# beta distribution probability density function proof

• Nov 11th 2008, 10:05 AM
JHUBmore
beta distribution probability density function proof
Hi
I'm having bit of trouble proving this fact.
Attachment 8657
can someone help me with the proof? or at least lead me thru it?

Thanks

-AC
• Nov 11th 2008, 10:52 AM
galactus
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.
• Nov 11th 2008, 02:54 PM
JHUBmore
Wow! that was quick!
Thanks for the help!