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I was trying to help a friend solve this and I am really troubled. The task was to show that $$\int_0^1x^t(1-x)^{s-t}dx=\frac{t!(s-t)!}{(s+1)!}$$ using the Gamma distribution properties.

What I did was initially to try and show that is $$\frac{t!(s-t)!}{(s+1)!}$$ this times the cumulative fuction of a Gamma multiplying/dividing the right things. What I could not think of is how to appear the exponential that I need in order to create the cumulative.

Am I wrong? How can I prove this? Also Gamma integrates from zero to infinity and in this case $$x\in (0,1)$$.

I am trying to use the formula: $$\text{Gamma}(\alpha,\lambda)\implies\frac{{\lambda}^{\alpha}}{\Gamma(\alpha)}\cdot x^{\alpha-1}e^{-\lambda x}$$

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    $\begingroup$ The Beta and Gamma functions satisfy the relationship $$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ $\endgroup$
    – Mark Viola
    Commented Jan 20, 2016 at 2:54
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    $\begingroup$ Thank you, I did not know it. Very interesting. However, I am looking for an elementary way of proving this, probably without using this formula. $\endgroup$
    – Chenmath
    Commented Jan 20, 2016 at 2:59
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    $\begingroup$ @Chenmath There are two different Gamma distribution parametrizations. What is the PDF of the Gamma distribution you are taught in your class? $\endgroup$ Commented Jan 20, 2016 at 2:59
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    $\begingroup$ Do you know the beta distribution? $\endgroup$
    – Em.
    Commented Jan 20, 2016 at 3:02
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    $\begingroup$ I also wonder about the Beta distribution. It would be much more useful in this case. $\endgroup$ Commented Jan 20, 2016 at 3:03

2 Answers 2

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I thought that it would be instructive to present a very efficient approach that relies on the well-known relationship between the Beta function and the Gamma function

$$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

where

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$

and

$$\begin{align} \Gamma(x)&=\int_0^\infty t^{x-1}e^{-t}\,dt\\\\ &=(x-1)! \end{align}$$

Here, the integral of interest is

$$\begin{align} \int_0^1 x^t(1-x)^{s-t}\,dx&=B(t+1,s-t+1)\\\\ &=\frac{\Gamma(t+1)\Gamma(s-t+1)}{\Gamma(s+2)}\\\\ &=\frac{t!(s-t)!}{(s+1)!} \end{align}$$

and we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$
    – Mark Viola
    Commented Jan 25, 2016 at 3:15
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Using the Gamma function, if $s-t,t$ are nonnegative integers, then notice that \begin{align*} \int_0^1x^t(1-x)^{s-t}dx&=\frac{\Gamma(t+1)\Gamma(s-t+1)}{\Gamma(t+1+(s-t+1))}\\ &\qquad\cdot\int_0^1\frac{\Gamma(t+1+(s-t+1))}{\Gamma(t+1)\Gamma(s-t+1)}x^{(t+1)-1}(1-x)^{(s-t+1)-1}\,dx\\ &=\frac{\Gamma(t+1)\Gamma(s-t+1)}{\Gamma(s+2)}\cdot 1\\ &=\frac{t!(s-t+1)!}{(s+1)!} \end{align*} where the second integrand is the density of a Beta distribution.

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