Why Severity Moments Need Careful Parameter Checks
A severity model describes the size \(X\) of a loss, with \(X\geq 0\). Its mean measures the expected loss size, while its variance measures the spread around that mean. For actuarial calculations, knowing a model’s name is not enough: parameter conventions differ, and a formula for a moment is useful only when that moment exists.
This tutorial calculates raw moments, means, and variances for uniform, exponential, gamma, Pareto, lognormal, and Weibull losses. The examples use hypothetical loss amounts. Unless otherwise stated, amounts are in the units named in the problem; variances are in squared units.
Moment Formulas and Parameter Conventions
The table gives common parameterizations and the corresponding first two raw moments. For continuous models, \(X\) has the stated distribution and the parameters satisfy the conditions listed. A scale parameter stretches the loss axis; an exponential rate parameter instead appears in the exponent as a multiplier of loss.
| Model and parameters | \(E[X]\) | \(E[X^2]\) and variance | Moment condition |
|---|---|---|---|
| Uniform on \([a,b]\), \(a<b\) | \((a+b)/2\) | \(E[X^2]=(a^2+ab+b^2)/3\); \(\operatorname{Var}(X)=(b-a)^2/12\) | All positive integer moments are finite |
| Exponential with rate \(\lambda>0\) | \(1/\lambda\) | \(E[X^2]=2/\lambda^2\); \(\operatorname{Var}(X)=1/\lambda^2\) | All positive integer moments are finite |
| Gamma with shape \(\alpha>0\), scale \(\theta>0\) | \(\alpha\theta\) | \(E[X^2]=\alpha(\alpha+1)\theta^2\); \(\operatorname{Var}(X)=\alpha\theta^2\) | All positive integer moments are finite |
| Pareto I with minimum \(\theta>0\), shape \(\alpha>0\) | \(\alpha\theta/(\alpha-1)\) | \(E[X^2]=\alpha\theta^2/(\alpha-2)\); \(\operatorname{Var}(X)=\alpha\theta^2/((\alpha-1)^2(\alpha-2))\) | \(E[X^r]\) is finite iff \(\alpha>r\) |
| Pareto II (Lomax) with scale \(\theta>0\), shape \(\alpha>0\) | \(\theta/(\alpha-1)\) | \(E[X^2]=2\theta^2/((\alpha-1)(\alpha-2))\); \(\operatorname{Var}(X)=\alpha\theta^2/((\alpha-1)^2(\alpha-2))\) | \(E[X^r]\) is finite iff \(\alpha>r\) |
| Lognormal, \(\ln X\sim N(\mu,\sigma^2)\) | \(\exp(\mu+\sigma^2/2)\) | \(E[X^2]=\exp(2\mu+2\sigma^2)\); \(\operatorname{Var}(X)=\exp(2\mu+\sigma^2)(\exp(\sigma^2)-1)\) | All positive integer moments are finite |
| Weibull with shape \(k>0\), scale \(\lambda>0\) | \(\lambda\Gamma(1+1/k)\) | \(E[X^2]=\lambda^2\Gamma(1+2/k)\); variance is \(E[X^2]-\{E[X]\}^2\) | All positive integer moments are finite |
Here \(\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt\) for \(z>0\), with \(\Gamma(z+1)=z\Gamma(z)\). For a gamma model, the scale convention means the density is proportional to \(x^{\alpha-1}e^{-x/\theta}\), not \(e^{-\theta x}\). The latter uses a rate parameter and has scale \(1/\theta\).
For Pareto I, the support begins at \(\theta\), with survival function \(P(X>x)=(\theta/x)^\alpha\) for \(x\geq\theta\). For Pareto II, the support begins at zero, with survival function \(P(X>x)=(\theta/(\theta+x))^\alpha\) for \(x\geq0\). Their different supports and scales matter: do not transfer a formula from one Pareto parameterization to the other.
Why the Raw-Moment and Pareto Formulas Hold
For any model with a finite second moment, the variance identity follows by expanding the square around the mean:
The Pareto I condition can be seen directly from its density \(f_X(x)=\alpha\theta^\alpha x^{-\alpha-1}\), for \(x\geq\theta\). For \(r>0\), integrate \(x^r f_X(x)\). The resulting power integral converges at infinity only when \(r-\alpha-1<-1\), or \(\alpha>r\).
If \(\alpha\leq r\), that integral diverges, so the \(r\)th moment is infinite; substituting those parameter values into the finite-moment expression is not valid. The same threshold applies to Pareto II: its density has a tail proportional to \(x^{-\alpha-1}\), so the integral for the \(r\)th moment has the same convergence requirement.
