Question

If for a poisson variable $X$, $P(X=1)=2.P(X=2)$, then the parameter $\lambda$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

For Poisson's distribution, $P\left(X\right)$ is given by
$P\left(X\right)=\frac{e^{-\lambda }.{\lambda }^x}{x!}$
Hence $P(X=1)=\frac{e^{-\lambda }.{\lambda }^1}{1!}$.
Similarly $P(X=2)=\frac{e^{-\lambda }.{\lambda }^2}{2!}$.
Given that, $P(X=1)=2P(X=2)$
$\Rightarrow \frac{e^{-\lambda }.{\lambda }^1}{1!}=2.\frac{e^{-\lambda }.{\lambda }^2}{2!}$
Simplifying we get $\lambda =1$