Stop-Loss Pricing from Grouped Loss Data

Why Grouped Loss Data Need an Assumption

A stop-loss payment with attachment point $d$ is $(S-d)_+=\max(S-d,0)$, where $S$ is aggregate loss. Its expected value is the pure premium for that cover. If the observed aggregate losses are available only as counts in intervals, the exact losses within each interval are unknown. In particular, an interval that straddles $d$ does not tell us how many losses exceed $d$, or by how much.

Grouped data therefore do not, by themselves, determine an exact stop-loss premium. A pricing estimate requires a stated assumption about how losses are distributed within each interval. This tutorial develops the uniform-within-bin assumption and compares it with a grouped-midpoint approximation. The techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let the observed aggregate losses be summarized by disjoint finite intervals $[a_j,b_j)$, with $n_j$ observations in interval $j$ and total $N=\sum_j n_j$. The interval weight is $p_j=n_j/N$. Under a uniform-within-bin assumption, the loss conditional on belonging to interval $j$ is uniform on $[a_j,b_j)$. The grouped-data estimate of the expected stop-loss payment is the weighted sum of the conditional expected payments in the intervals.

Calculating the Uniform-Within-Bin Estimate

For a bin $[a,b)$, the conditional expected payment depends on where the attachment $d$ lies. If the whole bin is below the attachment, its payment is zero. If the whole bin is above the attachment, the conditional mean loss is the bin midpoint, so the expected payment is that midpoint less $d$. If the attachment falls inside the bin, only the portion above $d$ contributes.

E[(S-d)_+\mid a\leq S<b]= \begin{cases} 0, & d\geq b,\\[4pt] \dfrac{(a+b)/2-d}{1}, & d\leq a,\\[8pt] \dfrac{(b-d)^2}{2(b-a)}, & a<d<b. \end{cases}

The denominator $1$ in the second case simply emphasizes that this is the conditional payment for the bin, not yet weighted by how many observations fall there. In applications, multiply each bin’s conditional expected payment by $p_j$, then add across bins. The result is in the same units as the loss intervals.

Find each interval weight.
Divide its count by the total count. Check that all weights sum to one.

Locate the attachment.
Classify each bin as entirely below, entirely above, or straddling $d$.

Calculate conditional payments.
Use the appropriate case of the formula for each interval.

Weight and sum.
Multiply each conditional payment by its interval weight and add the contributions.

Why the Straddling-Bin Formula Holds

Suppose $a<d<b$ and $S\mid(a\leq S<b)$ is uniform. Its density within the bin is $1/(b-a)$. Losses from $a$ through $d$ have no stop-loss payment; losses above $d$ pay $s-d$. Integrating that payment over the part of the bin that contributes gives

E[(S-d)_+\mid a\leq S<b] =\frac{1}{b-a}\int_d^b (s-d)\,ds =\frac{(b-d)^2}{2(b-a)}.

This derivation also explains why simply multiplying the bin probability by its midpoint’s payment can be inaccurate: the midpoint may lie below $d$, even though some losses in that bin exceed $d$. The formula accounts for those losses and their average excess, conditional on the uniform assumption.

Worked Example: An Attachment Inside One Bin

A hypothetical portfolio has 100 observed aggregate losses, measured in thousands of dollars. The counts are 20 in $[0,10)$, 30 in $[10,20)$, 30 in $[20,30)$, and 20 in $[30,40)$. Estimate the expected payment per portfolio under a stop-loss cover with attachment $d=25$, assuming losses are uniform within each bin. Also compare the result with the grouped-midpoint approximation.

Method and weights. The total count is $N=100$, so the four weights are $0.20, 0.30, 0.30,$ and $0.20$. They sum to $1.00$. All calculations are in thousands of dollars.

Bins below the attachment. The first two bins end at 10 and 20, both below $d=25$, so their conditional payments and weighted contributions are zero.

Straddling bin $[20,30)$. Here $a=20$, $b=30$, and $b-a=10$. The conditional expected payment is $ (30-25)^2/[2(10)]=25/20=1.25 $. Its weighted contribution is $0.30(1.25)=0.375$ thousand dollars.

Bin above the attachment $[30,40)$. The conditional midpoint is $(30+40)/2=35$, so the conditional expected payment is $35-25=10$ thousand dollars. Its weighted contribution is $0.20(10)=2.000$ thousand dollars.

Uniform estimate. Adding the four weighted contributions gives $0+0+0.375+2.000=2.375$ thousand dollars, or $\$2,375.00$ per portfolio.

Midpoint comparison. Representing each bin by its midpoint gives payments $0,0,0,$ and $35-25=10$ thousand dollars. The resulting estimate is $0.20(10)=2.000$ thousand dollars, or $\$2,000.00$. It is $\$375.00$ lower than the uniform estimate because it assigns no payment to the portion of $[20,30)$ above the attachment. The midpoint estimate is an approximation under a different within-bin representation, not a correction to the uniform calculation.

Check. The only positive contributions are from bins containing losses above 25. Their total is $2.375$ thousand dollars, as calculated. The common mistake to avoid is treating the whole straddling bin as either entirely below or entirely above the attachment.

Comparing Interpolation Choices

The uniform-within-bin approach spreads each interval’s probability evenly across its width. Equivalently, within that interval, the cumulative distribution increases linearly with loss. A grouped-midpoint approximation instead places all the interval’s weight at its midpoint. The two methods agree on the bin midpoint as the conditional mean, but need not agree on the expected stop-loss payment: the payment function changes at $d$, so knowing only the mean does not determine its expectation.

