Exercise on Random Variables

Question

I´m struggling with a random variable exercise of a book I´m reading. Anyone has an idea of how to approach this problem?

Thanks in advance :)

Answer 1

$$E(X)=\int^{1}_{0}\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}x(x^{a-1}(1-x)^{b-1})=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int^{1}_{0}x^{a}(1-x)^{b-1}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\left(\frac{\Gamma(a+1)\Gamma(b)}{\Gamma((a+b)+1)}\right)\int_{0}^{1}\left(\frac{\Gamma((a+b)+1)}{\Gamma(a+1)\Gamma(b)}\right)x^{(a+1)-1}(1-x)^{b-1}=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\left(\frac{\Gamma(a+1)\Gamma(b)}{\Gamma((a+b)+1)}\right)=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\left(\frac{a\Gamma(a)\Gamma(b)}{(a+b)\Gamma(a+b)}\right)=\frac{a}{a+b}$$

Answer 2

More directly: $EX = {B(a+1,b) \over B(a,b)}= { \Gamma(a+1) \Gamma(b) \over \Gamma(a+b+1)} { \Gamma(a+b) \over \Gamma(a) \Gamma(b) } = { a\over a+b } $ (using the fact that $\Gamma$ satisfies the functional equation $\Gamma(x+1) = x \Gamma(x)$).