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Let $X$ be a random variable with a Poisson distribution with mean $\lambda$. I am trying to show that $$ \mathbb{E}[\lambda f(X + 1) - X f(X)] = 0. $$

I have tried using linearity of expectation and writing out the pmf of the Poisson random variable, but I haven't been able to solve it. Could someone give me a hint?

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$$\mathbb E \left[\lambda f(X+1) - Xf(X)\right] = \sum_{k\ge 0} \frac{e^{-\lambda}\lambda^k}{k!}\left(\lambda f(k+1) - kf(k)\right) = \sum_{k\ge 0} \left(\frac{e^{-\lambda}\lambda^{k+1}}{k!}f(k+1) \right) - \sum_{k\ge 0} \left(\frac{e^{-\lambda}\lambda^{k}}{k!}kf(k) \right) = = \sum_{k\ge 1} \left(\frac{e^{-\lambda}\lambda^{k}}{(k-1)!}f(k) \right) - \sum_{k\ge 1} \left(\frac{e^{-\lambda}\lambda^{k}}{(k-1)!}f(k) \right) = 0$$

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  • $\begingroup$ Thanks, I didn't see that before. $\endgroup$ – Quanta Apr 16 '18 at 1:38

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