Why Compare Tails and Moments?
Two severity models can have the same mean and still imply very different risk at high loss amounts. A model’s tail describes how quickly the probability of a large loss decreases; its moments summarize features such as average loss and variability. For actuarial work, it is important to know both: a finite mean does not ensure a finite variance, and finite moments do not necessarily imply a finite moment-generating function.
The comparisons below concern positive continuous severity random variables. For each family, the parameters are stated explicitly because parameterizations differ, particularly for the gamma and Weibull distributions.
Five Severity Families
Use the following parameterizations throughout. The exponential distribution has rate \(\lambda>0\); the gamma distribution has shape \(\alpha>0\) and rate \(\beta>0\); the lognormal distribution has \(\ln X\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\); the Pareto Type I distribution has minimum \(\theta>0\) and tail index \(\alpha>0\); and the Weibull distribution has shape \(k>0\) and scale \(\lambda>0\).
| Family | Survival function | Positive-moment existence | Tail behavior |
|---|---|---|---|
| Exponential, rate \(\lambda\) | \(e^{-\lambda x}\), \(x\geq 0\) | All \(r>0\) | Exponential decay |
| Gamma, shape \(\alpha\), rate \(\beta\) | \(1-F_X(x)\) | All \(r>0\) | Exponential decay, with a power factor |
| Lognormal, \(\ln X\sim N(\mu,\sigma^2)\) | \(1-\Phi((\ln x-\mu)/\sigma)\), \(x>0\) | All \(r>0\) | Slower than any exponential tail; faster than any power tail |
| Pareto Type I, minimum \(\theta\), index \(\alpha\) | \((\theta/x)^\alpha\), \(x\geq\theta\) | Exactly when \(r<\alpha\) | Power-tail decay |
| Weibull, shape \(k\), scale \(\lambda\) | \(e^{-(x/\lambda)^k}\), \(x\geq0\) | All \(r>0\) | Compared with exponential: slower if \(k<1\), equal in form if \(k=1\), faster if \(k>1\) |
For a gamma distribution, the survival function has an exponential factor \(e^{-\beta x}\) multiplied asymptotically by a power of \(x\). The lognormal tail has a different pattern: it decreases more slowly than \(e^{-cx}\) for every fixed \(c>0\), yet more quickly than \(x^{-q}\) for every fixed \(q>0\). Thus it is not a power-tail model like Pareto, even though its upper tail is heavier than an exponential tail.
The Weibull comparison in the table is with an exponential tail: when \(k<1\), the survival probability decreases more slowly than any exponential with a fixed positive rate, while for \(k>1\) it decreases faster than any such exponential. All positive power moments nevertheless exist for every Weibull shape \(k>0\).
Moment Formulas and the Pareto Threshold
For positive \(r\), the raw moments make the existence distinctions precise. In these formulas, \(\Gamma(\cdot)\) is the gamma function. The exponential distribution is a gamma distribution with shape one, but its moment expression is shown separately for clarity.
The Pareto formula has a strict condition. Its density is \(f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}\) for \(x\geq\theta\). Multiplying by \(x^r\) and integrating gives
The integral converges only if the exponent \(r-\alpha-1\) is less than \(-1\), or equivalently \(r<\alpha\). When \(r<\alpha\), its value is \(\theta^{r-\alpha}/(\alpha-r)\), giving \(E[X^r]=\alpha\theta^r/(\alpha-r)\). At \(r=\alpha\), the integral is proportional to \(\int_\theta^\infty x^{-1}\,dx\), which diverges; for \(r>\alpha\), it also diverges. The boundary is therefore not included.
Worked Comparisons
The moment comparison is most informative when the models describe similar average losses. The first example holds the mean fixed and shows how much the variance can still differ.
