Deriving Pareto Moment Thresholds

Why Pareto Moments Need a Threshold

For a heavy-tailed severity model, calculating an integral is not enough: the integral must converge. A Pareto Type I loss can have a finite mean but an infinite variance, or even an infinite mean. The shape parameter determines which of these outcomes applies. This matters whenever a calculation or model summary uses an expected loss, a variance, or a higher-order moment.

Use the Pareto Type I parameterization with scale $\theta>0$ and shape $\alpha>0$. The loss $X$ has support $x\geq\theta$, density $f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}$, and survival function $S_X(x)=(\theta/x)^\alpha$ on that support. No interest or mortality assumptions are involved: this is a severity model for a single loss.

Definition: For a positive real order $r$, the $r$th raw moment of $X$ is $E[X^r]$, when the expectation is finite. Under this Pareto Type I parameterization, the moment exists as a finite number exactly when $0<r<\alpha$. In that case, $E[X^r]=\alpha\theta^r/(\alpha-r)$.

The condition $r<\alpha$ is part of the result, not an optional technical detail. If $r\geq\alpha$, the moment integral diverges and $E[X^r]=\infty$. In particular, the mean is finite exactly when $\alpha>1$, and the second raw moment is finite exactly when $\alpha>2$.

Deriving the Moment Threshold

For a nonnegative continuous loss, the raw moment is the integral of $x^r$ against its density over the support. Substituting the Pareto density gives an improper integral. Its behavior as the upper limit tends to infinity determines whether the moment exists.

E[X^r] =\int_{\theta}^{\infty}x^r\alpha\theta^\alpha x^{-(\alpha+1)}\,dx =\alpha\theta^\alpha\int_{\theta}^{\infty}x^{r-\alpha-1}\,dx.

When $r\ne\alpha$, integrate first to a finite upper limit $b$, then examine the limit as $b\to\infty$:

\alpha\theta^\alpha\int_{\theta}^{b}x^{r-\alpha-1}\,dx =\frac{\alpha\theta^\alpha}{r-\alpha} \left(b^{r-\alpha}-\theta^{r-\alpha}\right).

If $r<\alpha$, then $r-\alpha<0$, so $b^{r-\alpha}\to0$. The remaining value is $\alpha\theta^\alpha\theta^{r-\alpha}/(\alpha-r)=\alpha\theta^r/(\alpha-r)$, which is finite. If $r>\alpha$, then $b^{r-\alpha}\to\infty$, so the integral diverges. At the boundary $r=\alpha$, the power in the integrand is $-1$, and the integral is logarithmic:

\alpha\theta^\alpha\int_{\theta}^{b}\frac{1}{x}\,dx =\alpha\theta^\alpha\log\left(\frac{b}{\theta}\right) \longrightarrow\infty.

Thus the boundary case also diverges. Together, the three cases prove that a positive-order moment is finite exactly below the shape parameter.

Mean and Variance

Set $r=1$ in the general moment formula when $\alpha>1$. This gives the mean. Set $r=2$ when $\alpha>2$ to obtain the second raw moment, then subtract the squared mean to obtain the variance.

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

E[X^2]=\frac{\alpha\theta^2}{\alpha-2},\qquad \alpha>2.

The variance formula therefore requires $\alpha>2$, which ensures both the mean and second raw moment are finite:

\begin{aligned} \operatorname{Var}(X) &=E[X^2]-\big(E[X]\big)^2\\ &=\frac{\alpha\theta^2}{\alpha-2} -\left(\frac{\alpha\theta}{\alpha-1}\right)^2\\ &=\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)}. \end{aligned}

For the last simplification, take the common denominator $(\alpha-2)(\alpha-1)^2$. The numerator becomes $\alpha\theta^2(\alpha-1)^2-\alpha^2\theta^2(\alpha-2)$, which equals $\alpha\theta^2\{(\alpha^2-2\alpha+1)-(\alpha^2-2\alpha)\}=\alpha\theta^2$. This also shows why the variance is positive when $\alpha>2$.

Theorem: For Pareto Type I severity with $\theta>0$ and $\alpha>0$, $E[X^r]$ is finite exactly for $0<r<\alpha$. The mean is $\alpha\theta/(\alpha-1)$ only for $\alpha>1$. The variance is $\alpha\theta^2/((\alpha-1)^2(\alpha-2))$ only for $\alpha>2$. When the relevant condition fails, do not use the finite-moment formula.

Worked Example: Mean, Variance, and a Boundary Moment

A hypothetical portfolio uses a Pareto Type I model for individual claim severity, with $\theta=\$300$ and $\alpha=4$. Find the first three raw moments, the variance, and determine whether the fourth raw moment is finite.

Method and conditions. The distribution is Pareto Type I with the stated scale and shape. There is no interest or mortality assumption. Since $1,2,3<4$, the moment formula applies to each of the first three raw moments.

For $r=1$, $E[X]=4(300)/(4-1)=1200/3=\$400$. For $r=2$, $E[X^2]=4(300^2)/(4-2)=4(90{,}000)/2=\$180{,}000$. For $r=3$, $E[X^3]=4(300^3)/(4-3)=4(27{,}000{,}000)=108{,}000{,}000$.

Because $\alpha=4>2$, the variance exists. Using the second raw moment, $\operatorname{Var}(X)=180{,}000-(400)^2=180{,}000-160{,}000=20{,}000$ dollars squared. The direct formula checks this result: $4(300^2)/((4-1)^2(4-2))=360{,}000/(9\cdot2)=20{,}000$.

