Compare Severity Models Using Contract-Modified Data

Why Contract-Modified Data Require Care

Severity-model comparison is only meaningful when each candidate model is fitted to what the data actually reveal. A claim recorded at a policy cap does not generally tell us that the ground-up loss equals the cap; it tells us that the loss reached at least that amount. Similarly, a database containing only claims above a deductible is a selected sample, not an unrestricted sample from the severity distribution.

These distinctions affect both parameter estimates and the model comparison that follows. This tutorial considers continuous ground-up loss $X$, compares candidate severity models using censored or deductible-truncated observations, and then shows how fitted models imply different expected payments for a hypothetical reinsurance layer. The methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: An observation is right-censored at cap $u$ when the data identify that $X\ge u$, but do not reveal the exact value above $u$. Its likelihood contribution is the survival probability $S_X(u)$, not the density $f_X(u)$. A sample is left-truncated at deductible $d$ when only losses known to exceed $d$ enter the sample. Conditional on inclusion, an exact observation $x>d$ contributes $f_X(x)/S_X(d)$. If that included claim is also censored at $u>d$, its contribution is $S_X(u)/S_X(d)$.

Likelihoods Must Match the Observation Process

For independent observations, the likelihood is the product of the contributions for the observed records. Taking logarithms turns this product into a sum and makes maximization more convenient. For deductible-truncated data, the denominator $S_X(d)$ is essential: it accounts for the fact that a claim enters the data only if it exceeds the deductible. For capped data, a record at the cap contributes a survival probability because the actual loss may be larger.

If the sample is truncated at $d$ and capped at $u$, an exact ground-up loss $x$, where $d<x<u$, contributes $f_X(x)/S_X(d)$. A record known only to have reached the cap contributes $S_X(u)/S_X(d)$. These formulas presume that the observed sample includes only losses exceeding $d$, that the cap record represents $X\ge u$, and that the proposed model describes ground-up loss. If the data instead record payments rather than ground-up losses, first identify what interval or event each payment represents; a payment should not automatically be treated as an exact loss.

Worked Example: Comparing Exponential and Pareto Models With Capped Claims

A hypothetical portfolio has four observed ground-up claims. Two are exact losses of $200 and $250, while two are recorded at a cap of $300 and are known only to be at least $300. Compare an exponential model with mean $\theta$ against a Pareto Type I model with known minimum $100$ and unknown shape $\alpha$. Use AIC and BIC, counting one estimated parameter in each model.

Exponential fit. Its density and survival function are $f_X(x)=\theta^{-1}e^{-x/\theta}$ and $S_X(x)=e^{-x/\theta}$. The two exact losses contribute densities; the two capped records contribute survival probabilities. Thus

L(\theta)=\theta^{-2} \exp\left(-\frac{200+250+300+300}{\theta}\right).

There are $r=2$ exact observations and the sum of all observed amounts, including the two censoring thresholds, is $1{,}050$. The log-likelihood is $\ell(\theta)=-2\ln\theta-1{,}050/\theta$. Differentiating gives $-2/\theta+1{,}050/\theta^2=0$, so $\hat{\theta}=1{,}050/2=\$525$. Substitution gives $\ell(\hat{\theta})=-2\ln(525)-2\approx-14.5268$.

Pareto fit. For $x\ge100$, the model has $S_X(x)=(100/x)^\alpha$ and $f_X(x)=\alpha 100^\alpha/x^{\alpha+1}$. The exact losses contribute densities and the cap records contribute survival probabilities. The log-likelihood can be written

\ell(\alpha)=2\ln\alpha-2\ln100-\ln(200/100)-\ln(250/100) -\alpha\left[\ln(200/100)+\ln(250/100)+2\ln(300/100)\right].

The bracketed sum is $\ln2+\ln2.5+2\ln3\approx3.806662$. Differentiating and setting the derivative to zero gives $2/\alpha-3.806662=0$, so $\hat{\alpha}=2/3.806662\approx0.5254$. Substitution gives $\ell(\hat{\alpha})\approx-14.1070$.

For $k$ estimated parameters and maximized log-likelihood $\ell$, $AIC=2k-2\ell$, and $BIC=k\ln n-2\ell$, where $n$ is the number of observed records. Here $n=4$ and $k=1$ for each candidate. The exponential model has $AIC\approx31.05$ and $BIC\approx30.44$. The Pareto model has $AIC\approx30.21$ and $BIC\approx29.60$. Both criteria favor Pareto for these data, although the differences are small and the sample is very limited.

