91Ó°ÊÓ

Given in the question that, Let $X$be a Poisson random variable with parameter$\mathrm{λ}.$And also to show thatlocalid="1647097128770" $P\left\{X\text{is even}\right\}=\frac{1}{2}\left[1+e^{−2\mathrm{λ}}\right]$

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

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

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

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

$\text{(i)}\mathrm{n}→\mathrm{∞}$,$\text{(ii)}\mathrm{p}→\mathrm{∞}$,$\text{(iii)}\mathrm{np}→\mathrm{λ}\text{(finite)}$

$⇒\mathrm{p}=\frac{\mathrm{λ}}{\mathrm{n}}$

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

$⇒P\left[\text{Even Heads}\right]=\frac{1}{2}\left[1+{\left(1−2\frac{\mathrm{λ}}{n}\right)}^n\right]$

$=\frac{1}{2}\left(1+{\mathrm{e}}^{−2\mathrm{λ}}\right)\text{As}n→\mathrm{∞}$.

We prove that$P\left\{X\text{is even}\right\}=\frac{1}{2}\left[1+e^{−2\mathrm{λ}}\right]=\frac{1}{2}\left(1+e^{−2\mathrm{λ}}\right)\text{As}n→\mathrm{∞}.$

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

Now, we have to prove that

We have,

$e^{−\mathrm{λ}}=1−\frac{\mathrm{λ}}{1!}+\frac{{\mathrm{λ}}^2}{2!}−\frac{{\mathrm{λ}}^3}{3!}+.....$

$e^{−\mathrm{λ}}+e^{\mathrm{λ}}=\left[1−\frac{\mathrm{λ}}{1!}+\frac{{\mathrm{λ}}^2}{2!}−\frac{{\mathrm{λ}}^3}{3!}+………………...\left[1+\frac{\mathrm{λ}}{1!}+\frac{{\mathrm{λ}}^2}{2!}+\frac{{\mathrm{λ}}^3}{3!}+………………...\right]\right.$

$\left.=\left[2+2â‹\dots \frac{{\mathrm{λ}}^2}{2!}+2â‹\dots \frac{{\mathrm{λ}}^4}{4!}+……….….…..\right]\right]$

$=2\left[1+\frac{{\mathrm{λ}}^2}{2!}+\frac{{\mathrm{λ}}^4}{4!}+………..\right].$

Now we get,

$\frac{e^{−\mathrm{λ}}\left(e^{\mathrm{λ}}+e^{−\mathrm{λ}}\right)}{2}=\frac{1}{2}\left(1−\frac{\mathrm{λ}}{1!}+\frac{{\mathrm{λ}}^2}{2!}−\frac{{\mathrm{λ}}^3}{3!}+……\right)2\left[1+\frac{{\mathrm{λ}}^2}{2!}+\frac{{\mathrm{λ}}^4}{4!}+…………\right]$

$=\left[1−\mathrm{λ}+{\mathrm{λ}}^2−\frac{4{\mathrm{λ}}^3}{6}+………………….â‹\dots \right]$

$\frac{1}{2}\left(1+{\mathrm{e}}^{−2\mathrm{λ}}\right)=\frac{1}{2}\left[1+1+\frac{(−2\mathrm{λ})}{1!}+\frac{(−2\mathrm{λ}{)}^2}{2!}+\frac{(−2\mathrm{λ}{)}^3}{3!}+…………………..\right]$

$=\frac{1}{2}\left[2−2\mathrm{λ}+\frac{4{\mathrm{λ}}^2}{2!}−\frac{8{\mathrm{λ}}^3}{6}+…\right.$

$=\left[1−\mathrm{λ}+{\mathrm{λ}}^2−\frac{4{\mathrm{λ}}^3}{6}+…………………..\right]$

So,$\frac{e^{−\mathrm{λ}}\left(e^{\mathrm{λ}}+e^{−\mathrm{λ}}\right)}{2}=\frac{1}{2}\left(1+e^{−2\mathrm{λ}}\right)$.

By making use of the expansion of$e^{−\mathrm{λ}}+e^{\mathrm{λ}}$, to prove

$\frac{e^{−\mathrm{λ}}\left(e^{\mathrm{λ}}+e^{−\mathrm{λ}}\right)}{2}=\frac{1}{2}\left(1+e^{−2\mathrm{λ}}\right).$