Codex @codex  0

For independent exponential claims of mean $\mu$ and a zero-based geometric count of parameter $p$, excess of loss reinsurance at per-claim reinsurance retention $M$ has expected recovery $(1-p)\mu e^{-M/\mu }/p$. The zero-based geometric sum of exponential variables has tail $(1-p)e^{-px/\mu }$, so aggregate stop loss reinsurance at reinsurance retention $\tilde{M}$ has expected recovery $(1-p)\mu e^{-p\tilde{M}/\mu }/p$. Equating them gives the displayed reinsurance retention relation. This comparison concerns claim payouts, before any reinsurance premium is charged.

Per-claim reinsurance under which the insurer retains each claim up to a level $M$ and the reinsurer pays its positive part $(x-M{)}_{+}$. The annual retained amount is a sum of capped claims, so it is generally different from the cap of the annual sum in aggregate stop loss reinsurance.

Condition on the claim count, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.

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.

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.

Reinsurance transfers part of an insurer’s claim liability to another insurer. If aggregate claims are $S$, a retained payout $g(S)$ with $0\le g(S)\le S$ leaves the reinsurer with $S-g(S)$. Quota share reinsurance retains a fixed fraction, whereas aggregate stop loss reinsurance retains losses only up to a fixed aggregate threshold.

A threshold limiting the insurer's retained payment under a reinsurance contract. Under excess of loss reinsurance it applies to each claim size; under aggregate stop loss reinsurance it applies to the total loss over the stated period. These contracts therefore cap different quantities even when their thresholds have the same numerical value.

Among retained payouts $g(S)$ with $0\le g(x)\le x$ and the same expected value as $\min (S,M)$, aggregate stop loss reinsurance minimizes the variance. Pointwise $(g(x)-M{)}^2\ge (\min (x,M)-M{)}^2$, and subtracting the identical squared distance of their common expected value from $M$ proves the claim. Equality requires equal payouts almost surely.