Introduction
Severity comparisons can go wrong even when every formula is evaluated correctly. A finite variance does not guarantee a rapidly decaying tail, a high density at one loss amount does not mean a high probability of exceeding it, and a mixture’s tail can be governed by a component that is rare in the population. These distinctions matter whenever a model is being selected or a large-loss probability is being assessed.
The central quantity for comparing large losses is the survival function, \(S_X(x)=P(X>x)=1-F_X(x)\). A tail comparison should specify whether it concerns a particular threshold or behavior as the threshold becomes very large. Those are different questions: two survival curves can cross at a finite threshold even when one eventually decays more slowly.
Moments Are Not a Tail-Decay Ranking
A moment describes an average of powers of the loss, such as \(E[X^r]\). Its finiteness is useful, but it is not by itself a measure of how quickly the far tail decays. One common precise distinction is whether the moment-generating function \(E[e^{tX}]\) is finite for some \(t>0\). In this convention, a severity is called light-tailed if it has a finite positive exponential moment; it is called heavy-tailed if \(E[e^{tX}]=\infty\) for every \(t>0\). Terminology can vary across contexts, so a comparison should state the criterion being used.
For a Pareto Type I severity with scale \(x_m>0\) and shape \(\alpha>0\), the survival function is \(S_X(x)=(x_m/x)^\alpha\) for \(x\ge x_m\). Its \(r\)th raw moment is finite exactly when \(r<\alpha\), and then \(E[X^r]=\alpha x_m^r/(\alpha-r)\). Thus a Pareto model can have a finite mean and variance while having no finite third moment. It has no finite positive exponential moment.
A lognormal severity with \(\ln X\sim N(\mu,\sigma^2)\), where \(\sigma>0\), has all positive raw moments finite: \(E[X^r]=\exp(r\mu+r^2\sigma^2/2)\). Yet its positive exponential moments are infinite. The reason is that in the lognormal tail integral, the term \(tX\) eventually grows faster than \((\ln X)^2\), so multiplying the lognormal density by \(e^{tX}\) makes the integral diverge for every \(t>0\). Therefore, “all moments are finite” does not imply “light-tailed” under the exponential-moment criterion.
Worked Example: Finite Moments Do Not Mean a Light Tail
Method. Check moments using the distribution’s moment conditions, then compare survival probabilities at a stated threshold. All losses below are hypothetical and measured in units of one thousand dollars. There is no interest or mortality assumption because this is a single-loss severity comparison.
Let \(X\) be Pareto Type I with \(x_m=1\) and \(\alpha=2.5\). Let \(Y\) be lognormal with \(\mu=0\) and \(\sigma^2=1\). Compare their moments and their survival probabilities at \(x=e^4\). The models are not mean-matched; the calculation illustrates distinctions in tail behavior, not a controlled pricing comparison.
Pareto moments. The first moment is finite because \(1<2.5\): \(E[X]=2.5/(2.5-1)=5/3=1.6667\). The second moment is finite because \(2<2.5\): \(E[X^2]=2.5/(2.5-2)=5\). Hence \(\operatorname{Var}(X)=5-(5/3)^2=5-25/9=20/9=2.2222\). The third moment is infinite because \(3\not<2.5\).
Lognormal moments. The mean is \(E[Y]=e^{1/2}\approx1.6487\). The second raw moment is \(E[Y^2]=e^2\approx7.3891\), so \(\operatorname{Var}(Y)=e^2-e\approx7.3891-2.7183=4.6708\). Every positive raw moment is finite, since \(E[Y^r]=e^{r^2/2}\) for this choice of parameters.
Survival probabilities at \(x=e^4\). For the Pareto model, \(S_X(e^4)=(1/e^4)^{2.5}=e^{-10}\approx0.0000454\). For the lognormal model, \(S_Y(e^4)=P(\ln Y>4)=P(Z>4)\approx0.0000317\), where \(Z\sim N(0,1)\). At this particular threshold, the Pareto exceedance probability is larger.
Interpretation and check. The lognormal has every positive raw moment finite, but it does not have a finite positive exponential moment. At this threshold its survival probability is smaller than the Pareto’s; that does not make it light-tailed. In fact, the Pareto tail eventually exceeds the lognormal tail: writing \(y=\ln x\), the lognormal survival has leading logarithmic decay \(-y^2/2\), whereas the Pareto log survival is \(-2.5y\). As \(y\) grows, the quadratic term makes the lognormal tail decay faster than this Pareto tail. The calculation avoids the mistake of using moment finiteness as a tail ranking.
