Question

If $X$ is a Poisson variate such that $P(X=2)=9P(X=4)$ then mean and variance of $X$ are

• A. $1,1$
• B. $2,2$
• C. $\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$
• D. $\frac{2}{3},\frac{2}{3}$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Poisson's distribution implies
$P(X=n)=\frac{{\lambda }^n{e}^{-\lambda }}{n!}$
Where mean=variance=$\lambda$
Hence $P(x=2)=9P(x=4)$
$9\frac{{\lambda }^4{e}^{-\lambda }}{4!}=\frac{{\lambda }^2{e}^{-\lambda }}{2!}$
$9{\lambda }^2=12$
${\lambda }^2=\frac{4}{3}$

$\lambda =\frac{2}{\sqrt{3}}$
Hence mean $=$ variance $=$ $\frac{2}{\sqrt{3}}$