Atoms at Zero and Treaty Limits

Why a Continuous Loss Model Can Produce Point Masses

A loss amount may have a continuous distribution, but treaty payments need not. Under a per-claim excess-of-loss treaty, every loss below the attachment produces the same recovery: zero. Every sufficiently large loss produces the same recovery: the full layer limit. Those ranges of loss values are each mapped to a single payment amount, creating probability masses even when the original loss has no atoms.

The retained payment has a related feature. While the loss moves through the reinsured layer, the insurer's retained payment stays at the attachment. This creates a plateau in the retained-payment function and, when losses fall in that range with positive probability, an atom at the attachment. Correctly identifying which endpoints belong to each event is essential, especially for discrete or mixed loss distributions.

Definition: Let $X\geq 0$ be the loss on one claim. A per-claim excess-of-loss treaty has attachment $d\geq 0$ and layer limit $u>0$, where $u$ is the maximum amount recoverable from the reinsurer on that claim. The ceded recovery is $$ R=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ The insurer's retained payment per loss is $Y^L=X-R$. The recovery is zero for $X\leq d$, increases with the loss between $d$ and $d+u$, and is capped at $u$ for $X\geq d+u$.

Here $d+u$ is the exhaustion point: a loss at or above this amount generates the full layer recovery. In the notation $F_X(x)=P(X\leq x)$, use $F_X(a-)=P(X<a)$ for the left limit at $a$. The left limit matters when the loss distribution has probability at a treaty boundary.

The Three Key Probability Masses

The treaty recovery is zero precisely when the loss is at or below the attachment. It is the full limit precisely when the loss is at or above the exhaustion point. Between those two ranges, recovery is strictly between zero and $u$.

\begin{aligned} P(R=0)&=P(X\leq d)=F_X(d),\\ P(0<R<u)&=P(d<X<d+u)=F_X((d+u)-)-F_X(d),\\ P(R=u)&=P(X\geq d+u)=1-F_X((d+u)-). \end{aligned}

The retained payment is equal to $d$ for every loss in the closed interval from $d$ to $d+u$, including both endpoints. For losses below $d$, the retained payment equals the loss; for losses above $d+u$, it equals $X-u$. Therefore,

P(Y^L=d)=P(d\leq X\leq d+u)=F_X(d+u)-F_X(d-).

This is the mass at the retained-loss plateau. It is not generally the same event as full recovery: the retained payment equals $d$ for losses throughout the layer, whereas full recovery occurs for losses at or above the exhaustion point. The two events overlap at $X=d+u$, if there is an atom there.

Why the Plateau Creates an Atom

The point mass follows from how the treaty maps losses to retained payments. Inside the layer, the ceded amount is $X-d$, so subtracting it from $X$ leaves exactly $d$. Many distinct loss values therefore produce the same retained amount.

Below or at attachment.
If $X\leq d$, then $R=0$, so $Y^L=X$. The retained amount reaches $d$ at the boundary $X=d$.

Within the layer.
If $d<X<d+u$, then $R=X-d$, and $Y^L=X-(X-d)=d$. At $X=d+u$, recovery has reached $u$ and retained payment is still $d$.

Above exhaustion.
If $X>d+u$, then $R=u$, so $Y^L=X-u>d$. The retained payment rises again beyond the plateau.

Consequently, the event $Y^L=d$ is exactly $d\leq X\leq d+u$. Its probability is $F_X(d+u)-F_X(d-)$, because $F_X(d+u)$ includes losses at the upper endpoint while $F_X(d-)$ excludes losses at the lower endpoint. If $X$ is continuous, both endpoint probabilities are zero and the expression simplifies to $F_X(d+u)-F_X(d)$. If $X$ is discrete or mixed, the endpoint terms must be retained.

