Why Compare Two Stop-Loss Estimates?
A stop-loss contract with attachment \(d\) pays the amount by which an aggregate loss \(X\) exceeds \(d\). Its expected payment per observation is \(E[(X-d)_+]\), where \((z)_+=\max(z,0)\). An insurer can estimate this quantity directly from observed aggregate losses or calculate it from a fitted probability model. The calculations may differ even when both use the same attachment.
The empirical estimate summarizes the observations actually collected. The parametric estimate uses a model for the distribution of aggregate losses, with parameters estimated from data or specified from other information. Comparing the estimates helps separate two possible explanations for a difference: the observed sample may be atypical, or the model may describe the upper tail differently from the data. Neither estimate is automatically the “correct” one.
These are expected payments, not premiums including expenses, profit, or risk loads. A meaningful comparison requires the observations and model to refer to the same exposure unit, time period, coverage definition, and attachment. For example, an annual aggregate loss should not be compared with a model for monthly aggregate loss without first aligning the time basis.
The Empirical Estimate
For each observed aggregate loss, subtract the attachment and replace any negative result with zero. Add the resulting payments and divide by the total number of observed periods. Periods with no payment remain in the denominator: the target is the average payment across all periods, not the average payment conditional on a payment occurring.
If the observations are independent and identically distributed draws from the aggregate-loss distribution, the empirical estimate is an unbiased estimator of \(E[(X-d)_+]\). To see why, define \(Z_j=(X_j-d)_+\). Each \(Z_j\) has expectation \(E[(X-d)_+]\), so linearity of expectation gives $$ E\left[\frac{1}{m}\sum_{j=1}^{m} Z_j\right] =\frac{1}{m}\sum_{j=1}^{m}E[Z_j] =E[(X-d)_+]. $$ This justification does not mean a finite-sample estimate must be close to the target. A small sample can contain unusually few or unusually many large losses. The estimate can also be unrepresentative if the observations are not comparable—for example, if exposure levels or coverage terms changed across periods.
Worked Example: Calculating an Empirical Stop-Loss Estimate
A hypothetical portfolio has six observed annual aggregate losses, in thousands of dollars: \(0,4,9,12,18,25\). Estimate the annual stop-loss payment at an attachment of \(10\) thousand dollars.
Method. Calculate the payment \((x_j-10)_+\) for each year, then average all six payments.
The six payments, in thousands, are $$ (0-10)_+=0,\quad (4-10)_+=0,\quad (9-10)_+=0, $$ $$ (12-10)_+=2,\quad (18-10)_+=8,\quad (25-10)_+=15. $$ Their total is \(0+0+0+2+8+15=25\) thousand dollars. Therefore, $$ \widehat{\Pi}_{\mathrm{emp}}(10)=\frac{25}{6} =4.1667\text{ thousand dollars}. $$ Rounded to two decimal places, the estimate is \(\$4,166.67\) per year.
Check. The only positive payments are \(2,8,15\) thousand dollars, whose sum is \(25\) thousand dollars; dividing by six observed years gives the stated average. A common mistake is to divide by three, the number of years with positive payment. That would estimate the average payment conditional on exceeding the attachment, not the stop-loss payment per year.
The Parametric Estimate
A parametric estimate starts with a specified aggregate-loss model. In a compound model, aggregate loss is the sum of a random number of individual losses. The frequency distribution describes the number of losses, and the severity distribution describes the amount of each loss. After fitting or specifying the parameters, calculate the expected excess above the attachment under that model.
For a discrete aggregate-loss distribution with possible values \(s\) and probabilities \(P_{\widehat{\theta}}(X=s)\), the calculation is $$ \widehat{\Pi}_{\mathrm{par}}(d) =\sum_s (s-d)_+P_{\widehat{\theta}}(X=s). $$ For a continuous distribution, the corresponding calculation is an expectation or integral under the fitted density. In either case, the model assigns probability to losses that may not appear in the observed sample. This can matter greatly when the attachment lies in the tail.