Compare Survival Probabilities, Not Density Heights
A density \(f_X(x)\) describes probability per unit of loss near \(x\); it is not the probability of a loss exceeding \(x\). For a continuous severity, \(P(x<X\le x+h)\) is approximately \(f_X(x)h\) for small positive \(h\), while \(S_X(x)\) accumulates probability over the entire range above \(x\). Density values also carry units—the reciprocal of the loss unit—so comparing a density height with an exceedance probability is not meaningful.
Worked Example: A Higher Density Can Mean a Smaller Tail Probability
Method. Evaluate both densities at the threshold, but rank the models using their survival probabilities. Consider hypothetical single-loss severities measured in thousands of dollars. There is no interest or mortality assumption. Let \(A\) be uniform on \([9.9,10.1]\), and let \(B\) be exponential with mean \(20\), so its rate is \(\lambda=1/20=0.05\) per severity unit. Compare at \(x=10\).
Density values. The uniform density on its interval is \(f_A(10)=1/(10.1-9.9)=1/0.2=5\). For the exponential model, \(f_B(10)=0.05e^{-0.05(10)}=0.05e^{-0.5}\approx0.0303\). Thus \(A\) has the higher density at exactly 10.
Exceedance probabilities. By symmetry of the uniform interval around 10, \(S_A(10)=P(A>10)=0.5\). For the exponential model, \(S_B(10)=e^{-0.05(10)}=e^{-0.5}\approx0.6065\). Therefore \(B\), despite its much lower density at 10, has the greater probability of exceeding 10.
Check and interpretation. The probabilities are both between zero and one, and the exponential result exceeds the uniform result by about \(0.6065-0.5000=0.1065\). The uniform model concentrates probability in a narrow interval around 10 and has no losses above 10.1. The exponential model spreads probability over a much longer range, which produces the larger exceedance probability. The common mistake avoided is to treat the height of a density curve at the threshold as if it were the area to the right of that threshold.
Mixture Tails and Changing Component Weights
In a finite mixture, a severity is generated by first selecting a component and then drawing a loss from that component. If component \(j\) is selected with probability \(w_j\), where \(w_j\ge0\) and \(\sum_jw_j=1\), and its severity has survival function \(S_j(x)\), then the mixture survival function is \(S_{\mathrm{mix}}(x)=\sum_j w_jS_j(x)\). The mixture survival is the weighted average of the component survival probabilities—not the survival of a model obtained by averaging component parameters.
The chance that a loss exceeding \(x\) came from component \(j\) is also threshold-dependent. Applying conditional probability gives \(P(J=j\mid X>x)=w_jS_j(x)/S_{\mathrm{mix}}(x)\). A component can have a small population weight but a large share among extreme losses if its own survival probability decays more slowly.
Worked Example: A Rare Component Can Dominate Large Losses
Method. Average component survival probabilities, then use Bayes’ rule to find the component weights among exceedances. Consider a hypothetical two-component exponential severity mixture, with losses measured in thousands of dollars. There is no interest or mortality assumption. Component 1 has weight \(0.9\) and rate \(\lambda_1=1\) per severity unit; component 2 has weight \(0.1\) and rate \(\lambda_2=0.1\) per severity unit. Compare at \(x=5\).
Component survival probabilities. For an exponential distribution with rate \(\lambda\), \(S(x)=e^{-\lambda x}\). Thus \(S_1(5)=e^{-5}\approx0.0067379\), and \(S_2(5)=e^{-0.5}\approx0.6065307\).
Mixture survival. Weight the survival probabilities: \(S_{\mathrm{mix}}(5)=0.9(0.0067379)+0.1(0.6065307)\). The two contributions are \(0.0060641\) and \(0.0606531\), so \(S_{\mathrm{mix}}(5)=0.0667172\), or \(0.0667\) to four decimal places.
Component share among exceedances. The probability an exceedance came from component 2 is \(0.1(0.6065307)/0.0667172=0.0606531/0.0667172\approx0.9091\). Thus a component representing only 10% of the original population accounts for about 90.91% of losses above 5.