Worked Example: A Continuous Exponential Loss

A hypothetical insurer models a single claim loss as exponential with mean $1,000, so $X$ has rate $1/1000$ per dollar and $P(X>x)=e^{-x/1000}$ for $x\geq0$. A per-claim treaty has attachment $d=\$500$ and limit $u=\$1{,}000$. Find the probabilities of zero recovery, partial recovery, full recovery, and retained payment equal to the attachment. There is no interest or mortality assumption in this claim-payment problem.

Method. The exhaustion point is $d+u=\$1{,}500$. Because the exponential distribution is continuous, endpoint probabilities are zero.

Zero recovery. This occurs for $X\leq500$: $$ P(R=0)=1-e^{-500/1000}=1-e^{-0.5}=0.393469\ldots $$ Thus, to four decimal places, $P(R=0)=0.3935$.

Full recovery. This occurs for $X\geq1{,}500$: $$ P(R=1000)=e^{-1500/1000}=e^{-1.5}=0.223130\ldots $$ Thus, $P(R=1000)=0.2231$.

Partial recovery. This occurs for $500<X<1{,}500$: $$ P(0<R<1000)=e^{-0.5}-e^{-1.5} =0.606531\ldots-0.223130\ldots =0.383401\ldots $$ Thus, $P(0<R<1000)=0.3834$.

Retained payment at the attachment. The retained payment is $500 throughout the layer: $$ P(Y^L=500)=e^{-0.5}-e^{-1.5}=0.383401\ldots=0.3834. $$ The recovery categories check: $0.3935+0.3834+0.2231=1.0000$, to four decimal places. The plateau mass equals the partial-recovery probability here because the loss is continuous and the plateau is generated by losses strictly inside the layer.

Worked Example: Boundary Atoms in a Discrete Loss Model

A hypothetical equipment portfolio has a single-claim loss $X$ with the following distribution: $P(X=0)=0.10$, $P(X=400)=0.20$, $P(X=900)=0.25$, $P(X=1{,}400)=0.25$, and $P(X=2{,}000)=0.20$, with amounts in dollars. A treaty has $d=\$400$ and $u=\$1{,}000$. Find the probability of zero recovery, full recovery, partial recovery, and retained payment of exactly $400. No interest or mortality assumption is relevant.

The exhaustion point is $400+1{,}000=\$1{,}400$. At $X=400$, recovery is zero; at $X=1{,}400$, recovery is the full limit. These equality cases must not be lost by using strict inequalities.

Zero recovery: $X\leq400$, which includes losses 0 and 400. Therefore $P(R=0)=0.10+0.20=0.30$.

Full recovery: $X\geq1{,}400$, which includes losses 1,400 and 2,000. Therefore $P(R=1{,}000)=0.25+0.20=0.45$.

Partial recovery: $400<X<1{,}400$, which includes only the loss 900. Its recovery is $900-400=\$500$, so $P(0<R<1{,}000)=0.25$.

Retained payment of exactly $400: this occurs for $400\leq X\leq1{,}400$, so $$ P(Y^L=400)=0.20+0.25+0.25=0.70. $$ The recovery categories check: $0.30+0.25+0.45=1.00$. The plateau probability is $0.70$, not $0.25$: it includes the atoms at both treaty boundaries as well as the loss strictly inside the layer.

Worked Example: A Mixed Loss Distribution

For a hypothetical single claim, suppose $X$ has probability $0.20$ at zero and, with probability $0.80$, has an exponential distribution with mean $500. Thus, for $x\geq0$, $P(X>x)=0.80e^{-x/500}$. A per-claim treaty has $d=\$500$ and $u=\$500$. Calculate the three recovery probabilities and the mass at retained payment $500. There is no interest or mortality assumption.

The exhaustion point is $1,000. The atom at zero belongs to the zero-recovery event, and the remaining component is continuous.

