# Derivation of the third moment of Poisson distribution using Stein-Chen identity

(a) Use LOTUS to show that for $X \sim \operatorname{Pois}(\lambda)$ and any function g, $E(Xg(X)) = λE(g(X + 1))$. This is called the Stein-Chen identity for the Poisson.

(b) Find the third moment $E(X^3)$ for $X \sim \operatorname{Pois}(\lambda)$ by using the identity from (a) and a bit of algebra to reduce the calculation to the fact that $X$ has mean $\lambda$ and variance $\lambda$.

Only part b) is concerned. My solution

Let $g(X) = X^2$, then

\begin{align} E(X^3) &= \lambda E(g(X+1)) \\ &= \lambda E((X+1)^2) \\ &= \lambda (E(X^2) + 2E(X) + 1) \\ &= \lambda (\lambda+\lambda^2 + 2\lambda + 1)\\ &= \lambda^3 + 3\lambda^2 + \lambda \end{align}

However, from litarature I know that the third moment should be $\lambda$. What went wrong?

• The third central moment is $E[(X-\lambda)^3]=\lambda$. The third moment is given by your formula, which is correct.
– Ian
Commented Dec 20, 2014 at 14:28
• Oh, I thought that was the same thing. I'm just learning about that stuff. Thank you for the hint. Commented Dec 20, 2014 at 14:48
• @Dominik To help keep down the number of unanswered questions, could you please paste your correct solution down as an answer, and accept it. Thank you! Commented Dec 20, 2014 at 17:22
• @Sasha I did so, but I'll have to wait for 2 days to be able to accept it. Commented Dec 20, 2014 at 17:26

Let $g(X) = X^2$, then