Kurtosis of Poisson Distribution [closed]

I am trying to find the kurtosis of a Poisson distribution without using

$$\frac{E(X^4) - 4\mu E(X^3) +6\mu^2E(X^2) - 3\mu^4}{\sigma^4}$$

The professor did not teach this equation and will not. The same was the only way that I could learn using books or websites.

First, if your prof only taught the centered version, I assume once you prove the uncentered version, you can use it.

Second, to compute moments of a Poisson RV, it may help to use

$$X^2=X(X-1)+X\\ X^3=X(X-1)(X-2)+3X^2-2X\\ X^4=X(X-1)(X-2)(X-3)+6X^3-11X^2+6X\\$$

and so forth. Or differentiate the MGF or PGF (see J.G.'s response).

In general, the moments of Poisson($\lambda$) are given by Dobinski's formula,

$$E[X^k]=\sum_{i=0}^k \lambda^i{k\brace i},$$

where ${k\brace i}$ are Stirling numbers of the second kind.

The Poisson distribution's PGF $\Bbb E t^X=e^{-\mu}e^{t\mu}$ has $n$th derivative $\Bbb E\prod_{j=0}^{n-1}=\mu^n$ at $t=1$, i.e.$$\Bbb EX=\mu,\,\Bbb E(X^2-X)=\mu^2,\,\Bbb E(X^3-3X^2+2X)=\mu^3,\,\Bbb E(X^4-6X^3+11X^2-6X)=\mu^4.$$Before you start rearranging to compute each $\Bbb EX^n$ so you can ultimately compute$$\Bbb E(X^4-4\mu X^3+6\mu^2X^2-4\mu^3X+\mu^4),$$note the large-$\mu$ Normal approximation implies the kurtosis has $m\to\infty$ limit $3$, so the kurtosis must be of the form $3\mu^2+c\mu$ for some $c$ I leave you to find. Therefore, the coefficients of higher-order terms must cancel, so don't even bother tracking them (nor need you worry about the $\mu^2$ terms).