Second moment of Poisson distribution

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I know that the second moment of Poisson distribution is equal to square of it's rate parameter, and in general it's true for other moments. i can prove it algebraic by Taylor expansion but i'm looking for a intuitive proof. is there any combinatorial consideration at least for second moment?

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1 Answer

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It's not true. It's true for falling moments, i.e. $E[X(X-1)...(X-i+1)]=\mu^i$ and there is a combinatorial proof:

Hint

Prove $$E{X \choose i}=\frac{\mu^i}{i!}$$ instead of the original equation.

Use linearity of expectation.

Consider that you divide $[0,1]$ in $n$ parts of size $\frac{1}{n}$ and select each part with probability $\frac{\mu}{n}$. (just like Poisson derivation from Binomial distribution.)

So in general there would be $n \choose i$ sets which is selected with probability $\mu^i$.

You can achieve the desired result by a limitation.

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