Pareto Type II Severity And Its Moments

Introduction

The Pareto Type II model describes a nonnegative loss whose probability of exceeding a large amount declines according to a power law. Its scale parameter sets the units and general size of losses; its shape parameter controls the tail and determines which moments exist. These properties make careful identification of the parameterization essential before calculating probabilities, quantiles, or moments.

This tutorial uses a common actuarial parameterization of Pareto Type II, also called the Lomax distribution. In this convention, the model has support starting at zero and has scale parameter $\theta$ and shape parameter $\alpha$. Always check the stated survival function when interpreting parameter values: other texts may use different conventions.

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

The scale $\theta$ has the same units as $X$, such as dollars. The shape $\alpha$ is dimensionless. A larger $\alpha$ gives a more rapidly declining survival function. For $x<0$, $F_X(x)=0$ and $S_X(x)=1$. The density follows by differentiating the CDF, or equivalently by taking $-S_X'(x)$.

Probabilities and Quantiles

For a threshold $c\geq 0$, the probability that a loss exceeds $c$ is obtained directly from the survival function. The probability that it is at most $c$ is the CDF, and the probability that it falls between two thresholds is the difference of their CDF values. Since this is a continuous model, including or excluding an endpoint does not change these probabilities.

Worked Example: Exceedance and Interval Probabilities

A hypothetical portfolio has loss severity $X$ following a Pareto Type II model with $\theta=\$1{,}000$ and $\alpha=2$. Calculate (a) the probability a loss exceeds $\$1{,}000$, and (b) the probability a loss is between $\$1{,}000$ and $\$3{,}000$. Give probabilities to four decimal places.

Method. Use the survival function for the exceedance probability. For the interval, subtract the survival probability at the upper threshold from the survival probability at the lower threshold.

(a) Substitute $x=1{,}000$: $$ P(X>1{,}000)=\left(1+\frac{1{,}000}{1{,}000}\right)^{-2} =2^{-2}=\frac14=0.2500. $$

(b) At the lower threshold, $S_X(1{,}000)=0.2500$. At the upper threshold, $$ S_X(3{,}000)=\left(1+\frac{3{,}000}{1{,}000}\right)^{-2} =4^{-2}=\frac1{16}=0.0625. $$ Therefore, $$ P(1{,}000<X\leq 3{,}000) =S_X(1{,}000)-S_X(3{,}000) =\frac14-\frac1{16} =\frac3{16}=0.1875. $$

Check. The interval probability is positive and less than $P(X>1{,}000)=0.2500$, as it must be. A common mistake is to subtract the two survival probabilities in the reverse order, which would give a negative probability.

The $p$-quantile $x_p$, for $0<p<1$, is the value satisfying $F_X(x_p)=p$. Solving that equation gives a direct formula.

Result: For the Pareto Type II model, $$ x_p=\theta\left((1-p)^{-1/\alpha}-1\right),\qquad 0<p<1. $$

Worked Example: Finding and Verifying a Quantile

A hypothetical loss has $\theta=\$750$ and $\alpha=3$. Find the $26/27$-quantile, then verify it using the CDF. Report the amount to the nearest cent.

Method. Substitute $p=26/27$ into the inverse CDF. Here $1-p=1/27$, whose power $-1/3$ is exactly $3$.

$$ x_{26/27} =750\left(\left(\frac1{27}\right)^{-1/3}-1\right) =750(3-1) =\$1{,}500.00. $$

Verification. At $x=\$1{,}500$, $$ F_X(1{,}500) =1-\left(1+\frac{1{,}500}{750}\right)^{-3} =1-3^{-3} =1-\frac1{27} =\frac{26}{27}. $$ Thus the proposed amount has the required cumulative probability.

A common mistake is to use $p$, rather than $1-p$, inside the inverse survival expression. The quantile equation comes from $1-p=S_X(x_p)$, which is why the formula uses $(1-p)^{-1/\alpha}$.

