$E[Y^3]$ of a Poisson Distribution

Question

I have been able to find $E[Y^2]$ by simply solving for $E[Y^2]$ in the formula for variance

$Var[Y] = E[Y^2] - E[Y]^2 \rightarrow E[Y^2] = Var[Y] + E[Y^2]$

but how could I solve for $E[Y^3]$ where Y is a Poisson random variable.

Answer 1

Hint

Use the fact that $$\sum_{k=p}^\infty k^q\frac{\lambda ^k}{(k-p)!}=\sum_{k=p+1}^\infty k^{q-1}\frac{\lambda ^k}{(k-(p+1))!}+\sum_{k=p}^\infty k^{q-1}\frac{\lambda ^k}{(k-p)!}.$$

Answer 2

Hint:

The Poisson distribution has finite mgf

$$t \mapsto \exp (\lambda(e^t-1))$$

By differentiating this three times and evaluating in $t=0$, you get the desired expectation.