Pareto Severity: Moments and Their Existence

Introduction

Pareto models describe losses whose upper tails decrease as a power of the loss threshold. This tail behavior can make very large losses sufficiently influential that a model has an infinite mean or variance, even though every individual loss is finite. For actuarial work, calculating a moment formula is only part of the task: first check whether that moment exists.

There are multiple distributions called Pareto. This tutorial uses the Pareto Type II, also called the Lomax distribution, with a nonnegative loss and a positive scale parameter. Stating the parameterization matters: the survival formula and moment conditions below belong to this specific model.

Definition: A loss $X$ has a Pareto Type II distribution with scale $\theta>0$ and shape $\alpha>0$ if, for $x\geq 0$, $$ S_X(x)=P(X>x)=\left(\frac{\theta}{\theta+x}\right)^\alpha. $$ Its cumulative distribution function is $F_X(x)=1-S_X(x)$, and its density is $$ f_X(x)=\frac{\alpha\theta^\alpha}{(\theta+x)^{\alpha+1}},\qquad x\geq 0. $$ The scale $\theta$ has the same units as $X$; the shape $\alpha$ is dimensionless.

The model is continuous, so $P(X>x)=P(X\geq x)$ for any fixed threshold. Its survival function starts at $S_X(0)=1$ and decreases toward zero as $x$ increases. A larger scale makes losses larger on the same monetary scale, while the shape controls how quickly the tail decreases.

Survival Probabilities

For any nonnegative threshold $c$, substitute $x=c$ in the survival function. To find the probability that a loss lies in an interval, subtract survival probabilities: for $0\leq a<b$, $P(a<X\leq b)=S_X(a)-S_X(b)$. Continuity means the inclusion or exclusion of either fixed endpoint does not change this probability.

Worked Example: Two Tail Probabilities and Both Moments

A hypothetical portfolio models an individual claim severity $X$, in dollars, as Pareto Type II with $\theta=\$12{,}000$ and $\alpha=3$. Calculate the probabilities that a claim exceeds $\$24{,}000$ and $\$60{,}000$, then calculate the mean and variance and check their existence.

Method and tail probabilities. Use $S_X(x)=(\theta/(\theta+x))^\alpha$. At $\$24{,}000$, $$ P(X>\$24{,}000)=\left(\frac{12{,}000}{12{,}000+24{,}000}\right)^3 =\left(\frac{1}{3}\right)^3 =\frac{1}{27} =0.0370 $$ to four decimal places. At $\$60{,}000$, $$ P(X>\$60{,}000)=\left(\frac{12{,}000}{72{,}000}\right)^3 =\left(\frac{1}{6}\right)^3 =\frac{1}{216} =0.0046. $$

Mean. Since $\alpha=3>1$, the mean exists and is $$ E[X]=\frac{\theta}{\alpha-1} =\frac{\$12{,}000}{2} =\$6{,}000. $$ Variance. Since $\alpha=3>2$, the variance exists: $$ \operatorname{Var}(X) =\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)} =\frac{3(\$12{,}000)^2}{2^2(1)} =\$108{,}000{,}000. $$ The units of variance are dollars squared. The checks are that both moments are finite for this shape, the probabilities decrease with the threshold, and $0.0046<0.0370$.

Common mistake avoided: Substituting the shape into a moment expression without checking its existence condition can produce an invalid result. Here $\alpha>2$, so both formulas apply.

Why the Mean and Variance Have Shape Thresholds

The density follows by differentiating the cumulative distribution function. The $r$th raw moment, when its defining integral is finite, is $E[X^r]=\int_0^\infty x^r f_X(x)\,dx$. Set $u=1+x/\theta$, so $x=\theta(u-1)$, $dx=\theta\,du$, and $\theta+x=\theta u$. This gives the integral

E[X^r]=\alpha\theta^r\int_1^\infty (u-1)^r u^{-\alpha-1}\,du.

This substitution also makes the tail behavior visible. For large $u$, the factor $(u-1)^r$ grows at the same rate as $u^r$, so the integrand behaves like a constant times $u^{r-\alpha-1}$. The integral over an unbounded interval converges only when $r-\alpha-1<-1$, or $\alpha>r$. The following direct arguments establish the relevant first- and second-moment conditions, including divergence at the boundary.