Deriving the Raw Moments

For a nonnegative continuous loss, the tail-integral formula for a positive real power $r$ is $E[X^r]=r\int_0^\infty x^{r-1}S_X(x)\,dx$, provided the integral is finite. Applying it to the Pareto Type II survival function reveals the exact condition for moment existence.

Substitute the survival function.
For $r>0$, the candidate moment is $r\int_0^\infty x^{r-1}(1+x/\theta)^{-\alpha}\,dx$.

Scale the integration variable.
Let $u=x/\theta$, so $x=\theta u$ and $dx=\theta\,du$. The integral becomes $r\theta^r\int_0^\infty u^{r-1}(1+u)^{-\alpha}\,du$.

Evaluate and check convergence.
This is a beta integral, equal to $B(r,\alpha-r)=\Gamma(r)\Gamma(\alpha-r)/\Gamma(\alpha)$ when $\alpha>r$. The condition is necessary: at large $u$, the integrand behaves like $u^{r-1-\alpha}$, whose integral to infinity converges exactly when $r<\alpha$.

Using $r\Gamma(r)=\Gamma(r+1)$ gives the moment formula. If $\alpha\leq r$, the tail integral diverges, so the $r$th raw moment is infinite. This convergence check is part of the result, not an optional qualification.

Theorem: For $X$ with Pareto Type II parameters $\theta>0$ and $\alpha>0$, and for any $r>0$, $$ E[X^r]=\frac{\Gamma(r+1)\Gamma(\alpha-r)}{\Gamma(\alpha)}\theta^r \quad\text{if }\alpha>r. $$ If $\alpha\leq r$, then $E[X^r]=\infty$. In particular, $E[X]$ is finite exactly when $\alpha>1$, and the variance is finite exactly when $\alpha>2$.

For positive integers $k<\alpha$, the gamma-function expression simplifies to $$ E[X^k]=\frac{k!\,\theta^k}{(\alpha-1)(\alpha-2)\cdots(\alpha-k)}. $$ In particular, when $\alpha>1$, $E[X]=\theta/(\alpha-1)$. When $\alpha>2$, the second raw moment is $E[X^2]=2\theta^2/[(\alpha-1)(\alpha-2)]$, and subtracting the squared mean gives the variance below.

Result: The mean and variance are $$ E[X]=\frac{\theta}{\alpha-1},\qquad \alpha>1, $$ $$ \operatorname{Var}(X)=\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)},\qquad \alpha>2. $$ For $1<\alpha\leq2$, the mean is finite but the second raw moment and variance are infinite. For $0<\alpha\leq1$, the mean is infinite, so there is no finite variance.

To see the variance algebra, for $\alpha>2$ subtract $E[X]^2=\theta^2/(\alpha-1)^2$ from $E[X^2]=2\theta^2/[(\alpha-1)(\alpha-2)]$. Putting these over a common denominator gives $$ \operatorname{Var}(X) =\frac{\theta^2}{(\alpha-1)^2(\alpha-2)} \left(2(\alpha-1)-(\alpha-2)\right) =\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)}. $$ The result requires $\alpha>2$, because both the second raw moment and the mean must be finite for this calculation.

Using the Moment Conditions Correctly

The shape parameter, not the scale, determines whether a moment exists. Changing $\theta$ changes the numerical size of a finite moment, but cannot make an infinite moment finite. In particular, the conditions $\alpha>1$ for a finite mean and $\alpha>2$ for a finite variance must not be interchanged.

Worked Example: Mean and Variance Near the Existence Threshold

A hypothetical severity model has $\theta=\$1{,}000$ and $\alpha=2.5$. Calculate its mean and variance, and state whether each is finite. Report money to the nearest cent and variance to the nearest cent in dollars squared.

Method. Since $2.5>2$, both the mean and variance formulas apply. First calculate the mean: $$ E[X]=\frac{1{,}000}{2.5-1} =\frac{1{,}000}{1.5} =\$666.666\ldots =\$666.67. $$

Next, calculate the variance: $$ \operatorname{Var}(X) =\frac{(2.5)(1{,}000)^2}{(2.5-1)^2(2.5-2)} =\frac{2{,}500{,}000}{(1.5)^2(0.5)} =\frac{2{,}500{,}000}{1.125} =2{,}222{,}222.222\ldots $$ Thus the variance is approximately $\$2{,}222{,}222.22$ dollars$^2$, and it is finite.

