# gamma

• Aug 15th 2009, 03:08 AM
Katina88
gamma
I'm just wondering why in some sites they put the moment generating function of the gamma distribution is:

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

while in other sites it's

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

which one do i use to calculate the mean and variance?
• Aug 15th 2009, 07:05 AM
matheagle
Both are correct and yet they are different.
It depends on how you write your density.

Some write the density as ${1\over \Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}I(x>0)$
here $\mu=\alpha \beta$

and others write it as ${\beta^{\alpha}\over \Gamma(\alpha) }x^{\alpha-1}e^{-x\beta}I(x>0)$
here $\mu={\alpha\over \beta}$
• Aug 15th 2009, 04:24 PM
Katina88
thanks!
Um about the mean being the product of the two parameters.. Well when i did maths stats we proved that the mean for gamma was:
$E(X) = \alpha \beta$

But now in this other subject, I'm asked to show that the mean is
$E(X) = \frac{\alpha}{\beta}$

So which one is correct, since you said both of the densities should have the mean being the product of the two parameters?
• Aug 15th 2009, 05:02 PM
mr fantastic
Quote:

Originally Posted by Katina88
thanks!
Um about the mean being the product of the two parameters.. Well when i did maths stats we proved that the mean for gamma was:
$E(X) = \alpha \beta$

But now in this other subject, I'm asked to show that the mean is
$E(X) = \frac{\alpha}{\beta}$

So which one is correct, since you said both of the densities should have the mean being the product of the two parameters?

I'll bet you're working with the second of the two pdf's that matheagle posted. In which case, note that the beta in the second is the reciprocal of the beta in the first. So unsurprisingly the mean of the second will be the result you're being asked to prove.

You should find the mean for both as an exercise and to clarify your thoughts.