Pareto Severity Increased Limits Factors

Why Pareto Severity Matters for Increased Limits

An increased limits factor (ILF) compares expected claim payments at two per-claim limits. The factor is usually greater than one: a higher limit allows the insurer to pay more on large claims. How much more depends on the severity distribution’s tail. A model with relatively more probability in large losses can produce a greater increase in expected payment when the limit rises.

The Pareto Type II model is useful for examining this effect because its tail decreases as a power of loss size. Its shape parameter controls how heavy that tail is. This tutorial calculates ILFs from the model’s limited expected values (LEVs), with no deductible and no discounting. The hypothetical examples assume no mortality component; claim severity is modeled directly.

Definition: Let $X$ be a nonnegative ground-up claim severity with Pareto Type II parameters $\theta>0$ and $\alpha>0$. Its survival function is $S_X(x)=P(X>x)=(\theta/(\theta+x))^\alpha$ for $x\geq 0$. At per-claim limit $u$, the insurer’s payment is $\min(X,u)$, and the limited expected value is $\operatorname{LEV}(u)=E[\min(X,u)]$. For limits $0<L_1<L_2$, the increased limits factor from $L_1$ to $L_2$ is $\operatorname{ILF}(L_1,L_2)=\operatorname{LEV}(L_2)/\operatorname{LEV}(L_1)$, provided the denominator is positive.

Deriving the Pareto Limited Expected Value

For a nonnegative random variable, the expected amount paid up to a cap $u$ equals the area under its survival function from zero to $u$. One way to see this is that $\min(X,u)$ can be written as the integral, over $0\leq x\leq u$, of an indicator that equals one when $X>x$ and zero otherwise. Taking expectations converts that indicator to $P(X>x)=S_X(x)$.

\operatorname{LEV}(u) =E[\min(X,u)] =\int_0^u S_X(x)\,dx =\int_0^u \left(\frac{\theta}{\theta+x}\right)^\alpha dx.

For $\alpha\neq 1$, integrate using $t=\theta+x$. The integrand becomes $\theta^\alpha t^{-\alpha}$, whose antiderivative is $\theta^\alpha t^{1-\alpha}/(1-\alpha)$. Evaluating at $t=\theta$ and $t=\theta+u$ gives the result below. When $\alpha=1$, the integral is logarithmic instead.

\operatorname{LEV}(u)= \begin{cases} \displaystyle \frac{\theta}{\alpha-1} \left[1-\left(\frac{\theta}{\theta+u}\right)^{\alpha-1}\right], & \alpha\neq 1,\\[1.2em] \displaystyle \theta\ln\left(1+\frac{u}{\theta}\right), & \alpha=1. \end{cases}

For $\alpha>1$, the uncapped mean severity is finite and equals $E[X]=\theta/(\alpha-1)$. The formula for a finite limit remains valid for every $\alpha>0$, including $\alpha\leq 1$, even though the uncapped mean is then infinite. A finite limit caps each payment, so its expected value is finite. Consequently, an ILF between finite positive limits can still be calculated when $\alpha\leq 1$.

Key result: For $\alpha\neq 1$, calculate the ILF by dividing the higher-limit LEV by the basic-limit LEV: $$ \operatorname{ILF}(L_1,L_2)= \frac{1-\left(\dfrac{\theta}{\theta+L_2}\right)^{\alpha-1}} {1-\left(\dfrac{\theta}{\theta+L_1}\right)^{\alpha-1}}. $$ For $\alpha=1$, use $$ \operatorname{ILF}(L_1,L_2)= \frac{\ln(1+L_2/\theta)}{\ln(1+L_1/\theta)}. $$

Calculating Factors from Limited Expected Values

The safest calculation sequence is to find each limited expected value first, then divide. This makes the payment basis visible and helps catch errors in the numerator or denominator. If the expected claim frequency is the same at both limits, multiplying each LEV by that frequency gives expected loss cost per exposure; the frequency cancels from the factor.

Worked Example: A Factor from a Basic Limit of $1,800

A hypothetical liability portfolio is modeled with Pareto Type II severity, $\alpha=3$ and $\theta=\$1{,}200$. Find the ILF from a $\$1{,}800$ limit to a $\$4{,}800$ limit. There is no deductible, no discounting, and no mortality assumption.

Method. Calculate each limit’s LEV using the Pareto formula, then take the higher-limit LEV divided by the basic-limit LEV. Since $\alpha-1=2$, the common multiplier is $\theta/(\alpha-1)=\$600$.

At $L_1=\$1{,}800$, the survival-function ratio in the formula is $1{,}200/(1{,}200+1{,}800)=1{,}200/3{,}000=2/5$. Thus,

\operatorname{LEV}(1{,}800) =600\left[1-\left(\frac25\right)^2\right] =600\left(1-\frac4{25}\right) =600\left(\frac{21}{25}\right) =\$504.