Check. The second raw moment is $$ E[X^2]=\frac{2(1{,}000)^2}{(1.5)(0.5)} =2{,}666{,}666.666\ldots $$ and $E[X]^2=444{,}444.444\ldots$. Their difference is $2{,}222{,}222.222\ldots$, agreeing with the variance formula.

Common exam trap. Calling this variance infinite because $\alpha$ is “close to 2” is wrong: $2.5>2$, so the variance exists. The boundary matters; it is not enough to say that a Pareto-type distribution has a heavy tail.

Worked Example: A Finite Mean but Infinite Variance

A hypothetical loss model has $\theta=\$800$ and $\alpha=1.5$. Determine whether the mean and variance exist, and calculate the mean if finite.

Method. Check the shape condition separately for each moment. Since $1.5>1$, the mean exists and is $$ E[X]=\frac{800}{1.5-1} =\frac{800}{0.5} =\$1{,}600.00. $$ But $1.5\leq2$, so the second raw moment is infinite. Therefore the variance is infinite; the finite mean does not imply a finite variance.

Check. The conditions predict exactly this combination: $\alpha>1$ gives a finite mean, while $\alpha\leq2$ rules out a finite second moment. A common mistake is to treat the variance formula as valid for every $\alpha>1$; its denominator includes $\alpha-2$, and its derivation requires $\alpha>2$.

Common Mistakes and Practical Checks

Mixing up scale and shape. The scale $\theta$ carries the loss units; the shape $\alpha$ controls tail decay and moment existence.
Using the wrong tail expression. Under this parameterization, $S_X(x)=(1+x/\theta)^{-\alpha}$, not $(1+x)^{-\alpha}$. The ratio $x/\theta$ is dimensionless.
Forgetting the existence condition. Do not report a mean without $\alpha>1$, or a variance without $\alpha>2$.
Reversing an interval subtraction. For $a<b$, $P(a<X\leq b)=S_X(a)-S_X(b)$, which is nonnegative because the survival function is decreasing.
Rounding too early. Keep exact fractions or unrounded values through intermediate calculations, then round the final result to the requested precision.

Key takeaway: For Pareto Type II, use $S_X(x)=(1+x/\theta)^{-\alpha}$ and $x_p=\theta((1-p)^{-1/\alpha}-1)$. The $r$th positive raw moment is finite exactly when $\alpha>r$; in particular, the mean requires $\alpha>1$ and the variance requires $\alpha>2$. This tutorial is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical loss model, use the stated Pareto Type II parameterization. Probabilities are rounded to four decimal places and monetary amounts to the nearest cent.

A loss has $\theta=\$500$ and $\alpha=2$. Find $P(X>\$500)$ and $P(X\leq\$1{,}500)$. (Answer: $0.2500$ and $0.9375$, respectively.)
A loss has $\theta=\$800$ and $\alpha=4$. Find its $15/16$-quantile and verify the survival probability at that amount. (Answer: $\$800.00$; $P(X>\$800)=1/16=0.0625$.)
A loss has $\theta=\$400$ and $\alpha=2$. Find the probability that it is greater than $\$400$ but no greater than $\$1{,}200$. (Answer: $3/16=0.1875$.)
A loss has $\theta=\$1{,}200$ and $\alpha=3$. Calculate its mean and variance, and state whether both are finite. (Answer: Mean $=\$600.00$; variance $=\$1{,}080{,}000.00$ dollars$^2$; both are finite.)
A loss has $\theta=\$1{,}000$ and $\alpha=1.5$. Determine whether its mean and variance are finite and calculate the mean if it exists. (Answer: Mean is finite and equals $\$2{,}000.00$; variance is infinite.)

What You'll Learn