If the mean of a poisson variate $X$ is 1 , then $\sum_{r=0}^{\infty}|r-1| P(X=r)=$

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Aicharya · Accepted Answer

It is given that mean of a poisson variate $X$ is $\lambda=\mathbf{l}$, $\begin{aligned} \because P(X & =r)=\frac{e^{-\lambda} \lambda^r}{r !}=\frac{e^{-1}}{r !} \ \therefore & \sum_{r=0}^{\infty}|r-1| P(X=r)=\sum_{r=0}^{\infty}|r-1| \frac{e^{-1}}{r !} \ & =e^{-1}\left[\frac{1}{0 !}+\frac{0}{1 !}+\frac{1}{2 !}+\frac{2}{3 !}+\ldots \ldots \ldots . .\right] \ & =e^{-1}\left(1+\frac{1}{2 !}+\frac{2}{3 !}+\ldots \ldots . . .\right)=e^{-1}(1+1)=\frac{2}{e} \end{aligned}$ Hence, option (c) correct.

If the mean of a poisson variate \(X\) is 1 , then \(\sum_{r=0}^{\infty}|r-1| P(X=r)=\)

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