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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.

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.

For excess of loss reinsurance with retention $M$, the direct insurer caps its payment at $M$ and the reinsurer pays the excess. ThusLet $I_i=1_{\{X_i>M\}}$ and $\alpha =\mathbb{P}(X_i>M)=1-F_{X_1}(M)\in (0,1)$. These are independent Bernoulli random variables, independent of $N$. Since $N_R={\sum }_{i=1}^N{I}_i$, its conditional law given $N=n$ is a binomial distribution. Conditioning on $N$ therefore gives the Bernoulli thinning identityHere the probability generating function may be evaluated on $0\le z\le 1$, and the identity extends wherever the transforms converge.

The positive reinsurer marks have the conditional law $W\stackrel{d}{=}{X}_1-M\mid X_1>M$. HenceThe independent-mark representation is justified as follows. Conditional on any fixed count and pattern of exceedance indicators, the positive excesses are independent with this same conditional law; that law depends neither on the number of exceedances nor on their locations. Averaging over the indicator pattern thus leaves the positive mark sequence independent of $N_R$. We may extend it by additional independent copies if needed, giving $S_R={\sum }_{j=1}^{N_R}{W}_j$ with the stated independence.

Now write $q=1-p$. For exponential distribution claim sizes of mean $\mu$,Thus $W$ is exponential with the original mean $\mu$, by memorylessness of the exponential distribution. The original zero-based geometric distribution count has $G_N(z)=p/(1-qz)$. ConsequentlyThis is the geometric count under Bernoulli thinning, withFinally use the Laplace transform of a nonnegative random variable, keeping the mass at zero. For $t\ge 0$,The last expression is the transform of a mixture distribution: mass $p_R$ at zero and, with probability $q_R$, an exponential distribution of rate $p_R/\mu$. Uniqueness of the Laplace transform identifies this zero-based geometric sum of exponential variables:Equivalently, $\mathbb{P}(S_R=0)=p_R$ and its positive density is $q_R(p_R/\mu )e^{-p_{R}s/\mu }$. Therefore the requested exceedance probability isThe prefactor is essential: the aggregate is zero whenever no claim crosses the retention.