For $r=4$, the order equals $\alpha$, so the fourth moment diverges logarithmically; it is not finite. Check: the first three orders are below 4 and have finite values, while the boundary order 4 does not. The common mistake avoided here is assuming that once several moments are finite, the next one must also be finite.

Worked Example: Finite Mean but Infinite Variance

For a hypothetical individual-loss model, suppose $X$ is Pareto Type I with $\theta=\$600$ and $\alpha=1.5$. Determine whether the mean and variance exist, and calculate the mean if it is finite.

Method and conditions. There is no interest or mortality assumption. The mean is finite when $\alpha>1$, and the variance requires $\alpha>2$. Here $1<1.5<2$, so the mean exists but the second raw moment does not.

Calculate the mean: $E[X]=1.5(600)/(1.5-1)=900/0.5=\$1{,}800$. For the second moment, its order is $r=2$, which is greater than $\alpha=1.5$. The moment threshold therefore gives $E[X^2]=\infty$, and the variance is infinite.

This conclusion can also be checked in the integral: the integrand for the second moment is proportional to $x^{2-1.5-1}=x^{-0.5}$, whose integral from 600 to infinity diverges. Check: the finite mean is consistent with $1<1.5$, but finite variance would require $2<1.5$, which is false. The common mistake avoided is inferring finite variance merely from a finite mean.

Worked Example: The Logarithmic Boundary at Shape Two

A hypothetical severity model is Pareto Type I with $\theta=\$1{,}000$ and $\alpha=2$. Find the mean, decide whether the variance exists, and explain the boundary calculation for the second moment.

Method and conditions. There is no interest or mortality assumption. Since $\alpha=2>1$, the mean is finite: $E[X]=2(1{,}000)/(2-1)=\$2{,}000$. The second raw moment would require $r=2<\alpha$, but here $r=\alpha=2$; hence it is not finite.

At the boundary, the integrand for $E[X^2]$ simplifies to $2(1{,}000)^2/x$. Integrating to $b$ gives $2{,}000{,}000\log(b/1{,}000)$, which tends to infinity as $b\to\infty$. Therefore $\operatorname{Var}(X)=\infty$, despite the finite mean. Check: the mean condition is met, and the second-moment condition is not. The common mistake avoided is substituting $\alpha=2$ into a variance formula whose denominator contains $\alpha-2$; that formula is valid only for $\alpha>2$.

Common Exam Traps and Practical Interpretation

The shape parameter is a threshold for the order of the moment. It is not enough to check whether the parameter itself is positive, and the equality case must not be treated as finite. A useful summary is:

Shape range	Mean	Variance	Reason
$0<\alpha\leq1$	Infinite	Infinite	The first moment is at or above the threshold
$1<\alpha\leq2$	Finite	Infinite	The first moment is below the threshold, but the second is not
$\alpha>2$	Finite	Finite	Both the first and second moments are below the threshold

One tempting but incorrect calculation illustrates the issue. With $\alpha=1.5$, directly inserting the parameter into the finite-variance expression gives a denominator containing $\alpha-2=-0.5$, and produces a negative number. A variance cannot be negative. The correction is not to take an absolute value or to alter the formula: the formula's condition $\alpha>2$ fails, so the variance is infinite.

Likewise, the general moment expression $\alpha\theta^r/(\alpha-r)$ must not be used at $r=\alpha$, where its denominator is zero. The integral shows what happens at that boundary: it grows like a logarithm without bound. For $r>\alpha$, it diverges even faster, as a positive power of the upper integration limit.

These thresholds matter because a model may support some summaries but not others. A finite expected claim does not guarantee a finite variance, and even a finite variance does not guarantee that every higher-order moment exists. When a requested moment is beyond the threshold, report that it is infinite rather than presenting a formal expression outside its conditions. This moment analysis is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical Pareto Type I model, identify the requested finite moments and calculate any requested finite values.

A severity has $\theta=\$400$ and $\alpha=5$. Find its mean and variance. (Answer: $E[X]=\$500$; $\operatorname{Var}(X)=\$16{,}666.67$ dollars squared.)
A severity has $\theta=\$250$ and $\alpha=1.8$. Decide whether the mean and variance are finite, and calculate the mean if it exists. (Answer: Mean finite, $E[X]=\$562.50$; variance infinite.)
A severity has $\theta=\$1{,}200$ and $\alpha=2$. Determine whether the moments of orders 1, 2, and 3 are finite, and calculate the first moment. (Answer: Order 1 is finite with $E[X]=\$2{,}400$; orders 2 and 3 are infinite.)
A severity has $\theta=\$1{,}000$ and $\alpha=3.5$. Calculate its mean, second raw moment, and variance, and classify its moment of order 3.5. (Answer: $E[X]=\$1{,}400$; $E[X^2]=\$2{,}333{,}333.33$; $\operatorname{Var}(X)=\$373{,}333.33$ dollars squared; the order-3.5 moment is infinite.)
A severity has $\theta=\$900$ and $\alpha=2.2$. Calculate its mean and variance, then decide whether the moment of order 2.2 is finite. (Answer: $E[X]=\$1{,}650$; $\operatorname{Var}(X)=\$6{,}187{,}500$ dollars squared; the order-2.2 moment is infinite.)

Shape range	Mean	Variance	Reason
\(0<\alpha\leq1\)	Infinite	Infinite	The first moment is at or above the threshold
\(1<\alpha\leq2\)	Finite	Infinite	The first moment is below the threshold, but the second is not
\(\alpha>2\)	Finite	Finite	Both the first and second moments are below the threshold

What You'll Learn