Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 40 / 2

For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

Under quota share reinsurance, $S_I=\alpha S$ and $S_R=(1-\alpha )S$. Write $v=\mathrm{Var}(S)=\lambda \mathbb{E}[X^2]$. ThenThe variance-minimizing quota share split is thereforeFor positive claims and $\lambda >0$ this minimizer is unique. If the aggregate has zero variance, every retained fraction gives the same zero objective.

Set $\overline{F}(M)=1-F_X(M)$. The compound Poisson distribution variance identity and the per-claim excess of loss reinsurance payouts giveApply differentiation under the integral sign. The endpoint terms $M^2{f}_X(M)$ in the first two terms cancel, while the last integrand is zero at its lower endpoint. Equivalently, differentiate the two payout squares inside their expected values; the finite second moment supplies a dominating integrable function. This gives the total variance stationary condition for excess of lossHence the specified equality makes $g^{\prime }(M)=0$. When $\overline{F}(M)>0$, its interpretation is $M=m_X(M)$, where the mean residual life is $m_X(M)=\mathbb{E}[X-M\mid X>M]$.

For the exponential distribution of mean $\mu$, $\overline{F}(M)=e^{-M/\mu }$ andThus $g^{\prime }(M)=2\lambda e^{-M/\mu }(M-\mu )$ is negative below $\mu$ and positive above it. The variance-minimizing exponential retention is the unique global minimizerFor an explicit value, put $m=M/\mu$. The capped claim moments and the excess-claim second moment giveThe sign argument establishes global minimality, rather than just stationarity.

Let $v=\mathrm{Var}(S)>0$ and $v_I=\mathrm{Var}(S_I)$. The variance-matching quota share reinsurance in this part has the nonnegative retained fraction ${\alpha }^{\ast }=\sqrt{v_I/v}$. For claimwise retentions $0\le h(x)\le x$, its existence within $[0,1]$ follows from $v_I=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]=v$. The comparison below only needs the matching contract specified in the question.

Because $S_R=S-S_I$, the variance and covariance identities giveThe Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le \sqrt{v{v}_I}={\alpha }^{\ast }v$. Therefore the quota share comparison at matched retained variance isThe last expression is the sum of variances under the matching quota share reinsurance. HenceIf $v=0$, the given matching contract forces $v_I=0$, so all relevant totals are constant and equality holds. For $v>0$, equality in the bound is equivalent to $S_I-\mathbb{E}[S_I]={\alpha }^{\ast }(S-\mathbb{E}[S])$ almost surely.