Find the mean, variance and $f(21)$ of this if it is a p.g.f. $$G(t) = t^{11}e^{9t-9}$$
Find the mean, variance and $f(21)$
Please help me with find $f(21)$ as easily as possible.
For the mean and variance:
$$G'(t) = 11t^{10}e^{9t-9}+9t^{11}e^{9t-9}$$
$$G^{(2)}(t) = 11(10) t^9e^{9t-9} + 11(9)t^{10}e^{9t-9} + 9(11) t^{10}e^{9t-9} + 9(9)t^{11}e^{9t-9}$$
$$\mu = G'(1) = 11+9 = 20$$
$$\mu_{(2)} = G^{(2)}(1) = 110+99 +99 + 81 = 389$$
$$\mu_2' = \mu_{(2)} + \mu = 389 + 20 = 409$$
$$\sigma^2= \mu_2 = \mu_2'-\mu^2 = 409 - 20^2 = 9$$
Now for $f(21)$
$$f(21) = \frac{G^{(21)}(0)}{21!}$$
Okay now surely there must be a better way of finding the $21^{st}$ derivative other than deriving $21$ times but I don't know how I would go about finding the $r^{th}$ factorial moment for this one.

This answered was inspired by the examples mentioned here.
If $V$ is $\lambda$-Poisson-distributed then: $$G_{V}\left(t\right)=\mathbb{E}t^{V}=\sum_{n=0}^{\infty}t^{n}e^{-\lambda}\frac{\lambda^{n}}{n!}=e^{-\lambda}\sum_{n=0}^{\infty}\frac{\left(\lambda t\right)^{n}}{n!}=e^{\lambda t-\lambda}$$
For $\lambda=9$ we find $G_V(t)=e^{9t-9}$
So: $$G_{11+V}\left(t\right)=\mathbb{E}t^{11+V}=t^{11}e^{9t-9}$$
and: $$f(21)=P(11+V=21)=P(V=10)=e^{-9}\frac{9^{10}}{10!}$$
Moreover: $$\mathbb E(11+V)=11+\mathbb EV=11+9=20$$
and:$$\text{Var}(11+V)=\text{Var}V=9$$
