# Thread: Moment Generating Function Gamma Distrib

1. ## Moment Generating Function Gamma Distrib

$\displaystyle \displaystyle\frac{1}{\sqrt{7}\Gamma\left(\frac{1} {2}\right)}\lim_{b\to\infty}\int_0^b\left(x^{-1/2}\exp\left[-x\left(\frac{1}{7}-t\right)\right]\right) \ dx$

$\displaystyle \displaystyle\frac{1}{\sqrt{7}\Gamma\left(\frac{1} {2}\right)}\lim_{b\to\infty}\int_0^b\left(x^{-1/2}\exp\left[\frac{-x}{\frac{1}{\left(\frac{1}{7}-t\right)}}\right]\right) \ dx$

$\displaystyle \displaystyle\alpha=\frac{1}{2} \ \beta=\frac{1}{\frac{1}{7}-t}$

$\displaystyle \displaystyle =\frac{1}{\sqrt{7}\Gamma\left(\frac{1}{2}\right)}\ left(\frac{1}{\frac{1}{7}-t}\right)^{1/2}\Gamma\left(\frac{1}{2}\right)$

Correct?

2. Originally Posted by dwsmith
$\displaystyle \displaystyle\frac{1}{\sqrt{7}\Gamma\left(\frac{1} {2}\right)}\lim_{b\to\infty}\int_0^b\left(x^{-1/2}\exp\left[-x\left(\frac{1}{7}-t\right)\right]\right) \ dx$

$\displaystyle \displaystyle\frac{1}{\sqrt{7}\Gamma\left(\frac{1} {2}\right)}\lim_{b\to\infty}\int_0^b\left(x^{-1/2}\exp\left[\frac{-x}{\frac{1}{\left(\frac{1}{7}-t\right)}}\right]\right) \ dx$

$\displaystyle \displaystyle\alpha=\frac{1}{2} \ \beta=\frac{1}{\frac{1}{7}-t}$

$\displaystyle \displaystyle =\frac{1}{\sqrt{7}\Gamma\left(\frac{1}{2}\right)}\ left(\frac{1}{\frac{1}{7}-t}\right)^{1/2}\Gamma\left(\frac{1}{2}\right)$

Correct?
After simplification, does it agree with the general result given here: Gamma distribution - Wikipedia, the free encyclopedia ?

3. It does.

$\displaystyle \displaystyle\frac{1}{(1-7t)^{1/2}}$

But when I take the second derivative as t = 0 and subtract that from the mean squared, I don't obtain the correct variance.

4. Originally Posted by dwsmith
It does.

$\displaystyle \displaystyle\frac{1}{(1-7t)^{1/2}}$

But when I take the second derivative as t = 0 and subtract that from the mean squared, I don't obtain the correct variance.
I do You should try again.