Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 35 (Actuarial Statistics)

Condition on the claim count. The empty random sum of independent claims is zero, and independence gives $\mathbb{E}[e^{tS}\mid N=n]=M_X(t{)}^n$. Therefore the random-sum transform identity iswherever the composition is finite. This follows directly from the law of total expectation, rather than replacing the random count by its mean.

Put $q=1-p$. For the specified negative binomial distribution, the negative binomial series gives $G_N(z)=[p/(1-qz){]}^k$. A rate-$\lambda$ exponential distribution has moment-generating function $M_X(t)=\lambda /(\lambda -t)$. HenceThe last expression is exactly the moment-generating function of a compound binomial distribution with independent count and severitiesThe moment-generating function exists in a neighborhood of zero, so its uniqueness identifies the distributions. In particular, the new claim sizes have mean $1/(p\lambda )$, not $1/\lambda$, and both aggregate distributions have an atom $p^k$ at zero. This is the negative-binomial sum of exponential claims representation with the severity rate explicitly retained.

Conditional on $N=n\ge 1$, the aggregate has Erlang distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$ on $s>0$. Repeated integration by parts of its tail givesIndeed, after substituting $v=\lambda s$, the integral ${\int }_{\lambda x}^{\infty }{e}^{-v}{v}^{n-1}dv/(n-1)!$ reduces recursively to the corresponding integral with exponent $n-2$, terminating at $e^{-\lambda x}$. Applying the law of total probability yields the original infinite mixture distribution tailApplying the same argument to the compound binomial distribution gives instead the finite Erlang-mixture tail for a negative-binomial exponential aggregate:The two tails are equal. The binomial representation replaces an infinite sum by a finite exact calculation, eliminating the need to choose a claim-count truncation and control its omitted tail. The inner sums can be evaluated recursively, using $a_{j+1}=a_j(p\lambda x)/(j+1)$ for their exponential-weighted terms.

Write the surplus in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with premium rate $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the positive solution ofFor the exponential distribution of mean $\mu$, $M_X(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. Cancelling the nonzero root $R$ in its equation gives $\lambda \mu /(1-\mu R)=c$, and henceIt is strictly between zero and $1/\mu$, as required by the transform domain.

The ultimate ruin probability is $\psi (u)=\mathbb{P}({\inf }_{t\ge 0}{U}_t<0)$. The Lundberg inequality statesRuin can occur only at a claim arrival. Thus the event defining the finite-claim ruin probability ${\psi }_n(u)$ is contained in the ultimate ruin event, giving ${\psi }_n(u)\le \psi (u)\le e^{-Ru}$ for every $n\ge 1$ and $u>0$.

For the explicit calculations, take $\mu =1$, put $d=1+\rho =c/\lambda$ and $h=d+1=2+\rho$, and let $T$ be the first arrival time. It has rate-$\lambda$ exponential distribution, independent of the severity. Ruin on the first claim is the event $X_1>u+cT$, soFor ruin by the second claim, condition on whether the first claim ruins the insurer or leaves capital $u+cT-x$. The independent future arrivals and severities have the same law as a fresh classical risk model. ThereforeThe two terms in the first line refer to disjoint ruin events, so there is no double counting.

For a direct check of the Lundberg inequality, now $R=1-1/d$. The ratio ${\psi }_1(u)/e^{-Ru}=h^{-1}{e}^{-u/d}$ is less than one. For the second finite-claim ruin probability, put $A=h^{-1}+d{h}^{-3}$ and $B=h^{-2}$. Its ratio is $e^{-u/d}(A+Bu)$. Sincethe derivative of this ratio is $e^{-u/d}[B-(A+Bu)/d]<0$ for $u\ge 0$, and its value at zero is $A<1$. This verifies both required bounds directly from the calculated probabilities.

In the Bühlmann model, one policy has a random risk parameter $\Theta$. Conditional on $\Theta$, its annual observations $X_i$ are independent and identically distributed random variables, with mean $m(\Theta )$ and variance $v(\Theta )$. Put $m_0=\mathbb{E}[m(\Theta )]$, let $v=\mathbb{E}[v(\Theta )]$ be the expected process variance, and let $a=\mathrm{Var}(m(\Theta ))$ be the variance of hypothetical means. Assume the required second moments are finite.

The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ under mean squared error. After choosing the intercept to make the estimate unbiased, write it as $m_0+{\sum }_{i=1}^n{b}_i(X_i-m_0)$. If $B={\sum }_i{b}_i$, the law of total variance and conditional independence give its errorFor fixed $B$, the Cauchy-Schwarz inequality minimizes the last sum at $b_i=B/n$. Differentiating $a(1-B{)}^2+v{B}^2/n$ gives $B=na/(na+v)$. Thus the credibility factor and credibility premium areThe ratio form assumes $a>0$; when $a=0$ and $v>0$, the conditional mean is constant and $Z=0$. Predicting $X_{n+1}$ rather than its conditional mean adds the constant $v$ to the error and leaves the optimal affine estimate unchanged.

Here the latent intensity has a Pareto distribution with shape three and lower bound one. For $r<3$, its moments are $\mathbb{E}[{\Theta }^r]=3{\int }_1^{\infty }{\theta }^{r-4}d\theta =3/(3-r)$. ConsequentlyThe equality for $v$ uses the Poisson distribution conditional variance $v(\Theta )=\Theta$. There are two years of experience, so $Z=1/2$ and the observed sample mean is $k/2$. The Poisson credibility with a shape-three Pareto intensity is thereforeIn particular it equals $2$ when $k=5$.

For the exact Bayesian credibility estimate, the total count conditional on $\Theta =\theta$ has Poisson distribution of mean $2\theta$. Multiplication of its likelihood by the prior distribution density gives a Bayesian posterior density proportional to ${\theta }^{k-4}{e}^{-2\theta }$ on $\theta >1$. At $k=5$, its normalizing integral and first-moment integral areHence the next-year predictive expected value isThe Bühlmann credibility premium optimizes over affine estimates, whereas the Bayesian posterior mean optimizes over all square-integrable estimates. This explains why the two credibility estimates need not agree for this nonconjugate prior.

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.