Introduction
Choosing a severity family is a model-selection decision, not merely a matter of finding a distribution with a convenient formula. A proposed model should be compatible with the loss characteristics that matter: its mean and variability, its skewness, how its hazard changes with loss size, and how quickly its tail decays. These features can rule out a family or make it plausible, even before any parameter estimation is attempted.
This tutorial focuses on that qualitative and moment-based choice. It does not fit a model to a data set. A family may be compatible with the stated features without being uniquely determined by them; different families can share selected moments or similar behavior over a limited range.
What Each Family Typically Implies
Parameter conventions must be fixed before comparing formulas. Here, an exponential loss with rate \(\lambda>0\) has mean \(1/\lambda\). A gamma loss has shape \(\alpha>0\) and scale \(\theta>0\). A lognormal loss is specified by \(\ln X\sim N(\mu,\sigma^2)\), so \(\mu\) and \(\sigma^2\) describe the logarithm, not \(X\). A Type I Pareto loss has minimum \(x_m>0\) and shape \(\alpha>0\). A Weibull loss has shape \(k>0\) and scale \(\eta>0\).
| Family | Useful risk-feature implications | Important qualification |
|---|---|---|
| Exponential | Mean \(1/\lambda\), variance \(1/\lambda^2\), coefficient of variation 1, skewness 2, and constant hazard \(\lambda\). | It is restrictive when variability or hazard changes systematically with loss size. |
| Gamma | Mean \(\alpha\theta\), variance \(\alpha\theta^2\), skewness \(2/\sqrt{\alpha}\). Its hazard increases for \(\alpha>1\), is constant for \(\alpha=1\), and decreases for \(\alpha<1\). | Its tail has exponential decay, modified by a power factor. |
| Lognormal | Positive, right-skewed losses; all positive integer moments are finite. Its tail decays more slowly than an exponential tail. | Its hazard is not globally increasing: it eventually decreases toward zero. |
| Type I Pareto | Power-law survival \(S_X(x)=(x_m/x)^\alpha\) for \(x\ge x_m\). The mean is finite only for \(\alpha>1\), and the variance only for \(\alpha>2\). | Higher moments can be infinite even when the mean and variance exist. |
| Weibull | Hazard increases for \(k>1\), is constant for \(k=1\), and decreases for \(k<1\). | For \(k<1\), its tail is heavier than an exponential tail but still decays faster than a Pareto power tail. |
| Finite mixture | Can represent heterogeneous groups and can have a changing hazard, including a decreasing hazard from exponential components. | Its overall moments and hazard depend on both component behavior and mixture weights. |
A decreasing hazard alone does not identify one family: it can occur for a Weibull model with \(k<1\), some gamma models, a Pareto model, or a suitable mixture. Likewise, right skewness is common across several positive-loss families. Use combinations of features rather than a single visual or qualitative clue.
Moment Patterns as a Screening Tool
Moments provide useful restrictions. For example, the exponential family requires the coefficient of variation, \(\sqrt{\operatorname{Var}(X)}/E[X]\), to equal 1 and the skewness to equal 2. The gamma family allows different coefficients of variation and skewness, but the two are linked through its shape parameter.
The relationship follows directly from the gamma raw moments. For a positive integer \(r\), \(E[X^r]=\theta^r\Gamma(\alpha+r)/\Gamma(\alpha)\). Applying \(\Gamma(z+1)=z\Gamma(z)\) gives \(E[X]=\alpha\theta\), \(E[X^2]=\alpha(\alpha+1)\theta^2\), and \(E[X^3]=\alpha(\alpha+1)(\alpha+2)\theta^3\). Therefore,
The third central moment is \(E[(X-E[X])^3]=E[X^3]-3E[X]E[X^2]+2E[X]^3=2\alpha\theta^3\). Dividing by \(\operatorname{Var}(X)^{3/2}=(\alpha\theta^2)^{3/2}\) gives skewness \(2/\sqrt{\alpha}\). This linked pattern is a diagnostic, not proof that the gamma family is the only possible model.
Worked Example: Screening a Model With Mean, Variance, and Skewness
A hypothetical portfolio is described by a positive severity with mean 8, variance 64, and skewness 2. The risk analyst also reports an approximately constant hazard over the range of interest. Decide whether an exponential or gamma model is compatible with these features, and explain whether the moment evidence identifies a unique family.
Method. Check the implications of each candidate against all stated features, using the rate parameter for the exponential and shape-scale parameters for the gamma.
Exponential check. The mean requires \(\lambda=1/8\). Its variance is \(1/\lambda^2=64\), and its skewness is 2. Its hazard is the constant \(\lambda=0.125\). All four features agree.
Gamma check. From the mean and variance, the implied shape and scale are \(\alpha=E[X]^2/\operatorname{Var}(X)=8^2/64=1\) and \(\theta=\operatorname{Var}(X)/E[X]=64/8=8\). Its skewness is \(2/\sqrt{1}=2\), and shape 1 gives a constant hazard. This gamma specification is exactly the exponential distribution with rate \(1/8\), since a gamma distribution with shape 1 and scale 8 is exponential with that rate.
