From a Severity Model to an Exposure Curve
An exposure curve summarizes how expected losses change as the available policy limit changes. When ground-up claim data are sparse at high severities, a fitted parametric severity model can provide a smooth survival function and a basis for calculating limited expected losses. The benefit comes with a responsibility: the resulting curve depends on the chosen distribution, the quality of the data, and the range over which the model is used.
This tutorial fits exponential and Pareto severity models and uses their survival functions to calculate limited expected losses. All claim amounts in the examples are hypothetical and are stated in thousands of dollars. Assume a consistent ground-up loss basis: claims have been adjusted, as needed, to represent the same coverage and valuation basis before fitting.
The curve \(G(u)\) is nondecreasing and lies between zero and one. It approaches one as the limit grows without bound when \(E[X]\) is finite. An increased limits factor is not the same as \(G(u_2)\): it compares limited losses at two limits, while \(G(u_2)\) compares limited loss with unlimited expected loss.
Limited Expected Loss from the Survival Function
For any nonnegative loss, the limited expected loss has a useful integral representation. For a fixed loss amount \(X=x\), the amount \(\min(x,u)\) is the length of the interval from zero to \(u\) over which the indicator \(I(x>t)\) equals one. Therefore,
Taking expectations and interchanging expectation and integration gives the survival-function formula. The interchange is valid here because the integrand is nonnegative.
This result explains why the entire severity distribution below the limit matters: \(L(u)\) is the area under the survival curve from zero to \(u\). It also shows that, wherever the survival function is continuous, the slope of \(L(u)\) is \(S_X(u)\). A model with an excessively heavy tail can materially increase estimates at high limits.
Fitting and Using an Exponential Severity Model
An exponential model with rate \(\lambda>0\) has density \(f_X(x)=\lambda e^{-\lambda x}\) and survival function \(S_X(x)=e^{-\lambda x}\) for \(x\geq 0\). Its mean is \(E[X]=1/\lambda\). For independent, complete observations \(x_1,\ldots,x_n\) from this model, the maximum-likelihood estimate is \(\hat{\lambda}=n/\sum_j x_j=1/\bar{x}\). The model assumes a constant hazard over severity: conditional on exceeding any amount, the remaining severity has the same exponential distribution. That assumption may be unsuitable for claim data with a heavier tail or changing hazard.
Integrating the exponential survival function gives the limited expected loss and, after dividing by the model mean, the exposure curve:
Worked Example: Fitting an Exponential Model to Complete Claims
A hypothetical insurer has four complete ground-up claims of \(2,4,6,\) and \(8\) thousand dollars. Fit an exponential model by maximum likelihood, then calculate limited expected losses at limits of \(5\) and \(10\) thousand dollars, the exposure-curve values at those limits, and the increased limits factor from \(5\) to \(10\).
Method and estimate. The observations are complete and measured in thousands, so their total is \(2+4+6+8=20\) thousand and their count is \(n=4\). Thus \(\hat{\lambda}=4/20=0.2\) per thousand, and the fitted mean is \(1/0.2=5\) thousand.
Limit \(u=5\). The fitted limited expected loss is \(L(5)=5(1-e^{-0.2(5)})=5(1-e^{-1})\). Since \(e^{-1}=0.3678794\), this is \(5(0.6321206)=3.1606\) thousand. The exposure-curve value is \(G(5)=3.1606/5=0.6321\).
Limit \(u=10\). Here \(L(10)=5(1-e^{-0.2(10)})=5(1-e^{-2})\). Since \(e^{-2}=0.1353353\), this is \(5(0.8646647)=4.3233\) thousand. The exposure-curve value is \(G(10)=4.3233/5=0.8647\).
Increased limits factor and check. The factor from \(5\) to \(10\) thousand is \(L(10)/L(5)=4.3233/3.1606=1.3679\). Using the unrounded expressions gives \((1-e^{-2})/(1-e^{-1})=1+e^{-1}=1.3679\). The factor exceeds one, as expected when the limit increases, and the exposure-curve values are both between zero and one. Rounding is to four decimal places for rates, curve values, and the factor, and to four decimal places in thousands for the limited losses. The common mistake this method avoids is treating the exposure-curve value at the higher limit as the increased limits factor.
