Hello, I was looking through the materials and I saw that the question on page 14 chapter 20 had claims to be a normal distribution iwth 0.7p and 2.0p. However when they derived the distribution of S(1) in particular the variance is 100*(2*5000)^2. How did they derive this? The above result doesn't match the formula where Var(S) = lambda*m_2 where m_2 is the second moment of the claims distribution? Warm regards, Brian
The formula you've given for Var(S) is only valid when the claims arise according to a Poisson process. In this question the total claims at the end of year 1 is normally distributed, without any reference to the number of claims arising. The 100*(2*5,000)^2 comes from (number of policies sold) * (standard deviation of total claim amount)^2.
Thanks for your reply Steve, I had an additional question in section 6.2 following that, when considering the claims below and above e, how can E[e^(rX)|X>e] be equal to e^(re)?
Oh, I see! It's not the case that E[e^(rX)|X>e] and e^(re) are equal, the inequality sign between them is important. Notice that MGF(r) = E[exp(r*X)], therefore E[exp(r*X) | X>epsilon] > MGF(r). This is because the expectation given that X is greater than epsilon must be greater than the expectation which includes the smaller values of X. The Core Reading has already established that MGF(r) >= exp(r*epsilon)*pi, so this leads to: E[exp(r*X) | X>epsilon] > MGF(r) >= exp(r*epsilon)*pi as required.
