Pareto Type I Severity And Its Tail

Introduction

The Pareto Type I model is used for positive severities when a model with a minimum possible loss and a slowly decreasing upper tail is appropriate. Its parameters have direct interpretations: the minimum parameter sets the lower endpoint of possible losses, while the shape parameter controls how quickly exceedance probabilities decrease as the threshold rises.

For this tutorial, the main task is to calculate probabilities such as $P(X>x)$ and $P(a<X\leq b)$. The support matters: a threshold below the minimum loss has a different probability interpretation from one above it. All loss scenarios and data below are hypothetical.

Definition: A loss $X$ has a Pareto Type I distribution with minimum parameter $x_m>0$ and shape parameter $\alpha>0$ if its cumulative distribution function is $$ F_X(x)= \begin{cases} 0, & x<x_m,\\ 1-\left(\dfrac{x_m}{x}\right)^\alpha, & x\geq x_m. \end{cases} $$ The loss is continuous and has support $[x_m,\infty)$. Thus $X$ cannot be less than $x_m$, and there is no probability mass at any single point.

Deriving the Density and Survival Function

For $x<x_m$, the cumulative distribution is zero because such a loss is outside the support. At and above $x_m$, the cumulative probability is one minus the probability of exceeding $x$. This gives the survival function $S_X(x)=P(X>x)$ directly. Differentiating the cumulative distribution on the support gives the density.

Start with the cumulative distribution.
For $x\geq x_m$, $F_X(x)=1-(x_m/x)^\alpha$.

Take its complement.
Since $S_X(x)=1-F_X(x)$, the probability of exceeding a threshold on the support is $(x_m/x)^\alpha$.

Differentiate to obtain the density.
The derivative of $1-x_m^\alpha x^{-\alpha}$ is $\alpha x_m^\alpha x^{-\alpha-1}$.

Result: The survival function and density are $$ S_X(x)= \begin{cases} 1, & x<x_m,\\ \left(\dfrac{x_m}{x}\right)^\alpha, & x\geq x_m, \end{cases} \qquad f_X(x)= \begin{cases} \alpha x_m^\alpha x^{-\alpha-1}, & x\geq x_m,\\ 0, & x<x_m. \end{cases} $$

The density formula follows from the derivative because, for $x>x_m$, $$ \frac{d}{dx}\left[1-x_m^\alpha x^{-\alpha}\right] =\alpha x_m^\alpha x^{-\alpha-1}. $$ It also integrates to one: $$ \int_{x_m}^{\infty}\alpha x_m^\alpha x^{-\alpha-1}\,dx =\left[-x_m^\alpha x^{-\alpha}\right]_{x_m}^{\infty} =0-(-1)=1. $$ This verifies that the density accounts for all probability. At $x=x_m$, the survival function equals one. This is consistent with the model: there is no mass at $x_m$, and losses are at least $x_m$.

For a threshold $x\geq x_m$, the key calculation is $$ P(X>x)=S_X(x)=\left(\frac{x_m}{x}\right)^\alpha. $$ For $x<x_m$, $P(X>x)=1$, since every possible loss exceeds that threshold. The cumulative probability $P(X\leq x)$ is the complement of the survival probability. For two thresholds $a<b$, use the difference in survival probabilities: $$ P(a<X\leq b)=S_X(a)-S_X(b). $$ The piecewise definition of $S_X$ handles thresholds below the minimum without requiring a separate extrapolation.

Worked Tail-Probability Calculations

Worked Example: Probability Between Two Loss Thresholds

A hypothetical specialty-equipment portfolio is modelled by a Pareto Type I loss $X$, measured in dollars, with $x_m=\$10{,}000$ and $\alpha=2$. Calculate $P(X>\$20{,}000)$, $P(\$20{,}000<X\leq \$50{,}000)$, and $P(X>\$50{,}000)$. Check that the results are consistent.

Method. Both thresholds are above the minimum, so apply $S_X(x)=(x_m/x)^\alpha$.

At $\$20{,}000$, $$ S_X(20{,}000)=\left(\frac{10{,}000}{20{,}000}\right)^2 =(0.5)^2=0.25. $$ Therefore $P(X>\$20{,}000)=0.2500$.

At $\$50{,}000$, $$ S_X(50{,}000)=\left(\frac{10{,}000}{50{,}000}\right)^2 =(0.2)^2=0.04. $$ Therefore $P(X>\$50{,}000)=0.0400$. The probability between the thresholds is the first exceedance probability minus the second: $$ P(20{,}000<X\leq 50{,}000)=0.25-0.04=0.21. $$

Check. The probability above $\$20{,}000$ is partitioned into the probability between $\$20{,}000$ and $\$50{,}000$, plus the probability above $\$50{,}000$: $0.21+0.04=0.25$. The calculation is consistent. It avoids the common mistake of adding the two exceedance probabilities instead of subtracting them for an interval.

Worked Example: A Threshold Below the Minimum

A hypothetical equipment-breakdown severity $X$, in dollars, follows a Pareto Type I model with $x_m=\$12{,}000$ and $\alpha=3$. Calculate $P(X\leq \$9{,}000)$, $P(X>\$24{,}000)$, and $P(\$9{,}000<X\leq \$60{,}000)$.

