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Computing the MGF of the Gamma Distribution

Matthew

Posted: Submitted by Matthew on 15 December 2008 - 5:22pm.

Joined: 2007-11-03
Posts: 12

I'm having trouble computing the MGF of the Gamma( $\alpha$ , $\beta$ ) distribution from first principles. I can do it by repeated integration by parts if $\alpha$ is an integer like so:
Take $t<1/\beta$

$M_X(t) = \int_{0}^{\infty}\frac{x^{\alpha-1}e^{-x/\beta}e^{tx}}{\Gamma(\alpha)\beta^{\alpha}}dx$

$= \int_{0}^{\infty}\frac{x^{\alpha-1}e^{x(t-1/\beta)}}{(\alpha-1)!\beta^{\alpha}}dx$

$= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\left[\frac{x^{\alpha-n}e^{x(t-1/\beta)}}{(\alpha-n)!\beta^{\alpha-n}(\beta t - 1)^n}\right]_0^{\infty} + (-1)^{\alpha}\int_{0}^{\infty}\frac{e^{x(t-1/\beta)}}{\beta(\beta t - 1)^{\alpha - 1}}dx$

$=\left[\frac{e^{x(t-1/\beta)}}{(1 - \beta t)^{\alpha}}\right]_0^{\infty}$

$=\frac{1}{(1-\beta t)^{\alpha}}$

but I can't see what to do for general $\alpha>0$ . Can anyone help me with this?

dsmccormick

Posted: 16 December 2008 - 7:08pm

Joined: 2006-10-06
Posts: 75

There's a trick or two to be had, which to be fair I had to get off Mathworld, but once you do it, it falls out relatively easily. So, we have

$M_X (t) = \int_0^\infty \frac{x^{\alpha-1} e^{-x/\beta} e^{tx}}{\Gamma(\alpha) \beta^\alpha} \, \mathrm{d}x.$

Suppose $t < 1/\beta$ . Let $y = (1/\beta - t)x$ , so $\mathrm{d}y = (1/\beta - t) \mathrm{d}x$ . Using this substitution, we obtain:

$M_X (t) = \int_0^\infty \left( \frac{y}{1/\beta - t} \right)^{\alpha-1} \frac{ e^{-y}}{\Gamma(\alpha) \beta^\alpha} \, \frac{\mathrm{d}y}{(1/\beta - t)}$

$= \int_0^\infty \left( \frac{1}{\beta ( 1/\beta - t)} \right)^{\alpha} \frac{1}{\Gamma(\alpha)} y^{\alpha - 1} e^{-y} \, \mathrm{d}y$

Since the first two terms do not depend on $y$ we can take them out of the integral to obtain

$M_X (t) = \left( \frac{1}{\beta ( 1/\beta - t)} \right)^{\alpha} \frac{1}{\Gamma(\alpha)} \int_0^\infty y^{\alpha - 1} e^{-y} \, \mathrm{d}y$

Now, by definition $\Gamma(\alpha) = \int_0^\infty s^{\alpha - 1} e^{-s} \, \mathrm{d}s$ , so the integral on the right cancels with the $\Gamma(\alpha)$ on the bottom line and the fraction reduces to

$M_X(t) = \left( \frac{1}{1 - \beta t} \right)^{\alpha}$

as required. Hope that helps!

(Wow, that's the first time having done my second year essay on the gamma function has actually paid off...)

Dave McCormick

Matthew

Posted: 16 December 2008 - 10:59pm

Joined: 2007-11-03
Posts: 12

Thanks.

Ramildinho

Posted: 30 December 2008 - 10:37pm

Joined: 2008-01-26
Posts: 6

thanks

jonloton0098

Posted: 15 February 2010 - 9:39am

Joined: 2010-02-13
Posts: 1

Oh no, I am very afraid of math. I am very weak in math. But I have tried to solve the problem. But sorry.

You shouldn't have moved into check.

Colin Mkz

Posted: 26 May 2010 - 9:56am

Joined: 2010-05-26
Posts: 1

I wish I had seen this before my stats exam... sigh..

mzingdolow

Posted: 24 April 2012 - 5:48am

Joined: 2012-04-24
Posts: 4