Distribution of a Layered Payment

From Ground-Up Loss to Layer Payment

A layer pays only for the part of a loss between an attachment point and an exhaustion point. Its payment distribution therefore differs from the ground-up loss distribution in two important ways: losses at or below attachment produce a payment of zero, and sufficiently large losses all produce the same maximum payment. The probability distribution must account for both of these concentrations of probability, as well as payments strictly between the endpoints.

Let $X$ be a nonnegative ground-up loss, let $a\geq 0$ be the layer attachment point, and let $b>a$ be the exhaustion point. The layer width is $b-a$. For this tutorial, the layer pays the portion of the loss above $a$, subject to a maximum payment of $b-a$. Write the payment per loss as $Y^L$.

Definition: The payment from a layer attaching at $a$ and exhausting at $b$ is $$ Y^L=\min\bigl((X-a)_+,\,b-a\bigr), \qquad (z)_+=\max(z,0). $$ Equivalently, $Y^L=0$ when $X\leq a$, $Y^L=X-a$ when $a<X<b$, and $Y^L=b-a$ when $X\geq b$.

The convention at the boundaries matters. A loss exactly equal to $a$ produces no payment, while a loss exactly equal to $b$ produces the full layer payment. In particular, if the ground-up distribution has probability at $b$, that probability belongs to the mass at the layer cap. Use $F_X(x)=P(X\leq x)$, and write $F_X(b-)=P(X<b)$ for the left limit of the distribution function.

The Distribution and Its Probability Masses

Set $L=b-a$, the maximum layer payment. The layer payment is supported on $[0,L]$. Its distribution function at zero is the probability that the ground-up loss does not exceed attachment. For a payment strictly below the cap, the loss must have been exactly the payment amount above attachment, or less.

F_{Y^L}(y)=P(Y^L\leq y)= \begin{cases} 0, & y<0,\\ F_X(a+y), & 0\leq y<L,\\ 1, & y\geq L. \end{cases}

At $y=0$, this gives $P(Y^L=0)=F_X(a)$. At the cap, the probability mass is the jump in the payment distribution at $L$, not the value of the distribution function at $L$, which is one. The cap mass is $P(X\geq b)$, including any ground-up probability at exactly $b$.

P(Y^L=0)=F_X(a), \qquad P(Y^L=L)=P(X\geq b)=1-F_X(b-).

Between zero and the cap, ground-up probability is carried across without changing its size: a loss of $a+y$ produces a payment of $y$. Thus, for $0<y<L$, any point mass at payment $y$ equals the ground-up point mass at $a+y$. If the ground-up distribution has a density, the density of the payment on the interior is shifted by $a$.

P(Y^L=y)=P(X=a+y),\quad 0<y<L; \qquad f_{Y^L}(y)=f_X(a+y),\quad 0<y<L

The density statement describes the continuous part only. If $X$ is mixed or discrete, its atoms must be listed separately; a density does not describe those point masses. It is also useful to report the positive-payment part of the distribution without renormalizing it. For $0<y<L$, the probability of a strictly positive payment no greater than $y$ is $F_X(a+y)-F_X(a)$. At the cap, the corresponding probability is $P(X>a)$. This positive-payment component has total probability $P(X>a)$, which need not equal one.

P(0<Y^L\leq y)=F_X(a+y)-F_X(a),\quad 0<y<L; \qquad P(0<Y^L\leq L)=1-F_X(a).

Why the Distribution Formula Holds

For any $y$ strictly between zero and $L$, a payment no greater than $y$ occurs precisely when the ground-up loss is no greater than $a+y$. Losses at or below $a$ pay zero, and losses between $a$ and $a+y$ pay an amount between zero and $y$. Losses above $a+y$ pay more than $y$, because $a+y<b$. Therefore the two events are identical:

\{Y^L\leq y\}=\{X\leq a+y\}, \qquad 0\leq y<L.

Taking probabilities gives $F_{Y^L}(y)=F_X(a+y)$ below the cap. At $y=0$, the event is $X\leq a$, yielding the zero-payment mass. At $y=L$, every possible layer payment is at most $L$, so the distribution function is one. The cap mass is the probability entering at $L$: $P(X\geq b)$. This event-based argument also shows why the formulas apply to discrete and mixed losses, not just continuous ones.

