Find mean and variance using Moment generating function of the negative binomial. I was asked to derive the mean and variance for the negative binomial using the moment generating function of the negative binomial.  
However i am not sure how to go about using the formula to go out and actually solve for the mean and variance. 
 A: The moment generating function of a random variable $X$ is defined by
$$  M_X(t) = E(e^{tX}) =
\begin{cases}
\sum_i e^{tx_i}p_X(x_i),  & \text{(discrete case)} \\
\\
\int_{-\infty}^{\infty} e^{tx}f_X(x)dx, & \text{(continuous case)}
\end{cases}
$$
If we express $e^{tX}$ formally and take expectation
$M_X(t) = E(e^{tX}) = 1 + tE(X) + \frac{t^2}{2!}E(X^2)+...+
\frac{t^k}{k!}E(X^k)+...$
and the $k$th moment of $X$ is given by
$E(X^k) = M_X^{(k)}(0) \:\:\:\:\:\:k = 1, 2...$
$M_X^{(k)}(0) = \frac{d^k}{dt^k} M_X(t) |_{t=0}$

For the negative binomial we have the moment generating function
$M_X(t)=E(e^{tX})=(pe^t)^r [1-(1-p)e^t]^{-r}$
and we want to calculate (writing $M_X(t)=M(t)$ from here on)
$\mu = E[X] = M'(0)$
$\sigma^2= E[X^2] - (E[X])^2 = M''(0)-[M'(0)]^2$

$M'(t)= (1-p) r e^t (p e^t)^r (1-(1-p) e^t)^{-r-1}+p r e^t (p e^t)^{r-1} (1-(1-p) e^t)^{-r}$
$M'(0) \Rightarrow e^t = e^0 =1$
$M'(0)= (1-p) r (p)^r (1-(1-p))^{-r-1}+p r (p)^{r-1} (1-(1-p))^{-r}$
$M'(0)= (1-p) r (p)^r (p)^{-r-1}+p r (p)^{r-1} (p)^{-r}$
$M'(0)= (1-p) r (p)^{-1}+p r (p)^{-1}$
$M'(0)= \frac{(1-p)r}{p} + r$
$M'(0)= \frac{(1-p)r}{p} + \frac{pr}{p}$
$M'(0)= \frac{(1-p)r + pr}{p}$
$M'(0)= \frac{r}{p}$
$\therefore \mu = \frac{r}{p}$

$\sigma^2= E[X^2] - (E[X])^2 = M''(0)-[M'(0)]^2$
$M''(t) = r(pe^t)^r(-r-1)[1-(1-p)e^t]^{-r-2}[-(1-p)e^t]+r^2(pe^t)^{r-1}(pe^t)[1-(1-p)e^t]^{-r-1}$ 
$M''(0) = r(p)^r(-r-1)[1-(1-p)]^{-r-2}[-(1-p)]+r^2(p)^{r-1}(p)[1-(1-p)]^{-r-1}$ 
and then it is just to simplify this and use the formula for the variance.
A: The key idea is that if $M_X(t) = \operatorname{E}[e^{tX}]$ is the moment generating function for a random variable $X$, then we have $$\left[\frac{d^k M_X(t)}{dt^k} \right]_{t=0} = \operatorname{E}[X^k], \quad k = 0, 1, 2, \ldots.$$  That is to say, the $k^{\rm th}$ derivative of $M_X(t)$ with respect to $t$, evaluated at $t = 0$, is equal to the $k^{\rm th}$ raw moment of $X$, whenever such moments exist.  So if you are given the negative binomial MGF, all you need to do to calculate $\operatorname{E}[X]$ is to take the derivative of the MGF, and evaluate it at $t = 0$.
To get the variance, recall that $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2,$$ so you would calculate the second derivative $M''_X(0)$ at $t = 0$ and subtract the square of the previous result.
