Question

A financial analyst has found out that, policyholders are four times as likely to file two
claims as to file four claims. If the number of claims led has a Poisson distribution,
what is the expected value of the number of claims led?

Answer 1

Let $X=$ the number of claims: $X\sim Po(\lambda)$

$$P(X=x)={e^{-\lambda}\cdot \lambda^x\over x!}, \lambda>0$$

Given $P(X=2)=4P(X=4)$

$$P(X=2)={e^{-\lambda}\cdot \lambda^2\over 2!}=4P(X=4)=4{e^{-\lambda}\cdot \lambda^4\over 4!}$$

We want to find $\lambda$ for the Poisson RV

$${ \lambda^2\over 2}=4\cdot{ \lambda^4\over 24}, \lambda>0$$

$$\lambda^2=3, \lambda>0$$

So $\lambda=\sqrt{3}.$

The expected value of the number of claims led is $\sqrt{3}.$