If the mean of a Poisson variate X is 1,then | vedclass

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(C) Given that the mean of a Poisson variate $X$ is $\lambda = 1$. The probability mass function is $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \fra

Vedclass · Accepted Answer

$\frac{2}{e}$

Vedclass · Answer

(C) Given that the mean of a Poisson variate $X$ is $\lambda = 1$. The probability mass function is $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \frac{e^{-1}}{r!}$. We need to calculate $\sum_{r=0}^{\infty} |r-1| P(X=r)$. $\sum_{r=0}^{\infty} |r-1| \frac{e^{-1}}{r!} = e^{-1} \left[ |0-1| \frac{1}{0!} + |1-1| \frac{1}{1!} + |2-1| \frac{1}{2!} + |3-1| \frac{1}{3!} + \dots \right]$ $= e^{-1} \left[ 1 \cdot 1 + 0 \cdot 1 + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{6} + \dots \right]$ $= e^{-1} \left[ 1 + 0 + \frac{1}{2} + \frac{1}{3} + \dots \right]$ Note that $\sum_{r=0}^{\infty} |r-1| P(X=r) = E[|X-1|]$. For a Poisson distribution with $\lambda=1$,$E[|X-1|] = \sum_{r=0}^{\infty} |r-1| \frac{e^{-1}}{r!} = e^{-1} \left( \sum_{r=0}^{\infty} r \frac{1}{r!} - \sum_{r=0}^{\infty} \frac{1}{r!} + P(X=0) \right)$ is not the direct path. Calculating the sum: $e^{-1} [ |0-1| + |1-1|\frac{1}{1!} + |2-1|\frac{1}{2!} + |3-1|\frac{1}{3!} + \dots ] = e^{-1} [ 1 + 0 + \frac{1}{2} + \frac{2}{6} + \dots ] = e^{-1} [ 1 + 0 + 0.5 + 0.333 + \dots ] = \frac{2}{e}$. Thus,the correct option is $C$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

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

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