Fitting a Pareto Tail
A Pareto Type I model with threshold \(x_m>0\) and shape parameter \(\alpha>0\) has support \(x\geq x_m\), survival function \(S_X(x)=(x_m/x)^\alpha\) for \(x\geq x_m\), and survival equal to one below \(x_m\). If all observations are complete and independently drawn from this distribution with known threshold \(x_m\), the maximum-likelihood estimate is
This follows from the density \(f_X(x)=\alpha x_m^\alpha x^{-(\alpha+1)}\): the log-likelihood, apart from terms not involving \(\alpha\), is \(n\ln(\alpha)-\alpha\sum_j\ln(x_j/x_m)\). Differentiating with respect to \(\alpha\), setting the derivative to zero, and solving yields the estimate above. For this model \(E[X]=\alpha x_m/(\alpha-1)\) only when \(\alpha>1\); if \(\alpha\leq1\), the unlimited mean is infinite, so the normalized exposure curve \(L(u)/E[X]\) is not defined as an ordinary finite ratio.
When \(u\geq x_m\), integration of the Pareto survival function gives
For \(\alpha>1\), this expression tends to \(x_m+x_m/(\alpha-1)=E[X]\) as \(u\) tends to infinity. For \(u<x_m\), the loss always exceeds the limit, so \(L(u)=u\). At \(\alpha=1\), the integral from \(x_m\) to \(u\) is \(x_m\ln(u/x_m)\); the limited loss remains finite at any finite limit, even though the unlimited mean is infinite.
Worked Example: Fitting a Pareto Model and Building Curve Values
Suppose a hypothetical portfolio has four complete ground-up claims, in thousands of dollars, of \(10,20,20,\) and \(40\). Assume a Pareto Type I model with known threshold \(x_m=10\). Estimate the shape parameter, then calculate the mean, limited expected losses at \(20\) and \(40\), and the exposure-curve values at those limits.
Fit the parameter. The logarithmic ratios are \(\ln(10/10)=0\), \(\ln(20/10)=\ln2\), \(\ln(20/10)=\ln2\), and \(\ln(40/10)=\ln4=2\ln2\). Their sum is \(4\ln2=2.7725887\). Therefore, \(\hat{\alpha}=4/(4\ln2)=1/\ln2=1.4427\), using the unrounded value in subsequent calculations. Since \(\hat{\alpha}>1\), the fitted unlimited mean is finite.
Calculate the fitted mean. With \(\hat{\alpha}=1/\ln2\), the shape excess is \(\hat{\alpha}-1=(1-\ln2)/\ln2=0.4426950\). Thus \(E[X]=10\hat{\alpha}/(\hat{\alpha}-1)=10/(1-\ln2)=32.5889\) thousand.
Calculate \(L(20)\). The coefficient \(x_m/(\hat{\alpha}-1)\) is \(10( \ln2)/(1-\ln2)=22.5889\) thousand. Also, \((10/20)^{\hat{\alpha}-1}=2^{-(1/\ln2-1)}=2/e=0.7357589\). Hence \(L(20)=10+22.5889(1-0.7357589)=15.9689\) thousand, using unrounded parameter values in the calculation. The corresponding curve value is \(G(20)=15.9689/32.5889=0.4900\).
Calculate \(L(40)\). At this limit, \((10/40)^{\hat{\alpha}-1}=4^{-(1/\ln2-1)}=4/e^2=0.5413411\). Therefore, \(L(40)=10+22.5889(1-0.5413411)=20.3606\) thousand. The corresponding curve value is \(G(40)=20.3606/32.5889=0.6248\).
Check. The values satisfy \(10<L(20)<L(40)<E[X]\): \(10<15.9689<20.3606<32.5889\). The exposure-curve values increase with the limit and remain below one. All money amounts are rounded to four decimal places in thousands, and the parameter and curve values to four decimal places. A key mistake to avoid is rounding \(\hat{\alpha}\) or the exponent too early; retain the fitted value through the calculations and round only the reported results.
Right-Censored Claims and Model Fitting
A claim recorded only as exceeding a threshold \(c\) is right-censored: its exact severity is unknown, but the observation establishes \(X>c\). Under a parametric model, an exact observation contributes its density to the likelihood, while a right-censored observation contributes its survival probability. This treatment assumes the censoring mechanism is noninformative for severity, conditional on the model and the information used to set the censoring threshold. A policy limit that suppresses the observed amount is one possible source of censoring.
For independent exponential observations, let \(r\) be the number of exact observations, let their amounts be \(x_1,\ldots,x_r\), and let the remaining observations be censored at amounts \(c_1,\ldots,c_s\). The likelihood is proportional to \(\lambda^r\exp[-\lambda(\sum_jx_j+\sum_kc_k)]\). Differentiating its log with respect to \(\lambda\) gives the estimate
A censored claim contributes its known amount at risk to the denominator but does not count as an exact event in the numerator. This estimator depends on the exponential assumption; other severity distributions require their own likelihood and may not yield such a simple closed form.
