TITLE: Derivation of the third moment of Poisson distribution using Stein-Chen identity
QUESTION [8 upvotes]: (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?

REPLY [8 votes]: I'm answering my own question to keep the number of unanswered questions low.
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}