Conclusion and check. Both descriptions are compatible, but they do not give two distinct models here: the gamma specification reduces to the exponential. The calculated mean is \(\alpha\theta=8\), the variance is \(\alpha\theta^2=64\), and the skewness is 2. The common mistake is to treat “gamma” and “exponential” as disjoint alternatives without checking the gamma shape-one case.
Hazard Shape and Tail Behavior
The hazard can distinguish models whose means and variances happen to be similar. Under the Weibull model, \(S_X(x)=\exp[-(x/\eta)^k]\), so \(h_X(x)=(k/\eta)(x/\eta)^{k-1}\). The power \(k-1\) determines whether the hazard decreases, stays constant, or increases. A gamma hazard also changes with shape, but it does not generally have the same exact functional form as a Weibull hazard.
Tail behavior adds another dimension. A Pareto survival function declines as a power of \(x\), whereas lognormal survival declines more slowly than any exponential survival as \(x\) grows. A gamma tail has exponential decay with a polynomial factor; a Weibull tail has exponential-of-a-power form. Consequently, a Weibull with \(k<1\) decays more slowly than an exponential but faster than a Pareto power tail. “Heavy-tailed” should be interpreted relative to a stated comparison, not as a synonym for “right-skewed.”
Worked Example: Selecting a Pareto Model From Tail and Moment Conditions
A hypothetical large-loss layer is modeled with a Type I Pareto severity having minimum loss \(x_m=3\) and shape \(\alpha=2.5\). Check whether its mean and variance are finite, determine whether its third moment is finite, and calculate \(P(X>12)\).
Method. Use the Type I Pareto survival function and moment conditions. For \(\alpha>1\), \(E[X]=\alpha x_m/(\alpha-1)\); for \(\alpha>2\), \(\operatorname{Var}(X)=\alpha x_m^2/[(\alpha-1)^2(\alpha-2)]\). The \(r\)th raw moment exists only when \(\alpha>r\).
Mean. Since \(2.5>1\), \(E[X]=2.5(3)/(2.5-1)=7.5/1.5=5\).
Variance. Since \(2.5>2\), \(\operatorname{Var}(X)=2.5(3^2)/[(1.5)^2(0.5)] =22.5/(2.25\cdot0.5)=22.5/1.125=20\).
Third moment. It is not finite, because the condition \(\alpha>3\) fails: \(2.5\) is not greater than 3. Thus finite mean and variance do not imply finite third moment.
Tail probability. Since \(12\ge3\), \(P(X>12)=(3/12)^{2.5}=(1/4)^{2.5}=1/32=0.03125\), or 0.0313 to four decimal places. The check is \(1/4^{2.5}=1/(16\cdot2)=1/32\). A common arithmetic trap is to report 0.0361; that is not the value of this power. The Pareto model matches the stipulated power-law tail, while its moment conditions also meet the stated finite-mean and finite-variance requirements.
Worked Example: Recognizing an Increasing Weibull Hazard
For a hypothetical equipment-repair severity, the stated hazard is \(h_X(x)=2x/100\) for \(x>0\). Among exponential, Weibull, and Type I Pareto candidates, identify the model consistent with this exact hazard. Then calculate its survival probability at 10 and 20, its mean, and its variance.
Method. Compare the functional form of the stated hazard with each candidate, then use the implied Weibull parameters. For a Weibull severity, \(h_X(x)=(k/\eta)(x/\eta)^{k-1}\).
Identify the parameters. The stated hazard is \(0.02x\), a linear increasing function. Taking \(k=2\) and \(\eta=10\) in the Weibull hazard gives \((2/10)(x/10)=0.02x\), exactly as specified. An exponential hazard is constant, while a Type I Pareto hazard decreases over its support, so neither has this exact hazard.
Survival calculations. The Weibull survival function is \(S_X(x)=\exp[-(x/\eta)^k]\). Hence \(S_X(10)=\exp[-(10/10)^2]=e^{-1}=0.3679\) and \(S_X(20)=\exp[-(20/10)^2]=e^{-4}=0.0183\), to four decimal places. As a conditional check, \(P(X>20\mid X>10)=S_X(20)/S_X(10)=e^{-3}=0.0498\).
Moments. For Weibull shape 2 and scale 10, \(E[X]=\eta\Gamma(1+1/k)=10\Gamma(3/2)=5\sqrt{\pi}\approx8.8623\). Also, \(\operatorname{Var}(X)=\eta^2\{\Gamma(1+2/k)-[\Gamma(1+1/k)]^2\} =100(1-\pi/4)\approx21.4602\). The variance is positive, and the survival probabilities decrease as loss size increases, as required. The common mistake is to infer “Weibull” from any increasing hazard; here the stronger evidence is the exact linear hazard, which implies shape 2.
When a Mixture Is More Plausible
A single severity family may be inadequate when the risk population combines distinct types of exposure. A finite mixture first selects a component and then generates a loss from that component. Its overall mean is the weighted average of component means, but its variance also includes variation between those means. A mixture can therefore have substantially more variability than either component alone, and its hazard can change as the remaining losses become increasingly dominated by one component.
