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Please help on how would we solve the summation while deriving mean deviation about mean for Poisson Distribution.

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  • $\begingroup$ Is this the "mean absolute deviation" or the "mean square deviation"? $\endgroup$
    – mlc
    Apr 29 '17 at 11:12
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For the absolute mean deviation of $X \sim Poi(\lambda)$,

$$\mathbf{E}_\lambda|X - \lambda| = 2\frac{e^{-\lambda}\lambda^{[\lambda]+1}}{[\lambda]!} = 2 \lambda \mathbf{P}_\lambda(X = [\lambda])$$

where $[\lambda]$ is the integer part of $\lambda$. Here is a quick reference for this kind of results, and here is the original proof if you are interested.

Ramasubban, T. A. "The mean difference and the mean deviation of some discontinuous distributions." Biometrika 45.3/4 (1958): 549-556.

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