Search any question & find its solution
Question:
Answered & Verified by Expert
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)
Options:
Solution:
1825 Upvotes
Verified Answer
The correct answer is:
\(\frac{2}{\mathrm{e}}\)
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.
\(\begin{aligned}
& \text {so } \sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=\sum_{x=0}^{\infty}|x-1| \frac{e^{-1}}{x !} \\
& =\frac{1}{e}\left[\frac{1}{0 !}+\sum_{x=1}^{\infty} \frac{x-1}{x !}\right]=\frac{1}{e}\left[1+\sum_{x=1}^{\infty} \frac{1}{(x-1) !}-\sum_{x=1}^{\infty} \frac{1}{x !}\right]
\end{aligned}\)
\(=\frac{1}{e}[1+e-(e-1)]=\frac{1}{e}(1+e-e+1)=\frac{2}{e}\)
Hence, option (3) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.