a) $E(X) =$ ?
b) $E(X^2) = $ ?
c) $Var(X) = $ ?
My thoughts:
I know that moment-generating function for $\Gamma(k,\theta) = ( 1 - t\theta)^{-k}$ for $t < \frac{1}{\theta}$.
I also know that $E(X) = k\theta $ and $ Var(X) = k\theta^2$ (I think?)
Also I may need to use the idea that $E(X^n) = M_x^{(n)} (0)$ ?
But I am not really sure how to do these things using $M_X(t)$... Any help is greatly appreciated!
You have written down what needs to be done, now it is a matter of doing it.
For $E(X)$, we differentiate $M_X(t)$, that is, $(1-t\theta)^{-k}$, once and evaluate the derivative at $t=0$.
The first derivative of $M_X(t)$ is $(-\theta)(-k)(1-t\theta)^{-k-1}$. At $t=0$ this is $k\theta$, so $E(X)=k\theta$.
Differentiate again. The second derivative of $M_X(t)$ is $k(k+1)\theta^2(1-t\theta)^{-k-2}$. Thus $E(X^2)=k(k+1)\theta^2$.
For the variance, use the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. This simplifies to $k\theta^2$.