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(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?

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    The third central moment is $E[(X-\lambda)^3]=\lambda$. The third moment is given by your formula, which is correct. – Ian Dec 20 '14 at 14:28
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    Oh, I thought that was the same thing. I'm just learning about that stuff. Thank you for the hint. – NoBackingDown Dec 20 '14 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! – Sasha Dec 20 '14 at 17:22
  • @Sasha I did so, but I'll have to wait for 2 days to be able to accept it. – NoBackingDown Dec 20 '14 at 17:26

1 Answers1

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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}

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