LER for Pareto Severities

Why the Pareto Model Needs a Parameterization

Pareto severity models can describe losses with a substantial upper tail, but the name “Pareto” is used for more than one parameterization. A formula for the loss elimination ratio (LER) is therefore meaningful only after the model and its parameters have been specified. Here, $X$ is a nonnegative ground-up loss, and we use the Lomax, also called Pareto Type II, parameterization. The deductible is an ordinary deductible applied to each loss.

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

Under an ordinary deductible $d\geq 0$, the amount eliminated from a ground-up loss is $\min(X,d)$, while the insurer payment per loss is $(X-d)_+=\max(X-d,0)$. Thus the LER is the expected eliminated amount divided by the expected ground-up loss. This ratio requires a finite, positive expected ground-up loss, so the Pareto shape must satisfy $\alpha>1$.

Definition: For a nonnegative severity with $0<E[X]<\infty$, the limited expected value at deductible $d$ is $\operatorname{LEV}(d)=E[\min(X,d)]$. The loss elimination ratio is $$ \operatorname{LER}(d)=\frac{\operatorname{LEV}(d)}{E[X]}. $$ For an ordinary deductible, the expected insurer payment is $E[(X-d)_+]=E[X]-\operatorname{LEV}(d)$.

Deriving the Pareto Limited Expected Value

For a nonnegative random variable, the expected value of a limited loss can be found by integrating its survival function up to the limit. The identity follows by expressing the limited loss as an area: for each outcome $X$, the length of the interval from 0 to $\min(X,d)$ is $\min(X,d)$. Equivalently, $\min(X,d)=\int_0^d \mathbf{1}_{\{X>x\}}\,dx$, apart from an endpoint that does not affect the integral. Taking expectations gives the survival integral.

Write the limited-loss identity.
$E[\min(X,d)]=\int_0^d S_X(x)\,dx$ for a nonnegative severity.

Insert the Lomax survival function.
$E[\min(X,d)]=\int_0^d \left(\frac{\theta}{\theta+x}\right)^\alpha dx$.

Integrate.
For $\alpha\ne 1$, an antiderivative is $\frac{\theta^\alpha(\theta+x)^{1-\alpha}}{1-\alpha}$. Evaluating at 0 and $d$ gives the limited expected value.

Divide by the mean to obtain LER.
When $\alpha>1$, the Pareto mean is $\theta/(\alpha-1)$, so the scale cancels from the ratio after expressing the deductible relative to $\theta$.

\operatorname{LEV}(d) =\frac{\theta}{\alpha-1} \left[1-\left(\frac{\theta}{\theta+d}\right)^{\alpha-1}\right], \qquad \alpha>0,\ \alpha\ne 1,

The displayed limited expected value formula is valid for $\alpha>0$, $\alpha\ne 1$; when $0<\alpha<1$, the numerator and the bracket are both negative, leaving a positive value. At $\alpha=1$, integrating $\theta/(\theta+x)$ gives $\operatorname{LEV}(d)=\theta\ln(1+d/\theta)$. For $\alpha>1$, integrating the survival function over all $x\geq 0$ gives the finite mean $E[X]=\theta/(\alpha-1)$. If $0<\alpha\leq1$, the mean is infinite, so the LER as defined here is not a ratio of finite expected losses.

\operatorname{LER}(d) =1-\left(\frac{\theta}{\theta+d}\right)^{\alpha-1}, \qquad \alpha>1,

The same result can be checked using the insurer payment. Its expected value is the remaining survival area above the deductible:

E[(X-d)_+] =\int_d^\infty S_X(x)\,dx =\frac{\theta}{\alpha-1} \left(\frac{\theta}{\theta+d}\right)^{\alpha-1}, \qquad \alpha>1.

Adding this expected payment to $\operatorname{LEV}(d)$ recovers $E[X]$. This is a useful check because the deductible splits each ground-up loss into the eliminated amount and the insurer-paid amount.

Worked Pareto LER Calculations

Worked Example: Deductible Equal to the Scale

A hypothetical portfolio has individual ground-up losses following a Lomax distribution with $\theta=\$2{,}000$ and $\alpha=3$. An ordinary deductible is $d=\$2{,}000$. Find the mean loss, the limited expected value, the LER, and the expected insurer payment per loss.

Method. Use the Pareto mean and limited expected value, then divide the limited expected value by the mean. This keeps the ratio of expected losses distinct from the probability of exceeding the deductible.

First, $\alpha=3>1$, so the mean is finite:

E[X]=\frac{\theta}{\alpha-1} =\frac{\$2{,}000}{3-1} =\$1{,}000.

Next, $\theta+d=\$4{,}000$, so $\theta/(\theta+d)=2{,}000/4{,}000=1/2$. The limited expected value is

\operatorname{LEV}(d) =\$1{,}000\left[1-\left(\frac12\right)^2\right] =\$1{,}000\left(1-\frac14\right) =\$750.

Therefore, $\operatorname{LER}(d)=\$750/\$1{,}000=0.75$, or 75.00%. The expected insurer payment is $\$1{,}000-\$750=\$250$ per loss. As a check, the expected eliminated and retained amounts sum to the $\$1{,}000$ ground-up mean. The probability of exceeding the deductible is $(1/2)^3=0.125$, not 0.75; that probability is not the LER.

Worked Example: A Fractional Shape Parameter

A hypothetical severity has a Lomax distribution with $\theta=\$1{,}200$ and $\alpha=2.5$. The policy has an ordinary deductible of $d=\$1{,}200$. Calculate the LER and both expected loss components, rounding money to cents and the ratio to four decimal places.

