Discrete Losses and Layer Recoveries
When claim severity takes only specified values, a reinsurance layer can be evaluated directly at each possible loss. The main care is at the treaty boundaries: a loss exactly at the attachment point produces no recovery, while a loss exactly at the top of the layer exhausts it. If either boundary is an atom of the severity distribution, using a strict inequality where an inclusive one is needed can change a probability or an expectation.
This tutorial considers a single per-claim excess-of-loss layer applied to a nonnegative loss \(X\). The layer attaches at \(a\) dollars and has width \(b\) dollars. Thus the layer covers losses above \(a\), up to a maximum payment of \(b\). All examples use hypothetical one-period claim losses. There is no discounting or mortality assumption; the stated discrete probability table fully specifies the severity distribution.
The recovery is zero for \(X\le a\), increases dollar for dollar when \(a<X<a+b\), and equals \(b\) for \(X\ge a+b\). In particular, a claim at the attachment is not a positive recovery, but a claim at the exhaustion point receives the full layer limit. The amount \(b\) is the layer width, not the upper boundary itself.
Direct Calculation From a Discrete Table
Suppose \(X\) takes values \(x_1,\ldots,x_n\), with \(P(X=x_j)=p_j\), where \(p_j\ge0\) and \(\sum_jp_j=1\). Calculate the recovery and retention for each severity, then weight each amount by its probability. This method requires no continuous-density approximation and handles atoms at either boundary directly.
The boundary events can be stated precisely:
- Positive recovery occurs exactly when \(X>a\), so \(P(C>0)=P(X>a)\).
- Full recovery occurs exactly when \(X\ge a+b\), so \(P(C=b)=P(X\ge a+b)\).
- Strictly partial recovery occurs exactly when \(a<X<a+b\).
If \(P(C>0)>0\), the expected recovery conditional on positive recovery is \(E[C\mid C>0]=E[C]/P(C>0)\). This works because \(C=0\) whenever the event \(C>0\) does not occur.
Worked Example: A Severity Atom at the Attachment
A hypothetical insurer has one-claim severity distribution \(P(X=0)=0.10\), \(P(X=100)=0.20\), \(P(X=200)=0.25\), \(P(X=350)=0.25\), and \(P(X=600)=0.20\), with amounts in dollars. A per-claim layer attaches at \(a=\$200\) and has width \(b=\$300\). Find the expected recovery, the probability of positive recovery, the expected recovery conditional on positive recovery, and the expected retained loss. No interest or mortality adjustment applies.
Method. Apply \(C=\min((X-200)_+,300)\) and \(T=X-C\) separately at each severity. The attachment is \(\$200\); the exhaustion point is \(200+300=\$500\).
| Severity \(x\) | Probability | Recovery \(C\) | Retention \(T=x-C\) |
|---|---|---|---|
| \$0 | 0.10 | \$0 | \$0 |
| \$100 | 0.20 | \$0 | \$100 |
| \$200 | 0.25 | \$0 | \$200 |
| \$350 | 0.25 | \$150 | \$200 |
| \$600 | 0.20 | \$300 | \$300 |
At \(X=\$200\), the loss is exactly at attachment, so recovery is zero. At \(X=\$350\), recovery is \(350-200=\$150\). At \(X=\$600\), the amount above attachment exceeds the layer width, so recovery is capped at \(\$300\).
The expected recovery is $$ E[C]=0.10(0)+0.20(0)+0.25(0)+0.25(150)+0.20(300) =37.50+60.00=\$97.50. $$
Positive recovery occurs for \(X=350\) or \(X=600\), giving \(P(C>0)=0.25+0.20=0.4500\). Therefore, $$ E[C\mid C>0]=\frac{97.50}{0.4500}=\$216.67, $$ rounded to the nearest cent.
The expected retention is $$ E[T]=0.10(0)+0.20(100)+0.25(200)+0.25(200)+0.20(300) =20+50+50+60=\$180.00. $$ As a check, \(E[X]=0.20(100)+0.25(200)+0.25(350)+0.20(600)=\$277.50\), and \(E[T]+E[C]=180.00+97.50=\$277.50\). All amounts are in dollars; expectations are reported to the cent and probabilities to four decimals.
Common mistake avoided: Treating the atom at \(X=a\) as a positive recovery would incorrectly add \(0.25\) to the positive-recovery probability. The formula requires \(X>a\), not \(X\ge a\).