At $L_2=\$4{,}800$, the ratio is $1{,}200/(1{,}200+4{,}800)=1/5$. Therefore,

\operatorname{LEV}(4{,}800) =600\left[1-\left(\frac15\right)^2\right] =600\left(\frac{24}{25}\right) =\$576.

The factor is $576/504=8/7=1.142857\ldots$, or 1.1429 to four decimal places. As a check, the higher-limit LEV exceeds the basic-limit LEV, and the resulting factor is greater than one.

Common mistake avoided. The ILF is not the ratio of excess losses above the two limits. It compares insurer payments capped at each limit, so the relevant quantities are $\operatorname{LEV}(L_1)$ and $\operatorname{LEV}(L_2)$.

Worked Example: Converting an ILF to Expected Loss Cost

A hypothetical book of small commercial policies has claim count $N\sim\operatorname{Poisson}(0.04)$ claims per policy-year, independent of severities. Each claim severity has a Pareto Type II distribution with $\alpha=2$ and $\theta=\$1{,}000$. Calculate the ILF from a $\$2{,}000$ limit to a $\$6{,}000$ limit and the expected capped claim cost per policy-year at each limit. Assume no deductible, discounting, or mortality.

Method. For $\alpha=2$, the LEV formula reduces to $1{,}000[1-1{,}000/(1{,}000+u)]$. Expected capped claim cost per policy-year is expected claim count times LEV.

At $u=\$2{,}000$, the ratio is $1{,}000/3{,}000=1/3$, so

\operatorname{LEV}(2{,}000) =1{,}000\left(1-\frac13\right) =\$666.666\ldots.

The expected capped claim cost per policy-year is $0.04\times \$666.666\ldots=\$26.666\ldots$, or $\$26.67$ to the nearest cent.

At $u=\$6{,}000$, the ratio is $1{,}000/7{,}000=1/7$, so

\operatorname{LEV}(6{,}000) =1{,}000\left(1-\frac17\right) =\frac{6{,}000}{7} =\$857.142857\ldots.

The expected capped claim cost is $0.04\times \$857.142857\ldots=\$34.285714\ldots$, or $\$34.29$ to the nearest cent. Dividing the two LEVs gives

\operatorname{ILF}(2{,}000,6{,}000) =\frac{6{,}000/7}{2{,}000/3} =\frac97 =1.2857 \quad\text{to four decimal places.}

As a check, the ratio of the unrounded expected costs is also $34.285714\ldots/26.666666\ldots=9/7$. The displayed rounded dollar costs should not be used to recompute the factor, because rounding each cost first can slightly change the ratio.

Common mistake avoided. The claim frequency multiplies each capped severity to produce a cost per policy-year, but it does not change the ILF when frequency is the same at both limits. The factor is based on the ratio of like-for-like costs.

How Tail Heaviness Changes the Factor

A smaller Pareto shape parameter generally corresponds to a heavier tail. This means that, relative to the distribution’s scale, more probability remains in large losses. A higher limit can therefore add more expected payment relative to the payment under a lower limit. To make a fair comparison of tail shapes, it is helpful to hold the mean severity and the two limits fixed rather than merely holding $\theta$ fixed. Since $E[X]=\theta/(\alpha-1)$, the scale parameter must change when $\alpha$ changes and the mean is held constant.

Worked Example: Comparing Two Tail Shapes at the Same Mean

Consider two hypothetical Pareto Type II severity models, each with mean severity $\$1{,}200$. Model A has $\alpha=2$, $\theta=\$1{,}200$; Model B has $\alpha=3$, $\theta=\$2{,}400$. Find each model’s ILF from a $\$1{,}200$ limit to a $\$4{,}800$ limit. Assume no deductible, discounting, or mortality.

Method. Verify the means, calculate both LEVs for each model, and compare the ratios. The smaller shape parameter in Model A gives the heavier tail in this equal-mean comparison.

For Model A, $E[X]=1{,}200/(2-1)=\$1{,}200$. At the basic limit, the ratio $\theta/(\theta+L_1)$ is $1/2$. At the higher limit, it is $1{,}200/6{,}000=1/5$. Thus,

\operatorname{LEV}_A(1{,}200) =1{,}200\left(1-\frac12\right) =\$600, \qquad \operatorname{LEV}_A(4{,}800) =1{,}200\left(1-\frac15\right) =\$960.

Therefore, $\operatorname{ILF}_A=960/600=1.6000$.

