# Gamma Distribution

• Apr 7th 2011, 03:26 PM
dwsmith
Gamma Distribution
Expected Value

$\displaystyle\text{E}[X]=\frac{1}{\Gamma\left(\frac{2}{5}\right)20^{2/5}}\lim_{b\to\infty}\int_0^bx^{2/5}e^{-x/20}dx$

$\displaystyle\Gamma\left(\frac{2}{5}\right)=\lim_{ b\to\infty}\int_0^bx^{-3/5}e^{-x}dx$

How do I solve this Gamma function?
• Apr 7th 2011, 04:00 PM
pickslides
Have a go with this,

It is known that

$\displaystyle\Gamma\left(\frac{1}{5}\right)\approx 4.591$

And $\displaystyle\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)}= \int_0^1 t^{x-1}(1-t)^{y-1}~dt$

so $\displaystyle \Gamma \left(\frac{2}{5}\right) = \Gamma \left(\frac{1}{5}+\frac{1}{5}\right)=\frac{\Gamma \left(\frac{1}{5}\right)\Gamma \left(\frac{1}{5}\right)}{ \int_0^1 t^{\frac{-4}{5}}(1-t)^{\frac{-4}{5}}~dt}$

Not sure how valid this will be given my time away from the properties governing this function.
• Apr 7th 2011, 06:27 PM
mr fantastic
Quote:

Originally Posted by dwsmith
Expected Value

$\displaystyle\text{E}[X]=\frac{1}{\Gamma\left(\frac{2}{5}\right)20^{2/5}}\lim_{b\to\infty}\int_0^bx^{2/5}e^{-x/20}dx$

$\displaystyle\Gamma\left(\frac{2}{5}\right)=\lim_{ b\to\infty}\int_0^bx^{-3/5}e^{-x}dx$

How do I solve this Gamma function?

Why can't you leave it as $\Gamma \left(\frac{2}{5}\right)$?
• Apr 7th 2011, 09:17 PM
dwsmith
Quote:

Originally Posted by mr fantastic
Why can't you leave it as $\Gamma \left(\frac{2}{5}\right)$?

I don't know if I can. I was just going to solve everything if I could.
• Apr 7th 2011, 09:28 PM
pickslides