^r(1-x)^sf(x)\,\mathrm dx)$ can be expressed as a Beta function (times a constant) if you will only substitute the expression for $f(x)$ which is of the form $x^i(1-x)^j$ for $0 \leq x \leq 1$. (The constant itself can be expressed in terms of a beta function as noted in @did's comment) – Dilip Sarwate Sep 18 '12 at 13:04
up vote 2 down vote accepted

This is essentially an exercise in integration. If $X$ has a Beta($\alpha$,$\beta$) distribution then the expected value you're trying to calculate can be written as:

$$ \int_{0}^{1} x^r (1-x)^s \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}x^{\alpha -1}(1-x)^{\beta -1} dx = \int_{0}^{1}\frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )}x^{r+\alpha -1}(1-x)^{s+\beta -1} dx $$

The trick is to multiply by constants so that the integrand is another beta density, which we know integrates to 1. The constants required will all be $\Gamma$ functions:

$$ E(X^r (1-X)^s ) = \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )} \cdot \frac{\Gamma (\alpha+r )\Gamma (\beta+s )}{\Gamma (\alpha +\beta+r+s )}\int_{0}^{1} \underbrace{\frac{\Gamma (\alpha +\beta+r+s )}{\Gamma (\alpha+r )\Gamma (\beta+s )} x^{r+\alpha -1}(1-x)^{s+\beta -1}}_{{\rm Beta}(\alpha+r, \beta+s) \ \ {\rm density}}dx $$

So the integral equals 1, since it's a probability density. Therefore,

$$ E(X^r (1-X)^s ) = \frac{\Gamma (\alpha +\beta )}{\Gamma (\alpha )\Gamma (\beta )} \cdot \frac{\Gamma (\alpha+r )\Gamma (\beta+s )}{\Gamma (\alpha +\beta+r+s )} $$

I hope this helps!

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Thanks mate, yeah this helps heaps! – Stewart Sep 18 '12 at 21:47

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