I want to calculate $EX$ of beta distribution given by certain formula (if necessary i will post it) i am stuck with an integral of this sort: $$\int_0^1\frac{\Gamma (a+b+1)}{\Gamma(a+1)\Gamma(b)}x^a(1-x)^{b-1}dx=\frac{a}{a+b}$$ I know the solution from the wikipedia but i have no idea why this integral works in such a manner?
$\endgroup$1 Answer
$\begingroup$Hint: Use facts about the beta distribution, and about the gamma function. You know the beta$(\alpha,\beta)$ density integrates to 1, which means for all $\alpha>0$ and $\beta>0$: $$ \int_0^1 x^{\alpha-1}(1-x)^{\beta-1}={\Gamma(\alpha)\Gamma(\beta)\over \Gamma (\alpha+\beta)}.\tag1 $$ Evaluate your integral by applying (1) with a suitable choice of $\alpha$ and $\beta$. Also, the gamma function satisfies $$\Gamma(t+1)=t\Gamma(t)$$
$\endgroup$