Method. First compare each threshold with the support boundary $\$12{,}000$. Since $X\geq \$12{,}000$, no loss can be at or below $\$9{,}000$, so $P(X\leq \$9{,}000)=0$.

For $\$24{,}000$, which is above the minimum, $$ P(X>\$24{,}000)=\left(\frac{12{,}000}{24{,}000}\right)^3 =(0.5)^3=0.1250. $$ At $\$60{,}000$, $$ S_X(60{,}000)=\left(\frac{12{,}000}{60{,}000}\right)^3 =(0.2)^3=0.0080. $$ Because losses cannot be below $\$12{,}000$, the event $X>\$9{,}000$ has probability one. Thus $$ P(9{,}000<X\leq 60{,}000) =1-S_X(60{,}000) =1-0.0080=0.9920. $$

Check. The result says that $0.8\%$ of losses exceed $\$60{,}000$, while $99.2\%$ are above $\$9{,}000$ and at most $\$60{,}000$. These probabilities sum to one. The calculation avoids incorrectly substituting $\$9{,}000$ into the above-minimum survival formula.

Worked Example: Keeping the Loss Unit Consistent

A hypothetical model records severity in thousands of dollars. Let $X$ have a Pareto Type I distribution with $x_m=5$ (thousand dollars) and $\alpha=2$. Calculate the probability that a loss is between $\$15{,}000$ and $\$50{,}000$, the probability it exceeds $\$50{,}000$, and the probability it is at most $\$15{,}000$.

Method. Since the model unit is thousands of dollars, use the numerical thresholds 15 and 50, not 15,000 and 50,000. The minimum is 5 in the same unit.

The exceedance probabilities are $$ S_X(15)=\left(\frac{5}{15}\right)^2=\frac{1}{9}, \qquad S_X(50)=\left(\frac{5}{50}\right)^2=\frac{1}{100}. $$ Therefore, $$ P(15<X\leq 50)=\frac{1}{9}-\frac{1}{100} =\frac{100-9}{900} =\frac{91}{900} =0.1011. $$ Also, $P(X>50)=1/100=0.0100$, and $$ P(X\leq 15)=1-S_X(15)=1-\frac{1}{9} =\frac{8}{9}=0.8889. $$

Check. Using unrounded values, $91/900+1/100+8/9=91/900+9/900+800/900=1$. Rounded to four decimal places, the three probabilities are $0.1011$, $0.0100$, and $0.8889$, which also sum to $1.0000$. The calculation avoids mixing dollar amounts with amounts measured in thousands.

Common Exam Traps

The Pareto formula is short, but several errors can produce plausible-looking answers. Check the support and units before calculating, then verify the probability interpretation.

Using the tail formula below the minimum. The expression $(x_m/x)^\alpha$ is the survival formula for $x\geq x_m$, not for every positive $x$. If $x<x_m$, the correct survival probability is one and the cumulative probability is zero.
Reversing the ratio. For a threshold above the minimum, use $x_m/x$, not $x/x_m$. Since $x_m/x<1$, a positive shape parameter gives an exceedance probability between zero and one.
Subtracting interval probabilities in the wrong order. For $a<b$, $S_X(a)\geq S_X(b)$, so the interval probability is $S_X(a)-S_X(b)$, a nonnegative quantity.
Changing units for only one quantity. The threshold and minimum must use the same monetary unit before forming their ratio.
Confusing $P(X>x)$ with $P(X\leq x)$. They are complements. For $x\geq x_m$, $P(X\leq x)=1-(x_m/x)^\alpha$, whereas $P(X>x)=(x_m/x)^\alpha$.

For example, if $x_m=10$, $\alpha=2$, and the threshold is $x=20$, the incorrect reversed ratio gives $(20/10)^2=4$, which cannot be a probability. The correction is $(10/20)^2=0.25$, which lies between zero and one and decreases as the threshold rises.

Key takeaway: For a Pareto Type I loss, first check whether the threshold is below or above $x_m$. At or above the minimum, calculate the exceedance probability as $(x_m/x)^\alpha$; for an interval, subtract the survival probability at the upper endpoint from that at the lower endpoint.

Check Your Understanding

For each hypothetical Pareto Type I model, identify whether the thresholds lie on the support and calculate the requested probabilities. Give probabilities to four decimal places.

A model has $x_m=\$4{,}000$ and $\alpha=2$. Find $P(X>\$8{,}000)$ and $P(X\leq \$8{,}000)$. (Answer: $0.2500$ and $0.7500$)
A model has $x_m=\$20{,}000$ and $\alpha=3$. Find $P(X\leq \$10{,}000)$, $P(X>\$40{,}000)$, and $P(\$10{,}000<X\leq \$40{,}000)$. (Answer: $0.0000$, $0.1250$, and $0.8750$)
A model has $x_m=\$6{,}000$ and $\alpha=2$. Find $P(\$12{,}000<X\leq \$30{,}000)$ and verify using the two survival probabilities. (Answer: $0.2100$; the survival probabilities are $0.2500$ and $0.0400$)
A model measures losses in thousands of dollars, with $x_m=10$ and $\alpha=2$. Find $P(\$30{,}000<X\leq \$50{,}000)$, using consistent units. (Answer: $0.0711$)
A model has $x_m=\$25{,}000$ and $\alpha=1.5$. Find $P(\$50{,}000<X\leq \$100{,}000)$. (Answer: $0.2286$)

What You'll Learn