# How can we efficiently find the fourth moment of a Poisson distribution?

Suppose we have $$X\sim \textrm{Poisson}(\lambda)$$ and we know that moment generating function $$M(t)=\mathbb{E}(e^{tX})$$. How do we use the moment generating function property $$M^k(0)=\mathbb{E}(X^k)$$ to derive, let's say, $$\mathbb{E}(X^4)$$? It seems impossibly tedious for me to differentiate the moment generating function all the way down. Is there a faster way for this?

#### Get the raw moments via the factorial moments

As with many other discrete distributions with simple factorial moments, obtaining the high-order raw moments is simplest when done through the factorial moments. For the Poisson distribution, the factorial moments have an extremely simple form:

\begin{align} \mathbb{E}((X)_r) &= \sum_{x=0}^\infty (x)_r \cdot \text{Pois}(x|\lambda) \\[6pt] &= \sum_{x=r}^\infty (x)_r \cdot \text{Pois}(x|\lambda) \\[6pt] &= \sum_{x=r}^\infty (x)_r \cdot \frac{\lambda^x}{x!} \cdot \exp(-\lambda) \\[6pt] &= \sum_{x=r}^\infty \frac{\lambda^x}{(x-r)!} \cdot \exp(-\lambda) \\[6pt] &= \lambda^r \sum_{x=r}^\infty \frac{\lambda^{x-r}}{(x-r)!} \cdot \exp(-\lambda) \\[6pt] &= \lambda^r \sum_{x=0}^\infty \frac{\lambda^x}{x!} \cdot \exp(-\lambda) \\[6pt] &= \lambda^r \sum_{x=0}^\infty \text{Pois}(x|\lambda) \\[12pt] &= \lambda^r. \\[6pt] \end{align}

Using the relationship between the raw moments and the factorial moments, we get the following general formula for the raw moments of the Poisson distribution:

\begin{align} \mathbb{E}(X^k) &= \sum_{r=0}^k S(k,r) \cdot \mathbb{E}((X)_r) \\[6pt] &= \sum_{r=0}^k S(k,r) \cdot \lambda^r, \\[6pt] \end{align}

where the terms $$S(k,r)$$ denote the Stirling numbers of the second kind. The fourth raw moment for the Poisson distribution is therefore given by:

$$\mathbb{E}(X^4) = \lambda^4 + 6 \lambda^3 + 7 \lambda^2 + \lambda.$$

• Yep, factorial moments are usually easier with Poisson and binomial (and a couple of other distributions) Commented Jan 8 at 4:06
• +1 -- applied upon reading the first line of this answer.
– whuber
Commented Jan 8 at 14:52

I don't think there's any genuinely easy way. The factorial moments are straightforward, as are the cumulants, but then it's messy combinatorial stuff to recover the moments (raw moments or central moments) for arbitrary $$k$$.

The cumulant generating function is the log of the moment generating function, so $$\kappa(t)=\log \exp\lambda(e^t-1))=\lambda(e^t-1)$$

All the derivatives of this are easy: for any $$k\geq 1$$ $$\frac{d^k\kappa(t)}{dt^k}=\lambda e^t,$$ so the cumulants are all $$\lambda$$. You can express the (central) moments in terms of the cumulants using Bell polynomials, which can be looked up rather than you needing to derive them.

$$\newcommand{\E}{\mathbb{E}}$$ You can also use the Stein-Chen identity. If $$X\sim\mathrm{Po}(\lambda)$$, then $$\mathbb{E}(Xg(X))=\lambda \mathbb{E}(g(X+1))$$where $$g$$ is any suitable function for which both expectations exist and $$g(0)$$ is defined. The derivation of this identity boils down to reindexing the series you get when you calculate the expectation on the left side.

If we apply this to $$g(x)=x^3$$, then we get $$\mathbb{E}(X^4)=\lambda\mathbb{E}[(X+1)^3],$$which you can then expand out and continue the process iteratively with $$g(x)=x^2$$, $$g(x)=x$$, etc. Here's what it looks like. It is (IMO) easier than computing the derivatives or factorial moments:\begin{align}\E(X^4)&=\lambda\E[(X+1)^3]\\&=\lambda\E[X^3+3X^2+3X+1]\\&=\lambda[\lambda\E[(X+1)^2]+3\lambda\E[X+1]+3\lambda+1]\\&=\lambda[\lambda\E[X^2+2X+1]+3\lambda(\lambda+1)+3\lambda+1]\\&=\lambda[\lambda[\lambda\E(X+1)+2\lambda+1]+3\lambda^2+6\lambda+1]\\&=\lambda[\lambda[\lambda(\lambda+1)+2\lambda+1]+3\lambda^2+6\lambda+1]\\&=\lambda[\lambda[\lambda^2+3\lambda+1]+3\lambda^2+6\lambda+1]\\&=\lambda[\lambda^3+6\lambda^2+7\lambda+1]\\&=\lambda^4+6\lambda^3+7\lambda^2+\lambda.\end{align}

If you already have the moment generating function, differentiating a few times will show you the pattern given in wikipedia: Touchard polynomials. They satisfy $$E[X^n] = \lambda(1 + \tfrac{d}{d\lambda})E[X^{n-1}].$$

A simple way to calculate $$E[X^4] = e^{-\lambda}\sum_{k=0}^\infty \frac{k^4\lambda^k}{k!}$$ is by differentiating:

$$\frac{d^4}{d\lambda^4} e^{\lambda} = e^\lambda = \sum_{k=0}^\infty \frac{(k^4 - 6k^3 + 11k^2 - 6k)\lambda^{k-4}}{k!}\\ = \frac{e^{\lambda}}{\lambda^4}(E[X^4] - 6E[X^3] + 11E[X^2] - 6E[X])$$ This gives $$E[X^4] = \lambda^4 + 6E[X^3] - 11E[X^2] + 6E[X]$$ and, in the same way, $$E[X^3] = \lambda^3 + 3E[X^2] - 2E[X] = \lambda^3 + 3\lambda^2 + \lambda$$ since $$E[X^2] = \lambda + \lambda^2$$. Thus

$$E[X^4] = \lambda^4 + 6\lambda^3 + 18\lambda^2 + 6\lambda - 11\lambda^2 - 5\lambda = \lambda^4 + 6\lambda^3 + 7\lambda^2 + \lambda.$$

Find the factorial moments, i.e. E[X], E[X(X-1)], E[X(X-1)(X-2)], etc. until you get to a moment containing the 4th power.

• This answer doesnt explain the solution well and others have already noted getting the factorial moments. It would be better if this answer was more detailed. Commented Jan 8 at 1:14