In a poisson distribution with unit mean, $\sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=(\bar{x}$ is the mean of the distribution)

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Aicharya · Accepted Answer

For the unit mean, $\bar{x}=1$ $\begin{aligned} & \text {so } \sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=\sum_{x=0}^{\infty}|x-1| \frac{e^{-1}}{x !} \ & =\frac{1}{e}\left[\frac{1}{0 !}+\sum_{x=1}^{\infty} \frac{x-1}{x !}\right]=\frac{1}{e}\left[1+\sum_{x=1}^{\infty} \frac{1}{(x-1) !}-\sum_{x=1}^{\infty} \frac{1}{x !}\right] \end{aligned}$ $=\frac{1}{e}[1+e-(e-1)]=\frac{1}{e}(1+e-e+1)=\frac{2}{e}$ Hence, option (3) is correct.

In a poisson distribution with unit mean, \(\sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=(\bar{x}\) is the mean of…

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