Introduction
An annual aggregate excess-of-loss treaty responds to the insurer’s total loss over a defined period, rather than to each claim separately. Its effect on the distribution of retained aggregate loss is determined by a simple transformation: the insurer retains the loss below the attachment, retains a fixed amount across the treaty’s layer, and retains only the excess above the layer limit. That fixed retained amount can create a probability atom, even when aggregate loss itself is continuous.
This tutorial develops the transformation from a known annual aggregate-loss distribution to the insurer’s retained aggregate-loss distribution. The focus is on the full distribution—not on deriving an aggregate model from claim frequency and severity.
The retained amount can be written directly as
Thus losses in the treaty layer all leave the insurer retaining \(a\); above exhaustion, the insurer again bears each additional dollar of loss. The amount \(u\) is the layer limit, not the exhaustion point. The exhaustion point is \(a+u\).
The Retained-Loss Distribution
Write \(F_X(x)=P(X\le x)\) for the CDF of aggregate loss and \(F_R(r)=P(R\le r)\) for the CDF of retained aggregate loss. The transformation gives a particularly compact result.
This formula applies whether \(X\) is discrete, continuous, or mixed. It does not assume a density, and it includes any probability masses at the attachment or exhaustion points. Because \(X\ge 0\), the first case also gives \(F_R(r)=0\) for \(r<0\).
Why the CDF Formula Holds
Below the retained amount \(a\), only aggregate losses below \(a\) can produce retained losses below \(a\): losses in the treaty layer produce exactly \(a\), while losses above exhaustion produce at least \(a\). More generally, for any \(r<a\), the condition \(R\le r\) is equivalent to \(X\le r\). For \(r\ge a\), all losses through exhaustion produce \(R\le r\); above exhaustion, \(R=X-u\), so the additional condition is \(X-u\le r\), or \(X\le r+u\). Together these events are exactly \(X\le r+u\).
If \(X\le a\), then \(R=X\). If \(X>a\), then \(R\ge a\). Therefore \(R\le r<a\) holds exactly when \(X\le r\), giving \(F_R(r)=F_X(r)\).
Every \(X\le a+u\) produces \(R\le a\le r\). For \(X>a+u\), retained loss is \(X-u\), which is at most \(r\) exactly when \(X\le r+u\). The combined event is \(X\le r+u\).
Taking probabilities in the two cases gives the stated piecewise CDF. In particular, \(F_R(a)=F_X(a+u)\), which accounts for all aggregate-loss outcomes that leave exactly \(a\) retained.
Atoms and Continuous Portions
The retained distribution may contain probability mass at \(a\). The exact mass is
The closed interval is important: \(X=a\) is retained in full, while \(X=a+u\) exhausts the layer and also leaves exactly \(a\) retained. The atom is positive only if this interval has positive probability under the distribution of \(X\). If \(X\) has a density \(f_X\), its size is \(\int_a^{a+u} f_X(x)\,dx\); that integral can be zero if the model assigns no probability to the interval.
Other atoms can also occur. Any atom of \(X\) at a value below \(a\) remains at that value; an atom above \(a+u\) shifts down by \(u\). For a model with a density and no other discrete component, the continuous density of \(R\), away from the attachment, is \(f_X(r)\) for \(0<r<a\) and \(f_X(r+u)\) for \(r>a\). The probability in the interval \([a,a+u]\) is concentrated at \(R=a\), not spread over that interval.
Worked Example: Transforming a Discrete Aggregate Distribution
A hypothetical insurer’s annual aggregate loss \(X\), in thousands of dollars, has the following distribution. An annual aggregate excess-of-loss treaty attaches at \(a=100\) thousand dollars and has limit \(u=80\) thousand dollars. Find the distribution of retained aggregate loss \(R\), and calculate \(P(R\le 100)\).
Method. The layer exhausts at \(a+u=100+80=180\). Apply the piecewise transformation to each possible value of \(X\).
| Aggregate loss \(X\) | \(P(X)\) | Retained loss \(R\) |
|---|---|---|
| 0 | 0.10 | 0 |
| 80 | 0.20 | 80 |
| 120 | 0.15 | 100 |
| 180 | 0.25 | 100 |
| 260 | 0.30 | 180 |
For \(X=0\) and \(X=80\), the loss does not exceed the attachment, so \(R=X\). For \(X=120\) and \(X=180\), the loss lies in the layer or at exhaustion, so \(R=100\). For \(X=260\), the recovery is the full limit of 80 and \(R=260-80=180\).
Combining outcomes that map to the same retained value gives \(P(R=0)=0.10\), \(P(R=80)=0.20\), \(P(R=100)=0.15+0.25=0.40\), and \(P(R=180)=0.30\). The probabilities sum to \(0.10+0.20+0.40+0.30=1.00\), as required.
Since \(R\le100\) at retained values 0, 80, and 100, \(P(R\le100)=0.10+0.20+0.40=0.70\). This also agrees with the CDF formula: \(F_R(100)=F_X(100+80)=F_X(180)=0.70\). Common mistake avoided: treating the limit \(80\) as the exhaustion point would incorrectly place \(X=180\) above the layer; the correct exhaustion point is \(a+u=180\).