Worked Example: Fitting an Exponential Model with One Censored Claim
A hypothetical dataset contains exact claims of \(2,3,\) and \(6\) thousand dollars, plus one claim known only to exceed \(4\) thousand dollars. Assume independent exponential severities and noninformative right censoring. Estimate the model and calculate the limited expected loss and exposure-curve value at \(5\) thousand dollars.
Form the likelihood estimate. There are \(r=3\) exact claims. Their total is \(2+3+6=11\) thousand; the censored observation contributes \(4\) thousand of known severity exposure. The total in the denominator is \(11+4=15\) thousand. Thus \(\hat{\lambda}=3/15=0.2\) per thousand, and the fitted mean is \(1/0.2=5\) thousand.
Calculate the limited loss. The fitted exponential survival function gives \(L(5)=(1-e^{-0.2(5)})/0.2=5(1-e^{-1})=5(0.6321206)=3.1606\) thousand. The exposure-curve value is \(G(5)=3.1606/5=0.6321\).
Check and interpret. The limited expected loss is below both the limit of \(5\) thousand and the fitted unlimited mean of \(5\) thousand. The fitted curve value is therefore between zero and one. Report the limited loss in thousands and round the final loss, rate, and curve value to four decimal places. The common mistake is to treat the censored amount as an exact claim: doing so would use four exact claims and a total of \(15\), incorrectly producing \(\hat{\lambda}=4/15\). The correct numerator counts only the three exact claims.
Practical Checks Before Using a Parametric Curve
A fitted curve is a model-based estimate, not a direct observation of losses at every possible limit. Check that the fitted distribution is compatible with the data and the intended application before using it for pricing. In particular, a Pareto tail can imply much greater high-limit losses than a thin-tailed exponential model. Sparse observations at large severities make the tail parameter especially influential.
- Check support and units. A Pareto Type I model with threshold \(x_m\) applies only at or above that threshold. Keep the claim amounts, thresholds, and limits in the same units.
- Check whether the unlimited mean exists. For Pareto Type I, the mean is finite only when \(\alpha>1\). Limited expected loss at a finite \(u\) can still be calculated when \(\alpha\leq1\), but a conventional normalized exposure curve cannot use an infinite mean as its denominator.
- Represent censoring correctly. An exact loss contributes a density; a right-censored loss contributes a survival probability. Treating a limit value as an exact ground-up claim generally distorts the fit.
- Check curve behavior. For nonnegative losses, \(L(u)\) must not decrease as \(u\) increases. When the finite-mean exposure curve applies, \(G(u)\) must also be nondecreasing and no greater than one.
- Keep the loss basis consistent. An increased limits factor is meaningful only when the numerator and denominator use compatible severity definitions, claim adjustments, and coverage terms.
These checks are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. The central calculation is compact—fit a model, integrate its survival function, and normalize or take a ratio—but the quality of the answer depends on choosing the right likelihood, retaining consistent units, and stating where the fitted model is being extrapolated.
Check Your Understanding
Work each problem using the stated model and units. Round monetary answers to four decimal places in thousands and curve values or factors to four decimal places.
- Three complete hypothetical claims are \(1,3,\) and \(4\) thousand dollars. Fit an exponential model by maximum likelihood and calculate \(L(4)\). (Answer: \(\hat{\lambda}=0.3750\) per thousand; \(L(4)=2.0717\) thousand.)
- A Pareto Type I model has \(x_m=8\) thousand and \(\alpha=2.5\). Calculate \(L(16)\) and \(G(16)\). (Answer: \(L(16)=11.4477\) thousand; \(G(16)=0.8586\).)
- Four complete claims are \(12,12,24,\) and \(48\) thousand dollars, and a Pareto Type I model has known \(x_m=12\). Calculate the maximum-likelihood estimate of \(\alpha\), determine whether the unlimited mean is finite, and calculate that mean if it is. (Answer: \(\hat{\alpha}=4/\ln 8=1.9236\); the mean is finite and equals approximately \(24.9930\) thousand.)
- Under an exponential model, exact claims are \(3,5,\) and \(7\) thousand dollars and one claim is censored at \(5\) thousand. Assuming noninformative censoring, estimate \(\lambda\), then calculate \(L(10)\). (Answer: \(\hat{\lambda}=0.1500\) per thousand; \(L(10)=5.1791\) thousand.)
- A Pareto Type I model has \(x_m=5\) thousand and \(\alpha=3\). Calculate \(L(10)\), \(L(20)\), and the increased limits factor from \(10\) to \(20\). (Answer: \(L(10)=6.8750\) thousand; \(L(20)=7.3438\) thousand; factor \(=1.0682\).)