Pareto Severity ILF Calculation

Why Pareto Severity Matters for Increased Limits

An increased limits factor (ILF) compares expected claim payments at two per-claim limits. The severity distribution matters because increasing the limit adds payments only when a claim exceeds the lower limit. A distribution with a heavy tail can therefore assign materially different expected payments to higher layers than a light-tailed model does.

This tutorial uses the Pareto Type II severity distribution, also called the Lomax distribution. Naming the type is important: Pareto distributions have multiple parameterizations, and their formulas are not interchangeable. Assume $X$ is a nonnegative ground-up claim severity, measured in dollars, with scale parameter $\theta>0$ and shape parameter $\alpha>0$. There is no deductible or aggregate limit in the calculations below.

Definition: A Pareto Type II severity has survival function $S_X(x)=P(X>x)=(1+x/\theta)^{-\alpha}$ for $x\geq 0$, where $\theta>0$ is the scale and $\alpha>0$ is the shape. Its density is $f_X(x)=(\alpha/\theta)(1+x/\theta)^{-(\alpha+1)}$. The uncapped mean exists only when $\alpha>1$, in which case $E[X]=\theta/(\alpha-1)$.

The survival function has a polynomial tail: for large $x$, $S_X(x)$ decreases approximately in proportion to $x^{-\alpha}$. An exponential survival function decreases exponentially instead. This slower tail decay is why Pareto models are often considered for claim severities with substantial potential for large losses. The shape parameter controls the tail: smaller $\alpha$ means a heavier tail, holding $\theta$ fixed.

Limited Expected Loss Under Pareto Type II

For a per-claim limit $u>0$, the insurer’s payment on a claim is $\min(X,u)$. Define the limited expected loss at that limit by $L(u)=E[\min(X,u)]$. For any nonnegative severity, this expectation can be calculated by integrating the survival function from zero to the limit.

Theorem: For Pareto Type II severity with scale $\theta>0$, shape $\alpha>0$, and per-claim limit $u>0$, $L(u)=\theta\{1-(1+u/\theta)^{1-\alpha}\}/(\alpha-1)$ when $\alpha\ne1$, and $L(u)=\theta\ln(1+u/\theta)$ when $\alpha=1$. These limited expectations are finite for every finite $u$, including when $\alpha\leq1$ and the uncapped mean is infinite.

The formula follows from the identity $\min(X,u)=\int_0^u \mathbf{1}_{\{X>x\}}\,dx$. Taking expectations gives $E[\min(X,u)]=\int_0^u P(X>x)\,dx$; the equality is valid because the integrand is nonnegative. Substituting the Pareto survival function and integrating gives the result:

\begin{aligned} L(u) &=\int_0^u \left(1+\frac{x}{\theta}\right)^{-\alpha}\,dx\\ &=\theta\int_1^{1+u/\theta} t^{-\alpha}\,dt\\ &=\frac{\theta}{1-\alpha}\left[\left(1+\frac{u}{\theta}\right)^{1-\alpha}-1\right], \qquad \alpha\ne1. \end{aligned}

Here the substitution is $t=1+x/\theta$, so $dx=\theta\,dt$. When $\alpha=1$, the integrand in the substituted integral is $1/t$, giving $L(u)=\theta\ln(1+u/\theta)$. The logarithmic expression is not optional at $\alpha=1$: substituting $\alpha=1$ into the formula with denominator $\alpha-1$ produces an indeterminate form.

For a positive base limit $b$, the ILF at limit $u\geq b$ is the ratio of limited expected losses:

\operatorname{ILF}(u;b)=\frac{L(u)}{L(b)}.

This ratio applies to the same severity basis and coverage terms at both limits. If claim frequency is unchanged, expected claim counts multiply both limited expected losses by the same amount and cancel in the ratio. The ILF is not itself a premium indication when expenses, expenses varying by limit, or other rating adjustments differ between the two options.

Worked Pareto ILF Calculations

Worked Example: Shape Two, From a $1,000 Base Limit to $5,000

A hypothetical coverage has ground-up claim severity $X$ following Pareto Type II with $\theta=\$1{,}000$ and $\alpha=2$. Calculate the limited expected losses at per-claim limits of $1,000 and $5,000, the ILF from $1,000 to $5,000, and the expected payment in the layer above $1,000 and up to $5,000. Use the formula for $\alpha\ne1$.

At the $1,000 limit: $u/\theta=1{,}000/1{,}000=1$, and $1-\alpha=-1$. Therefore,

L(1{,}000) =\frac{1{,}000}{2-1}\left[1-(1+1)^{-1}\right] =1{,}000\left(1-\frac12\right) =\$500.00.

At the $5,000 limit: $u/\theta=5$, so

L(5{,}000) =\frac{1{,}000}{2-1}\left[1-(1+5)^{-1}\right] =1{,}000\left(1-\frac16\right) =\frac{5{,}000}{6} =\$833.33.

The ILF is $833.333\ldots/500=1.6666\ldots$, or 1.667 to three decimal places. The expected payment in the layer above $1,000 and up to $5,000 is the difference in limited expected losses: $833.333\ldots-500=\$333.33$. Check: the base-limit payment plus that layer payment is $500+333.333\ldots=833.333\ldots$, matching the $5,000 limited expected loss. The method avoids treating the ILF as the ratio of the limits, which here would incorrectly give 5.

Worked Example: Comparing Two Limits With Shape Three

For a hypothetical book with Pareto Type II severity parameters $\theta=\$1{,}200$ and $\alpha=3$, calculate the ILF from a $2,400 base limit to a $6,000 limit, and calculate the expected payment in the layer between them. Use the same ground-up severity basis at both limits.

