Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 35 / 4

Mix the conditional Poisson distribution over the rate-$\nu$ exponential distribution intensity. The gamma integral giveswhere $p=\nu /(1+\nu )$ and $q=1/(1+\nu )$. Thus the annual claim count has the zero-based geometric distribution, with probabilities $\mathbb{P}(N=0)=p$, $\mathbb{P}(N=1)=pq$ and $\mathbb{P}(N\ge 2)=q^2$.

For the intended Markov chain calculation, take these annual counts to be independent. Order the states by discounts $0,\alpha ,\beta$. From either of the first two states, a positive count returns the policy to zero, while a zero count raises it by one level. From the top, two or more claims send it to zero, exactly one sends it to $\alpha$, and no claim keeps it at $\beta$. Hence the transition matrix isWrite its stationary distribution as $({\pi }_0,{\pi }_{\alpha },{\pi }_{\beta })$. The final two balance equations give $q{\pi }_{\beta }=p{\pi }_{\alpha }$ and ${\pi }_{\alpha }=p{\pi }_0+pq{\pi }_{\beta }$. It follows that ${\pi }_{\alpha }=p{\pi }_0/[q(1+p)]$ and ${\pi }_{\beta }=p^2{\pi }_0/[q^2(1+p)]$. Normalizing yields the three-level discount equilibrium with geometric annual counts:The Markov chain is irreducible for $0<p,q<1$ and has a self-loop, so it is aperiodic and this equilibrium is the limiting distribution. Weighting the three premiums $c,c(1-\alpha ),c(1-\beta )$ givesThere is a necessary qualification to the temporal model: the one-year mixture alone does not specify independence between years. A persistent claim intensity can destroy the discount Markov property. If one fixed $\Lambda$ is drawn for the policyholder and annual counts are conditionally independent given it, then histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the top after three years but give different next-year no-claim probabilities. The first history is three zero counts, so its posterior intensity has exponential rate $\nu +3$ and next no-claim probability $(\nu +3)/(\nu +4)$. The second is a positive count followed by two zero counts; its posterior density is proportional to $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$, giving next no-claim probability $(\nu +2)/(\nu +4)$. Thus the displayed matrix and premium require independent annual marginal sampling, for example a fresh intensity each year. They are not valid for a persistent unobserved policy intensity without conditioning on that intensity.

Under the persistent-intensity interpretation, one can instead condition on $\lambda$. Put $a_0=e^{-\lambda }$ and $b_1=\lambda e^{-\lambda }$. The conditional transition matrix has rows $(1-a_0,a_0,0)$, $(1-a_0,0,a_0)$ and $(1-a_0-b_1,b_1,a_0)$. Its stationary distribution isThe corresponding population limiting expected value of the premium is $c{\int }_0^{\infty }[1-\alpha {\pi }_{\alpha }(\lambda )-\beta {\pi }_{\beta }(\lambda )]\nu e^{-\nu \lambda }d\lambda$. This gives the correct alternative when the intensity persists, and makes the extra assumption behind the requested marginal matrix explicit.

Compare future premium differences, since the current premium is already fixed. From the highest discount, making no claim keeps all future premiums at $c(1-\beta )$. One submitted claim raises next year's premium to $c(1-\alpha )$; if there are no further events, the year after returns to $c(1-\beta )$. The extra premium is therefore $c(\beta -\alpha )$. Claiming saves payment of the loss itself, so the first claim is advantageous precisely when $L_1>c(\beta -\alpha )$. For the unit-rate exponential distribution loss,Now condition on the first claim already having been submitted. If the second loss is not claimed, next year's discount is $\alpha$ and the following year's is $\beta$. Claiming the second loss instead makes these two discounts $0$ and $\alpha$, after which both paths are back at $\beta$. Thus the additional premium caused by this second claim isThis is the incremental second-claim threshold at the highest discount level: compare the second decision to the already-chosen one-claim path, not to a no-claim path. The second claim is advantageous exactly when $L_2>c\beta$. The losses are independent, so conditioning on the first claim does not change the exponential distribution of $L_2$. HenceEquality at a threshold has probability zero. The infinitely many premiums common to both paths cancel; all nonzero differences occur in the next two years, so no divergent sum is used. These thresholds use the undiscounted comparison in the question, with no interest rate or deductible introduced.