For example, the Pareto I mean requires \(\alpha>1\), and its variance requires \(\alpha>2\). A model with \(1<\alpha\leq2\) has a finite mean but infinite variance. More generally, for any of the models here, calculate \(E[X^2]\) and \(E[X]\) first, check that both are finite, and only then subtract the squared mean.
Worked Severity Moment Calculations
Worked Example: Comparing Uniform and Exponential Losses
A hypothetical review considers two loss models, with each loss measured in thousands of dollars. Model A is uniform on \([0,12]\). Model B is exponential with rate \(\lambda=1/4\) per thousand dollars. Calculate the mean, second raw moment, and variance for each model, then compare their means and variances.
Method. Use the uniform endpoints for Model A and the exponential rate formulas for Model B. The exponential parameter is a rate, so its mean is \(1/\lambda\), not \(\lambda\).
For Model A, \(a=0\) and \(b=12\). Thus $$ E[X]=\frac{0+12}{2}=6,\qquad E[X^2]=\frac{0^2+0(12)+12^2}{3}=\frac{144}{3}=48. $$ The variance is \(48-6^2=48-36=12\), in squared thousands of dollars.
For Model B, \(\lambda=1/4\). Therefore $$ E[X]=\frac{1}{1/4}=4,\qquad E[X^2]=\frac{2}{(1/4)^2}=\frac{2}{1/16}=32. $$ Its variance is \(32-4^2=32-16=16\), also in squared thousands of dollars.
Model A has the larger mean, \(6\) rather than \(4\) thousand dollars. Model B has the larger variance, \(16\) rather than \(12\) squared thousands of dollars. All results are exact, so no rounding is needed. As a check, each variance is positive and equals the second raw moment minus the squared mean.
Common mistake avoided: treating the exponential rate \(1/4\) as the mean would reverse the scale. With rate parameterization, the mean is its reciprocal.
Worked Example: Gamma and Pareto Models With Different Moment Conditions
Three hypothetical claim models are stated in thousands of dollars: (A) gamma with shape \(\alpha=3\) and scale \(\theta=2\); (B) Pareto I with minimum \(\theta=3\) and shape \(\alpha=5\); and (C) Pareto II with scale \(\theta=2\) and shape \(\alpha=5\). For each, calculate the mean and variance, and determine whether the third raw moment is finite.
Method. Apply each model’s own parameterization, check the relevant shape threshold, and use \(E[X^2]-\{E[X]\}^2\). In all three models the third moment is finite: gamma moments are finite for positive integer orders, and both Pareto shapes satisfy \(5>3\).
For Model A, the gamma mean and variance are $$ E[X]=\alpha\theta=3(2)=6,\qquad \operatorname{Var}(X)=\alpha\theta^2=3(2^2)=12. $$ Its second raw moment is \(12+6^2=48\). The third moment is finite.
For Model B, Pareto I has \(\alpha=5>2\), so its variance exists. Its mean and second raw moment are $$ E[X]=\frac{5(3)}{5-1}=\frac{15}{4},\qquad E[X^2]=\frac{5(3^2)}{5-2}=15. $$ Consequently, $$ \operatorname{Var}(X)=15-\left(\frac{15}{4}\right)^2 =\frac{240}{16}-\frac{225}{16} =\frac{15}{16}. $$ The third raw moment is finite because \(5>3\).
For Model C, Pareto II also has \(\alpha=5\). Its mean and second raw moment are $$ E[X]=\frac{2}{5-1}=\frac12,\qquad E[X^2]=\frac{2(2^2)}{(5-1)(5-2)}=\frac{8}{12}=\frac23. $$ Thus $$ \operatorname{Var}(X)=\frac23-\left(\frac12\right)^2 =\frac{8}{12}-\frac{3}{12}=\frac{5}{12}. $$ Its third raw moment is finite because \(5>3\). The results are exact and need no rounding.
Check. The second raw moments reconstructed from mean squared plus variance are \(36+12=48\), \(225/16+15/16=15\), and \(1/4+5/12=2/3\), respectively.
Common mistake avoided: Pareto I and Pareto II are not interchangeable. Model B’s minimum loss is 3, whereas Model C’s support begins at zero; using one model’s mean formula for the other gives an incorrect result.
Worked Example: Lognormal and Weibull Moments
A hypothetical severity analyst considers two models, with losses measured in dollars. For Model L, \(\ln X\sim N(0,\ln 4)\). For Model W, \(X\) is Weibull with shape \(k=1/2\) and scale \(\lambda=1\). Calculate each mean, second raw moment, and variance.
