Deriving Increased Limits Factors from Severity

Introduction

A higher per-claim policy limit increases the insurer’s payment only on losses that reach beyond the lower limit. The effect therefore depends not just on the ratio of the two limits, but on the severity distribution: how often losses occur at different sizes. An increased limits factor (ILF) quantifies the change in expected capped loss between a basic limit and a higher limit.

This tutorial derives the factor from the severity distribution. The setting is a per-claim limit with no deductible. Deductibles, expenses, and other policy features are not included in the calculation.

Definition: Let $X$ be the nonnegative amount of a claim before application of a policy limit. At limit $u$, the insurer’s payment per loss is $Y^L_u=\min(X,u)$. The limited expected value at $u$ is $E[Y^L_u]=E[\min(X,u)]$. For a basic limit $u_0$ and an increased limit $u_1$, where $u_1\geq u_0>0$, the increased limits factor is $$ \operatorname{ILF}(u_1;u_0)=\frac{E[\min(X,u_1)]}{E[\min(X,u_0)]}, $$ provided the denominator is positive. It expresses expected loss capped at the increased limit relative to expected loss capped at the basic limit.

Deriving Expected Capped Loss

For any fixed nonnegative claim amount $x$, the capped payment at limit $u$ can be written as the length of the interval from zero to the smaller of $x$ and $u$. Equivalently, it is the accumulated indicator of whether the loss exceeds each threshold below the limit:

\min(x,u)=\int_0^u \mathbf{1}(x>t)\,dt.

Taking expectations gives the survival-function formula. Since the integrand is nonnegative, expectation and integration can be interchanged:

E[\min(X,u)] =\int_0^u E[\mathbf{1}(X>t)]\,dt =\int_0^u P(X>t)\,dt =\int_0^u S_X(t)\,dt.

This derivation works for nonnegative severities whether or not their distribution is continuous. In particular, it handles point masses without requiring a density. When $X$ has atoms, the probability at an exact threshold does not affect the integral’s value; it affects the survival function only at individual points, which have zero length in the integral.

Result: The expected payment per loss at limit $u$ is the area under the severity survival function from 0 to $u$: $$ E[Y^L_u]=\int_0^u S_X(t)\,dt. $$ Consequently, $$ \operatorname{ILF}(u_1;u_0) =\frac{\int_0^{u_1}S_X(t)\,dt}{\int_0^{u_0}S_X(t)\,dt}. $$

A second useful expression is $E[\min(X,u)]=E[X\mathbf{1}(X\leq u)]+uP(X>u)$. It separates payments on losses at or below the limit from the capped payments on losses above it. The survival-integral expression is often more direct when the survival function is given.

From Expected Loss per Claim to an ILF

An ILF compares expected capped severity. If the claim frequency is the same at both limits, the same factor also compares expected capped loss cost per exposure. Let $\lambda$ be the expected number of claims per exposure unit, with claim severity distributed as $X$ at either limit. Then expected capped loss cost at limit $u$ is $\lambda E[\min(X,u)]$. Thus:

\frac{\lambda E[\min(X,u_1)]}{\lambda E[\min(X,u_0)]} =\frac{E[\min(X,u_1)]}{E[\min(X,u_0)]}.

The cancellation requires the same frequency basis and severity distribution in numerator and denominator. If the two contracts or groups being compared have different claim frequencies or severity distributions, their loss-cost ratio is not necessarily the ILF defined here.

Worked Example: Two Limits for a Discrete Severity Model

A hypothetical insurer models claim severity $X$, in thousands of dollars, as 20 with probability 0.50, 60 with probability 0.30, and 150 with probability 0.20. Find the ILF for a 100-thousand-dollar limit relative to a 50-thousand-dollar basic limit. Use the discrete expected-payment method.

Method and mistake avoided. Calculate the payment at each limit for each possible loss, then probability-weight those payments. Do not use the ratio of the limits, $100/50$: the payment increase depends on how much probability lies above each limit.