Check. The Pareto estimate is positive and its survival probability decreases as loss increases. The likelihood used two density contributions and two survival contributions in each model. The common mistake avoided here is treating the capped records as exact $300 losses.

Why Truncation Changes the Likelihood

For an exact loss observed only because it exceeded $d$, the conditional density is obtained from the definition of conditional probability. For a small interval of width $dx$ around $x>d$, the probability of a loss in that interval, conditional on exceeding $d$, is approximately $f_X(x)dx/S_X(d)$. Dividing by $dx$ gives the conditional density. The same reasoning for an observation known only to exceed $u>d$ gives conditional probability $P(X\ge u\mid X>d)=S_X(u)/S_X(d)$.

f_{X\mid X>d}(x)=\frac{f_X(x)}{S_X(d)},\quad x>d, \qquad P(X\ge u\mid X>d)=\frac{S_X(u)}{S_X(d)},\quad u>d.

This conditioning is not a correction for a deductible payment formula. It describes which ground-up losses are represented in a selected data set. If every claim, including claims below the deductible, is present in the records, the sample is not truncated in this way. The likelihood must follow the actual recording and inclusion rules.

Worked Example: Fitting an Exponential Model Above a Deductible

A hypothetical data extract contains only claims exceeding a $100 deductible. Two ground-up losses are exact: $200 and $300. A third claim is known to have reached a $400 cap, but its ground-up amount is not known. Assume $X$ is exponential with mean $\theta$. Estimate $\theta$ from these selected observations.

The exact observations contribute $f_X(200)/S_X(100)$ and $f_X(300)/S_X(100)$. The capped claim contributes $S_X(400)/S_X(100)$. Since $S_X(x)=e^{-x/\theta}$, each contribution can be expressed using the excess above the deductible. The likelihood is

L(\theta)=\theta^{-2} \exp\left(-\frac{(200-100)+(300-100)+(400-100)}{\theta}\right) =\theta^{-2}e^{-600/\theta}.

There are two exact observations, so the log-likelihood is $\ell(\theta)=-2\ln\theta-600/\theta$. Its derivative is $-2/\theta+600/\theta^2$. Setting this equal to zero gives $\hat{\theta}=600/2=\$300$. The estimate is finite because there is at least one exact observation; with no exact observations, this likelihood would increase toward its supremum as $\theta$ tends to infinity, so there would be no finite maximum-likelihood estimate.

Now calculate the expected payment for a hypothetical per-claim excess-of-loss layer attaching at $a=\$300$ with limit $L=\$300$. Under this fitted exponential model, the payment is $\min((X-a)_+,L)$, where $(z)_+=\max(z,0)$. Its expectation is the area under the survival curve from the attachment to the top of the layer:

E[\min((X-300)_+,300)] =\int_{300}^{600}e^{-x/300}\,dx =300(e^{-1}-e^{-2}) \approx \$69.76.

Check. The $400 cap contributes an excess of $300 above the $100 selection threshold, not an exact $400 loss. The fitted scale is $300, and the layer payment is less than its $300 maximum.

From a Fitted Severity Model to a Layer Cost

A lower AIC or BIC indicates a preferred balance of fit and parameter count among the candidates being compared; it does not itself measure reinsurance cost. Once a severity model is chosen, expected layer payment is a separate calculation. For a ground-up loss $X$, layer attachment $a$, and limit $L$, the per-claim payment is $\min((X-a)_+,L)$. For a nonnegative loss with survival function $S_X$, its expectation equals the integral of $S_X(x)$ across the layer.

Result: For a per-claim layer attaching at $a$ with limit $L$, $$ E[\min((X-a)_+,L)]=\int_a^{a+L}S_X(x)\,dx. $$ This is a per-claim expected payment under the specified ground-up severity model. It is not an aggregate stop-loss expectation, and it does not include claim frequency.

To justify the identity, write the layer payment $W=\min((X-a)_+,L)$. For any nonnegative random variable bounded by $L$, $E[W]=\int_0^L P(W>t)\,dt$. For $0\le t<L$, the event $W>t$ is the event $X>a+t$. Therefore $E[W]=\int_0^L S_X(a+t)\,dt$. Changing variables to $x=a+t$ gives the stated integral. For a continuous severity distribution, using $>$ rather than $\ge$ does not change the integral.