Method. For the lognormal model, use the variance of the underlying normal variable as \(\sigma^2=\ln4\). For the Weibull model, substitute the shape into the gamma-function moment formula. Both models have finite moments of all positive integer orders.
For Model L, $$ E[X]=\exp\left(0+\frac{\ln4}{2}\right)=2,\qquad E[X^2]=\exp(2(0)+2\ln4)=16. $$ Hence \(\operatorname{Var}(X)=16-2^2=12\) dollars squared.
For Model W, the \(r\)th raw moment is \(\lambda^r\Gamma(1+r/k)\). With \(\lambda=1\) and \(k=1/2\), $$ E[X]=1\cdot\Gamma(1+2)=\Gamma(3)=2,\qquad E[X^2]=1^2\Gamma(1+4)=\Gamma(5)=24. $$ Therefore \(\operatorname{Var}(X)=24-2^2=20\) dollars squared. The exact results are \(E[X]=2\), \(\operatorname{Var}(X)=12\) for Model L and \(E[X]=2\), \(\operatorname{Var}(X)=20\) for Model W; no rounding is needed.
Check. Both second raw moments are at least their squared means: \(16\geq4\) and \(24\geq4\). The resulting variances are positive.
Common mistake avoided: the lognormal parameter \(\sigma^2\) is the variance of \(\ln X\), not its standard deviation. The Weibull scale is \(\lambda\), and it is raised to the power \(r\) in the \(r\)th moment.
A Reliable Calculation Sequence
A consistent sequence reduces errors when parameters or moment conditions are unfamiliar. In particular, do not calculate a variance by memory alone when the model’s parameter convention is uncertain.
Record the support and whether each parameter is a rate, scale, shape, or endpoint. Distinguish Pareto I from Pareto II and a gamma scale from a gamma rate.
For Pareto models, verify \(\alpha>1\) for a mean and \(\alpha>2\) for a variance. For the other models listed here, positive integer moments are finite under their stated parameter restrictions.
Find \(E[X]\) and \(E[X^2]\) using the matching model formulas. Preserve exact values during the calculation.
Compute \(E[X^2]-\{E[X]\}^2\). Confirm that the result is nonnegative and that adding the variance to the squared mean recovers \(E[X^2]\).
Common Errors and Their Corrections
- Using the second raw moment as the variance. If \(E[X^2]=48\) and \(E[X]=6\), the variance is \(48-36=12\), not 48.
- Ignoring a moment-existence condition. A Pareto model with \(\alpha=2\) does not have a finite variance. The variance formula cannot be used at \(\alpha=2\).
- Mixing up rate and scale. An exponential model with rate \(\lambda\) has mean \(1/\lambda\). A gamma model with scale \(\theta\) has mean \(\alpha\theta\).
- Using a generic “Pareto” formula without specifying the type. Pareto I has a positive minimum-loss parameter; Pareto II (Lomax) begins at zero. Their moment formulas differ.
- Using \(\sigma\) where \(\sigma^2\) is specified. In \(\ln X\sim N(\mu,\sigma^2)\), the second normal parameter is the variance of the log loss.
Check Your Understanding
For each hypothetical loss model, calculate the requested moments and check the stated existence condition.
- A loss is uniform on \([4,16]\) dollars. Find its mean, second raw moment, and variance. (Answer: \(E[X]=10\), \(E[X^2]=112\), \(\operatorname{Var}(X)=12\) dollars squared.)
- A loss is exponential with rate \(\lambda=1/3\) per thousand dollars. Find its mean and variance. (Answer: \(E[X]=3\) thousand dollars; \(\operatorname{Var}(X)=9\) squared thousands of dollars.)
- A loss is gamma with shape \(\alpha=2\) and scale \(\theta=3\). Find its mean, second raw moment, and variance. (Answer: \(E[X]=6\), \(E[X^2]=54\), \(\operatorname{Var}(X)=18\).)
- A loss follows Pareto I with minimum \(\theta=4\) and shape \(\alpha=3\). Find its mean, second raw moment, variance, and determine whether its third raw moment is finite. (Answer: \(E[X]=6\), \(E[X^2]=48\), \(\operatorname{Var}(X)=12\); the third raw moment is infinite because \(\alpha\not>3\).)
- A loss is lognormal with \(\ln X\sim N(0,\ln2)\). Find its mean, second raw moment, and variance. (Answer: \(E[X]=\sqrt{2}\), \(E[X^2]=4\), \(\operatorname{Var}(X)=2\).)
- A loss is Weibull with shape \(k=1\) and scale \(\lambda=3\). Find its mean and variance using the gamma-function formula. (Answer: \(E[X]=3\), \(\operatorname{Var}(X)=9\).)