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

Answer:

1.3125

Step-by-step explanation:

Given that our random variable $X$ follows a Poisson distribution$P(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda$

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

$P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}$

#since

$4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda$

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