Expectation of Poisson variate

Question

How do I prove that $E[xe^{-kx}] = λe^{-λ\left(1-e^{-k}\right) - k}$

where $x$ is a Poisson variate.

I know that if $x$ is a Poisson variate with mean $m$, then $E[e^{-px}] = e^{-m\left(1-e^{-p}\right)}$ but how did we get this identity?

Answer 1

We have \begin{align*}E(xe^{-kx}) &= \sum_{t=0}^{\infty}te^{-kt}P(x=t) \\ &= \sum_{t=0}^{\infty}te^{-kt}\frac{e^{-\lambda}\lambda^t}{t!} \\ &= \sum_{t=1}^{\infty}\frac{te^{-kt-\lambda}\lambda^t}{t!} \qquad \text{(as the $t=0$ term is $0$)} \\ &= \lambda e^{-\lambda}\sum_{t=1}^{\infty}\frac{e^{-kt}\lambda^{t-1}}{(t-1)!} \\ &= \lambda e^{-\lambda-k}\sum_{t=0}^{\infty}\frac{e^{-kt}\lambda^t}{t!} \qquad \text{(reindexing by $t \mapsto t+1$)} \\ &= \lambda e^{-\lambda-k}\sum_{t=0}^{\infty}\frac{\left(\lambda e^{-k}\right)^t}{t!} \\ &= \lambda e^{-\lambda-k}e^{\lambda e^{-k}} \qquad \text{(by the Taylor series for $e^x$)} \\ &= \lambda e^{-\lambda\left(1-e^{-k}\right)-k}\end{align*} as required. $\qquad \rule{0.7em}{0.7em}$