Worked Examples

Worked Example: An Exponential Loss in a Layer

A hypothetical insurer covers a ground-up claim $X$, measured in thousands of dollars, with $X\sim\operatorname{Exponential}(\lambda=0.1)$. A layer attaches at $a=2$ and exhausts at $b=7$, also in thousands of dollars. Find the zero-payment mass, the cap mass, the interior density, and the positive-payment probability up to a payment of 3. Assume no interest or mortality adjustment is relevant; this is a single-loss severity calculation.

Method. Apply the layer distribution formulas to the exponential ground-up distribution. Its distribution function and survival function are $F_X(x)=1-e^{-0.1x}$ and $P(X>x)=e^{-0.1x}$, for $x\geq0$. The layer width is $L=7-2=5$.

Zero payment. The payment is zero for $X\leq2$, so $P(Y^L=0)=F_X(2)=1-e^{-0.2}=0.1813$, to four decimal places.

Payment at the cap. The payment is 5 when $X\geq7$. The exponential distribution is continuous, so $P(X\geq7)=P(X>7)=e^{-0.7}=0.4966$, to four decimal places.

Interior density. For $0<y<5$, the density is shifted from the ground-up density: $f_{Y^L}(y)=f_X(2+y)=0.1e^{-0.1(2+y)}$. This is density on the interior; it does not include either endpoint mass.

Positive payment no greater than 3. Since $3<5$, this event is $2<X\leq5$. Therefore $P(0<Y^L\leq3)=F_X(5)-F_X(2)=e^{-0.2}-e^{-0.5}=0.2122$, to four decimal places.

Check and trap avoided. The interior probability is $P(2<X<7)=e^{-0.2}-e^{-0.7}=0.3221$. Adding the three components gives $0.1813+0.3221+0.4966=1.0000$, allowing for rounding. A common error is to use $P(X>7)$ for the cap mass without checking for an atom at 7. Here that shortcut is valid only because the specified exponential distribution is continuous.

Worked Example: A Discrete Loss With Boundary Atoms

A hypothetical equipment policy has ground-up loss $X$, in thousands of dollars, with the following distribution: $P(X=0)=0.10$, $P(X=3)=0.15$, $P(X=4)=0.20$, $P(X=6)=0.25$, $P(X=10)=0.12$, and $P(X=13)=0.18$. A layer attaches at $a=4$ and exhausts at $b=10$. Find the complete distribution of $Y^L$. Interest and mortality assumptions are not applicable to this single-loss calculation.

Method. Map each ground-up loss through $Y^L=\min((X-4)_+,6)$, then combine probabilities whenever different losses give the same payment.

The layer width is $10-4=6$. For $X=0,3,$ or $4$, the loss is at or below attachment and the payment is zero. The combined zero-payment mass is $0.10+0.15+0.20=0.45$.

For $X=6$, the payment is $6-4=2$, so $P(Y^L=2)=0.25$. For $X=10$ or $X=13$, the payment is capped at 6. Thus $P(Y^L=6)=0.12+0.18=0.30$.

The complete distribution is $P(Y^L=0)=0.45$, $P(Y^L=2)=0.25$, and $P(Y^L=6)=0.30$, with zero probability at all other payment amounts.

Check and trap avoided. The masses sum to $0.45+0.25+0.30=1.00$, as required. In particular, $X=4$ contributes to zero and $X=10$ contributes to the cap. Using $P(X>10)=0.18$ as the cap mass would incorrectly omit the probability $0.12$ at exhaustion.

Worked Example: A Mixed Ground-Up Distribution

For a hypothetical property portfolio, suppose ground-up loss $X$, in thousands of dollars, has probability $0.20$ at zero, probability $0.10$ at 3, and a continuous component with total probability $0.70$ that is uniform on $(0,10)$. A layer attaches at $a=2$ and exhausts at $b=8$. Determine the zero-payment mass, the cap mass, the interior density, and the point mass at payment 1. Interest and mortality assumptions are not applicable.

