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I was reading the following about the 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 ?

Thanks in advance for your help.

Regards,

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    $\begingroup$ Hint: Use twice, once for $c=a$ and once for $c=a+b$, the identity $$\Gamma (n+c)= \Gamma (c)\prod _{ k=0 }^{ n-1 }(c+k).$$ $\endgroup$ – Did Oct 19 '15 at 17:04
  • $\begingroup$ Hmmm, I wasn't aware of this identity. Thanks so much, Did. Always impressive. $\endgroup$ – XCoder Oct 19 '15 at 17:18
  • $\begingroup$ Simply the iteration of $\Gamma(k+c)=(k-1+c)\Gamma(k-1+c)$, $n$ times. $\endgroup$ – Did Oct 19 '15 at 19:03
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Using the fundamental identity $\Gamma(k+c)=(k-1+c)\Gamma(k-1+c)$, one gets:

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

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