Q: In a group of policies, the monthly number of claims for a single policy has a Poisson distribution with parameter λ, where λ is a random variable with the following density: f(λ) = 2 exp { -2 λ } ; λ > 0 Find the moment generating function for the aggregate claims distribution if the claims have a gamma distribution with mean 2 and variance 2. The solution uses a method for calculation which is greek to me (see attachment). a) Can I use Ms(t) = Mn (log Mx(t)) ? Please explain steps to solve if YES. b) If NO , Why ? c) Where to find the method used for solving (as in attachment) in the core reading material ?
You could use the formula. The challenge here is what is the MGF of the number of claims given that f(n) = 2/[3^(n+1)] for n = 0, 1, 2, .... That is what the answer is trying to avoid calculating. This is not dealt with directly in the notes given that this is simply about mathematical manipulation of MGFs and PGFs as covered in CT3.
Explanation of Solution Ms(t) = E[e^(St)] = E{E[e^(St) |N]} = E[{E[e^(Xt)]}^N] This is the first steps in the acted notes on the MGF of an aggregate model. = Gn[Mx(t)] This follows because from the PGF formula Since {E[e^(Xt)]}^N = {Mx(t)}^N If we substitute Mx(t) by z E[{E[e^(Xt)]}^N] = E(z^N) resembling the formula for a PGF Hence E[{E[e^(Xt)]}^N] = Gn[Mx(t)] The PGF route was chosen because it is easier to calculate than the MGF The rest of it is just mathematical manipulation to get the PGF for N and then eventually the MGF of the aggregate model.
The first part is easy. The second part stumped me. They have used: Gx(t) = E[t^X] = Summation { t^X . P[N=n]} instead of Ms(t) = Mn(t) ...... Thanks for your reply.
