Actuarial Consequences Of Severity-Model Choice

Why Model Choice Has Actuarial Consequences

Two severity models can describe observed losses similarly while implying materially different future payments. A fitted model is not just a curve used to summarize data: it can determine a projected mean loss, an upper quantile, the probability of exceeding a threshold, or the expected cost of an insurance layer. Those quantities are often more directly relevant to pricing, reserving, and risk assessment than the model label itself.

Information criteria such as AIC and BIC compare candidate models using a balance of fit and complexity. Their rankings are useful, but they do not guarantee that the model ranked first is best for every actuarial quantity. The appropriate interpretation depends on the purpose of the analysis and on how important the part of the severity distribution is for that purpose.

Definition: For a nonnegative loss $X$, the mean severity is $E[X]$; the $p$-quantile is $q_p=\inf\{x:F_X(x)\ge p\}$; and the exceedance probability at threshold $c$ is $P(X>c)=S_X(c)$. These summarize different features of the distribution and need not rank candidate models in the same order.

Throughout, severity models describe individual losses in dollars. Unless stated otherwise, assume losses are independent draws from the specified severity distribution. Any frequency assumption used to calculate an aggregate expected payment will be stated separately. A model-based quantity is a consequence of the assumed distribution, not proof that the distribution is correct.

Translate a Fitted Model Into Quantities

For a continuous severity model with distribution function $F_X$, the mean is calculated from the distribution or its survival function, when finite. The quantile $q_p$ solves $F_X(q_p)=p$ when the distribution is continuous and strictly increasing. The tail probability at $c$ is $S_X(c)=1-F_X(c)$. These formulas can be used directly only when the model’s stated support includes the relevant loss values and the required moment exists.

For example, an exponential model with rate $\lambda>0$ has $S_X(x)=e^{-\lambda x}$ for $x\ge0$, mean $1/\lambda$, and $p$-quantile $-\ln(1-p)/\lambda$. A Lomax model with scale $\theta>0$ and shape $\alpha>0$ has $S_X(x)=(1+x/\theta)^{-\alpha}$ for $x\ge0$. Its mean is $\theta/(\alpha-1)$ only when $\alpha>1$; its $p$-quantile is $\theta[(1-p)^{-1/\alpha}-1]$. Using the Lomax mean formula when $\alpha\le1$ would incorrectly assign a finite mean to a distribution whose mean is infinite.

Worked Example: Similar Means, Different Tail Probabilities

A hypothetical insurer fits two models to individual claim severities. Model E is exponential with rate $\lambda=0.001$ per dollar. Model L is Lomax with $\theta=\$1{,}000$ and $\alpha=2$. Suppose their reported AIC values are 812.4 for Model E and 814.0 for Model L. Compare their means, 95th percentiles, and probabilities of a loss exceeding $\$5{,}000$. State what the AIC ranking does and does not establish.

Method. Use the distribution-specific formulas, checking the Lomax mean condition before applying it.

Means. For Model E, $E[X]=1/0.001=\$1{,}000$. For Model L, $\alpha=2>1$, so $E[X]=\theta/(\alpha-1)=1{,}000/(2-1)=\$1{,}000$. The models have the same mean.

95th percentiles. For Model E, $q_{0.95}=-\ln(0.05)/0.001=2{,}995.732$, or $\$2{,}995.73$ to the nearest cent. For Model L, $q_{0.95}=1{,}000[(0.05)^{-1/2}-1]$. Since $(0.05)^{-1/2}=\sqrt{20}=4.472136$, this gives $q_{0.95}=\$3{,}472.14$ to the nearest cent.

Exceedance probabilities. For Model E, $P(X>5{,}000)=e^{-0.001(5{,}000)}=e^{-5}=0.0067$ to four decimals. For Model L, $P(X>5{,}000)=(1+5{,}000/1{,}000)^{-2}=6^{-2}=1/36=0.0278$ to four decimals.

The AIC is lower for Model E, so it is preferred by this reported criterion among these two candidates. Yet Model L implies a higher 95th percentile and more than four times the probability of exceeding $\$5{,}000$. AIC does not establish which model is more suitable for a particular tail-sensitive decision; that requires considering the purpose and relevant evidence for the tail.

Check. Both models give a $\$1{,}000$ mean, but the computed quantiles and tail probabilities are different. That is consistent: equality of means does not imply equality of distributions.

