5.3 Practice Problems
Problem 5.3.1. The number of claims \(N\) in a month is Poisson with mean \(4\), and claim sizes are independent of \(N\) and of each other, each exponential with mean \(500\). Find the mean and variance of the monthly total, and state what proportion of the variance is due to uncertainty in the number of claims.
Show solution
Solution. For the exponential with mean \(\mu =500\), \(\sigma =500\), so \(E\left (X^{2}\right ) = \sigma ^{2}+\mu ^{2} = 500{,}000\).
By the compound Poisson results, \[E(S) = \lambda \mu = 4(500) = 2{,}000 ,\] \[\Var (S) = \lambda E\left (X^{2}\right ) = 4(500{,}000) = 2{,}000{,}000 ,\] so the standard deviation is \(\$1{,}414\).
Splitting the general formula, \[\Var (S) = \underbrace {\lambda \sigma ^{2}}_{1{,}000{,}000} + \underbrace {\Var (N)\mu ^{2}}_{1{,}000{,}000},\] so exactly half the variance comes from not knowing how many claims there will be. The two halves are equal here because the exponential has \(\sigma =\mu \); for a less variable claim size the count term would dominate still further.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.