Question

If m is the variance of P.D., then the ratio of sum of the terms in odd places to the sum of the terms in even places is

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

Answer 1

The correct option is C $\coth m$

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 = [ P$(0;\mu )+P(2;\mu )+P(4;\mu )+.........$ ]

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

= $e^{-\mu }[1+\frac{{\mu }^2}{2!}+\frac{{\mu }^4}{4!}+.......]$ --------------------- (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 (m) is equal to mean ($\mu$) in Poisson distribution ]
Similarly Sum of the terms in even places = $e^{-m}\sinh m$
The ratio of sum of the terms in odd places to the sum of the terms in even places is = $\frac{e^{-m}\cosh m}{e^{-m}\sinh m}=\coth m$