Pareto Type I Severity

Introduction

Some loss distributions allow for very large claims, even when such claims are relatively uncommon. The Pareto Type I model is a simple continuous severity model with a power-law tail: its survival probability decreases as a power of the loss amount. That makes it useful for practicing tail-probability and quantile calculations, and for seeing directly how a distribution’s parameters determine whether its mean or variance is finite.

This tutorial develops the Pareto Type I survival function and density, then uses them to calculate exceedance probabilities, quantiles, and moments. The distribution has a lower bound but no finite upper bound. The shape parameter governs how quickly its tail decreases; in particular, it determines which moments exist.

Definition: A loss $X$ has a Pareto Type I distribution with scale parameter $\theta>0$ and shape parameter $\alpha>0$ if its support is $x\geq\theta$, and its survival function on that support is $$ S_X(x)=P(X>x)=\left(\frac{\theta}{x}\right)^\alpha. $$ For $x<\theta$, $S_X(x)=1$. Equivalently, $F_X(x)=0$ for $x<\theta$, and $F_X(x)=1-(\theta/x)^\alpha$ for $x\geq\theta$. The parameter $\theta$ has the same units as $X$; $\alpha$ is dimensionless.

Survival Function and Density

The scale $\theta$ is the smallest possible loss. In particular, $P(X\geq\theta)=1$, and for any $x\geq\theta$, the probability that a loss exceeds $x$ is $(\theta/x)^\alpha$. A larger $\alpha$ makes that probability decrease more rapidly as $x$ increases, holding $\theta$ fixed.

For $x\geq\theta$, differentiating the CDF gives the density. Because the model is continuous, the probability of a loss equaling any one specified value is zero; thus $P(X>x)=P(X\geq x)$.

f_X(x)=\frac{d}{dx}F_X(x) =\frac{\alpha\theta^\alpha}{x^{\alpha+1}},\qquad x\geq\theta.

The density is zero below $\theta$. As a quick check that it defines a probability distribution, integrate it over its support:

\int_\theta^\infty \frac{\alpha\theta^\alpha}{x^{\alpha+1}}\,dx =\alpha\theta^\alpha\left[\frac{-x^{-\alpha}}{\alpha}\right]_\theta^\infty =1.

This check uses $\alpha>0$, so $x^{-\alpha}$ tends to zero as $x$ tends to infinity. The formula does not describe a Pareto Type I loss below its scale: below $\theta$, the CDF is zero and the survival function is one.

Worked Example: Tail Probability, Interval Probability, and a Quantile

A hypothetical portfolio models individual claim severity $X$, in dollars, by a Pareto Type I distribution with scale $\theta=\$1{,}200$ and shape $\alpha=3$. Find the probability that a claim exceeds $\$2{,}400$, the probability that it is between $\$1{,}200$ and $\$2{,}400$, and the $87.5$th percentile.

Method. Use the survival function for the exceedance probability, subtract CDF values for the interval probability, and solve the CDF equation for the percentile.

First, $2{,}400\geq\theta$, so the Pareto survival formula applies: $$ P(X>2{,}400)=\left(\frac{1{,}200}{2{,}400}\right)^3 =\left(\frac12\right)^3 =\frac18=0.1250. $$

Since all losses are at least $\$1{,}200$, $F_X(1{,}200)=0$. Also, $F_X(2{,}400)=1-P(X>2{,}400)=7/8$. There is no point mass at either endpoint, so $$ P(1{,}200\leq X\leq2{,}400)=F_X(2{,}400)-F_X(1{,}200) =\frac78-0=\frac78=0.8750. $$

For a percentile $x_p$, solve $F_X(x_p)=p$. At $p=0.875=7/8$, $$ 1-\left(\frac{1{,}200}{x_p}\right)^3=\frac78, \qquad \left(\frac{1{,}200}{x_p}\right)^3=\frac18, \qquad x_p=1{,}200\left(\frac18\right)^{-1/3}=\$2{,}400. $$

The answers are $0.1250$, $0.8750$, and $\$2{,}400$. As a check, the $87.5$th percentile is exactly the point below which the interval probability accumulates. Common mistake avoided: the interval probability is not the tail probability at $\$2{,}400$; it is the difference of the CDF values at the two endpoints.

Quantiles and the Shape of the Tail

For $0<p<1$, define the $p$th quantile $x_p$ by $F_X(x_p)=p$. The Pareto Type I CDF is continuous and strictly increasing on its support, so this equation has a unique solution. Rearranging it gives

x_p=\frac{\theta}{(1-p)^{1/\alpha}},\qquad 0<p<1.

The complement $1-p$ appears because $F_X(x)=1-S_X(x)$: a loss at the $p$th quantile has exceedance probability $1-p$. The formula also shows that quantiles scale in direct proportion to $\theta$. As $p$ approaches one, the quantile grows without bound, consistent with the model having no finite upper endpoint.

For a general threshold $c$, the exceedance probability is $$ P(X>c)= \begin{cases} 1, & c<\theta,\\ (\theta/c)^\alpha, & c\geq\theta. \end{cases} $$

The two cases matter. If a proposed threshold is below the scale, every possible loss exceeds it. Applying the power formula below the support would incorrectly give a probability greater than one.

Moments and When They Exist

The $r$th raw moment is $E[X^r]$, when the expectation is finite. For positive $r$, calculate it by integrating $x^r$ against the density. This integral converges only if the resulting power of $x$ is integrable at infinity.

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

When $r<\alpha$, the exponent $r-\alpha-1$ is less than $-1$, so the integral converges. Evaluating it gives

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

If $r=\alpha$, the integral contains $x^{-1}$ and diverges logarithmically. If $r>\alpha$, the integral also diverges. Therefore the moment formula must not be used when $r\geq\alpha$: the moment is infinite, not the value obtained by substituting into the expression.

