Monotonicity of LER in the Deductible

Why Monotonicity Matters

An ordinary deductible makes the insured responsible for the first portion of each ground-up loss. As the deductible rises, the insurer pays less on any given loss, but a larger portion of the ground-up loss is eliminated from the insurer’s payment. The loss elimination ratio (LER) measures the expected eliminated loss as a fraction of the expected ground-up loss.

The key question here is how that ratio behaves as the deductible changes. The result follows from a pointwise property of limited losses: for every fixed loss amount, increasing the deductible cannot decrease the amount of loss below the deductible. Taking expectations preserves that ordering. This argument applies to discrete, continuous, and mixed severity distributions, provided the mean loss is finite and positive.

Definition: Let $X$ be a nonnegative ground-up loss with $0<E[X]<\infty$, and let $d\geq 0$ be an ordinary deductible. The limited expected loss at $d$ is $E[\min(X,d)]$. The ordinary-deductible loss elimination ratio is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]}. $$ The insurer’s payment per loss is $(X-d)_+=\max(X-d,0)$, so the loss eliminated from the insurer’s payment is $\min(X,d)$.

This definition concerns an ordinary deductible applied to the ground-up loss $X$. It does not include a policy limit, coinsurance, or a franchise deductible. The denominator is the expected ground-up loss, not the expected insurer payment.

Limited Expected Loss and the Survival Function

For a nonnegative loss, the limited expected loss can also be calculated as an area under the survival function $S_X(x)=P(X>x)$. For any fixed value $X$, the quantity $\min(X,d)$ is the length of the interval from zero up to the smaller of $X$ and $d$. Equivalently, it is the integral of an indicator that equals one while $x<X$, over $0\leq x\leq d$.

\min(X,d)=\int_0^d \mathbf{1}_{\{X>x\}}\,dx

Taking expectations and interchanging expectation and integration for this nonnegative integrand gives the limited expected loss formula:

E[\min(X,d)] =\int_0^d E[\mathbf{1}_{\{X>x\}}]\,dx =\int_0^d S_X(x)\,dx.

The survival integral is valid whether $X$ has a density, probability masses, or both. In particular, it avoids the need to differentiate a limited expected loss formula or assume a continuous severity distribution.

Why the LER Is Nondecreasing

Let $0\leq d_1<d_2$. For every possible value $x\geq0$, $\min(x,d_2)\geq\min(x,d_1)$: raising the cap cannot lower the capped value. Therefore the expected limited loss at $d_2$ is at least as large as the expected limited loss at $d_1$. Since $E[X]$ is a fixed positive denominator, the same ordering holds for the LER.

Theorem: If $X\geq0$ and $0<E[X]<\infty$, then $\operatorname{LER}(d)$ is nondecreasing for $d\geq0$. More precisely, for $0\leq d_1<d_2$, $$ \operatorname{LER}(d_2)-\operatorname{LER}(d_1) =\frac{1}{E[X]}\int_{d_1}^{d_2}S_X(x)\,dx \geq0. $$ The increase is strict exactly when $\int_{d_1}^{d_2}S_X(x)\,dx>0$. Equivalently, for $d_2>d_1$, it is strict when $P(X>d_1)>0$.

To see the equivalence in the strictness condition, if $P(X>d_1)>0$, then there is positive probability of losses exceeding $d_1$. For a positive interval immediately to the right of $d_1$, the survival function is consequently positive, so its integral over $[d_1,d_2]$ is positive. Conversely, if $P(X>d_1)=0$, then $X\leq d_1$ with probability one, and increasing the deductible beyond $d_1$ cannot change any limited loss.

Thus, the ratio may increase strictly over part of the deductible range and later become constant. For a bounded loss distribution, once the deductible reaches the largest possible loss, the limited expected loss equals $E[X]$, and the LER is one for all larger deductibles. A plateau can also begin earlier if there is no probability of losses above the deductible. These conclusions concern the LER, not the insurer’s expected payment, which moves in the opposite direction as the deductible rises.

State the loss model and denominator.
Confirm that $X\geq0$ and $0<E[X]<\infty$. Calculate $E[X]$, the expected ground-up loss.

Calculate the limited expected loss.
Use $E[\min(X,d)]$, by summing over outcomes for a discrete model or integrating the survival function when that is convenient.