The comparison is most consequential when the attachment lies inside a bin. It can also matter when bins have different widths: a wider straddling bin can contain a larger range of losses that exceed the attachment. Interval weights come from counts, not widths. Width affects the conditional calculation; it does not replace the observed frequency.

Worked Example: Unequal Bin Widths and a Per-Policy Price

A hypothetical insurer groups 100 annual aggregate losses, measured in thousands of dollars, as follows: 30 losses in $[0,2)$, 40 in $[2,6)$, and 30 in $[6,10)$. Estimate the average stop-loss payment per policy at $d=4$, first assuming uniform losses within each bin and then using the grouped-midpoint approximation. State the corresponding expected payment for all 100 policies.

Weights. The interval weights are $30/100=0.30$, $40/100=0.40$, and $30/100=0.30$; they sum to one.

First bin $[0,2)$. It is below $d=4$, so its contribution is zero under either method.

Straddling bin $[2,6)$. Its width is $6-2=4$, and $b-d=6-4=2$. Under the uniform assumption, the conditional expected payment is $2^2/[2(4)]=4/8=0.5$ thousand dollars. Weighting gives $0.40(0.5)=0.20$ thousand dollars per policy.

Third bin $[6,10)$. This bin is entirely above $d$. Its conditional midpoint is $(6+10)/2=8$, so its conditional expected payment is $8-4=4$ thousand dollars. The weighted contribution is $0.30(4)=1.20$ thousand dollars per policy.

Uniform estimate. The sum is $0+0.20+1.20=1.40$ thousand dollars, or $\$1,400.00$ per policy. For 100 policies, the expected total payment is $100(\$1,400.00)=\$140,000.00$.

Midpoint comparison and check. The three bin midpoints are 1, 4, and 8. Their payments are 0, 0, and 4 thousand dollars, respectively, so the weighted midpoint estimate is $0.30(0)+0.40(0)+0.30(4)=1.20$ thousand dollars per policy. It gives $\$120,000.00$ for 100 policies, $\$20,000.00$ below the uniform estimate. The difference is exactly the uniform contribution from the straddling bin, which the midpoint representation assigns zero payment.

Common mistake avoided. The counts are not probabilities until divided by 100, and the per-policy estimate is not yet the total for 100 policies. Weight first, then multiply by the number of policies if a portfolio total is required.

Sensitivity and Limits of Grouped Data

The uniform estimate is a model-based estimate, not a value identified exactly by the grouped counts. If the true losses cluster near the lower end of a straddling bin, its stop-loss contribution may be smaller than the uniform estimate; if they cluster near the upper end, it may be larger. A midpoint approximation provides one alternative, but it too imposes a particular representation of the data. When the estimated price matters materially, report the assumption and, where practical, show how the estimate changes under a reasonable alternative.

Finite intervals are essential to the formulas above. An open-ended interval such as $[b,\infty)$ cannot be treated as a uniform bin: there is no finite upper endpoint, and a uniform distribution over an infinite interval is not a probability distribution. The grouped count in that tail does not reveal the average excess loss there. A finite upper bound supported by the data or an explicit tail model is needed before estimating its stop-loss contribution.

Common Exam Traps

Using counts without normalizing. For an expected payment per observation, each bin’s contribution is weighted by $n_j/N$, not by $n_j$. Multiplying by counts gives a total across observations only if that is the intended quantity.
Using the midpoint in a straddling bin as if it settled the question. The midpoint approximation can miss payments above $d$ when the midpoint is at or below $d$. Under a uniform assumption, use the partial-bin formula.
Weighting by interval width. The observed count determines the bin’s probability. Width appears in the conditional uniform formula, not as a substitute for frequency.
Applying the formula to an open-ended bin. Without a finite endpoint or tail assumption, the grouped data alone do not specify the stop-loss payment in that bin.
Mixing units. If intervals are in thousands of dollars, the result is in thousands of dollars until it is converted. State whether the result is per policy or for the whole group.

Key takeaway: For grouped aggregate losses, calculate a conditional stop-loss payment in each interval, weight it by that interval’s observed proportion, and sum. The uniform-within-bin estimate handles an attachment inside an interval explicitly; comparison with a midpoint approximation makes the effect of the interpolation assumption visible.

Check Your Understanding

Assume uniform losses within each finite interval unless a question asks for a midpoint comparison. Give expected payments in the units used for the loss intervals.

Grouped losses have probabilities $0.60$ in $[0,4)$ and $0.40$ in $[4,8)$. At $d=6$, what is the uniform-within-bin estimate? (Answer: $0.200$ loss units.)
There are 25 observations in $[0,10)$ and 75 in $[10,20)$. At $d=10$, what is the uniform estimate per observation? (Answer: $3.75$ loss units.)
There are 10, 20, and 20 observations in $[0,5)$, $[5,15)$, and $[15,25)$, respectively. At $d=8$, what is the uniform estimate per observation? (Answer: $5.78$ loss units.)
For 100 observations, counts are 30 in $[0,2)$, 40 in $[2,6)$, and 30 in $[6,10)$, with losses in thousands of dollars. At $d=4$, how much larger is the uniform estimate for all observations than the midpoint estimate? (Answer: $\$20,000.00$.)
A grouped table reports 12 observations in $[50,\infty)$. Can the uniform-within-bin formula estimate their expected stop-loss payment? What extra information is needed? (Answer: No; a finite upper endpoint or an explicit tail model is needed.)

What You'll Learn