Integer-Attachment Recursion for Stop-Loss Premiums

Why Integer Attachments Allow a Recursion

For an aggregate loss $S$, a stop-loss contract with attachment $d$ pays the amount by which $S$ exceeds $d$. When aggregate losses and attachments are integers, raising the attachment by one unit reduces the payment by either zero or one unit for each outcome. That simple outcome-by-outcome change gives a direct recursion for the expected payment.

This is useful when premiums are needed at several consecutive attachments. Instead of summing the payment separately at every attachment, calculate one starting premium and update it using the aggregate loss tail. The key is to use the correct tail: the update from $d$ to $d+1$ uses the strict probability $P(S>d)$.

Definition: Let $S$ be a nonnegative integer-valued aggregate loss with finite expected value. For a nonnegative integer attachment $d$, the undiscounted pure stop-loss premium is $$ \Pi(d)=E[(S-d)^+], \qquad (x)^+=\max(x,0). $$ It is the expected payment before expenses, profit loads, or discounting. All amounts in a calculation must use the same loss unit.

The Integer-Attachment Update

For any fixed integer outcome $S=s$, compare the payment at attachment $d$ with the payment at attachment $d+1$. If $s\leq d$, both payments are zero. If $s>d$, increasing the attachment by one reduces the payment by exactly one. Therefore the difference is an indicator of the event $S>d$.

(S-d)^+-(S-(d+1))^+ = \begin{cases} 1, & S>d,\\ 0, & S\leq d. \end{cases} = \mathbf{1}_{\{S>d\}}.

Taking expectations gives the recursion. The strict inequality matters: when $S=d$, neither contract pays anything, and that outcome does not reduce the payment when the attachment rises to $d+1$.

Theorem: If $S$ is nonnegative and integer-valued and $d$ is a nonnegative integer, then $$ \Pi(d+1)=\Pi(d)-P(S>d). $$ Equivalently, $P(S>d)=P(S\geq d+1)$. This one-unit update is specific to integer attachment steps and integer-valued losses; it should not be applied unchanged to noninteger attachments or to a continuous aggregate loss.

A convenient starting value is $\Pi(0)=E[S]$, since a contract attached at zero pays the entire nonnegative loss. Repeated updates then give

\Pi(d)=E[S]-\sum_{k=0}^{d-1}P(S>k) =\sum_{k=d}^{\infty}P(S>k).

The second equality follows because $E[S]=\sum_{k=0}^{\infty}P(S>k)$ for a nonnegative integer-valued $S$. The tail-sum form can also be used directly when the tail probabilities are provided. If $S$ has a finite maximum value, the premiums are zero at and above that maximum.

Computing Premiums from an Aggregate Probability Table

When a probability table gives the distribution of $S$, first find the strict tail at each integer attachment. A tail at $d$ includes probabilities at loss values strictly greater than $d$, not probabilities at $d$ itself. Then subtract that tail from the premium at $d$ to obtain the premium at $d+1$.

Worked Example: Updating Premiums from a Loss Table

A hypothetical aggregate contract has loss $S$, measured in thousands of dollars, with the following distribution. Find the pure premiums at attachments $0$ through $7$, in thousands of dollars.

$s$	0	2	4	7
$P(S=s)$	0.20	0.30	0.30	0.20

Method. Calculate $\Pi(0)=E[S]$, obtain each strict tail from the table, and apply the integer-attachment recursion. The probabilities sum to $0.20+0.30+0.30+0.20=1.00$, as required.

