LER as a Survival-Function Integral

Why the Survival Function Gives the Loss Elimination Ratio

An ordinary deductible does not eliminate the same dollar amount from every claim. For a loss below the deductible, the entire loss is eliminated; for a loss above it, the eliminated amount is exactly the deductible. The loss elimination ratio (LER) compares the expected dollars eliminated with the expected ground-up loss. A survival-function integral expresses both quantities as areas, making the ratio possible to calculate directly from the severity distribution.

Definition: Let $X$ be a nonnegative ground-up loss with $0<E[X]<\infty$, and let $d\geq 0$ be an ordinary deductible. The loss eliminated on a claim is $\min(X,d)$, and the insurer payment is $(X-d)_+=\max(X-d,0)$. The loss elimination ratio is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]}. $$ Here $S_X(x)=P(X>x)$ is the survival function of $X$.

The definition concerns expected loss amounts, not claim counts. The numerator is the expected amount removed by the deductible, while the denominator is the expected loss before coverage modification. The condition $0<E[X]<\infty$ ensures that the ratio is defined and finite. If $E[X]=0$, there is no positive expected ground-up loss to use as a denominator.

Deriving the Survival-Integral Formula

For a fixed nonnegative loss 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 area under an indicator that equals one while the integration point is below the loss and below the deductible. Taking expectations turns that indicator into a survival probability.

Write the eliminated amount as an integral.
For each realized $X$, the indicator $1_{\{X>x\}}$ equals one for $0\leq x<X$. Thus, up to an endpoint that does not affect the integral, $\min(X,d)=\int_0^d 1_{\{X>x\}}\,dx$.

Take expectations.
Since the integrand is nonnegative, expectation and integration can be interchanged. The expected indicator at $x$ is $P(X>x)=S_X(x)$, giving $E[\min(X,d)]=\int_0^d S_X(x)\,dx$.

Express the ground-up mean the same way.
Letting the upper limit extend to infinity gives $E[X]=\int_0^\infty S_X(x)\,dx$, provided the mean is finite.

\operatorname{LER}(d) = \frac{\displaystyle\int_0^d S_X(x)\,dx} {\displaystyle\int_0^\infty S_X(x)\,dx}.

This is the survival-integral form of the ordinary-deductible LER. The numerator is the area under the survival curve from zero to the deductible. The denominator is the entire area under that curve. The derivation does not require a density, so the identity also holds for severity distributions with atoms, as long as the nonnegative loss has finite positive mean. In a continuous model, whether the survival function is written using $P(X>x)$ or $P(X\geq x)$ does not change the integral.

Key result: For a nonnegative loss with finite positive mean, the expected amount eliminated by an ordinary deductible $d$ is $\int_0^d S_X(x)\,dx$, and the ground-up mean is $\int_0^\infty S_X(x)\,dx$. Their ratio is the LER.

Worked Examples

Worked Example: A Uniform Severity and a 40-Dollar Deductible

A hypothetical insurer models a claim severity $X$ as uniform on $[0,100]$ dollars. Find the expected ground-up loss, the expected amount eliminated by an ordinary deductible of $d=40$ dollars, and the LER. Use the survival-integral method.

Method and model. The model is $X\sim\operatorname{Uniform}(0,100)$. Its survival function is $S_X(x)=1-x/100$ for $0\leq x\leq100$, and zero for $x>100$. The mean is finite and positive.

Ground-up mean. Integrate the survival function over its full range:

E[X]=\int_0^{100}\left(1-\frac{x}{100}\right)dx =\left[x-\frac{x^2}{200}\right]_0^{100} =100-50 =50\text{ dollars}.

Expected eliminated amount. Integrate only up to $d=40$:

E[\min(X,40)] =\int_0^{40}\left(1-\frac{x}{100}\right)dx =\left[x-\frac{x^2}{200}\right]_0^{40} =40-\frac{1600}{200} =32\text{ dollars}.

LER and check.

\operatorname{LER}(40)=\frac{32}{50}=0.6400.

The result is between zero and one, and the eliminated expected amount, 32 dollars, is less than the 50-dollar ground-up mean. A common mistake is to use $P(X\leq40)=0.4000$ as the LER. That is the probability a loss falls below the deductible, not the expected fraction of dollars eliminated; the correct LER is $0.6400$.

Worked Example: Integrating a Piecewise Survival Function

For a hypothetical severity measured in dollars, suppose the survival function is $S_X(x)=1-x/200$ for $0\leq x\leq100$, $S_X(x)=\tfrac12(200-x)/100$ for $100<x\leq200$, and zero for $x>200$. Find the ground-up mean, the expected amount eliminated by a 150-dollar deductible, and the LER.

Check the model. The survival function starts at one, is nonincreasing, remains nonnegative, and reaches zero at 200. The two expressions agree at 100: both give $0.5$. Thus the piecewise definition is consistent at the join.

