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Random variable $\xi$ has Poisson distribution with parameter $\lambda$. Compute $E(\xi \lambda^\xi)$

Poisson distribution is: $P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda}$

Then we can compute:

$$E(\xi \lambda^\xi) = \sum_{k=0}^\infty k\; P[k=\xi]=\sum_{k=0}^\infty \frac{k\, \lambda^k \, \lambda^k \, e^{-\lambda}}{k!} = e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^{2k}}{(k-1)!}$$

What can I do with the sum?

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    You have run into a common, mildly annoying problem in algebraic manipulation of factorials: $k/k!$ is $(k-1)!$ if $k \geq 1$ but it is just $0$ if $k=0$. You need to change indices to fix that. Once you've done that, you will have a multiple of $\sum_{k=0}^\infty \lambda^{2k}/k!$. You should recognize that sum... – 2017-01-12
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    Ok, if I change indices, it should be this: $e^{-\lambda}(\sum_{k=1}^{\infty}\frac{\lambda^{2k}}{(k-1)!} + 1)$ right? – 2017-01-12
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    I cannot recognize it right now, but it is familiar to me. I will try to remember. – 2017-01-12
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    Typo: I meant $1/(k-1)!$. And no: the sum (without the $e^{-\lambda}$) is $\sum_{k=1}^\infty \lambda^{2k}/(k-1)!$. (The $k=0$ term is $0$, not $1$.) Now that you are starting at $k=1$ you can change indices to start at $k=0$ again. This makes the denominator be $k!$, but what does it do to the exponent in the numerator? – 2017-01-12

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