At the basic limit of 50, the three capped payments are $\min(20,50)=20$, $\min(60,50)=50$, and $\min(150,50)=50$. Therefore, in thousands of dollars:

E[\min(X,50)] =(0.50)(20)+(0.30)(50)+(0.20)(50) =10+15+10 =35.

At the increased limit of 100, the capped payments are 20, 60, and 100:

E[\min(X,100)] =(0.50)(20)+(0.30)(60)+(0.20)(100) =10+18+20 =48.

The ILF is $48/35=1.371428\ldots$, or 1.371 to three decimal places. The increased limit produces expected capped loss per claim of 48 thousand dollars, compared with 35 thousand at the basic limit. As a check, the factor exceeds 1 because raising the limit cannot reduce any claim payment, and it is less than the limit ratio of 2 because not every claim payment doubles.

Worked Example: Applying the Survival-Function Integral

A hypothetical severity model is $X\sim\operatorname{Uniform}(0,120)$, with amounts measured in thousands of dollars. Find the ILF for a limit of 80 relative to a basic limit of 40. Also find the corresponding capped loss costs if expected claim frequency is 2.4 claims per exposure unit.

Method and mistake avoided. Integrate the survival function to obtain limited expected values, then multiply each by the same frequency for loss costs. Do not multiply the ILF by frequency; the common frequency cancels in the ratio.

For $0\leq t\leq120$, the survival function is $S_X(t)=(120-t)/120$. Therefore, for a limit $u$ within the support:

E[\min(X,u)] =\int_0^u \frac{120-t}{120}\,dt =u-\frac{u^2}{240}.

At the basic limit $u=40$, the limited expected value is $40-40^2/240=40-6.6667=33.3333$ thousand dollars. At the increased limit $u=80$, it is $80-80^2/240=80-26.6667=53.3333$ thousand dollars. Keeping full precision for the ratio:

\operatorname{ILF}(80;40) =\frac{53.3333\ldots}{33.3333\ldots} =1.600.

The capped loss costs are $2.4(33.3333\ldots)=80$ thousand dollars per exposure unit at the basic limit and $2.4(53.3333\ldots)=128$ thousand dollars per exposure unit at the increased limit. Their ratio is $128/80=1.600$, checking the result. The limited expected values and loss costs are reported to four decimal places in thousands where needed; the factor is reported to three decimal places.

Worked Example: A Nonlinear Survival Function and Frequency Check

For a hypothetical portfolio, severity $X$, in thousands of dollars, has survival function $S_X(t)=(1-t/100)^2$ for $0\leq t\leq100$, and $S_X(t)=0$ for $t>100$. Find the ILF for a limit of 80 relative to a limit of 50. If expected frequency is 0.04 claims per exposure unit, also calculate both capped loss costs.

Method and mistake avoided. Integrate the specified survival function at each limit and use the same frequency for both loss costs. Avoid treating the survival value at the limit as the expected payment; the expected capped loss is the area over the whole interval from zero to the limit.

For $u\leq100$:

E[\min(X,u)] =\int_0^u (1-t/100)^2\,dt =u-\frac{u^2}{100}+\frac{u^3}{30000}.

At $u=50$, this gives $50-2500/100+125000/30000=50-25+4.1667=29.1667$ thousand dollars. At $u=80$, it gives $80-6400/100+512000/30000=80-64+17.0667=33.0667$ thousand dollars. Using unrounded values, the factor is:

\operatorname{ILF}(80;50) =\frac{33.0666\ldots}{29.1666\ldots} =\frac{992}{875} =1.133714\ldots \approx 1.134.

The capped loss costs are $0.04(29.1666\ldots)=1.1667$ thousand dollars and $0.04(33.0666\ldots)=1.3227$ thousand dollars per exposure unit. Their ratio, using unrounded values, is $1.322666\ldots/1.166666\ldots=1.133714\ldots$, confirming the ILF. Money is shown to four decimal places in thousands, and the factor is rounded to three decimal places.

