Q. 4.17 - A First Course in Probability

Question

Let X be a Poisson random variable with parameter λ.

•  (a) Show that$P\left\{X\text{ is even }\right\}=\frac{1}{2}\left[1+e^{-2\lambda }\right]$by using the result of Theoretical Exercise 4.15 and the relationship between Poisson and binomial random variables.
• (b) Verify the formula in part (a) directly by making use of the expansion of $e^{-\lambda }+e^{\lambda }$

Step-by-Step Solution

Answer

a. Show that $P\left\{X\text{ is even }\right\}=\frac{1}{2}\left[1+e^{-2\lambda }\right]=\frac{1}{2}\left(1+e^{-2\lambda }\right)\text{ As }n\to \mathrm{\infty }$

b. By making use of the expansion of  $e^{-\lambda }+e^{\lambda }$, to prove$\frac{e^{-\lambda }\left(e^{\lambda }+e^{-\lambda }\right)}{2}=\frac{1}{2}\left(1+e^{-2\lambda }\right).$

Step 1: Given Information (Part-a)

Given in the question that, Let $X$be a Poisson random variable with parameter$\lambda .$And also to show that$P\left\{X\text{ is even }\right\}=\frac{1}{2}\left[1+e^{-2\lambda }\right]$

Step 2: Poisson distribution is a limiting case of Binomial distribution (Part-a)

$\mathrm{P}\left[\text{ even heads }\right]=\frac{1}{2}\left[1+(q-p{)}^n\right]$

$=\frac{1}{2}\left[1+(1-p-p{)}^n\right]$

$=\frac{1}{2}\left[1+(1-2p{)}^n\right]----------\left(1\right)$

Poisson distribution is a limiting case of Binomial distribution under the following conditions

$\text{ (i) }\mathrm{n}\to \mathrm{\infty }$,$\text{ (ii) }\mathrm{p}\to \mathrm{\infty }$,$\text{ (iii) }\mathrm{np}\to \lambda \text{ (finite) }$

$\Rightarrow \mathrm{p}=\frac{\lambda }{\mathrm{n}}$

Substitute $\mathrm{p}=\frac{\lambda }{\mathrm{n}}\mathrm{in}\left(1\right)$

$\Rightarrow P\left[\text{ Even Heads }\right]=\frac{1}{2}\left[1+{\left(1-2\frac{\lambda }{n}\right)}^n\right]$

$=\frac{1}{2}\left(1+{\mathrm{e}}^{-2\lambda }\right)\text{ As }n\to \mathrm{\infty }$.

Step 3: Final Answer (Part-a)

We prove that $P\left\{X\text{ is even }\right\}=\frac{1}{2}\left[1+e^{-2\lambda }\right]=\frac{1}{2}\left(1+e^{-2\lambda }\right)\text{ As }n\to \mathrm{\infty }.$

Step 4: Given Information (Part-b)

Given in the question that, Let $X$be a Poisson random variable with parameter$\lambda$ and$e^{-\lambda }+e^{\lambda }.$

Step 5: Expansion of the Equation (Part-b)

Now, we have to prove that 

$\frac{e^{-\lambda }\left(e^{\lambda }+e^{-\lambda }\right)}{2}=\frac{1}{2}\left(1+{\mathrm{e}}^{-2\lambda }\right)$

We have,

$e^{\lambda }=1+\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}+....$

$e^{-\lambda }=1-\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}-\frac{{\lambda }^3}{3!}+.....$

$e^{-\lambda }+e^{\lambda }=\left[1-\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}-\frac{{\lambda }^3}{3!}+\dots \dots \dots \dots \dots \dots ...\left[1+\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}+\dots \dots \dots \dots \dots \dots ...\right]\right.$

$\left.=\left[2+2\cdot \frac{{\lambda }^2}{2!}+2\cdot \frac{{\lambda }^4}{4!}+\dots \dots \dots .\dots .\dots ..\right]\right]$

$=2\left[1+\frac{{\lambda }^2}{2!}+\frac{{\lambda }^4}{4!}+\dots \dots \dots ..\right].$

Step 6: Prove the Equation (Part-b)

Now we get,

$\frac{e^{-\lambda }\left(e^{\lambda }+e^{-\lambda }\right)}{2}=\frac{1}{2}\left(1-\frac{\lambda }{1!}+\frac{{\lambda }^2}{2!}-\frac{{\lambda }^3}{3!}+\dots \dots \right)2\left[1+\frac{{\lambda }^2}{2!}+\frac{{\lambda }^4}{4!}+\dots \dots \dots \dots \right]$

$=\left[1-\lambda +{\lambda }^2-\frac{4{\lambda }^3}{6}+\dots \dots \dots \dots \dots \dots \dots .\cdot \right]$

$\frac{1}{2}\left(1+{\mathrm{e}}^{-2\lambda }\right)=\frac{1}{2}\left[1+1+\frac{(-2\lambda )}{1!}+\frac{(-2\lambda {)}^2}{2!}+\frac{(-2\lambda {)}^3}{3!}+\dots \dots \dots \dots \dots \dots \dots ..\right]$

$=\frac{1}{2}\left[2-2\lambda +\frac{4{\lambda }^2}{2!}-\frac{8{\lambda }^3}{6}+\dots \right.$

$=\left[1-\lambda +{\lambda }^2-\frac{4{\lambda }^3}{6}+\dots \dots \dots \dots \dots \dots \dots ..\right]$

So,$\frac{e^{-\lambda }\left(e^{\lambda }+e^{-\lambda }\right)}{2}=\frac{1}{2}\left(1+e^{-2\lambda }\right)$.

Step 7: Final Answer (Part-b)

By making use of the expansion of$e^{-\lambda }+e^{\lambda }$, to prove

$\frac{e^{-\lambda }\left(e^{\lambda }+e^{-\lambda }\right)}{2}=\frac{1}{2}\left(1+e^{-2\lambda }\right).$