# What do we mean when we say a gamma function has index v?

• Feb 4th 2009, 07:51 AM
alakazam
What do we mean when we say a gamma function has index v?
What is the density function of a gamma distribution with mean u and index v? I know this may be a very simple question, but there are so many parametrisations of the gamma distribution that it's confusing me. Thanks!

(I'm trying to show that the variance of a gamma distribution with mean u and index v is u^2 / v. I think this should be fairly manageable if I knew what the distribution was in the first place, and I haven't managed to find it elsewhere.)
• Feb 4th 2009, 02:03 PM
Laurent
Quote:

Originally Posted by alakazam
What is the density function of a gamma distribution with mean u and index v? I know this may be a very simple question, but there are so many parametrisations of the gamma distribution that it's confusing me. Thanks!

(I'm trying to show that the variance of a gamma distribution with mean u and index v is u^2 / v. I think this should be fairly manageable if I knew what the distribution was in the first place, and I haven't managed to find it elsewhere.)

There are various definitions of the parameters of a gamma distribution, but there are always two parameters, and since you know (given the answer) both the mean and the variance, you have two equations, and these allow you to identify the parameters and thus the distribution.

For instance, if you define a "Gamma distribution with scale parameter $\displaystyle \lambda$ and index $\displaystyle \alpha$" to be the distribution with density $\displaystyle \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}$ on $\displaystyle \mathbb{R}_+$, then you can compute the mean (it is $\displaystyle \frac{\alpha}{\lambda}$) and the variance (it is $\displaystyle \frac{\alpha}{\lambda^2}$), so that you must have $\displaystyle \frac{\alpha}{\lambda}=u$ and $\displaystyle \frac{\alpha}{\lambda^2}=\frac{u^2}{v}$.

Solving this, you get $\displaystyle \lambda=\frac{v}{u}$ and $\displaystyle \alpha=v$. This answers your question. By the way, the "index" you were given happens to be the "index" of my definition. This could have allowed to find $\displaystyle \lambda$ from the only knowledge of the mean.