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If the mean of a poisson variate \(X\) is 1 , then \(\sum_{r=0}^{\infty}|r-1| P(X=r)=\)
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Verified Answer
The correct answer is:
\(\frac{2}{e}\)
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.
\(\begin{aligned}
\because P(X & =r)=\frac{e^{-\lambda} \lambda^r}{r !}=\frac{e^{-1}}{r !} \\
\therefore & \sum_{r=0}^{\infty}|r-1| P(X=r)=\sum_{r=0}^{\infty}|r-1| \frac{e^{-1}}{r !} \\
& =e^{-1}\left[\frac{1}{0 !}+\frac{0}{1 !}+\frac{1}{2 !}+\frac{2}{3 !}+\ldots \ldots \ldots . .\right] \\
& =e^{-1}\left(1+\frac{1}{2 !}+\frac{2}{3 !}+\ldots \ldots . . .\right)=e^{-1}(1+1)=\frac{2}{e}
\end{aligned}\)
Hence, option (c) correct.
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