In particular, the mean exists only if $\alpha>1$:

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

The second raw moment exists only if $\alpha>2$. When it does, subtracting the square of the mean gives

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

Consequently, for $1<\alpha\leq2$, the mean is finite but the variance is infinite. For $0<\alpha\leq1$, even the mean is infinite. The restriction on $\alpha$ must be checked before reporting a moment.

Worked Example: Mean and Variance With a Noninteger Shape

A hypothetical claim model has Pareto Type I severity $X$, measured in thousands of dollars, with scale $\theta=5$ and shape $\alpha=2.5$. Calculate the mean and variance in dollars, and check whether both moments exist.

Method. Compare each moment order with $\alpha$, then apply the raw-moment formulas in thousands of dollars.

For the mean, $1<2.5$, so it is finite: $$ E[X]=\frac{2.5(5)}{2.5-1} =\frac{12.5}{1.5} =\frac{25}{3} =8.333333\ldots\text{ thousand dollars}. $$ Thus the mean is $\$8{,}333.33$, rounded to the nearest cent.

For the second raw moment, $2<2.5$, so it is finite: $$ E[X^2]=\frac{2.5(5^2)}{2.5-2} =\frac{62.5}{0.5} =125\text{ thousand-dollars squared}. $$ The squared mean is $$ \big(E[X]\big)^2=\left(\frac{25}{3}\right)^2 =\frac{625}{9} =69.444444\ldots\text{ thousand-dollars squared}. $$ Therefore, $$ \operatorname{Var}(X)=125-\frac{625}{9} =\frac{500}{9} =55.555555\ldots\text{ thousand-dollars squared}. $$ Since one thousand-dollars squared equals $1{,}000{,}000$ dollars squared, the variance is $\$55{,}555{,}555.56$ dollars squared, rounded to the nearest cent. Its standard deviation is $\sqrt{500/9}$ thousand dollars, or approximately $\$7{,}453.56$.

The results are consistent with the model: $\alpha=2.5$ exceeds both moment orders 1 and 2, so both mean and variance are finite. Common mistake avoided: comparing $\alpha$ with the scale would not determine moment existence; only the shape parameter is relevant.

Using the Model Carefully

Pareto Type I calculations require tracking both the scale and the shape. The scale sets the lower endpoint and the units of loss. The shape controls the tail and determines moment existence. Before calculating, check that a threshold is on or above the support, that a quantile uses the tail probability $1-p$, and that a moment order $r$ is strictly less than $\alpha$.

A useful direct check is to compare an answer to its probability bounds. A survival probability must lie between zero and one; for $c<\theta$, it must equal one. A quantile for a high percentile must be large relative to the scale. These checks often expose using $p$ instead of $1-p$, or applying the support formula in the wrong region.

Key takeaway: For Pareto Type I severity, $S_X(x)=(\theta/x)^\alpha$ on $x\geq\theta$, and $x_p=\theta/(1-p)^{1/\alpha}$. The $r$th positive raw moment is finite exactly when $r<\alpha$, in which case $E[X^r]=\alpha\theta^r/(\alpha-r)$.

Worked Example: A High Quantile and an Infinite Variance

A hypothetical specialty-equipment loss $X$, in dollars, follows a Pareto Type I distribution with scale $\theta=\$800$ and shape $\alpha=2$. Find the $93.75$th percentile and the probability of exceeding $\$2{,}400$. Then determine whether the mean and variance are finite.

Method. Invert the CDF for the quantile, evaluate the survival function at the threshold, and compare moment orders with the shape.

For $p=0.9375=15/16$, the tail probability is $1-p=1/16$. Therefore, $$ x_{0.9375}=\frac{800}{(1/16)^{1/2}} =800(4) =\$3{,}200. $$ At $\$2{,}400$, which is above the scale, $$ P(X>2{,}400)=\left(\frac{800}{2{,}400}\right)^2 =\left(\frac13\right)^2 =\frac19 =0.1111 $$ to four decimal places.

The mean exists because $1<\alpha=2$, and $$ E[X]=\frac{2(800)}{2-1}=\$1{,}600. $$ The second raw moment does not exist: its order is $r=2$, equal to $\alpha$. Thus the variance is infinite, even though the mean is finite.

The percentile is above $\$2{,}400$, and its exceedance probability is $1/16$, as required by its definition. Common mistake avoided: at $\alpha=2$, the variance is not finite; the condition is strictly $\alpha>2$, not $\alpha\geq2$.

Check Your Understanding

For each problem, assume the stated loss follows a Pareto Type I distribution. Keep units consistent and check the support and moment conditions.

A hypothetical repair loss $X$, in dollars, has scale $\theta=\$600$ and shape $\alpha=2$. Find $P(X>\$1{,}200)$. (Answer: $0.2500$)
A hypothetical cargo loss has scale $\theta=\$1{,}000$ and shape $\alpha=4$. Find the probability that the loss is between $\$1{,}000$ and $\$2{,}000$, inclusive. (Answer: $15/16=0.9375$)
A hypothetical loss, measured in thousands of dollars, has scale $\theta=2$ and shape $\alpha=3$. Find its $87.5$th percentile. (Answer: $4$ thousand dollars, or $\$4{,}000$)
A hypothetical equipment loss has scale $\theta=\$900$ and shape $\alpha=1.5$. Determine its mean and whether its variance is finite. (Answer: Mean $=\$2{,}700$; variance is infinite)
A hypothetical claim severity has scale $\theta=\$3{,}000$ and shape $\alpha=3$. Find its $99$th percentile and variance. (Answer: $ \$3{,}000\cdot100^{1/3}\approx\$13{,}924.77$; variance $=\$6{,}750{,}000$ dollars squared)

What You'll Learn