Interpreting and Checking an ILF

For $u_1\geq u_0$, $\min(X,u_1)\geq\min(X,u_0)$ for every nonnegative claim amount. Taking expectations preserves that inequality, so the ILF is at least 1 when the denominator is positive. The expected capped loss cannot decrease when the limit rises.

The factor is not generally equal to $u_1/u_0$. That ratio would describe proportional growth in the limits, not in expected payments. The factor depends on the distribution of losses below, between, and above the two limits. A limit increase can have little effect if few claims reach the added layer. If every claim exceeds both limits, expected capped loss equals each limit and the ILF does equal the ratio of limits; that is a special case, not a general rule.

The difference $E[\min(X,u_1)]-E[\min(X,u_0)]$ is the expected payment in the layer above $u_0$ and up to $u_1$. In survival-function form, it is $\int_{u_0}^{u_1}S_X(t)\,dt$. This gives a useful check: the numerator must equal the denominator plus this nonnegative layer cost.

Specify the payment basis.
Confirm that the calculation concerns per-claim payments $\min(X,u)$, with no deductible or other adjustment included.

Calculate each limited expected value.
Use the severity probabilities directly, or integrate $S_X(t)$ from zero to the relevant limit.

Take the ratio in the stated order.
Divide the increased-limit expected payment by the basic-limit expected payment.

Check the result.
For a higher limit, the factor should be at least 1. If common frequency is used, the ratio of the two loss costs should match the ILF.

Common Exam Traps

Using the limit ratio. The ILF is a ratio of expected capped losses, not $u_1/u_0$, except in special severity situations.
Reversing the ratio. The increased-limit value belongs in the numerator and the basic-limit value in the denominator. A higher limit should not produce a factor below 1 under the stated assumptions.
Using only losses above the basic limit. The ILF numerator includes payments on all claims, including those below the basic limit. The layer cost is the difference between the two limited expected values, not the numerator itself.
Applying frequency inconsistently. Frequency cancels only when the same frequency basis applies at both limits. If computing loss costs, apply it to both limited expected values.
Mixing payment bases. The formulas here concern a per-claim limit with no deductible. A deductible changes the payment function, so it cannot be silently combined with this calculation.

Key takeaway: Derive an increased limits factor from the severity distribution by calculating $E[\min(X,u)]$ at each limit and dividing the increased-limit value by the basic-limit value. The survival-function integral $\int_0^u S_X(t)\,dt$ provides a general method for discrete or continuous severity models. This material is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Check Your Understanding

For each hypothetical severity model, calculate the requested ILF. Keep units consistent and check that the result is at least 1.

A severity is 10 thousand dollars with probability 0.40, 40 thousand with probability 0.40, and 90 thousand with probability 0.20. Find the ILF for a 60-thousand-dollar limit relative to a 25-thousand-dollar basic limit. (Answer: $32/19=1.684$, to three decimals.)
Severity is uniform on 0 to 150 thousand dollars. Find the ILF for a 100-thousand-dollar limit relative to a 50-thousand-dollar basic limit. (Answer: $1.600$.)
A severity in thousands has survival function $S_X(t)=(1-t/120)^2$ for $0\leq t\leq120$, and zero above 120. Find the ILF for a 90-thousand-dollar limit relative to a 30-thousand-dollar limit. (Answer: $63/37=1.703$, to three decimals.)
A severity is 20 thousand dollars with probability 0.70 and 80 thousand with probability 0.30. At frequency 0.12 claims per exposure unit, find the loss costs at limits 30 thousand and 70 thousand, then find the ILF. (Answer: 2.76 thousand and 4.20 thousand per exposure unit; ILF $35/23=1.522$, to three decimals.)
For any nonnegative severity distribution, explain why the expected payment in the layer above $u_0$ and up to $u_1$ can be added to the expected payment at $u_0$ to obtain the expected payment at $u_1$. (Answer: $\min(X,u_1)=\min(X,u_0)+[\min(X,u_1)-\min(X,u_0)]$ claim by claim; taking expectations gives the corresponding sum, and the layer expectation is $\int_{u_0}^{u_1}S_X(t)\,dt$.)

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