Form the ratio and compare deductibles.
Divide the limited expected loss by $E[X]$. For two deductibles, compare the limited expected losses, or use the survival integral over the interval between them.

Worked Calculations

Worked Example: Comparing Deductibles for a Discrete Loss

A hypothetical ground-up loss $X$ is $0$ with probability $0.20$, $\$60$ with probability $0.50$, and $\$150$ with probability $0.30$. Find the LER at deductibles of $\$40$, $\$100$, and $\$200$, and verify their ordering.

Method. For a discrete severity, calculate $E[X]$ and then sum the limited amounts $\min(X,d)$, each weighted by its probability. The distribution and probabilities are the same at each deductible.

First calculate the expected ground-up loss:

E[X]=0(0.20)+60(0.50)+150(0.30)=0+30+45=\$75.

At $d=\$40$, the limited losses are $\$0$, $\$40$, and $\$40$. Thus

E[\min(X,40)]=0(0.20)+40(0.50)+40(0.30)=0+20+12=\$32, \qquad \operatorname{LER}(40)=\frac{32}{75}=0.4267.

At $d=\$100$, the limited losses are $\$0$, $\$60$, and $\$100$. Therefore

E[\min(X,100)]=0(0.20)+60(0.50)+100(0.30)=0+30+30=\$60, \qquad \operatorname{LER}(100)=\frac{60}{75}=0.8000.

At $d=\$200$, every possible loss is at or below the deductible, so

E[\min(X,200)]=0(0.20)+60(0.50)+150(0.30)=\$75, \qquad \operatorname{LER}(200)=\frac{75}{75}=1.0000.

The values $0.4267$, $0.8000$, and $1.0000$ are in increasing order, as the theorem requires. As a check on the interpretation, at $d=\$40$ the expected insurer payment is $20(0.50)+110(0.30)=\$43$, and $\$32+\$43=\$75$, the expected ground-up loss.

Common mistake avoided. The amount $\$43$ is the expected insurer payment, not the expected eliminated loss. Using $43/75=0.5733$ as the LER would give the wrong ratio. The correct numerator is the limited expected loss, $\$32$, giving $32/75=0.4267$.

Worked Example: Using a Survival Integral for a Uniform Loss

Suppose a hypothetical loss is uniformly distributed on $[0,\$200]$. Find the LER at deductibles of $\$50$ and $\$100$. Show how the integral makes the ordering clear.

Model and mean. Under the uniform distribution on $[0,200]$, the density is $1/200$ on that interval, and the mean is

E[X]=\frac{0+200}{2}=\$100.

For $0\leq x\leq200$, the survival function is $S_X(x)=(200-x)/200$. Applying the limited expected loss representation for $d\leq200$,

E[\min(X,d)] =\int_0^d\frac{200-x}{200}\,dx =\left[x-\frac{x^2}{400}\right]_0^d =d-\frac{d^2}{400}.

At $d=\$50$, the limited expected loss is

50-\frac{50^2}{400}=50-\frac{2500}{400} =50-6.25=\$43.75, \qquad \operatorname{LER}(50)=\frac{43.75}{100}=0.4375.

At $d=\$100$, it is

100-\frac{100^2}{400}=100-\frac{10000}{400} =100-25=\$75, \qquad \operatorname{LER}(100)=\frac{75}{100}=0.7500.

The increase can also be checked directly from the theorem:

\operatorname{LER}(100)-\operatorname{LER}(50) =\frac{1}{100}\int_{50}^{100}\frac{200-x}{200}\,dx =\frac{75-43.75}{100} =0.3125>0.

The ratio is strictly increasing over this interval because the survival probability is positive throughout it. No rounding was needed before the final ratios.

Common mistake avoided. The limited expected loss is not $d$ times the probability that $X$ exceeds $d$. It accounts for losses both below and above $d$; the survival integral accumulates the survival probability across the entire interval from zero to $d$.

Worked Example: Detecting a Plateau in a Mixed Discrete Severity

A hypothetical loss is $\$25$ with probability $0.40$ and $\$100$ with probability $0.60$. Calculate the LER at $d=\$25$, $\$60$, $\$100$, and $\$140$. Identify where the ratio increases and where it becomes constant.

