How to easily calculate the expected value of powers of Poisson variable?

Question

For a statistics class the solution to one of the question starts with: "First note that, if $Y$ is a Poisson random variable with mean $\lambda$ then,
$\mathbb{E}[Y]=\lambda$,
$\mathbb{E}[Y^2]=\lambda^2+\lambda$,
$\mathbb{E}[Y^3]=\lambda^3+3\lambda^2+\lambda$,
$\mathbb{E}[Y^4]=\lambda^4+6\lambda^3+7\lambda^2+\lambda$.

I know the first one is a property of the Poisson distribution and the second one follows relatively easy from that but I was wondering how to quickly get the last two.

I know it is possible to compute these values with $$\sum_{k=0}^{\infty}k^3 e^{-\lambda}\frac{\lambda^k}{k!} \text{ and }\sum_{k=0}^{\infty}k^4 e^{-\lambda}\frac{\lambda^k}{k!}$$

But when I do that it takes quite some time and computational work. Because the solutions just said "note" it makes me think that there is a way faster way to do this or that this is just a well known fact.

So my question is, is there a faster way to compute this then working out the sum?

Answer 1

Let's say you know $E(Y^j)$ for all $\ j\in \{1,2,\ldots, n-1\}.\ $ Then we can get $E(Y^n)$ in terms of $E(Y^{n-1}),\ E(Y^{n-2}),\ \dots,\ $ and $\ E(Y).\ $

$$ E(Y^n) = \sum y^nP(Y=y) = \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} k^n $$

$$ = 0 + \sum_{k=1}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} k^n$$

$$ = \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^{k+1}}{(k+1)!} (k+1)^n$$

$$ = \lambda\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} (k+1)^{n-1}$$

$$ = \lambda \left( \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} \binom{n-1}{0} k^{n-1} + \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} \binom{n-1}{1} k^{n-2} + \ldots + \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} \binom{n-1}{n-1} k^{0} \right)$$

$$ = \lambda \left( \binom{n-1}{0} \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!}k^{n-1} + \binom{n-1}{1}\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} k^{n-2} + \ldots + \binom{n-1}{n-1}\sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} k^{0} \right)$$

$$ = \lambda \left( \binom{n-1}{0} E(Y^{n-1}) + \binom{n-1}{1}E(Y^{n-2}) + \ldots + \binom{n-1}{n-2}E(Y) + \binom{n-1}{n-1} 1 \right).$$

Answer 2

If $Y\sim \text{Poisson}(\lambda)$ $$\mathbb{E}[Y^n]=\sum_{k=0}^{n}\left\{n\atop k\right\}\lambda^k=B_n(\lambda)$$

Where:

• $\displaystyle\left\{n\atop k\right\}$ is the Stirling number of the $2$nd kind
• $B_n(x)$ is the Bell polynomial

You can find all the Stirling numbers here