For Model B, $E[X]=2{,}400/(3-1)=\$1{,}200$. The common multiplier is $2{,}400/2=\$1{,}200$. At the basic limit, the ratio is $2{,}400/3{,}600=2/3$, and at the higher limit it is $2{,}400/7{,}200=1/3$. Hence,

\operatorname{LEV}_B(1{,}200) =1{,}200\left[1-\left(\frac23\right)^2\right] =\$1{,}200\left(\frac59\right) =\$666.666\ldots, $$ $$ \operatorname{LEV}_B(4{,}800) =1{,}200\left[1-\left(\frac13\right)^2\right] =\$1{,}200\left(\frac89\right) =\$1{,}066.666\ldots.

Thus, $\operatorname{ILF}_B=(1{,}066.666\ldots)/(666.666\ldots)=1.6000$. In this particular comparison the factors are equal, even though the models have different shape parameters. This is a useful check against assuming that a smaller $\alpha$ must always produce a larger factor for every pair of limits. The factor also depends on how the limits relate to the scale.

For a comparison that does show a larger factor for the heavier tail, keep the same mean of $\$1{,}200$ but use limits of $\$600$ and $\$4{,}800$. Model A has ratios $2/3$ and $1/5$, giving LEVs $1{,}200(1-2/3)=\$400$ and $1{,}200(1-1/5)=\$960$. Its ILF is $960/400=2.4000$. Model B has ratios $2{,}400/3{,}000=4/5$ and $1/3$, giving LEVs $1{,}200[1-(4/5)^2]=\$432$ and $1{,}200[1-(1/3)^2]=\$1{,}066.666\ldots$. Its ILF is $1{,}066.666\ldots/432=2.4691$ to four decimal places. The heavier-tailed Model A does not have the larger factor here; Model B’s factor is slightly larger. This illustrates why comparisons should be computed rather than inferred from tail labels alone.

Common mistake avoided. A qualitative statement about heavier tails does not replace the calculation. An ILF is a ratio at particular limits, and its value depends jointly on $\alpha$, $\theta$, and both limits.

Important Checks and Special Cases

For $\alpha=1$, use the logarithmic LEV formula rather than substituting one into the formula containing $\alpha-1$. For example, with $\theta=\$1{,}000$, the LEV at $\$1{,}000$ is $1{,}000\ln 2$, while at $\$3{,}000$ it is $1{,}000\ln 4$. The ILF is $\ln 4/\ln 2=2$. The formula is finite because each payment is capped, although the uncapped mean severity is infinite.

An ILF should be at least one when $L_2>L_1>0$: for every possible loss, $\min(X,L_2)\geq\min(X,L_1)$, so its expectation cannot decrease. The factor equals one if the two limits are equal. These checks can reveal a reversed ratio or an arithmetic error, but they do not verify the model assumptions.

Use the correct payment function. Under a per-claim limit with no deductible, payment is $\min(X,u)$, not the amount in excess of $u$.
Keep the parameterization consistent. Here $\theta$ is the Pareto Type II scale in $S_X(x)=(\theta/(\theta+x))^\alpha$; other Pareto conventions may use different parameters.
Do not divide by the uncapped mean when the question asks for an ILF between limits. The required denominator is the LEV at the stated basic limit.
Check finiteness. The mean severity is finite only for $\alpha>1$, but a finite-limit LEV is finite for every $\alpha>0$.

Key takeaway: A Pareto increased limits factor is the ratio of two limited expected values under the same severity model. Calculate the LEV at each stated limit, use the logarithmic case when $\alpha=1$, and evaluate the effect of tail shape at the actual limits rather than relying on a qualitative rule.

Check Your Understanding

For each hypothetical model, assume Pareto Type II severity, no deductible, no discounting, and no mortality. Round factors to four decimal places where needed.

For $\alpha=2$, $\theta=\$800$, find the ILF from a $\$800$ limit to a $\$3{,}200$ limit. (Answer: 1.6000)
For $\alpha=3$, $\theta=\$1{,}500$, calculate the LEV at limits of $\$1{,}500$ and $\$4{,}500$, then find the ILF. (Answer: LEVs of $562.50 and $703.125; ILF = 1.2500)
For $\alpha=2$, $\theta=\$1{,}000$, calculate the ILF from a $\$1{,}000$ limit to a $\$5{,}000$ limit. (Answer: 1.6667)
For $\alpha=1.5$, $\theta=\$1{,}000$, calculate the ILF from a $\$1{,}000$ limit to a $\$3{,}000$ limit. Use the general $\alpha\neq1$ formula. (Answer: 1.7071)
For $\alpha=1$, $\theta=\$1{,}000$, calculate the ILF from a $\$1{,}000$ limit to a $\$3{,}000$ limit. (Answer: 2.0000)

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