Method. Treat the discrete atoms and continuous component separately, then combine their contributions where they map to the same payment. The continuous component has density $0.70/10=0.07$ on $(0,10)$, and the layer width is $L=8-2=6$.

Zero payment. The atom at zero contributes $0.20$. The continuous component contributes probability $0.70(2/10)=0.14$ on $(0,2]$. The atom at 3 is above attachment. Therefore $P(Y^L=0)=0.20+0.14=0.34$.

Cap payment. Only the continuous component has losses at or above 8; its probability there is $0.70(2/10)=0.14$. There is no atom at 8 or above. Thus $P(Y^L=6)=0.14$.

Interior density and atom. For $0<y<6$, the continuous component gives density $f_{Y^L}(y)=0.07$, since $f_X(2+y)=0.07$. The atom at $X=3$ maps to payment $Y^L=3-2=1$, so $P(Y^L=1)=0.10$. The density does not absorb this atom.

Check and trap avoided. The continuous component between attachment and exhaustion contributes $0.70(6/10)=0.42$. The total probability is $0.34+0.42+0.10+0.14=1.00$. Treating the entire distribution as continuous would miss the mass $0.10$ at payment 1 and would give an incomplete distribution.

Reading the Result Correctly

A layer generally produces a mixed payment distribution even when the ground-up loss is continuous: it creates a mass at zero and another at the layer cap. For a continuous ground-up model, there is no atom at an interior payment, but there is still a density between the endpoints. For a discrete ground-up model, interior payment atoms can occur, and distinct ground-up losses at or above $b$ all combine into the cap mass. A mixed model may have both interior atoms and an interior density.

The formulas also provide useful checks. The zero-payment mass is $F_X(a)$; the cap mass is $1-F_X(b-)$; and all ground-up probability strictly between $a$ and $b$ maps to interior payments. These three regions partition the ground-up outcomes. Thus the zero mass, all interior probability, and the cap mass must sum to one. If a calculation does not satisfy this check, a boundary atom may have been omitted or counted twice.

Key takeaway: A layer payment distribution is obtained by mapping $X\leq a$ to zero, shifting losses $a<X<b$ down by $a$, and mapping $X\geq b$ to the cap $b-a$. Keep the boundary probabilities explicit, especially when the ground-up distribution has atoms.

Check Your Understanding

For each problem, identify endpoint masses as well as any interior distribution. Answers are given so that the probability accounting can be checked.

A hypothetical loss is uniform on $(0,12)$, in thousands of dollars. A layer attaches at 3 and exhausts at 9. Find the zero mass, cap mass, and interior density. (Answer: zero mass $0.2500$, cap mass $0.2500$, density $1/12$ on $0<y<6$.)
A hypothetical loss has exponential distribution with rate $0.2$ per thousand dollars. A layer attaches at 1 and exhausts at 4. Give the zero mass, cap mass, and interior density. (Answer: zero mass $1-e^{-0.2}=0.1813$, cap mass $e^{-0.8}=0.4493$, density $0.2e^{-0.2(1+y)}$ on $0<y<3$.)
A discrete hypothetical loss takes values $2,5,8,11$, in thousands of dollars, with probabilities $0.20,0.30,0.10,0.40$, respectively. A layer attaches at 5 and exhausts at 10. State the payment distribution. (Answer: $P(Y^L=0)=0.50$, $P(Y^L=3)=0.10$, and $P(Y^L=5)=0.40$.)
For a layer attaching at $a$ and exhausting at $b$, suppose $F_X(a)=0.20$ and $F_X(b-)=0.75$. Find the zero mass, the cap mass, and total interior probability. (Answer: zero mass $0.20$, cap mass $1-0.75=0.25$, interior probability $0.75-0.20=0.55$.)
A hypothetical ground-up distribution has $P(X=2)=0.25$, $P(X=5)=0.20$, and $P(X=9)=0.55$. A layer attaches at 2 and exhausts at 7. Find all payment masses. (Answer: $P(Y^L=0)=0.25$, $P(Y^L=3)=0.20$, and $P(Y^L=5)=0.55$.)

The mapping from $X$ to $Y^L$ is a direct way to construct the full layer-payment distribution from the ground-up model. This distributional view is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

What You'll Learn