Give the mgf of a gamma random variable having mean $6$ and variance $12$, and then use the mgf to obtain the $3rd$ moment of the random variable.
I have that
\begin{align} \operatorname{Var}(X)&=\frac{\alpha}{\beta^2} \\ \operatorname{E}(X)&=\frac{\alpha}{\beta} \end{align}
From the given mean and variance, I obtain $\alpha=3$, $\beta=\frac{1}{2}$.
The Wikipedia Page says that the mgf of a gamma distribution is
$$\left(1-\frac{t}{\beta}\right)^{-\alpha}$$
So I have that the mgf in my case is
$$\left(1-2t\right)^{-3} I_{(-\infty, \beta)}(t)$$
since the Wiki page says $t \lt \beta$.
I think to find the $3rd$ moment, I would take the third derivative of this and then plug in $t=0$.
$$\frac{d^3}{dt^3}\left(1-2t\right)^{-3}=\frac{480}{(1-2t)^6}=480$$
Are these correct?