In a Poisson distribution with unit mean,calc | vedclass

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(C) For a Poisson distribution with unit mean,$\bar{x} = 1$. The probability mass function is $P(X=x) = \frac{e^{-1}}{x!}$.We need to calculate $\sum_

Vedclass · Accepted Answer

$\frac{2}{e}$

Vedclass · Answer

(C) For a Poisson distribution with unit mean,$\bar{x} = 1$. The probability mass function is $P(X=x) = \frac{e^{-1}}{x!}$.We need to calculate $\sum_{x=0}^{\infty} |x-1| \frac{e^{-1}}{x!}$.$= \frac{1}{e} \left[ |0-1| \frac{1}{0!} + |1-1| \frac{1}{1!} + |2-1| \frac{1}{2!} + |3-1| \frac{1}{3!} + \dots \right]$$= \frac{1}{e} \left[ 1 + 0 + \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots \right]$$= \frac{1}{e} \left[ 1 + \sum_{x=2}^{\infty} \frac{x-1}{x!} \right] = \frac{1}{e} \left[ 1 + \sum_{x=2}^{\infty} \frac{1}{(x-1)!} - \sum_{x=2}^{\infty} \frac{1}{x!} \right]$$= \frac{1}{e} \left[ 1 + (e-1) - (e-1-1) \right] = \frac{1}{e} [1 + e - 1 - e + 2] = \frac{2}{e}$.Thus,the correct option is $C$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

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

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