Check the first moment.
For $r=1$, the integrand is $(u-1)u^{-\alpha-1}$. If $\alpha>1$, its two terms can be integrated and give a finite result. If $0<\alpha\leq1$, then for $u\geq2$, $u-1\geq u/2$, so the integrand is at least $\tfrac12u^{-\alpha}$. The integral of this lower bound diverges for $0<\alpha\leq1$. Thus the mean is finite exactly when $\alpha>1$.

Check the second raw moment.
For $r=2$, the integrand is $(u-1)^2u^{-\alpha-1}$. If $\alpha>2$, each term after expanding $(u-1)^2$ has a convergent integral. If $0<\alpha\leq2$, then for $u\geq2$, $u-1\geq u/2$, and therefore $(u-1)^2u^{-\alpha-1}\geq\tfrac14u^{1-\alpha}$. The integral of this lower bound diverges when $0<\alpha\leq2$. Hence the second raw moment is finite exactly when $\alpha>2$.

Calculate the moments when the integrals converge.
For $\alpha>1$, integrating the first-moment integrand gives $E[X]=\theta/(\alpha-1)$. For $\alpha>2$, integrating the three terms in the second-moment integrand gives $E[X^2]=2\theta^2/((\alpha-1)(\alpha-2))$. Subtracting $E[X]^2$ gives the variance formula.

For clarity, the convergent integrations in Step 3 are

E[X]=\alpha\theta\left(\frac{1}{\alpha-1}-\frac{1}{\alpha}\right) =\frac{\theta}{\alpha-1},\qquad \alpha>1,

E[X^2]=\alpha\theta^2\left(\frac{1}{\alpha-2}-\frac{2}{\alpha-1}+\frac{1}{\alpha}\right) =\frac{2\theta^2}{(\alpha-1)(\alpha-2)},\qquad \alpha>2.

The second-moment calculation is valid as a term-by-term integration only for $\alpha>2$, when all three integrals converge. For $\alpha\leq2$, the lower-bound argument above establishes divergence without separating divergent terms or subtracting improper integrals.

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)},\qquad \alpha>2.

Theorem: For this Pareto Type II model, $E[X]$ is finite exactly when $\alpha>1$. The second raw moment is finite exactly when $\alpha>2$. Consequently, the variance is finite exactly when $\alpha>2$. When $1<\alpha\leq2$, the mean is finite but the variance is infinite. When $0<\alpha\leq1$, the mean is infinite, so there is no finite variance.

Reading the Existence Conditions in Practice

The mean and variance conditions are distinct. A shape of $1.5$, for example, permits a finite mean but not a finite variance. A shape of $1$ is the boundary at which the mean ceases to be finite; it is not a case where the mean formula gives an unusually large but valid number. More generally, do not use a moment formula at its boundary: the corresponding integral diverges there.

Worked Example: Finite Mean, Infinite Variance

A hypothetical equipment-loss model has scale $\theta=\$8{,}000$ and shape $\alpha=1.5$. Find $P(X>\$24{,}000)$, the mean, and whether the variance is finite.

Tail probability. The ratio is $\theta/(\theta+x)=8{,}000/(8{,}000+24{,}000)=1/4$. Therefore, $$ P(X>\$24{,}000)=\left(\frac14\right)^{1.5} =\frac{1}{8} =0.1250. $$

Mean. Since $1.5>1$, the mean exists: $$ E[X]=\frac{\$8{,}000}{1.5-1} =\frac{\$8{,}000}{0.5} =\$16{,}000. $$ Variance. Since $1.5\leq2$, the second raw moment diverges and the variance is not finite. The full set of conclusions is therefore a tail probability of $0.1250$, a finite mean of $\$16{,}000$, and an infinite variance.

Common mistake avoided: Applying the variance formula at $\alpha=1.5$ would put a negative value, $1.5-2=-0.5$, in its denominator and could suggest a nonsensical negative variance. The formula requires $\alpha>2$; here the variance is infinite, not negative.

