Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 34 / 2

For quota share reinsurance, each claim and therefore its aggregate are retained in the same proportion. For aggregate stop loss reinsurance, the insurer pays the aggregate up to the retention, and the reinsurer pays the excess. Thus the insurer's payouts areThe subscript $+$ denotes the positive part. The stop loss contract here applies to the annual aggregate, rather than separately to each claim.

The expected value and variance under quota share reinsurance follow by scaling the exponential distribution:The retained stop loss moments for an exponential aggregate follow from the payout $Y=\min (S,M)$, its survival function equals $e^{-x/{\mu }_S}$ for $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. Because $M>0$, the difference is actually positive. At equal retained expected value, aggregate stop loss reinsurance reduces the variance more than quota share reinsurance.

Write $Y=g(S)$ and $Z=\min (S,M)$. Their common expected value is $c$. Expanding about the retention gives the requested variance identityThe stop loss variance minimization principle follows from a pointwise comparison. For $0\le x\le M$, the constraint $0\le g(x)\le x$ impliesFor $x>M$, the squared distance of $\min (x,M)=M$ from $M$ is zero, so the same squared-distance comparison is immediate. ThereforeSubtracting the same $(M-c{)}^2$ proves optimality of the retained stop loss payout:Since $Z$ is bounded, its variance is finite; if $\mathbb{E}[Y^2]=\infty$, the inequality remains valid with infinite variance on the left. When it is finite, equality requires $g(S)=\min (S,M)$ almost surely, because the pointwise squared-distance inequality is strict whenever the two payouts differ.