A plug-in estimate treats the fitted parameter values as fixed when calculating the expectation. It therefore does not, by itself, include uncertainty from estimating those parameters. It also depends on the adequacy of the chosen frequency and severity distributions. A precise calculation under a poor tail model is not necessarily a reliable estimate.
Worked Example: Comparing an Empirical Estimate with a Fitted Frequency Model
In eight hypothetical periods, a risk has aggregate losses \(0,6,12,0,6,12,0,6\), in dollars. In each period there are two independent exposure units. Each unit produces a claim of exactly \(\$6\) with probability \(p\), and otherwise produces no claim. Thus \(N\sim\operatorname{Binomial}(2,p)\) and aggregate loss is \(X=6N\). Estimate the stop-loss payment at \(d=\$10\) empirically and using the fitted model. Estimate \(p\) by maximum likelihood from the observed counts.
Empirical method. The stop-loss payments for the observed losses are \(0,0,2,0,0,2,0,0\) dollars. Their sum is \(\$4\), so $$ \widehat{\Pi}_{\mathrm{emp}}(10)=\frac{4}{8}=\$0.50. $$
Fit the frequency parameter. Divide each aggregate loss by \(\$6\) to recover the observed counts \(N\): \(0,1,2,0,1,2,0,1\). The total number of claims is \(7\), out of \(8\times 2=16\) exposure units. The binomial maximum-likelihood estimate is $$ \widehat{p}=\frac{7}{16}=0.4375. $$
Calculate the model estimate. Aggregate loss can be \(0,\$6,\) or \(\$12\). At a \(\$10\) attachment, only \(X=\$12\) produces a positive payment, of \(\$2\). This occurs when both exposure units produce a claim: $$ \widehat{\Pi}_{\mathrm{par}}(10) =2P(N=2) =2\widehat{p}^{\,2} =2(0.4375)^2 =2(0.19140625) =\$0.3828125. $$ Rounded to the nearest cent, the model estimate is \(\$0.38\) per period. It is about \(\$0.12\) lower than the empirical estimate.
Check and interpretation. The model assigns probability \(0.19140625\) to the only outcome above the attachment, and multiplying that probability by its \(\$2\) payment gives \(\$0.3828125\). The observed sample contains two such outcomes in eight periods, a sample frequency of \(0.25\), compared with the fitted model probability of about \(0.1914\). This difference is consistent with sampling variation; it does not, on its own, establish that the binomial model is wrong. The common mistake is to average only the two positive observed payments, obtaining \(\$2\), rather than average across all eight periods.
How Tail Assumptions Change the Comparison
The attachment determines which part of the aggregate-loss distribution drives the estimate. When \(d\) is high, most observations may contribute zero, and the estimate depends on relatively few large outcomes or on the model’s probability of losses beyond \(d\). A model that assigns more probability to large losses can produce a larger stop-loss estimate even if its mean loss is the same as that of another model.
For a continuous loss with survival function \(S_X(x)=P(X>x)\), the expected stop-loss payment can also be written as $$ E[(X-d)_+]=\int_d^\infty S_X(x)\,dx, $$ provided the expectation is finite. This identity follows from writing \((X-d)_+=\int_d^\infty \mathbf{1}_{\{X>x\}}\,dx\), where the indicator is one when \(X>x\) and zero otherwise, and then taking expectations. It makes the tail dependence explicit: the estimate accumulates the survival probabilities above the attachment.
Worked Example: Comparing Tail Assumptions at the Same Mean
A hypothetical one-claim-per-period portfolio has eight observed claim amounts, in dollars: \(2,5,8,10,12,20,15,8\). Compare the empirical stop-loss estimate at \(d=\$10\) with two possible severity models, each having mean \(\$10\): an exponential model with mean \(\$10\), and a two-point model with claim amount \(\$5\) or \(\$15\), each with probability \(0.5\). Treat the two models as alternative tail assumptions, not as claims that both have been fitted to every feature of the sample.
