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The problem:

Let X have the Poisson distribution with parameter $\lambda$, where $\lambda >0$.

Calculate E[X!] for every possible value of $\lambda$,

So what I got so far:

$$E[X!] = \sum_0^\infty\frac{i!\cdot e^{-\lambda}\cdot\lambda^i}{i!} = e^{-\lambda} \sum_0^\infty\lambda^i=\frac{1}{e^\lambda\cdot(1-\lambda)}$$

Is this correct?

Help will be much appreciated.


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Well it holds for $\lambda < 1$ I believe... – homiee Apr 28 '13 at 11:11
yes that's correct – roger Apr 28 '13 at 11:17

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