Question

If $m$ is the variance of Poisson distribution, then sum of the terms in even 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 C $e^{-m}\sinh m$

If m is the variance of P. D, then P$(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$
Sum of the terms in even places = [ P$(1;\mu )+P(3;\mu )+P(5;\mu )+.........$ ]

= [$\frac{e^{-\mu }{\mu }^1}{1!}+\frac{e^{-\mu }{\mu }^3}{3!}+\frac{e^{-\mu }{\mu }^5}{5!}+.......$]

= $e^{-\mu }[\mu +\frac{{\mu }^3}{3!}+\frac{{\mu }^5}{5!}+.......]$ --------------------- (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 subtracting , we get
$e^x-e^{-x}=2[x+\frac{x^3}{3!}+\frac{x^5}{5!}+.....]$
$\frac{e^x-e^{-x}}{2}=[x+\frac{x^3}{3!}+\frac{x^5}{5!}+.....]$
So,
$\frac{e^{\mu }-e^{-\mu }}{2}=[x+\frac{{\mu }^3}{3!}+\frac{{\mu }^5}{5!}+.....]$
put this value in equation (1),
Sum of the terms in even places = $e^{-\mu }[\mu +\frac{{\mu }^3}{3!}+\frac{{\mu }^5}{5!}+.......]$
= $e^{-\mu }\left(\frac{e^{\mu }-e^{-\mu }}{2}\right)$
= $e^{-\mu }\sinh \mu$
= $e^{-m}\sinh m$ [ Variance (m) is equal to mean ($\mu$) in Poisson distribution ]