Why the Expectation and Boundary Rules Hold
For any realized severity \(x\), the layer first pays zero below or at attachment. Above attachment, the amount above \(a\) is \(x-a\); the layer pays that amount unless it exceeds its width \(b\), in which case payment is capped at \(b\). This gives the clipping formula \(\min((x-a)_+,b)\). Substituting each possible \(x_j\) and using the finite-distribution expectation rule yields the weighted sum for \(E[C]\).
The same pointwise calculation also proves the retained-loss identity. For every severity, $$ T=X-C. $$ Taking expectations on both sides and using linearity gives $$ E[T]=E[X]-E[C], \qquad E[X]=E[T]+E[C]. $$ This identity is both a useful shortcut and an important audit check. It does not mean the layer recovery can be subtracted from every severity as a constant; \(C\) varies with \(X\).
At the upper boundary, \(X=a+b\) gives \((X-a)_+=b\), so the recovery is \(b\). Consequently, full recovery uses the inclusive event \(X\ge a+b\). An atom at \(a+b\) belongs in the full-recovery probability, not the partial-recovery probability.
Worked Example: An Atom at the Exhaustion Point
A hypothetical portfolio has one-claim severity distribution \(P(X=50)=0.15\), \(P(X=100)=0.20\), \(P(X=250)=0.30\), \(P(X=400)=0.20\), and \(P(X=700)=0.15\), in dollars. A per-claim layer attaches at \(a=\$100\) with width \(b=\$300\). Find the probability of positive recovery, the probability of full recovery, the expected recovery, the conditional expected recovery given positive recovery, and the expected retention. There is no interest or mortality adjustment.
Method. The exhaustion point is \(100+300=\$400\). Apply the recovery formula to each table value.
| Severity \(x\) | Probability | Recovery \(C\) | Retention \(T\) |
|---|---|---|---|
| \$50 | 0.15 | \$0 | \$50 |
| \$100 | 0.20 | \$0 | \$100 |
| \$250 | 0.30 | \$150 | \$100 |
| \$400 | 0.20 | \$300 | \$100 |
| \$700 | 0.15 | \$300 | \$400 |
The positive-recovery probability is \(P(X>100)=0.30+0.20+0.15=0.6500\). The full-recovery probability is \(P(X\ge400)=0.20+0.15=0.3500\); the atom at exactly \(\$400\) is included.
The expected recovery is $$ E[C]=0.30(150)+0.20(300)+0.15(300) =45+60+45=\$150.00. $$ Thus $$ E[C\mid C>0]=\frac{150.00}{0.6500}=\$230.77, $$ rounded to the nearest cent. Expected retention is $$ E[T]=0.15(50)+0.20(100)+0.30(100)+0.20(100)+0.15(400) =7.50+20+30+20+60=\$137.50. $$ For verification, \(E[X]=0.15(50)+0.20(100)+0.30(250)+0.20(400)+0.15(700)=\$287.50\), which equals \(150.00+137.50\).
Common mistake avoided: Counting only losses strictly above \(\$400\) as full recoveries would omit the \(\$400\) atom and report \(0.1500\). A loss at the exhaustion point already generates the full \(\$300\) layer payment.
Retained-Loss Outcomes
A layer changes each severity value into a retained amount \(T=x-C\). Different severities can map to the same retained amount. For example, every severity between the attachment and exhaustion point has retention \(a\): the insurer retains the first \(a\) dollars, while the layer pays the excess. Therefore, construct the retained-loss distribution by mapping each original outcome and combining probabilities for any equal retained amounts.
The following example calculates recovery probabilities and the entire retained distribution. The probability table describes a hypothetical single claim, with no interest or mortality adjustment.
Worked Example: Constructing the Retained-Loss Distribution
A hypothetical severity \(X\) takes values \(0,80,150,250,350,500\) dollars with probabilities \(0.10,0.15,0.20,0.25,0.20,0.10\), respectively. A per-claim layer attaches at \(a=\$150\) and has width \(b=\$200\), so its exhaustion point is \(\$350\). Find the probabilities of positive, partial, and full recovery; the expected recovery; the conditional expected recovery given positive recovery; and the retained-loss distribution and mean.