Worked Example: Equal Means, Different Variances
A hypothetical analyst considers five candidate severities, each with mean \(2\) (in thousands of dollars). The models are exponential with rate \(\lambda=0.5\); gamma with shape \(\alpha=2\) and rate \(\beta=1\); Weibull with shape \(k=0.5\) and scale \(\lambda=1\); lognormal with \(\mu=\tfrac12\ln 2\) and \(\sigma^2=\ln 2\); and Pareto Type I with \(\theta=\tfrac43\) and tail index \(\alpha=3\). Calculate each variance and identify whether all second moments exist.
Method. Use \(E[X]\), \(E[X^2]\), and \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\). Each model’s parameterization is as stated above.
Exponential. The mean is \(1/\lambda=1/0.5=2\). Its second moment is \(\Gamma(3)/(0.5)^2=2/0.25=8\). Thus its variance is \(8-2^2=4\).
Gamma. Its mean is \(\alpha/\beta=2/1=2\). Its second moment is \(\Gamma(4)/(\Gamma(2)1^2)=6/1=6\). Its variance is \(6-4=2\).
Weibull. Its mean is \(1\cdot\Gamma(1+1/0.5)=\Gamma(3)=2\). Its second moment is \(1^2\Gamma(1+2/0.5)=\Gamma(5)=24\). Its variance is \(24-4=20\).
Lognormal. Its mean is \(\exp(\mu+\sigma^2/2)=\exp(\tfrac12\ln2+\tfrac12\ln2)=2\). Its second moment is \(\exp(2\mu+2\sigma^2)=\exp(\ln2+2\ln2)=8\), so its variance is \(8-4=4\).
Pareto. Since \(2<3\), its second moment exists and equals \(3(4/3)^2/(3-2)=16/3\). Its mean is \(3(4/3)/(3-1)=2\), so its variance is \(16/3-4=4/3\).
Check. All five means equal \(2\), but the variances are \(4,\ 2,\ 20,\ 4,\) and \(4/3\), respectively. All second moments exist here; the Pareto model’s tail index \(3\) exceeds the moment order \(2\). The calculation avoids the mistake of assuming that equal means imply similar upper-tail risk.
The Pareto threshold is particularly important for identifying when common summaries fail.
Worked Example: Testing Moments at the Pareto Boundary
A hypothetical excess-loss severity follows Pareto Type I with minimum \(\theta=2\) and tail index \(\alpha=2\). Determine whether the mean and variance exist, and calculate \(E[X^{1.5}]\).
Method. Compare each requested moment order with \(\alpha=2\). A moment is finite only for an order strictly below \(2\).
For the mean, \(r=1<2\), so it is finite: \(E[X]=2(2)/(2-1)=4\).
For the second raw moment, \(r=2=\alpha\), so it is not finite. Directly, the integrand in the moment integral is proportional to \(x^{2-2-1}=x^{-1}\), whose integral from \(2\) to infinity diverges. Therefore the variance is infinite; it is not valid to subtract the squared mean from a finite second moment.
For \(r=1.5\), which is below \(2\), \(E[X^{1.5}]=2(2^{1.5})/(2-1.5)=4\sqrt{2}/0.5=8\sqrt{2}\), approximately \(11.31\).
Check. The mean is \(4\), the variance is infinite, and the \(1.5\)th moment is finite at approximately \(11.31\). The boundary order diverges, even though every smaller positive order has a finite moment.
Tail Behavior Is Not Just a Moment Question
A finite set of moments does not completely describe how a model behaves at very high losses. For example, lognormal severities have every positive power moment finite, but their moment-generating function \(E[e^{tX}]\) is infinite for every \(t>0\). The same is true for Pareto severities. A moment-generating function weights large values exponentially, a substantially stronger requirement than finiteness of each power moment.
For comparison, a gamma severity with rate \(\beta\) has a finite moment-generating function for \(t<\beta\); an exponential severity with rate \(\lambda\) has one for \(t<\lambda\). A Weibull severity has a finite moment-generating function for all positive \(t\) if \(k>1\), for \(0<t<1/\lambda\) if \(k=1\), and for no positive \(t\) if \(0<k<1\). These conditions concern exponential moments, not the positive power moments listed earlier.