Expected Payments for a Deductible and Limit

A distributional difference can also affect the expected payment for a particular coverage. Let $d$ be a per-loss deductible and $u$ the maximum insurer payment per loss. The payment is $Y^L=\min((X-d)^+,u)$, where $(z)^+=\max(z,0)$. For a nonnegative loss with survival function $S_X$, its expected value is

E[Y^L]=\int_d^{d+u}S_X(x)\,dx.

This identity assumes $d\ge0$, $u\ge0$, and a well-defined nonnegative severity distribution. It concerns the insurer’s payment per loss, not the insured’s payment or the total loss. If the coverage has no payment limit, the upper integration endpoint is unbounded, and the integral may be infinite.

Derivation: For each fixed loss $X$, the payment $\min((X-d)^+,u)$ equals the length of the interval of thresholds $x$ in $[d,d+u]$ for which $X>x$. Equivalently, $Y^L=\int_d^{d+u}\mathbf{1}_{\{X>x\}}\,dx$. Taking expectations and interchanging expectation and integration for this nonnegative integrand gives $E[Y^L]=\int_d^{d+u}P(X>x)\,dx=\int_d^{d+u}S_X(x)\,dx$.

Worked Example: Two Models for a Limited Layer

A hypothetical portfolio has 100 claims, each subject to the same coverage: a $\$100$ deductible and a maximum insurer payment of $\$400$ per claim. Compare the expected total insurer payment under (i) an exponential severity model with scale $\$100$, and (ii) a Lomax model with $\theta=\$50$ and $\alpha=1.5$. Assume the stated 100 claims occur; no frequency model is needed.

Method. The payment per claim is $Y^L=\min((X-100)^+,400)$, so the relevant severity interval in the survival integral is from $d=\$100$ to $d+u=\$500$. Multiply the resulting expected payment per claim by 100.

(i) Exponential model. Scale $\$100$ means rate $\lambda=1/100$, so $S_X(x)=e^{-x/100}$. Therefore, $E[Y^L]=\int_{100}^{500}e^{-x/100}\,dx$. An antiderivative is $-100e^{-x/100}$, giving $E[Y^L]=100(e^{-1}-e^{-5})$. Using $e^{-1}=0.367879$ and $e^{-5}=0.006738$, the per-claim expectation is $100(0.361141)=\$36.1141$. For 100 claims, the expected total is $100(\$36.1141)=\$3{,}611.41$, to the nearest cent.

(ii) Lomax model. Here $S_X(x)=(1+x/50)^{-1.5}$. Thus $E[Y^L]=\int_{100}^{500}(1+x/50)^{-1.5}\,dx$. An antiderivative is $-100(1+x/50)^{-0.5}$, so $E[Y^L]=100(1/\sqrt{3}-1/\sqrt{11})$. Since $1/\sqrt{3}=0.577350$ and $1/\sqrt{11}=0.301511$, the per-claim expectation is $100(0.275839)=\$27.5839$. For 100 claims, the expected total is $100(\$27.5839)=\$2{,}758.39$, to the nearest cent.

The layer’s expected total payment is therefore higher under the exponential model in this comparison, even though the Lomax model has a heavier far tail. The layer covers losses only between $\$100$ and $\$500$ in the survival-integral calculation; it does not weight every part of the tail equally. As a check, each per-claim expectation is between $\$0$ and the $\$400$ payment cap, and each aggregate expectation is below the maximum possible $\$40{,}000$.

Common mistake avoided. A decimal-place error can inflate both aggregate results by a factor of ten. For example, the exponential calculation is $100\text{ claims}\times100(e^{-1}-e^{-5})=\$3{,}611.41$, not $\$36{,}114.15$. The Lomax calculation is $100\text{ claims}\times100(1/\sqrt3-1/\sqrt{11})=\$2{,}758.39$, not $\$27{,}583.89$. Keep the integral’s scale factor and the claim count separate, then multiply once by each.

A Criterion Ranking Is Not a Ranking for Every Purpose

An information criterion assigns a score to a fitted candidate model. A lower score supports that model under the criterion’s fit-and-complexity comparison, provided the candidates were fitted to comparable data and the criterion was calculated consistently. It does not directly compare the resulting premium, quantile, layer cost, or tail probability. Those require calculating the quantity under each candidate model.

Consequently, the model with the lower AIC can still imply a larger premium for a particular layer, while a model with a slightly higher AIC can imply a much larger extreme quantile. This is not a contradiction. The criterion and the actuarial quantity answer different questions. A decision-maker should state the target quantity, compare its implications across plausible models, and consider whether the data support the part of the distribution that drives the decision. A model’s extrapolation far beyond observed losses deserves particular caution.