Method. Apply \(C=\min((X-150)_+,200)\), then calculate \(T=X-C\). At \(X=150\), recovery is zero; at \(X=350\), the layer is exhausted and recovery is full.
| Severity \(x\) | Probability | Recovery \(C\) | Retention \(T\) |
|---|---|---|---|
| \$0 | 0.10 | \$0 | \$0 |
| \$80 | 0.15 | \$0 | \$80 |
| \$150 | 0.20 | \$0 | \$150 |
| \$250 | 0.25 | \$100 | \$150 |
| \$350 | 0.20 | \$200 | \$150 |
| \$500 | 0.10 | \$200 | \$300 |
Positive recovery occurs at \(250,350,\) and \(500\), so its probability is \(0.25+0.20+0.10=0.5500\). Partial recovery occurs only at \(250\), with probability \(0.2500\). Full recovery occurs at \(350\) and \(500\), with probability \(0.20+0.10=0.3000\).
The expected recovery and its conditional mean are $$ E[C]=0.25(100)+0.20(200)+0.10(200)=25+40+20=\$85.00, $$ $$ E[C\mid C>0]=\frac{85.00}{0.5500}=\$154.55, $$ rounded to the nearest cent.
Combining the outcomes that produce the same retention gives \(P(T=0)=0.10\), \(P(T=80)=0.15\), \(P(T=150)=0.20+0.25+0.20=0.65\), and \(P(T=300)=0.10\). These probabilities sum to \(1.00\). Hence $$ E[T]=0.10(0)+0.15(80)+0.65(150)+0.10(300) =0+12+97.50+30=\$139.50. $$ The original mean is \(E[X]=0.15(80)+0.20(150)+0.25(250)+0.20(350)+0.10(500)=\$224.50\); \(E[T]+E[C]=139.50+85.00=\$224.50\), as required. Probabilities are reported to four decimals and money to the nearest cent.
Common mistake avoided: Retentions at \(X=250\) and \(X=350\) are both \(\$150\), not \(\$250\) and \(\$350\). In the layer's covered range, the ceded amount is the severity above attachment.
Exam Checks and Common Traps
A reliable calculation separates three questions: what does the layer pay at each severity, which outcomes satisfy the event in the question, and how are those outcomes weighted? Do not infer a probability from an expected payment, or an expected payment from a probability alone. In particular, positive-recovery probability does not determine expected recovery unless the conditional payment amount is also known.
Before finalizing a table-based answer, check that the severity probabilities sum to one, that each recovery lies between zero and the layer width, and that each retention equals severity less recovery. Then compare \(E[X]\) with \(E[C]+E[T]\). This check catches arithmetic errors, including a retention value that was not recalculated after correcting a recovery.
Check Your Understanding
For each hypothetical one-claim discrete severity distribution below, there is no interest or mortality adjustment. Use dollars throughout and apply a per-claim layer. Each problem emphasizes a different boundary or expectation calculation.
- \(X\) is \(40,100,180,260\) with probabilities \(0.20,0.30,0.30,0.20\). A layer attaches at \(100\) with width \(100\). Find \(E[C]\), \(P(C>0)\), and \(P(C=100)\). (Answer: \(E[C]=\$44.00\), \(P(C>0)=0.5000\), \(P(C=100)=0.2000\).)
- \(X\) is \(0,75,150,250,450\) with probabilities \(0.10,0.20,0.30,0.25,0.15\). A layer attaches at \(150\) with width \(200\). Find \(E[C]\) and \(E[T]\). (Answer: \(E[C]=\$55.00\), \(E[T]=\$135.00\).)
- Use the distribution and layer in question 2. Find \(P(C>0)\) and \(E[C\mid C>0]\). (Answer: \(P(C>0)=0.4000\), \(E[C\mid C>0]=\$137.50\).)
- \(X\) is \(100,200,300,400\), each with probability \(0.25\). A layer attaches at \(100\) with width \(200\). Find the probabilities of zero, partial, and full recovery, and find \(E[C]\). (Answer: \(P(C=0)=0.2500\), \(P(0<C<200)=0.2500\), \(P(C=200)=0.5000\), \(E[C]=\$125.00\).)
- \(X\) takes \(0,100,200,300,500\), each with probability \(0.20\). A layer attaches at \(200\) with width \(100\). Find \(E[C]\), \(E[T]\), and \(P(T=200)\). (Answer: \(E[C]=\$40.00\), \(E[T]=\$180.00\), \(P(T=200)=0.4000\).)