The starting premium is $$ \Pi(0)=E[S]=(0)(0.20)+(2)(0.30)+(4)(0.30)+(7)(0.20) =0+0.60+1.20+1.40=3.20. $$ For example, $P(S>2)=P(S=4)+P(S=7)=0.30+0.20=0.50$. The full set of strict tails and updates is:

Attachment $d$	$P(S>d)$	Premium update	$\Pi(d)$
0	0.80	$\Pi(0)=3.20$	3.20
1	0.80	$\Pi(1)=3.20-0.80$	2.40
2	0.50	$\Pi(2)=2.40-0.80$	1.60
3	0.50	$\Pi(3)=1.60-0.50$	1.10
4	0.20	$\Pi(4)=1.10-0.50$	0.60
5	0.20	$\Pi(5)=0.60-0.20$	0.40
6	0.20	$\Pi(6)=0.40-0.20$	0.20
7	0.00	$\Pi(7)=0.20-0.20$	0.00

The entries use $P(S>0)=0.80$, $P(S>1)=0.80$, $P(S>2)=0.50$, $P(S>3)=0.50$, $P(S>4)=0.20$, $P(S>5)=0.20$, and $P(S>6)=0.20$. As a direct check, at attachment $4$, only the outcome $S=7$ produces a payment, so $\Pi(4)=(7-4)(0.20)=0.60$, matching the recursion. The common mistake to avoid is including $P(S=2)$ in $P(S>2)$; the strict tail at 2 is 0.50, not 0.80.

Using a Model to Obtain the Aggregate Tail

The recursion needs only the aggregate loss tail, not a particular way of constructing $S$. A model for claim count and claim severity can supply that tail. Once the aggregate distribution is known, the same update applies.

Worked Example: A Compound Binomial Aggregate Loss

A hypothetical portfolio has claim count $N\sim\operatorname{Binomial}(2,0.5)$. Each claim has a fixed severity of \$3, and there are no other payments. Thus $S=3N$. Find the pure stop-loss premiums at integer-dollar attachments from 0 through 6.

Method. First derive the aggregate loss probabilities from the binomial count model. Since $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$, the aggregate loss is 0, 3, or 6 dollars with those respective probabilities. The probabilities sum to 1, and $E[S]=3E[N]=3(2)(0.5)=\$3$, so $\Pi(0)=\$3$.

The strict tails are $P(S>0)=P(S>1)=P(S>2)=0.75$, because losses 3 and 6 exceed those attachments. Also, $P(S>3)=P(S>4)=P(S>5)=0.25$, because only loss 6 exceeds those attachments. Applying each update:

\begin{aligned} \Pi(1)&=3.00-0.75=2.25,\\ \Pi(2)&=2.25-0.75=1.50,\\ \Pi(3)&=1.50-0.75=0.75,\\ \Pi(4)&=0.75-0.25=0.50,\\ \Pi(5)&=0.50-0.25=0.25,\\ \Pi(6)&=0.25-0.25=0.00. \end{aligned}

Check directly at attachment $2$: the payment is \$1 when $S=3$, with probability 0.50, and \$4 when $S=6$, with probability 0.25. Thus $\Pi(2)=(1)(0.50)+(4)(0.25)=\$1.50$, as calculated. The common mistake is to treat $S$ as having the same distribution as $N$; severity must be applied to the count before calculating aggregate tails.

Starting at an Intermediate Attachment

A calculation does not have to begin at zero. If a premium at one attachment is known, subsequent premiums follow by subtracting the appropriate tails. This is especially efficient when a problem supplies a premium at a starting point and asks for a short sequence of higher attachments.

Worked Example: Advancing from a Given Attachment

A hypothetical aggregate loss, measured in thousands of dollars, has the distribution below. Find the premiums at attachments 2, 3, 4, and 5. State the values in thousands of dollars and verify the premium at attachment 2 directly.

$s$	0	1	2	4	5
$P(S=s)$	0.10	0.20	0.30	0.25	0.15

Method. The probabilities total $0.10+0.20+0.30+0.25+0.15=1.00$. The strict tails needed are $P(S>2)=0.25+0.15=0.40$, $P(S>3)=0.40$, and $P(S>4)=0.15$. To obtain the first requested premium, calculate $$ \Pi(2)=(4-2)(0.25)+(5-2)(0.15)=0.50+0.45=0.95. $$ Then update: $$ \begin{aligned} \Pi(3)&=0.95-0.40=0.55,\\ \Pi(4)&=0.55-0.40=0.15,\\ \Pi(5)&=0.15-0.15=0.00. \end{aligned} $$

All amounts are in thousands of dollars. At attachment 3, a direct check gives $(4-3)(0.25)+(5-3)(0.15)=0.25+0.30=0.55$, agreeing with the recursion. The common mistake is to subtract $P(S\geq 3)$ when moving from attachment 2 to 3; the correct probability is $P(S>2)=0.40$, which excludes the loss of 2 but includes losses 4 and 5.