Check and common mistake. The mixture survival lies between the component survivals, \(0.0067379\) and \(0.6065307\), as a weighted average should. A tempting but wrong calculation is to average the rates, obtaining \(0.9(1)+0.1(0.1)=0.91\), and then use \(e^{-0.91(5)}=e^{-4.55}\approx0.0106\). That is not the mixture survival: the mixture is not a single exponential with the average rate. The correct value is \(0.0667\), found by averaging the two component survival probabilities. At increasingly high thresholds, the slower-decaying component’s share typically grows; here its survival \(e^{-0.1x}\) decays more slowly than \(e^{-x}\).
A Reliable Tail-Comparison Process
Before concluding that one severity model has the “heavier” tail, identify what kind of comparison is being requested. A threshold probability, a comparison across every threshold, and an asymptotic tail-decay comparison are not interchangeable. The following sequence makes the distinction explicit.
Record the threshold if the question concerns a particular exceedance probability. If it asks about general tail behavior, state the criterion, such as existence of a positive exponential moment.
Compute \(S_X(x)=P(X>x)\) for an exceedance comparison. Do not substitute the density \(f_X(x)\) or infer tail behavior only from the mean or variance.
For a mixture, calculate the weighted sum of component survival probabilities. If asked which component generated an exceedance, calculate the threshold-specific conditional component weights.
Confirm that a mixture survival lies between its component survivals and that all probabilities are between zero and one. A comparison at one threshold does not prove a global ordering.
Common Exam Traps
- “All moments are finite, so the tail is light.” This does not follow. A lognormal severity has every positive raw moment finite but no finite positive exponential moment.
- “The density is higher here, so more losses exceed this amount.” Density is local; exceedance probability is the area to the right. Compare \(S_X(x)\) for tail questions.
- “The mixture’s rate is the weighted average rate.” A mixture of exponential distributions generally is not exponential. Average the component survival probabilities instead.
- “The larger tail at this threshold is larger everywhere.” A pointwise comparison at one \(x\) cannot establish first-order stochastic dominance. Check the full range or use a valid general argument.
- “A small component weight makes the component irrelevant to extreme losses.” The component’s conditional share among exceedances depends on both its original weight and its survival probability at the threshold.
Check Your Understanding
For each hypothetical severity model, identify the appropriate tail calculation and check the interpretation.
- A Pareto Type I severity has \(x_m=2\) and \(\alpha=4\). Find \(E[X^2]\), state whether the fourth raw moment is finite, and state whether any positive exponential moment is finite. (Answer: \(E[X^2]=8\); the fourth moment is infinite; no positive exponential moment is finite.)
- Severity \(A\) is uniform on \([2.9,3.1]\), while severity \(B\) is exponential with mean 4. Compare both densities at \(x=3\) and both exceedance probabilities there. Which model has the larger exceedance probability? (Answer: \(f_A(3)=5\), \(f_B(3)=0.25e^{-0.75}\approx0.1181\); \(S_A(3)=0.5000\), \(S_B(3)=e^{-0.75}\approx0.4724\); \(A\) has the larger exceedance probability.)
- A mixture has exponential components with weights \(0.8\) and \(0.2\), and rates \(0.5\) and \(0.1\), respectively. Find its survival probability at \(x=10\), then find the conditional share of exceedances from the slower component. (Answer: \(S_{\mathrm{mix}}(10)=0.8e^{-5}+0.2e^{-1}\approx0.0790\); the slower component’s conditional share is \(0.2e^{-1}/S_{\mathrm{mix}}(10)\approx0.9317\).)
- A lognormal severity has \(\ln X\sim N(0,1)\). Find its mean and variance, and explain whether finite values for both establish that it is light-tailed under the positive-exponential-moment definition. (Answer: \(E[X]=e^{1/2}\approx1.6487\), \(\operatorname{Var}(X)=e^2-e\approx4.6708\); no, it has no finite positive exponential moment.)
- A mixture has component weights \(0.95\) and \(0.05\). At a threshold \(x\), their survival probabilities are \(0.01\) and \(0.50\), respectively. Find the overall exceedance probability and the conditional share from component 2. (Answer: \(S_{\mathrm{mix}}(x)=0.95(0.01)+0.05(0.50)=0.0345\); component 2’s share is \(0.025/0.0345\approx0.7246\).)