Worked Example: Comparing Tails at a Common Threshold
Three hypothetical models each have mean \(10\): an exponential severity with rate \(\lambda=0.1\), a Weibull severity with shape \(k=0.5\) and scale \(\lambda=5\), and a Pareto Type I severity with minimum \(\theta=5\) and tail index \(\alpha=2\). Compare their probabilities of exceeding \(100\), then state what the comparison does and does not show.
Method. Substitute \(x=100\) in each survival function. First check the means: the exponential mean is \(1/0.1=10\); the Weibull mean is \(5\Gamma(3)=10\); and the Pareto mean is \(2(5)/(2-1)=10\).
Exponential. \(S_X(100)=e^{-0.1(100)}=e^{-10}\approx0.0000454\).
Weibull. \(S_X(100)=\exp[-(100/5)^{0.5}]=e^{-\sqrt{20}}\approx e^{-4.4721}\approx0.0114\).
Pareto. Since \(100\geq5\), \(S_X(100)=(5/100)^2=0.05^2=0.0025\).
Check. At \(100\), the exceedance probabilities rank Weibull, Pareto, then exponential. The Pareto variance is infinite because its tail index \(2\) does not exceed the second-moment order. The Weibull variance is finite despite its relatively large exceedance probability at this threshold. A single-threshold ranking is not a universal tail ranking: in this example the Pareto tail eventually exceeds the Weibull tail as \(x\) grows, since a power tail decays more slowly than \(e^{-\sqrt{x/5}}\). The common mistake is to infer eventual tail order from one threshold.
Common Exam Traps
- Including the Pareto boundary. The condition is \(r<\alpha\), not \(r\leq\alpha\). At \(r=\alpha\), the defining integral diverges logarithmically.
- Confusing finite power moments with a finite moment-generating function. The lognormal has all positive power moments, yet \(E[e^{tX}]\) is infinite for \(t>0\). One property does not imply the other.
- Calling every non-exponential tail a power tail. The lognormal tail is heavier than an exponential tail but still decreases faster than every inverse power \(x^{-q}\). Pareto has an actual power tail.
- Ranking tails from a single observation point. A survival probability at one selected \(x\) gives a pointwise comparison only. Tail classification concerns how survival probabilities behave as \(x\) increases.
- Ignoring the Weibull shape. All positive power moments exist for every \(k>0\), while the tail’s comparison with exponential decay changes at \(k=1\).
Check Your Understanding
For each problem, identify the relevant moment or tail condition before calculating.
- A gamma severity has shape \(\alpha=3\) and rate \(\beta=2\). Calculate its second raw moment and state whether it is finite. (Answer: finite, \(E[X^2]=3\).)
- A Weibull severity has shape \(k=2\) and scale \(\lambda=3\). Calculate its third raw moment in exact form. (Answer: finite, \(E[X^3]=\frac{81\sqrt{\pi}}{4}\).)
- A Pareto Type I severity has minimum \(\theta=2\) and tail index \(\alpha=1.8\). Determine whether its mean and variance exist, and calculate its \(1.5\)th moment. (Answer: mean and variance are infinite; \(E[X^{1.5}]=12\sqrt{2}\approx16.9706\).)
- A lognormal severity has \(\mu=\ln3\) and \(\sigma^2=\ln4\). Calculate its mean and variance, and state whether its moment-generating function is finite for positive \(t\). (Answer: mean \(6\), variance \(108\); its moment-generating function is infinite for every \(t>0\).)
- Compare the eventual tail behavior of a Weibull severity with \(k=0.75\) and an exponential severity. State whether all positive power moments of the Weibull severity exist. (Answer: the Weibull tail is heavier than exponential decay, and all its positive power moments are finite.)