Worked Example: A Continuous Model with an Atom at the Attachment
Suppose a hypothetical annual aggregate loss \(X\), in thousands of dollars, is uniformly distributed on \([0,400]\). An annual aggregate treaty attaches at \(a=100\) and has limit \(u=150\). Find the atom at \(R=100\), calculate \(P(R\le150)\), and describe the continuous part of the retained distribution.
The exhaustion point is \(a+u=100+150=250\). For a uniform distribution on \([0,400]\), the density is \(1/400=0.0025\) per thousand dollars. The interval \([100,250]\) has length \(150\), so
For \(0\le r<100\), the retained CDF is \(F_X(r)=r/400\). For \(r\ge100\), it is \(F_X(r+150)\), with the uniform CDF capped at 1 when \(r+150\ge400\). In particular,
Below 100, the retained distribution has density \(0.0025\). Above 100 and below 250, it also has density \(f_X(r+150)=0.0025\). At exactly 100, it has the additional atom of 0.375. The continuous probability below 100 is \(100/400=0.25\), the atom is \(0.375\), and the continuous probability above 100 is \(150/400=0.375\); these total 1.
Check. The CDF immediately to the left of 100 is \(100/400=0.25\), while \(F_R(100)=F_X(250)=250/400=0.625\). Their difference is \(0.625-0.25=0.375\), the atom calculated directly.
Worked Example: Tail Probability Under an Exponential Model
Suppose the hypothetical annual aggregate loss \(X\), in thousands of dollars, has an exponential distribution with rate \(\lambda=1/200\) per thousand dollars, so \(S_X(x)=P(X>x)=e^{-x/200}\) for \(x\ge0\). A treaty attaches at \(a=100\) and has limit \(u=200\). Find \(P(R=100)\) and \(P(R>150)\). Round probabilities to four decimal places.
The exhaustion point is \(100+200=300\). Since \(X\) is continuous, endpoint probabilities are zero, and
For \(r\ge a\), the CDF relationship gives \(P(R>r)=P(X>r+u)\). Taking \(r=150\) gives
Thus the answers, to four decimal places, are \(P(R=100)=0.3834\) and \(P(R>150)=0.1738\). Check. The tail event requires aggregate loss to exceed 350, not 150: once the layer is exhausted, retained loss is \(X-200\).
Common Exam Traps
- Using the limit as the exhaustion point. A layer attaching at \(a\) with limit \(u\) exhausts at \(a+u\), not at \(u\). Retained loss is \(a\) across the layer.
- Forgetting to combine probability at the plateau. The mass at \(R=a\) is \(P(a\le X\le a+u)\), including any probability atoms at either endpoint. In CDF notation, it is \(F_X(a+u)-F_X(a-)\), where \(F_X(a-)=P(X<a)\).
- Assuming every continuous model has an atom at the attachment. Continuity of \(X\) does not by itself make the atom positive. For example, if \(X\) is uniform on \([200,300]\), \(a=60\), and \(u=90\), then exhaustion is 150 and \(X\) always exceeds it. The retained loss is \(R=X-90\), which is uniform on \([110,210]\); there is no atom at 60 because \(P(60\le X\le150)=0\).
- Applying a per-claim treaty transformation to an annual total. Here the treaty’s attachment and limit are defined on annual aggregate loss \(X\). Transforming individual claim amounts first would describe a different treaty basis.
- Using the wrong tail threshold. For \(r\ge a\), \(P(R>r)=P(X>r+u)\), not \(P(X>r)\). The loss must exceed \(r+u\) for the insurer to retain more than \(r\) after the layer is exhausted.
Check Your Understanding
For each hypothetical annual aggregate-loss model, transform the distribution or probability using the stated attachment and limit. Treat monetary values as being in the same units throughout.
- \(X\) takes values \(0,50,90,140,210\) with probabilities \(0.10,0.20,0.25,0.25,0.20\), respectively. A treaty has \(a=80\) and \(u=60\). Give the distribution of \(R\). (Answer: \(P(R=0)=0.10,\ P(R=50)=0.20,\ P(R=80)=0.50,\ P(R=150)=0.20\).)
- \(X\) is uniform on \([0,500]\). A treaty has \(a=100\) and \(u=100\). Find the atom at the attachment and \(P(R\le150)\). (Answer: \(P(R=100)=0.2000,\ P(R\le150)=0.5000\).)
- \(X\) is exponential with rate \(1/100\). A treaty has \(a=50\) and \(u=100\). Find \(P(R=50)\) and \(P(R>200)\), to four decimal places. (Answer: \(P(R=50)=e^{-0.5}-e^{-1.5}=0.3834,\ P(R>200)=e^{-3}=0.0498\).)
- \(X\) is uniform on \([200,300]\). A treaty has \(a=60\) and \(u=90\). Describe the distribution of \(R\) and state whether there is an atom at \(a\). (Answer: \(R=X-90\) is uniform on \([110,210]\); \(P(R=60)=0\).)
- \(X\) is exponential with rate \(1/100\), and a treaty has \(a=100\), \(u=200\). Define the 0.95 quantile as \(\inf\{r:F_R(r)\ge0.95\}\). Find this quantile. (Answer: \(100\), since \(F_R(100)=F_X(300)=1-e^{-3}\approx0.9502\), while \(F_R(r)<0.95\) for \(r<100\).)