Question

lf $m$ is the variance of a Poisson Distribution, then the sum of the terms in odd places is:

• A. $e^{-m}$
• B. $e^{-m}\cosh m$
• C. $e^{-m}\sinh m$
• D. $e^{-m}\coth m$

Answer 1

The correct option is B $e^{-m}\cosh m$

If $m$ is the variance of P. D, then $P(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$
Sum of the terms in odd places $=$ $\left[P\right(0;\mu )+P(2;\mu )+P(4;\mu )+...]$

$=[$$\frac{e^{-\mu }{\mu }^0}{0!}+\frac{e^{-\mu }{\mu }^2}{2!}+\frac{e^{-\mu }{\mu }^4}{4!}+...]$

$=$ $e^{-\mu }[1+\frac{{\mu }^2}{2!}+\frac{{\mu }^4}{4!}+....]$ ----- (1)

Since $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$
$e^{-x}=1+\frac{(-x)}{1!}+\frac{{(-x)}^2}{2!}+\frac{{(-x)}^3}{3!}+\frac{{(-x)}^4}{4!}+....$

on adding , we get
$e^x+e^{-x}=2[1+\frac{x^2}{2!}+\frac{x^4}{4!}+.....]$
$\frac{e^x+e^{-x}}{2}=[1+\frac{x^2}{2!}+\frac{x^4}{4!}+.....]$
So,
$\frac{e^{\mu }+e^{-\mu }}{2}=[1+\frac{{\mu }^2}{2!}+\frac{{\mu }^4}{4!}+.....]$
put this value in equation (1),
Sum of the terms in odd places $=$ $e^{-\mu }[1+\frac{{\mu }^2}{2!}+\frac{{\mu }^4}{4!}+.......]$
$=$ $e^{-\mu }\left(\frac{e^{\mu }+e^{-\mu }}{2}\right)$
$=$ $e^{-\mu }\cosh \mu$
$=$ $e^{-m}\cosh m$ [Variance $\left(m\right)$ is equal to mean $\left(\mu \right)$ in Poisson distribution]