How do you find $E(X^3)$ of a Poisson Distribution?

Question

I know the proof to find the variance of a Poisson Distribution, and I tried to use that to find $E(X^3)$, but I can't get it to work. Any help would be great!!

Answer 1

$E(X^3)=e^{-\lambda}\sum_0^\infty \frac{k^3 \lambda^k}{k!}$.

$k^3=k(k-1)(k-2)+3k(k-1)+k$

so the sum can be changed to three sums

$\lambda^3\sum_3^\infty \frac{\lambda^{k-3}}{(k-3)!}+ 3\lambda^2\sum_2^\infty \frac{\lambda^{k-2}}{(k-2)!}+\lambda\sum_1^\infty \frac{\lambda^{k-1}}{(k-1)!}$

Multiply by $e^{-\lambda}$ and sum to get $\lambda^3+3\lambda^2+\lambda$.

Answer 2

Observe that $$E[e^{Xt}]=\sum_{k=0}^{\infty} e^{kt}P(X=k) = \sum_{k=0}^{\infty} e^{kt}e^{-\lambda } \frac{\lambda ^k}{k!} = e^{\lambda(e^t-1)} \quad (*). $$

Also $$\frac{d^3}{d^3t}E[e^{Xt}]|_{t=0} = \frac{d^3}{d^3t}(1+E[X]+\frac{1}{2!}t^2E[X^2]+\frac{1}{3!}t^3E[X^3]+ O(t^4X^4))|_{t=0}= E[X^3] \quad (**).$$ Combining (*) and (**) results

$$\frac{d^3}{d^3t}E[e^{Xt}]|_{t=0} = \frac{d^3}{d^3t}e^{\lambda(e^t-1)}|_{t=0}= \lambda e^{\lambda (e^t-1) + t} (1 + 3 \lambda e^t + \lambda^2 e^{2 t})|_{t=0}=\lambda+3\lambda^2+\lambda^3.$$