# Thread: beta distribution probability density function proof

1. ## beta distribution probability density function proof

Hi
I'm having bit of trouble proving this fact.

can someone help me with the proof? or at least lead me thru it?

Thanks

-AC

2. Prove $B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}$

$B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q) }$....[1]

Start with ${\Gamma}(p)=\int_{0}^{\infty}t^{p-1}e^{-t}dt$

Now, put $t=y^{2}$, and we get:

${\Gamma}(p)=2\int_{0}^{\infty}y^{2p-1}e^{-y^{2}}dy$....[2]

Similarly, (the dummy integration variable can be any letter):

${\Gamma}(p)=2\int_{0}^{\infty}x^{2q-1}e^{-x^{2}}dx$

Next, we multiply these 2 equations and change to polar coordinates:

${\Gamma}(p){\Gamma}(q)=4\int_{0}^{\infty}\int_{0}^ {\infty}x^{2q-1}y^{2p-1}e^{-(x^{2}+y^{2})}dxdy$

$=4\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}(rcos{\ theta})^{2q-1}(rsin{\theta})^{2p-1}e^{-r^{2}}rdrd{\theta}$

$=4\int_{0}^{\infty}r^{2p+2q-1}e^{-r^{2}}dr\int_{0}^{\frac{\pi}{2}}(cos{\theta})^{2q-1}(sin{\theta})^{2p-1}d{\theta}$....[3]

The r integral in [3] is $\frac{1}{2}{\Gamma}(p+q)$ by [2].

The ${\theta}$ in [3] is $\frac{1}{2}B(p,q)$ by

$B(p,q)=2\int_{0}^{\frac{\pi}{2}}(sin{\theta})^{2p-1}(cos{\theta})^{2q-1}d{\theta}$

Then, ${\Gamma}(p){\Gamma}(q)=4\cdot \frac{1}{2}{\Gamma}(p+q)\cdot \frac{1}{2}B(p,q)$ and [1] follows.

3. Wow! that was quick!
Thanks for the help!

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# proofing of beta distribution pdf

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