# kth Moment of Beta Distribution

$E({ X }^{ n })=\frac { \beta (n+a,b) }{ \beta (a,b) } =\frac { \Gamma (n+a)\Gamma (b)\Gamma (a+b) }{ \Gamma (n+a+b)\Gamma (a)\Gamma (b) } =\frac { \Gamma (n+a) }{ \Gamma (a) } \frac { \Gamma (a+b) }{ \Gamma (n+a+b) } =\prod _{ k=0 }^{ n-1 }{ \frac { a+k }{ a+b+k } }$

However, I cannot understand the last step:

How come:

$\frac { \Gamma (n+a) }{ \Gamma (a) } \frac { \Gamma (a+b) }{ \Gamma (n+a+b) } =\prod _{ k=0 }^{ n-1 }{ \frac { a+k }{ a+b+k } }$ ???

My thought is that:

$\frac { \Gamma (n+a) }{ \Gamma (n+a+b) } =\prod _{ k=0 }^{ n-1 }{ \frac { a+k }{ a+b+k } }$

But, in this case, what about:

$\frac { \Gamma (a+b) }{ \Gamma (a) }$ ?

Why is it gone ?

• Hint: Use twice, once for $c=a$ and once for $c=a+b$, the identity $$\Gamma (n+c)= \Gamma (c)\prod _{ k=0 }^{ n-1 }(c+k).$$ – Did Oct 19 '15 at 17:04
• Simply the iteration of $\Gamma(k+c)=(k-1+c)\Gamma(k-1+c)$, $n$ times. – Did Oct 19 '15 at 19:03
Using the fundamental identity $\Gamma(k+c)=(k-1+c)\Gamma(k-1+c)$, one gets:
$$\frac { \Gamma (n+a) }{ \Gamma (a) } \frac { \Gamma (a+b) }{ \Gamma (n+a+b) } =\frac { \Gamma (a)\prod _{ k=0 }^{ n-1 }{ a+k } }{ \Gamma (a) } \frac { \Gamma (a+b) }{ \Gamma (a+b)\prod _{ k=0 }^{ n-1 }{ a+b+k } } =\prod _{ k=0 }^{ n-1 }{ \frac { a+k }{ a+b+k } }$$