Empirical estimate. The eight excess payments are \(0,0,0,0,2,10,5,0\) dollars. Their sum is \(\$17\), giving $$ \widehat{\Pi}_{\mathrm{emp}}(10)=\frac{17}{8}=\$2.125, $$ or \(\$2.13\) to the nearest cent.
Exponential model. If \(X\) is exponential with mean \(\theta=\$10\), its survival function is \(S_X(x)=e^{-x/10}\) for \(x\geq 0\). Using the tail integral, $$ E[(X-10)_+]=\int_{10}^{\infty}e^{-x/10}\,dx =10e^{-1} \approx \$3.6788. $$ Thus the exponential-model estimate is \(\$3.68\) to the nearest cent.
Two-point model. A \(\$5\) claim has no excess at the \(\$10\) attachment, while a \(\$15\) claim has excess \(\$5\). Therefore, $$ E[(X-10)_+]=0.5(0)+0.5(5)=\$2.50. $$ The empirical estimate, the exponential estimate, and the two-point estimate are respectively \(\$2.13\), \(\$3.68\), and \(\$2.50\). Both models have mean claim amount \(\$10\), yet their stop-loss estimates differ because their probabilities of larger claims differ. The exponential estimate is also above the observed empirical estimate, but eight observations alone cannot determine whether this reflects ordinary sampling variation or a poor tail assumption.
Interpreting Differences Carefully
The comparison is most useful when each estimate is interpreted in light of how it was produced. The empirical estimate makes relatively few distributional assumptions, but can be unstable when the sample is small or the attachment is high. The parametric estimate can use the structure of a frequency-severity model and can assign probabilities to outcomes not seen in the sample, but its result depends on the model and fitted parameters.
| Feature | Empirical estimate | Parametric plug-in estimate |
|---|---|---|
| Calculation | Average observed excess payments over all periods | Expected excess payment under the fitted model |
| Main source of sensitivity | Which large losses happened to appear in the sample | Model form, fitted parameters, and modeled tail |
| Unobserved large losses | Cannot contribute unless they occur in the data | Can contribute if the model assigns them probability |
| Parameter uncertainty | Reflected indirectly in sample variation | Not included in a plug-in expectation unless assessed separately |
A large difference should prompt checks before a conclusion is drawn. Confirm that the same currency, exposure basis, period, and attachment were used. Verify that the empirical observations are aggregate losses under the same coverage represented by the model. Then inspect the fitted frequency and severity assumptions and consider whether the attachment is far enough into the tail that a small change in tail probabilities has a substantial effect.
A common exam trap is to treat the two estimates as interchangeable merely because both are called stop-loss estimates. They target the same expected payment only when they concern the same loss definition and population or process. One is a sample average; the other is an expectation under a specified model. Their numerical difference is information to investigate, not an arithmetic error by default.
Check Your Understanding
For each problem, use the stated attachment and keep the estimate on a per-period basis.
- Observed aggregate losses are \(\$3,\$7,\$11,\$16,\$22\). Calculate the empirical stop-loss estimate at \(d=\$12\). (Answer: \(\$2.80\))
- Let \(N\sim\operatorname{Binomial}(2,0.4)\), with a fixed claim amount of \(\$5\) per claim and aggregate loss \(X=5N\). Calculate the parametric stop-loss estimate at \(d=\$8\). (Answer: \(\$0.32\))
- Four observed annual aggregate losses are \(\$0,\$10,\$14,\$20\). Calculate the empirical stop-loss estimate at \(d=\$12\), and state why the two losses below the attachment remain in the denominator. (Answer: \(\$2.50\) per year; all four years are included because the estimate is average payment per year.)
- A single claim per period has an exponential severity distribution with mean \(\$8\). Calculate the stop-loss expectation at \(d=\$12\), rounded to the nearest cent. (Answer: \(8e^{-12/8}\approx \$1.79\))
- Two fitted severity models both have mean \(\$10\). At an attachment of \(\$10\), one model gives an expected excess of \(\$2.00\) and the other gives \(\$3.50\). Give one valid interpretation of the difference. (Answer: The models assign different probabilities or amounts to losses above the attachment; equal means do not imply equal stop-loss expectations.)