Worked Example: A Shape at the Mean Boundary

A hypothetical small-claim severity model has scale $\theta=\$5{,}000$ and shape $\alpha=1$. Calculate $P(\$5{,}000<X\leq\$20{,}000)$, then determine whether the mean and variance are finite.

Interval probability. First calculate the survival probabilities: $$ S_X(\$5{,}000)=\left(\frac{5{,}000}{5{,}000+5{,}000}\right)^1 =\frac12, $$ $$ S_X(\$20{,}000)=\left(\frac{5{,}000}{5{,}000+20{,}000}\right)^1 =\frac15. $$ Thus $$ P(\$5{,}000<X\leq\$20{,}000) =S_X(\$5{,}000)-S_X(\$20{,}000) =\frac12-\frac15 =\frac{3}{10} =0.3000. $$

Moment existence. Because $\alpha=1$ does not satisfy $\alpha>1$, the mean is infinite. The second raw moment is also infinite because $\alpha$ does not satisfy $\alpha>2$. There is no finite variance. The computed interval probability is still valid: a probability can be finite even when the mean is not.

Common mistake avoided: Substituting $\alpha=1$ into $\theta/(\alpha-1)$ does not produce a finite mean. The zero denominator marks the boundary of the formula’s validity; the defining integral diverges.

Common Exam Traps

Mixing Pareto parameterizations. This tutorial uses $S_X(x)=(\theta/(\theta+x))^\alpha$ for $x\geq0$. A formula for a different Pareto model may use another survival function or support. Check the stated model before substituting.
Ignoring existence conditions. Check $\alpha>1$ before using the mean formula and $\alpha>2$ before using the variance formula.
Confusing a finite mean with a finite variance. When $1<\alpha\leq2$, the mean exists but the variance does not.
Subtracting tail probabilities in the wrong order. For $a<b$, $P(a<X\leq b)=S_X(a)-S_X(b)$, which must be nonnegative.
Dropping units. The mean has the units of loss; the variance has squared units. Keep monetary amounts consistent when calculating $\theta^2$.

Key takeaway: For Pareto Type II severity, calculate tail probabilities directly from the stated survival function. Check the shape before calculating moments: the mean is finite only for $\alpha>1$, and the variance is finite only for $\alpha>2$. These are properties of the entire tail, not of any one observed loss.

The Pareto model is useful for examining how heavy tails affect severity summaries: two models can have valid survival probabilities while only one has a finite mean or variance. An actuarial analysis that relies on an average or a variance should therefore verify the moment condition before interpreting that quantity. Correct identification of the Pareto parameterization is equally important, since a threshold probability calculated under the wrong survival function can be materially incorrect. These checks are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical Pareto Type II model, use $S_X(x)=(\theta/(\theta+x))^\alpha$. Give probabilities to four decimal places and state whether each requested moment is finite.

A severity has $\theta=\$6$ and $\alpha=2.5$. Find $P(X>\$18)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $P(X>\$18)=0.0313$, $E[X]=\$4$, and $\operatorname{Var}(X)=\$80$; both moments are finite.)
A severity has $\theta=\$10$ and $\alpha=1.25$. Find $P(X>\$30)$, the mean, and whether the variance is finite. (Answer: $P(X>\$30)=0.1768$, $E[X]=\$40$, and the variance is infinite.)
A severity has $\theta=\$2$ and $\alpha=2$. Find $P(X>\$6)$, the mean, and whether the variance is finite. (Answer: $P(X>\$6)=0.0625$, $E[X]=\$2$, and the variance is infinite.)
A severity has $\theta=\$15$ and $\alpha=0.75$. Find $P(X>\$45)$ and determine whether its mean and variance are finite. (Answer: $P(X>\$45)=0.3536$; the mean and variance are not finite.)
A severity has $\theta=\$4$ and $\alpha=3$. Find $P(\$8<X\leq\$20)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $P(\$8<X\leq\$20)=7/216=0.0324$, $E[X]=\$2$, and $\operatorname{Var}(X)=\$12$.)

What You'll Learn