Question

Assume that a policyholder is four times more likely to file exactly two claims as to file exactly three claims. Assume also that the number X of claims of this policyholder is Poisson. Determine the expectation E(X²).

Answer 1

1.3125

Explanation:

Given that our random variable
`X` follows a Poisson distribution
`P(X=k)=(\lambda^k e^-^p)/(k!) \ \ \ \ \ \ \ \ p=\lambda`

Evaluate the formula at
`k=2,3:`

`P(X=2)=0.5\lambda^2e^(-\lambda)\\\\P(X=3)=(1)/(6)\lambda^2e^(-\lambda)`

#since

`4P(X=2)=P(X=3);\\\\0.5\lambda^2e^(-\lambda)=4*(1)/(6)\lambda^3e^(-\lambda)\\\\0.5\lambda^2=(2)/(3)\lambda^3\\\\0.75=\lambda`

The mean and variance of the Poisson distributed random variable is equal to
`\lambda`:

`\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75`

#By property variance:

`\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125`

The expectation is 1.3125