Method. Check the mean condition first, then use the exponent $\alpha-1$ in the LER formula. The deductible equals the scale, making $\theta/(\theta+d)=1/2$.

Since $2.5>1$, the ground-up mean exists:

E[X]=\frac{\$1{,}200}{2.5-1} =\frac{\$1{,}200}{1.5} =\$800.

The exponent in the LER is $\alpha-1=1.5$. Thus

\operatorname{LER}(d) =1-\left(\frac12\right)^{1.5} =1-\frac{1}{2\sqrt{2}} \approx 1-0.3535533906 =0.6464466094.

Rounded to four decimal places, the LER is 0.6464. The limited expected value is

\operatorname{LEV}(d) =\$800(0.6464466094) \approx \$517.16.

The expected insurer payment is $\$800-\$517.16=\$282.84$. The check is $\$517.16+\$282.84=\$800.00$. The fractional power matters: replacing $1.5$ by $2.5$ would use the survival probability at the deductible, not the LER.

Worked Example: Deductible Several Times the Scale

Suppose a hypothetical loss has $\theta=\$500$, $\alpha=4$, and an ordinary deductible $d=\$1{,}500$. Find the LER and the expected eliminated and retained amounts per loss. Give money to the nearest cent and the LER to four decimal places.

Method. Express the deductible relative to the scale: $d=3\theta$. Then $\theta/(\theta+d)=500/(500+1{,}500)=1/4$. The shape satisfies $\alpha>1$, so use the finite-mean LER formula.

The mean ground-up loss is

E[X]=\frac{\$500}{4-1}=\frac{\$500}{3}\approx \$166.67.

The LER is

\operatorname{LER}(d) =1-\left(\frac14\right)^3 =1-\frac{1}{64} =\frac{63}{64} =0.984375 \approx 0.9844.

The limited expected value is

\operatorname{LEV}(d) =\frac{\$500}{3}\left(\frac{63}{64}\right) =\frac{\$31{,}500}{192} =\$164.0625 \approx \$164.06.

The expected insurer payment is $\$500/3-\$164.0625=\$2.6041667$, or $\$2.60$ to the nearest cent. Before rounding, the two components sum to $\$166.6666667$, the ground-up mean. Rounding each component separately gives $\$164.06+\$2.60=\$166.66$; the one-cent difference from $\$166.67$ is only rounding.

Common Exam Trap: Confusing Probability and Expected Loss

A frequent error is to substitute the probability of a loss exceeding the deductible for the LER. Under this model,

P(X>d)=S_X(d)=\left(\frac{\theta}{\theta+d}\right)^\alpha, \qquad \operatorname{LER}(d)=1-\left(\frac{\theta}{\theta+d}\right)^{\alpha-1}.

These expressions answer different questions. The exceedance probability counts how often a loss is above the deductible. The LER measures the proportion of expected ground-up dollars eliminated by the deductible; it depends on the expected sizes of losses as well as their frequency. For the first worked example, the wrong calculation would report $P(X>d)=(1/2)^3=0.125$ as the LER. The correction is to use $1-(1/2)^{3-1}=0.75$.

Another important check is the shape condition. A finite deductible produces a finite limited expected value even when $0<\alpha\leq1$, but the ground-up mean is then infinite. Dividing by that mean does not produce the finite-mean LER defined in this tutorial. Do not use the simplified LER formula outside its condition $\alpha>1$.

Key takeaway: For the Lomax parameterization $S_X(x)=(\theta/(\theta+x))^\alpha$, the mean is $\theta/(\alpha-1)$ when $\alpha>1$, and the ordinary-deductible LER is $1-(\theta/(\theta+d))^{\alpha-1}$. The ratio describes expected dollars eliminated, not the probability that a loss exceeds the deductible.

Check Your Understanding

For each problem, assume the stated hypothetical ground-up loss follows the Lomax parameterization defined above and that the deductible is ordinary.

A severity has $\theta=\$900$, $\alpha=2$, and $d=\$900$. Find its mean, limited expected value, LER, and expected insurer payment. (Answer: mean $\$900$; limited expected value $\$450$; LER $0.5000$; expected insurer payment $\$450$.)
A severity has $\theta=\$1{,}500$, $\alpha=3$, and $d=\$1{,}500$. Find the LER and the expected eliminated and insurer-paid amounts. (Answer: LER $0.7500$; eliminated $\$562.50$; insurer-paid $\$187.50$.)
A severity has $\theta=\$600$, $\alpha=2$, and $d=\$1{,}800$. Find $P(X>d)$, the LER, and the expected insurer payment. (Answer: $P(X>d)=0.0625$; LER $0.7500$; expected insurer payment $\$150.00$.)
A severity has $\theta=\$800$, $\alpha=4$, and $d=\$800$. Find the finite mean, LER, limited expected value, and expected insurer payment, giving money to cents. (Answer: mean $\$266.67$; LER $0.8750$; limited expected value $\$233.33$; expected insurer payment $\$33.33$.)
A severity has $\theta=\$1{,}000$, $\alpha=1.5$, and $d=\$3{,}000$. Check the finite-mean condition, then find the mean, LER, limited expected value, and expected insurer payment. (Answer: the mean is finite; mean $\$2{,}000$; LER $0.5000$; limited expected value $\$1{,}000$; expected insurer payment $\$1{,}000$.)

What You'll Learn