The expected ground-up loss is

E[X]=25(0.40)+100(0.60)=10+60=\$70.

At $d=\$25$, both losses are limited to $\$25$:

E[\min(X,25)]=25(0.40)+25(0.60)=\$25, \qquad \operatorname{LER}(25)=\frac{25}{70}=0.3571.

At $d=\$60$, the limited amounts are $\$25$ and $\$60$:

E[\min(X,60)]=25(0.40)+60(0.60)=10+36=\$46, \qquad \operatorname{LER}(60)=\frac{46}{70}=0.6571.

At $d=\$100$, both losses are fully included in the limited amount:

E[\min(X,100)]=25(0.40)+100(0.60)=\$70, \qquad \operatorname{LER}(100)=1.0000.

At $d=\$140$, the limited amounts remain $\$25$ and $\$100$, so $E[\min(X,140)]=\$70$ and $\operatorname{LER}(140)=1.0000$. The LER increases strictly up to $\$100$, then plateaus: there is no probability of a loss greater than $\$100$. The ordering $0.3571<0.6571<1.0000=1.0000$ checks the result.

Common mistake avoided. A nondecreasing ratio need not be strictly increasing for every larger deductible. Once the deductible is at least the largest possible loss in this model, raising it further leaves every limited loss unchanged.

Common Interpretation Traps

Reversing the direction. A higher deductible reduces the insurer’s payment, but it increases or leaves unchanged the expected loss eliminated. The LER therefore cannot decrease.
Using the wrong numerator. The numerator is $E[\min(X,d)]$. The expected insurer payment $E[(X-d)_+]$ is a different quantity.
Assuming strict increase everywhere. Strict increase requires positive survival probability above the lower deductible. If no loss can exceed that deductible, the LER is already constant thereafter.
Assuming a continuous model. The monotonicity proof does not require a density. The limited-loss comparison and survival-integral representation also work with discrete and mixed severities.
Changing the denominator with the deductible. The LER denominator is the expected ground-up loss $E[X]$, which does not depend on $d$.

Key takeaway: For an ordinary deductible, $\operatorname{LER}(d)=E[\min(X,d)]/E[X]$. The limited expected loss is nondecreasing in $d$, so the LER is nondecreasing whenever $X$ has a finite, positive mean. Its increase is strict only while there remains positive probability of losses above the lower deductible.

Check Your Understanding

For each problem, use the ground-up loss as the denominator and the limited expected loss as the numerator. Give probabilities to four decimal places where needed.

A hypothetical ground-up loss is $\$0$, $\$50$, or $\$120$, with respective probabilities $0.20$, $0.50$, and $0.30$. Find $E[X]$ and the LER at a deductible of $\$80$. (Answer: $E[X]=\$61$; $\operatorname{LER}(80)=49/61\approx0.8033$.)
A hypothetical loss is uniform on $[0,\$150]$. Find the LER at deductibles of $\$30$ and $\$90$, and state whether the increase is strict. (Answer: $\operatorname{LER}(30)=0.3600$, $\operatorname{LER}(90)=0.8400$; the increase is strict.)
A hypothetical exponential loss has mean $\$80$. Find its LER at a deductible of $\$40$, using $E[\min(X,d)]=80(1-e^{-d/80})$. (Answer: $\operatorname{LER}(40)=1-e^{-0.5}\approx0.3935$.)
A hypothetical loss is $\$20$, $\$70$, or $\$160$, with respective probabilities $0.25$, $0.50$, and $0.25$. Calculate the LER at deductibles of $\$50$ and $\$100$, and verify the ordering. (Answer: $E[X]=\$80$; $\operatorname{LER}(50)=42.50/80=0.5313$, $\operatorname{LER}(100)=65/80=0.8125$, so the ratio increases.)
A hypothetical loss is $0$ with probability $0.30$, $\$40$ with probability $0.40$, and $\$90$ with probability $0.30$. Find the LER at deductibles of $\$40$ and $\$90$, and determine whether the increase is strict. (Answer: $E[X]=\$43$; $\operatorname{LER}(40)=28/43\approx0.6512$, $\operatorname{LER}(90)=1.0000$; the increase is strict because $P(X>40)=0.30>0$.)

What You'll Learn