Worked Example: Comparing Layer Payments From the Two Capped-Data Fits

Use the exponential and Pareto fits from the capped-claims example to compare the expected per-claim payment for a hypothetical layer attaching at $300 with a $300 limit. The layer covers the interval from $300 to $600.

Exponential model. With $\hat{\theta}=\$525$, the survival function is $e^{-x/525}$. Hence

E[W]_{\mathrm{Exp}} =\int_{300}^{600}e^{-x/525}\,dx =525\left(e^{-300/525}-e^{-600/525}\right) \approx \$129.05.

Pareto model. With minimum $100$ and $\hat{\alpha}\approx0.5254$, the survival function on this layer is $(100/x)^{\hat{\alpha}}$. Since $\hat{\alpha}\ne1$,

E[W]_{\mathrm{Par}} =\int_{300}^{600}(100/x)^{0.5254}\,dx =100\frac{6^{1-0.5254}-3^{1-0.5254}}{1-0.5254} \approx \$138.25.

The fitted Pareto model therefore gives an expected layer payment about $\$138.25-\$129.05=\$9.20$ higher per claim. The layer is bounded, so this expectation is finite even though the fitted Pareto shape is below one and the unbounded Pareto mean does not exist. This comparison illustrates why a model that is only modestly preferred by information criteria can still imply a material difference for a particular layer.

Check. Each calculation integrates the fitted survival function over the attachment-to-exhaustion interval, not over all positive losses. Both results are between zero and the $300 layer limit.

Common Mistakes and Practical Checks

A frequent error is to treat a capped record as an exact loss at the cap. In the first example, this incorrect approach would use all four amounts as exact exponential observations, producing $\hat{\theta}_{\mathrm{wrong}}=(200+250+300+300)/4=\$262.50$. The correct censored likelihood uses only the two exact density terms in the power of $\theta^{-1}$, and gives $\hat{\theta}=\$525$. A capped record contributes information through $S_X(u)$, not through $f_X(u)$.

Do not omit truncation conditioning. For a sample selected above $d$, use $f_X(x)/S_X(d)$ for exact observations. Using $f_X(x)$ as if the sample were unrestricted ignores the selection mechanism.
Keep the two mechanisms distinct. Truncation determines which claims appear; censoring limits what is known about a claim that appears.
Compare information criteria on the same data basis. Use the same observed records, observation units, and likelihood convention for all candidates. Count all freely estimated parameters in $k$.
Do not interpret an information criterion as a cost estimate. Calculate the relevant layer expectation under each fitted model when pricing or comparing a treaty.
Check what the record represents. A payment at a cap may imply a lower bound on ground-up loss, whereas a deductible-modified payment may correspond to a transformed value. Identify the observation event before writing its likelihood contribution.

Key takeaway: Fit severity models to the observed-data likelihood: densities for exact losses, survival probabilities for capped losses, and conditional contributions when the sample is selected above a deductible. Then compare models using a common information criterion and evaluate the layer payments implied by each fitted survival function.

Check Your Understanding

For each question, identify the observation mechanism before calculating.

A hypothetical portfolio has exact exponential losses of $80 and $120, plus one loss known only to have reached a $200 cap. With no deductible truncation, estimate the exponential mean by maximum likelihood. (Answer: $\hat{\theta}=\$200$.)
Only claims above a $50 deductible are included. Under an exponential model, the data contain exact losses of $90 and $150 and one claim censored at $250. Find the maximum-likelihood estimate of the mean. (Answer: $\hat{\theta}=\$170$.)
A Pareto Type I model has known minimum $100. There are exact losses of $200 and $400, and one loss censored at $500. Write the maximum-likelihood estimate of the shape parameter. (Answer: $\hat{\alpha}=2/(\ln2+\ln4+\ln5)\approx0.5422$.)
Two candidate models fitted to the same $n=20$ records have maximized log-likelihoods $-18.0$ with $k=2$, and $-16.5$ with $k=3$. Calculate both AIC values and identify the lower-AIC model. (Answer: $AIC_1=40.0$, $AIC_2=39.0$; the second model has the lower AIC.)
Under an exponential severity model with mean $\theta=\$250$, find the expected payment for a per-claim layer attaching at $250 with a $250 limit. (Answer: $250(e^{-1}-e^{-2})\approx\$58.14$.)

What You'll Learn