For $\alpha=3$, the formula becomes $L(u)=\theta[1-(1+u/\theta)^{-2}]/2$. At the $2,400 base limit, $u/\theta=2$:

L(2{,}400) =\frac{1{,}200}{2}\left[1-(1+2)^{-2}\right] =600\left(1-\frac19\right) =600\cdot\frac89 =\$533.33.

At the $6,000 limit, $u/\theta=5$:

L(6{,}000) =\frac{1{,}200}{2}\left[1-(1+5)^{-2}\right] =600\left(1-\frac1{36}\right) =600\cdot\frac{35}{36} =\$583.33.

The ILF is $(35/36)/(8/9)=35/32=1.09375$, or 1.094 to three decimal places. The layer expected payment is $583.333\ldots-533.333\ldots=\$50.00$. Check: multiplying the base limited expected loss by the ILF gives $533.333\ldots(35/32)=583.333\ldots$. The calculation avoids rounding the limited losses before taking their ratio.

Worked Example: Shape One and an Infinite Uncapped Mean

A hypothetical severity follows Pareto Type II with $\theta=\$500$ and $\alpha=1$. Calculate the ILF from a $500 base limit to a $4,500 limit. Explain whether the finite-limit calculation is valid even though the uncapped mean is not finite.

Because $\alpha=1$, use the logarithmic case. At the base limit, $u/\theta=500/500=1$:

L(500)=500\ln(1+1)=500\ln 2=\$346.57.

At the $4,500 limit, $u/\theta=4{,}500/500=9$:

L(4{,}500)=500\ln(1+9)=500\ln 10=\$1{,}151.29.

The ILF is $\ln 10/\ln 2=3.321928\ldots$, or 3.322 to three decimal places. Both limited expected losses are finite because the limits are finite. However, the uncapped mean does not exist: integrating the survival function to infinity gives $\int_0^\infty (1+x/500)^{-1}\,dx=\infty$. Thus a finite-limit ILF can be calculated, but it must not be interpreted as evidence that the uncapped expected severity is finite. This method avoids using the $\alpha>1$ mean formula at $\alpha=1$.

Interpreting Tail Effects and Checking the Model

For $\alpha>1$, $L(u)$ approaches $\theta/(\alpha-1)$ as $u$ increases without bound. For $\alpha\leq1$, $L(u)$ continues to increase without approaching a finite uncapped mean, even though each finite-limit value remains finite. This distinction matters when the pricing question concerns capped coverage: calculate the finite-limit expectation directly rather than substituting an infinite mean.

A smaller shape parameter gives a heavier tail when the scale is held fixed, but an ILF compares two limited losses, not just the tail at the higher limit. The base-limit expected loss also changes with the distribution. Consequently, one should not claim that every heavier-tailed model must produce a larger ILF for every pair of limits without specifying how the distributions are parameterized and compared. The layer expectation provides a direct check: for limits $u\geq b$, it is $L(u)-L(b)=\int_b^u S_X(x)\,dx$.

For a valid model and $u\geq b>0$, $L(u)\geq L(b)>0$, so the ILF is at least 1. Also, $L(u)\leq u$, because $\min(X,u)\leq u$. These simple checks can expose a parameter, unit, or arithmetic error. The values of $\theta$, $u$, and $b$ must use the same monetary unit.

Common exam trap: A limited expected loss is an integral of the survival function, not the limit multiplied by the survival probability at that limit. For $\theta=\$1{,}000$, $\alpha=2$, and $u=\$5{,}000$, the incorrect calculation $uS_X(u)=5{,}000(1+5)^{-2}=\$138.89$ is not $E[\min(X,u)]$. The correct calculation integrates the survival function: $L(5{,}000)=1{,}000[1-(1+5)^{-1}]=\$833.33$.

Pareto ILFs are useful when the assumed severity model needs to reflect substantial large-loss potential. They remain conditional on the chosen distribution, scale, shape, and policy terms; they are not empirical observations or guarantees about future losses. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, use the stated Pareto Type II model and calculate limited expected losses before forming an ILF. Round dollar amounts to the nearest cent and ILFs to three decimal places.

A hypothetical severity has $\theta=\$800$ and $\alpha=2$. Find $L(800)$, $L(3{,}200)$, and the ILF from $800 to $3,200. (Answer: $L(800)=\$400.00$, $L(3{,}200)=\$640.00$, ILF $=1.600$.)
A hypothetical liability severity has $\theta=\$1{,}500$ and $\alpha=3$. Find the limited expected losses at $1,500 and $4,500, then find the ILF. (Answer: $L(1{,}500)=\$562.50$, $L(4{,}500)=\$703.13$, ILF $=1.250$.)
A hypothetical severity has $\theta=\$250$ and $\alpha=1$. Find the ILF from a $250 base limit to a $2,250 limit, using the correct special case. (Answer: ILF $=\ln(10)/\ln(2)=3.322$; the limited expected losses are $\$173.29$ and $\$575.65$.)
For $\theta=\$1{,}000$ and $\alpha=2$, calculate the ILF from $2,000 to $6,000 and the expected payment in the layer between those limits. (Answer: $L(2{,}000)=\$666.67$, $L(6{,}000)=\$857.14$, ILF $=1.286$, layer expected payment $=\$190.48$.)
A hypothetical severity has $\theta=\$500$ and $\alpha=0.8$. Find the ILF from a $500 base limit to a $2,500 limit. State whether the uncapped mean exists. (Answer: $L(500)\approx\$371.75$, $L(2{,}500)\approx\$1{,}077.42$, ILF $\approx2.898$; the uncapped mean is infinite because $\alpha\leq1$.)

What You'll Learn