Ground-up mean. Split the area at the point where the formula changes:

\begin{aligned} E[X] &=\int_0^{100}\left(1-\frac{x}{200}\right)dx +\int_{100}^{200}\frac12\left(\frac{200-x}{100}\right)dx\\ &=\left[x-\frac{x^2}{400}\right]_0^{100} +\frac{1}{200}\left[200x-\frac{x^2}{2}\right]_{100}^{200}\\ &=(100-25)+\frac{1}{200}(20000-15000)\\ &=75+25=100\text{ dollars}. \end{aligned}

Expected eliminated amount. The deductible is 150, so the numerator includes all of the first segment and the part from 100 to 150 in the second:

\begin{aligned} E[\min(X,150)] &=\int_0^{100}\left(1-\frac{x}{200}\right)dx +\int_{100}^{150}\frac{200-x}{200}dx\\ &=75+\frac{1}{200} \left[200x-\frac{x^2}{2}\right]_{100}^{150}\\ &=75+\frac{18750-15000}{200}\\ &=75+18.75=93.75\text{ dollars}. \end{aligned}

LER and check. The ratio is $93.75/100=0.9375$. It is less than one, as required, and the eliminated amount is no greater than the ground-up mean. The key method is to split each integral at 100; using only one formula across the whole range would give the wrong area.

Worked Example: Proportional Inflation and a Matched Deductible

A hypothetical portfolio has historical severity $X$ with survival function $S_X(x)=1-x/100$ for $0\leq x\leq100$, and zero thereafter. Losses are trended by a factor of $1.20$, so current severity is $Y=1.20X$. The current ordinary deductible is 60 dollars. Calculate the current mean, expected amount eliminated, and LER.

Method and units. The historical amounts are in dollars, and the trend factor is dimensionless. The current deductible of 60 dollars corresponds to a historical amount of $60/1.20=50$ dollars. Since $Y=1.20X$, the current survival function is $S_Y(y)=S_X(y/1.20)$.

Current ground-up mean. First calculate the historical mean from its survival function, then apply the trend:

E[X]=\int_0^{100}\left(1-\frac{x}{100}\right)dx =100-50=50, \qquad E[Y]=1.20E[X]=1.20(50)=60\text{ dollars}.

Expected eliminated amount. In current dollars, the survival integral to 60 is

\begin{aligned} E[\min(Y,60)] &=\int_0^{60}S_X\left(\frac{y}{1.20}\right)dy\\ &=\int_0^{60}\left(1-\frac{y}{120}\right)dy\\ &=\left[y-\frac{y^2}{240}\right]_0^{60} =60-15=45\text{ dollars}. \end{aligned}

LER and check. Therefore $\operatorname{LER}(60)=45/60=0.7500$. Equivalently, the historical eliminated amount at deductible 50 is $50-50^2/200=37.5$ dollars, and trending gives $1.20(37.5)=45$ dollars. Both calculations agree. Applying the trend to severity but forgetting to express the deductible in the same price basis is a common error; here the current 60-dollar deductible corresponds to 50 historical dollars.

Interpreting the Areas and Avoiding Common Errors

The survival curve records the chance that a loss exceeds each threshold $x$. For every dollar of threshold below the deductible, the area accounts for the probability that the loss reaches that dollar. Adding those probabilities across thresholds gives expected dollars eliminated. Beyond the deductible, the remaining area corresponds to expected insurer payment:

E[(X-d)_+]=\int_d^\infty S_X(x)\,dx, \qquad E[X]=\int_0^d S_X(x)\,dx+\int_d^\infty S_X(x)\,dx.

These identities follow by applying the same indicator argument to the portions of the loss below and above $d$. They also provide a useful check: the expected eliminated amount plus the expected insurer payment must equal the ground-up mean. They do not say that the insurer pays a fixed amount $d$ on each claim; the payment varies with $X$.

Do not substitute a probability for a loss ratio. $P(X\leq d)$ counts claims below the deductible. The LER weights losses by dollars through the survival-area integrals.
Use the correct integration ranges. The eliminated amount uses $0$ to $d$; the insurer payment uses $d$ to infinity.
Use the same loss basis throughout. If severity is trended, the deductible and integration variable must be expressed in the corresponding price basis.
Check the denominator. The LER denominator is the ground-up expected loss, not the expected insurer payment or the probability of a claim.

Key takeaway: An ordinary deductible removes $\min(X,d)$ dollars from a loss $X$. Its expected eliminated amount is the area under $S_X$ up to $d$, and dividing this area by the full survival area gives the LER.

Check Your Understanding

For each problem, use the survival-integral definition. Amounts are in dollars unless stated otherwise.

A hypothetical severity is uniform on $[0,80]$, with ordinary deductible $d=20$. Find the ground-up mean and LER. (Answer: $E[X]=40$, $\operatorname{LER}=0.4375$.)
A severity has $S_X(x)=1-x/100$ for $0\leq x\leq100$, and zero thereafter. For $d=50$, calculate the expected eliminated amount and LER. (Answer: eliminated amount $=37.50$, $\operatorname{LER}=0.7500$.)
A severity has $S_X(x)=(1-x/60)^2$ for $0\leq x\leq60$, and zero thereafter. For $d=30$, calculate the mean and LER. (Answer: $E[X]=20$, $\operatorname{LER}=0.8750$.)
A severity has $S_X(x)=1-x/200$ for $0\leq x\leq100$, $S_X(x)=(200-x)/200$ for $100<x\leq200$, and zero thereafter. For $d=150$, find the expected eliminated amount and LER. (Answer: eliminated amount $=93.75$, $\operatorname{LER}=0.9375$.)
Historical severity has $S_X(x)=1-x/100$ for $0\leq x\leq100$, and zero thereafter. Current severity is $Y=1.10X$, and the current deductible is 55 dollars. Find the current mean, expected eliminated amount, and LER. (Answer: current mean $=55$, eliminated amount $=41.25$, $\operatorname{LER}=0.7500$.)

What You'll Learn