Tail Indexing and Practical Checks

The update can be read in two equivalent ways: subtract $P(S>d)$, or subtract $P(S\geq d+1)$. It is not the same as subtracting $P(S\geq d)$. When there is probability mass at $d$, that distinction changes the result. This is why tail tables should be labelled carefully and why a strict inequality should be retained in the recursion.

A reliable calculation has three checks:

Check the starting value. If starting at zero, $\Pi(0)$ must equal $E[S]$.
Check the tail event. For the update from $d$ to $d+1$, include only losses greater than $d$.
Check the endpoint. Once the attachment is at least the maximum possible aggregate loss, the premium must be zero.

The recursion is a calculation tool for expected stop-loss payments, not a change in the contract definition. It assumes a unit increase in attachment and integer-valued aggregate loss. For a noninteger attachment step, the payment reduction need not be exactly one on every outcome; for a continuous loss model, probability at a single point and the size of the attachment change must be handled using the appropriate tail or integral calculation.

Key takeaway: For nonnegative integer-valued $S$, one unit of additional attachment reduces the pure stop-loss premium by $P(S>d)$. Preserve the strict tail, calculate consistently in the stated loss unit, and verify at least one result using $E[(S-d)^+]$.

Check Your Understanding

Use the strict-tail recursion where it applies. Each problem is independent.

A hypothetical aggregate loss has $P(S=0)=0.25$, $P(S=1)=0.25$, and $P(S=3)=0.50$. Find $\Pi(0)$, $\Pi(1)$, $\Pi(2)$, and $\Pi(3)$. (Answer: $1.75,\ 1.00,\ 0.50,\ 0.00$, respectively.)
A loss has probabilities $P(S=0)=0.40$, $P(S=2)=0.40$, and $P(S=5)=0.20$. Find $\Pi(1)$ and $\Pi(3)$. (Answer: $\Pi(1)=1.20$ and $\Pi(3)=0.40$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.5)$ claims, each of fixed severity \$2, and $S=2N$. Find $\Pi(2)$ and $\Pi(4)$. (Answer: $\Pi(2)=\$1.25$ and $\Pi(4)=\$0.25$.)
Suppose $\Pi(0)=2.35$, and the strict tails at $d=0,1,2,3,4$ are $0.90, 0.70, 0.40, 0.20, 0.05$, respectively. Find $\Pi(3)$ and $\Pi(5)$. (Answer: $\Pi(3)=0.35$ and $\Pi(5)=0.10$.)
If $\Pi(2)=1.10$ and $\Pi(3)=0.65$ for a nonnegative integer-valued aggregate loss, what is $P(S>2)$? Can these two premiums alone determine $P(S=3)$? (Answer: $P(S>2)=0.45$; no, they do not determine $P(S=3)$.)

Attachment \(d\)	\(P(S>d)\)	Premium update	\(\Pi(d)\)
0	0.80	\(\Pi(0)=3.20\)	3.20
1	0.80	\(\Pi(1)=3.20-0.80\)	2.40
2	0.50	\(\Pi(2)=2.40-0.80\)	1.60
3	0.50	\(\Pi(3)=1.60-0.50\)	1.10
4	0.20	\(\Pi(4)=1.10-0.50\)	0.60
5	0.20	\(\Pi(5)=0.60-0.20\)	0.40
6	0.20	\(\Pi(6)=0.40-0.20\)	0.20
7	0.00	\(\Pi(7)=0.20-0.20\)	0.00

What You'll Learn