Zero recovery: this includes the atom at zero and all positive losses up to $500: $$ P(R=0)=0.20+0.80(1-e^{-500/500}) =0.20+0.80(1-e^{-1}) =0.705696\ldots=0.7057. $$

Full recovery: only the continuous component can reach $1,000 or more: $$ P(R=500)=0.80e^{-1000/500} =0.80e^{-2} =0.108268\ldots=0.1083. $$

Partial recovery: $$ P(0<R<500)=0.80(e^{-1}-e^{-2}) =0.80(0.367879-0.135335) =0.186035\ldots=0.1860. $$

Retained payment of $500: it occurs for losses from $500 through $1,000. There are no atoms at those positive endpoints, so $$ P(Y^L=500)=0.80(e^{-1}-e^{-2})=0.1860. $$ The three recovery probabilities sum to $0.705696+0.186035+0.108268=0.999999$, with the small discrepancy from 1 due to displayed intermediate rounding; using the unrounded exponential values, they sum exactly to 1.

Common Boundary Errors

A frequent mistake is to use strict inequalities for both endpoint events, as if the treaty paid zero only when $X<d$ and paid its full limit only when $X>d+u$. That omits any probability mass at $d$ or $d+u$. The treaty definition resolves the boundaries directly: at $X=d$, recovery is zero; at $X=d+u$, recovery is $u$.

For example, in the discrete example above, a mistaken zero-recovery calculation using $P(X<400)$ gives $0.10$, rather than the correct $P(X\leq400)=0.30$. A mistaken full-recovery calculation using $P(X>1{,}400)$ gives $0.20$, rather than the correct $P(X\geq1{,}400)=0.45$. The omitted probabilities are exactly the atoms at the treaty boundaries.

Another error is to use the full recovery event when calculating the retained plateau. The correct plateau event is $d\leq X\leq d+u$, not $X\geq d+u$. Losses above exhaustion have retained payment $X-u$, which exceeds $d$ unless the loss is exactly at the exhaustion point.

Key takeaway: For a per-claim excess-of-loss treaty, zero recovery corresponds to $X\leq d$, full recovery corresponds to $X\geq d+u$, and the retained payment equals $d$ exactly when $d\leq X\leq d+u$. Use the distribution's endpoint probabilities whenever $X$ can be discrete or mixed.

Check Your Understanding

For each hypothetical single-claim model, identify the relevant loss intervals first, then calculate the requested probabilities. Give probabilities to four decimal places where needed.

$X$ is uniform on $[0,2{,}000]$, in dollars. A treaty has $d=\$500$ and $u=\$500$. Find $P(R=0)$, $P(R=u)$, and $P(Y^L=500)$. (Answer: $0.2500,\ 0.5000,\ 0.2500$, respectively.)
A discrete loss has probabilities $0.10,0.20,0.30,0.25,0.15$ at losses $\$0,\$300,\$600,\$900,\$1{,}200$, respectively. For $d=\$300$, $u=\$600$, find $P(R=0)$, $P(R=u)$, $P(0<R<u)$, and $P(Y^L=300)$. (Answer: $0.3000,\ 0.4000,\ 0.3000,\ 0.7500$, respectively.)
$X$ is exponential with mean $800. A treaty has $d=\$400$ and $u=\$800$. Find the zero-recovery probability, full-recovery probability, and retained plateau mass. (Answer: $0.3935,\ 0.2231,\ 0.3834$, respectively.)
$X$ has probability $0.25$ at zero and, with probability $0.75$, is exponential with mean $1,000. For $d=\$1{,}000$, $u=\$500$, find $P(R=0)$, $P(R=u)$, and $P(Y^L=1{,}000)$. (Answer: $0.7241,\ 0.1673,\ 0.1086$, respectively.)
$X$ is exponential with mean $1,000, and a treaty has $d=\$500$, $u=\$1{,}000$. Given that recovery is positive, find the conditional probability that it is the full layer limit. (Answer: $P(R=1{,}000\mid R>0)=e^{-1}=0.3679$.)

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