Worked Example: Tail Quantile Versus Information-Criterion Preference

A hypothetical benefit program compares Model E, exponential with scale $\$100$, and Model L, Lomax with $\theta=\$50$ and $\alpha=1.5$. Suppose Model E has the lower AIC. Calculate each model’s 99th percentile and its probability of exceeding $\$600$. Explain why the criterion preference alone is insufficient if the decision concerns a high loss threshold.

Method. For the exponential model, use rate $\lambda=1/100$. For the Lomax model, the mean exists because $\alpha=1.5>1$, but the variance is infinite because $\alpha\le2$. The requested quantile and exceedance probability exist for both models.

99th percentiles. For Model E, $q_{0.99}=-100\ln(0.01)=\$460.52$ to the nearest cent. For Model L, $q_{0.99}=50[(0.01)^{-1/1.5}-1]=50(100^{2/3}-1)$. Since $100^{2/3}=21.544347$, this is $50(20.544347)=\$1{,}027.22$ to the nearest cent.

Exceedance probabilities. For Model E, $P(X>600)=e^{-600/100}=e^{-6}=0.0025$ to four decimals. For Model L, $P(X>600)=(1+600/50)^{-1.5}=13^{-1.5}=1/(13\sqrt{13})=0.0213$ to four decimals.

Model L implies a 99th percentile more than twice that of Model E and a substantially higher probability of exceeding $\$600$. Model E nevertheless has the lower AIC in the stated comparison. If the decision is driven by the 99th percentile or the $\$600$ threshold, the model-implied quantities are essential; the AIC ranking alone does not answer the decision question. The infinite Lomax variance is also a relevant warning if a variance-based summary is being considered.

Check. Substituting the calculated Lomax quantile into its survival function gives $0.01$, and substituting $\$600$ gives approximately $0.0213$. These agree with the requested quantile and tail probability.

Practical Interpretation and Exam Traps

A disciplined comparison separates three steps: identify the actuarial quantity of interest, calculate that quantity under each fitted model, and then interpret the information-criterion evidence alongside the quantity. A single summary such as the mean cannot stand in for the entire severity distribution. Likewise, a heavier far tail does not automatically mean a higher expected payment for every limited layer: the deductible, limit, and loss range covered matter.

Do not equate lower AIC with superiority for every use. It is a criterion-based comparison, not a guarantee about a chosen tail probability or contract payment.
Check moment conditions. For Lomax severity, $E[X]$ is finite only for $\alpha>1$, and $\operatorname{Var}(X)$ is finite only for $\alpha>2$.
Keep the layer endpoints straight. With deductible $d$ and maximum payment $u$, the interval in $E[Y^L]$ is $[d,d+u]$, not $[d,u]$.
Keep units and scaling explicit. A rate is expressed per dollar when losses are in dollars; a scale parameter is not a rate. In an aggregate expected payment, multiply the per-claim expectation by the number of claims once.
Do not overstate precision. Model-based outputs may be sensitive to parameter estimates and tail assumptions. Report the requested rounding, but interpret close numerical differences in light of that uncertainty.

Key takeaway: A severity-model choice has consequences through the quantities the model implies. Compare means, quantiles, exceedance probabilities, and coverage-specific expected payments directly; use information-criterion rankings as evidence about candidate models, not as substitutes for those calculations.

Check Your Understanding

For each problem, calculate the requested model implication and keep the target quantity distinct from any criterion ranking.

A hypothetical loss has an exponential distribution with rate $0.005$ per dollar. Find its mean and the probability that it exceeds $\$600$. (Answer: mean $\$200$; probability $e^{-3}=0.0498$.)
A hypothetical severity follows a Lomax distribution with $\theta=\$400$ and $\alpha=3$. Find the mean and 90th percentile. (Answer: mean $\$200$; 90th percentile $\$461.77$.)
An exponential model with scale $\$100$ is used for a policy with deductible $\$50$ and maximum insurer payment $\$150$. Find the expected payment for 20 claims. (Answer: $20[100(e^{-0.5}-e^{-2})]=\$942.39$, to the nearest cent.)
Two candidate models have AIC values 410.2 and 411.0. The lower-AIC model implies a 99th percentile of $\$900$; the other implies $\$1{,}400$. If a hypothetical decision is specifically based on the 99th percentile, what additional comparison is needed? (Answer: Compare the models’ evidence and suitability for that upper-tail quantity; the lower AIC alone does not establish which 99th-percentile implication is preferable.)

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