Worked Example: Identifying Heterogeneity in a Two-Component Mixture
A hypothetical service contract has two unobserved loss types. With probability 0.75, \(X\) is exponential with rate 1; with probability 0.25, \(X\) is exponential with rate 0.1. Find the overall mean and variance, and describe the mixture hazard and tail. Decide whether a single exponential with the same mean captures the stated variability.
Method. Apply the law of total expectation and the variance decomposition \(\operatorname{Var}(X)=E[\operatorname{Var}(X\mid C)]+\operatorname{Var}(E[X\mid C])\), where \(C\) denotes the selected component.
Mean. The component means are \(1/1=1\) and \(1/0.1=10\), so \(E[X]=0.75(1)+0.25(10)=0.75+2.5=3.25\).
Variance. The component variances are \(1\) and \(1/0.1^2=100\). The within-component contribution is \(E[\operatorname{Var}(X\mid C)]=0.75(1)+0.25(100)=25.75\). The between-component contribution is \(0.75(1-3.25)^2+0.25(10-3.25)^2 =0.75(5.0625)+0.25(45.5625) =3.796875+11.390625=15.1875\). Thus the overall variance is \(25.75+15.1875=40.9375\).
Hazard and tail. The survival and density are \(S_X(x)=0.75e^{-x}+0.25e^{-0.1x}\) and \(f_X(x)=0.75e^{-x}+0.025e^{-0.1x}\). Therefore \(h_X(0)=f_X(0)/S_X(0)=0.775\). As \(x\) grows, the slower-decaying component dominates both expressions, so the hazard tends to 0.1 and the survival tail is asymptotically proportional to \(e^{-0.1x}\). The hazard declines from its initial value toward 0.1.
Conclusion and check. A single exponential with mean 3.25 would have rate \(1/3.25\) and variance \(3.25^2=10.5625\), far below the mixture variance 40.9375. It does not capture the stipulated heterogeneity and variability. The mixture variance exceeds its mean squared, as the computed coefficient of variation is \(\sqrt{40.9375}/3.25\approx1.968\), rather than the exponential value 1.
Common Mistakes and a Practical Decision Process
Model choice is strongest when it proceeds from requirements to candidates, rather than from a familiar distribution name to a forced interpretation. The following sequence keeps the comparisons organized.
Confirm that the family permits the loss values and has finite moments wherever the problem requires them. For Pareto, compare the shape parameter with the order of each required moment.
For example, an exponential requires coefficient of variation 1 and skewness 2; a gamma links the two through its shape. Reject a candidate when its required pattern conflicts with the stated features.
A monotone hazard describes local conditional behavior; a tail comparison describes long-range decay. Neither should be substituted for the other.
A mixture may explain excess variation or a hazard that changes as the composition of surviving losses changes. Do not label an exponential mixture as Pareto merely because it has a longer tail than one exponential component.
Say that a family is ruled out, compatible, or particularly plausible. A few matched features do not establish uniqueness unless the information actually identifies the distribution.
- Mixing up gamma scale and rate. With the scale convention used here, the gamma mean is \(\alpha\theta\), not \(\alpha/\theta\).
- Assuming a decreasing hazard determines one family. Several families and mixtures can have decreasing hazards.
- Calling every right-skewed model heavy-tailed. Right skew is not the same as a power-law tail; state which tail comparison is intended.
- Checking only the mean. A family can match the mean while failing the variance, skewness, hazard, or required moment conditions.
- Confusing a plausible choice with a fitted or uniquely identified model. Risk features can narrow the options without determining a unique family.
These comparisons are useful because a severity model affects quantities beyond the average loss. Tail probabilities and the frequency of very large claims can be especially sensitive to the family selected. For professional analysis, the choice should therefore reflect the risk features that matter to the intended application, with uncertainty about the model made explicit. The selection principles here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each original scenario, identify the model implication and check the calculation or condition stated in the answer.
- A hypothetical severity has mean 6, variance 36, and skewness 2, with a constant hazard. Is an exponential model compatible? Give its rate. (Answer: Yes; \(\lambda=1/6\), and its variance and skewness are 36 and 2.)
- A gamma severity has shape \(\alpha=4\) and scale \(\theta=3\). Find its mean, variance, and skewness. (Answer: Mean \(12\), variance \(36\), skewness \(1\).)
- A Type I Pareto severity has minimum \(x_m=5\) and shape \(\alpha=3.5\). Find its mean and variance, and state whether its third moment is finite. (Answer: Mean \(7\), variance \(28/3\approx9.3333\), and the third moment is finite because \(3.5>3\).)
- A Weibull severity has shape \(k=0.5\) and scale \(\eta=4\). Is its hazard increasing or decreasing, and what is its mean? (Answer: Decreasing hazard; mean \(4\Gamma(3)=8\).)
- A mixture chooses an exponential rate of 0.5 with probability 0.8 and rate 0.05 with probability 0.2. Find the hazard at zero and its limiting value as \(x\) tends to infinity. (Answer: \(h_X(0)=0.8(0.5)+0.2(0.05)=0.41\); the limiting hazard is \(0.05\).)