Reinsurance Treaty: Thinning a Poisson Process When the Rate Is Not Constant
A reinsurer receives claims from a cedant under a quota-share treaty. Claims below £250,000 are classed as small and claims at or above £250,000 as large. In the last 52 weeks the cedant reported 624 claims, of which 187 were large. The weekly number of claims over those 52 weeks had mean 12.0 and variance 36.0. The pricing actuary proposes to model claim arrivals as a homogeneous Poisson process of rate 12 per week, to treat each claim as large with probability 0.30 independently of all else, and therefore to model large claims as a Poisson process of rate 3.6 per week that is independent of the small-claim process of rate 8.4 per week. The treaty runs for the four weeks of January. The actuary must give the probability of no large claims in that period, an approximate probability of 20 or more large claims, and must decide whether the proposed model is adequate for setting the treaty's risk margin, which is calibrated to a 99.5% VaR.
1. Under the actuary's assumptions, derive the distribution of the number of large claims in the four-week treaty period and compute the probability of none and the probability of 20 or more.
Read the worked answer
Thinning: if claims arrive as a Poisson process of rate $\lambda$ and each is marked large with probability $p$ independently, the large-claim process is Poisson with rate $p\lambda$. So $\lambda_L=0.30\times12=3.6$ per week, and over $t=4$ weeks $N_L\sim\text{Poisson}(14.4)$.
$$P(N_L=0)=e^{-14.4}=5.5\times10^{-7}.$$
For the upper tail, mean $=14.4$ and sd $=\sqrt{14.4}=3.795$. With a continuity correction, $z=(19.5-14.4)/3.795=1.344$, so $P(N_L\ge20)\approx1-\Phi(1.344)\approx0.090$, about 9%.
Two cautions. First, the normal approximation understates the far tail of a right-skewed Poisson; for a 99.5% VaR the normal quantile gives $14.4+2.576(3.795)+0.5\approx25$, but this should be confirmed against exact Poisson quantiles, since a Poisson tail at that level is heavier than the normal suggests. Second, all of this depends on the thinning probability being constant and independent of the total; the last task shows that assumption fails.
2. Given exactly 60 claims in the four weeks, state the conditional distribution of the number of large claims. Explain why this conditional result does not justify treating the two claim-type processes as independent if the underlying rate is random.
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Conditionally on the total $N=60$, each claim is independently large with probability 0.30, so $N_L\mid N=60\sim\text{Binomial}(60,0.30)$: mean 18, variance $60(0.30)(0.70)=12.6$, sd 3.55. This is the classic multinomial thinning result and it is exact under IID marks.
The independence of the two Poisson processes, however, follows only because the total rate $\lambda t$ is a fixed number. If the weekly rate is itself random - $\Lambda$ with mean 12 and variance 24, say - then marginally $N_L\mid\Lambda\sim\text{Poisson}(0.3\Lambda t)$ and $N_S\mid\Lambda\sim\text{Poisson}(0.7\Lambda t)$, and
$$\operatorname{Cov}(N_L,N_S)=0.3\times0.7\times t^2\operatorname{Var}(\Lambda)>0.$$
Large and small claims then move together: a week with an unusual total volume of claims carries more of both types. Treating them as independent processes gives the right means but understates the variance of the aggregate. For a reinsurer whose retention is breached only when both severity and frequency are high, this correlation is exactly the dependence that matters for the 99.5% VaR.
3. Assess the Poisson assumption using the reported mean and variance, identify a suitable alternative, and state the consequences for the risk margin.
Read the worked answer
Under a homogeneous Poisson process the weekly mean and variance are equal. Observed: mean 12.0, variance 36.0, a variance-to-mean ratio of 3. The sample variance of 52 observations has standard error roughly $\sigma^2\sqrt{2/51}=36(0.198)=7.1$, so 36 is about $(36-12)/7.1=3.4$ standard errors above 12 - clear overdispersion. Equivalently, the index of dispersion $36/12=3$ across 52 weeks is far in the tail of a $\chi^2_{51}/51$ distribution.
A negative binomial with mean $\mu$ and shape $r$ has variance $\mu+\mu^2/r$; setting $36=12+144/r$ gives $r=6$. Over four weeks the mean stays 14.4 but the variance becomes $14.4+14.4^2/6=49.0$, sd 7.0 instead of 3.8. A symmetric 99.5% quantile moves out by roughly $2.576(7.0-3.8)\approx8$ claims relative to the Poisson answer - on a treaty of this size, a material margin.
Consequences and advice: the point estimate of expected large claims is unaffected, but every tail-based number in the treaty (VaR, expected shortfall, reinstatement pricing) is understated. I would refit with an overdispersed frequency model, allow for the rate varying over time or by month (January weather), test whether large claims cluster within events - which would break independence of marks even after allowing for a random rate - and stress-test the tail rather than relying on a single calibrated model.