LER Using Limited Expected Values

Why Use a Limited Expected Value?

An ordinary deductible removes part of each ground-up loss: the entire loss is removed when it is below the deductible, while only the deductible amount is removed when the loss is larger. The expected amount removed can therefore be summarized by a limited expected value. This gives a direct route to the loss elimination ratio (LER) without first calculating the insurer’s payment for every possible loss.

Assume throughout that $X$ is a nonnegative ground-up loss with finite, positive mean, and that $d\geq 0$ is an ordinary per-loss deductible measured in the same monetary units and price basis as $X$. The deductible removes $\min(X,d)$ from a loss $X$. The ratio of the expected amount removed to the expected ground-up loss is the LER.

Definition: The limited expected value of $X$ at limit $d$ is $\operatorname{LEV}_X(d)=E[\min(X,d)]$. For an ordinary deductible $d$, the loss elimination ratio is $$ \operatorname{LER}(d)=\frac{\operatorname{LEV}_X(d)}{E[X]} =\frac{E[\min(X,d)]}{E[X]}. $$ The LER is a proportion of expected ground-up loss, not the proportion of claims eliminated.

For example, if a loss is $300 and the deductible is $500, the full $300 is eliminated. If a loss is $1,200, $500 is eliminated. Averaging these eliminated amounts over the severity distribution gives the numerator of the LER. The denominator is the mean of the ground-up severity, before applying the deductible.

Calculating the Limited Expected Value

For a nonnegative severity with distribution function $F_X$, define the survival function by $S_X(t)=P(X>t)=1-F_X(t)$. The limited expected value can be calculated using either function:

\operatorname{LEV}_X(d) =\int_0^d S_X(t)\,dt =d-\int_0^d F_X(t)\,dt.

The first integral uses only survival probabilities up to the deductible. The second uses the distribution function over that same interval. Either route gives the expected amount eliminated per loss. To calculate the LER, the denominator still requires the full ground-up mean; for a nonnegative loss with finite mean,

E[X]=\int_0^\infty S_X(t)\,dt.

Why the formula holds: For each realized loss $X$, the quantity $\min(X,d)$ is the length of the interval of values $t$ between 0 and $d$ for which $X>t$. Thus $\min(X,d)=\int_0^d \mathbf{1}_{\{X>t\}}\,dt$, where the indicator is 1 when the event holds and 0 otherwise. Taking expectations and integrating the event probability gives $E[\min(X,d)]=\int_0^d P(X>t)\,dt$. Since $P(X>t)=1-F_X(t)$, this also equals $d-\int_0^d F_X(t)\,dt$.

These formulas remain valid when the distribution has probability at particular loss amounts: the integral representation uses $P(X>t)$, and isolated endpoint values do not change the integral. In applications, check that $X$, $d$, and the mean in the ratio use consistent units. A deductible stated in current dollars cannot be combined directly with a severity distribution stated in historical dollars.

Worked Calculations

Worked Example: Using a Distribution Function

A hypothetical claim severity $X$, in thousands of dollars, has distribution function $F_X(x)=0$ for $x<0$, $F_X(x)=x^2/10{,}000$ for $0\leq x\leq100$, and $F_X(x)=1$ for $x>100$. Find the LER for an ordinary deductible of 40 (that is, $40,000). Assume the stated distribution applies to the severity being covered.

Method. Use the CDF form of the limited expected value, and calculate the ground-up mean over the full support. This avoids using the deductible-limited numerator as though it were the ground-up mean.

First calculate the numerator:

\begin{aligned} \operatorname{LEV}_X(40) &=40-\int_0^{40}\frac{x^2}{10{,}000}\,dx\\ &=40-\left[\frac{x^3}{30{,}000}\right]_0^{40}\\ &=40-\frac{64{,}000}{30{,}000}\\ &=40-\frac{32}{15} =\frac{568}{15} \approx 37.8667. \end{aligned}

Now calculate the denominator from the survival function $S_X(x)=1-x^2/10{,}000$ on $0\leq x\leq100$:

\begin{aligned} E[X] &=\int_0^{100}\left(1-\frac{x^2}{10{,}000}\right)\,dx\\ &=\left[x-\frac{x^3}{30{,}000}\right]_0^{100}\\ &=100-\frac{1{,}000{,}000}{30{,}000} =\frac{200}{3} \approx 66.6667. \end{aligned}

Therefore,

\operatorname{LER}(40) =\frac{568/15}{200/3} =0.568 =56.8\%.

The numerator is less than the ground-up mean, as it must be because $\min(X,40)\leq X$ for every loss. The LER is 56.8% of expected ground-up loss. The common mistake avoided here is to use $\operatorname{LEV}_X(40)$ as the denominator too; the denominator is $E[X]$, not the limited mean.

Worked Example: Integrating a Piecewise Survival Function

For a hypothetical severity $X$ measured in thousands of dollars, suppose $S_X(x)=1-x/50$ for $0\leq x\leq40$, $S_X(x)=(100-x)/300$ for $40<x\leq100$, and $S_X(x)=0$ for $x>100$. Find the LER at an ordinary deductible of 40. The survival function is continuous at 40, since both formulas give 0.2 there.

Method. Integrate the survival function to the deductible for the limited expected value, then over the entire support for the mean. This keeps the two integration ranges distinct.

\begin{aligned} \operatorname{LEV}_X(40) &=\int_0^{40}\left(1-\frac{x}{50}\right)\,dx\\ &=\left[x-\frac{x^2}{100}\right]_0^{40}\\ &=40-16=24. \end{aligned}

The tail contribution to the full mean is the area under the second part of the survival function. It is a triangle with base 60 and height 0.2, so its area is $60(0.2)/2=6$. The first part of the mean is the limited expected value just calculated. Therefore,

E[X]=24+6=30,\qquad \operatorname{LER}(40)=\frac{24}{30}=0.8=80.0\%.

As a check, the mean includes losses beyond 40, while the limited expected value does not accumulate any additional amount after 40. A common mistake is to stop the integral at 40 for both quantities, which would incorrectly make the denominator 24 and the ratio 100%.

Worked Example: Matching the Deductible to an Inflated Severity

A hypothetical ground-up severity $X$, in thousands of base-year dollars, has CDF $F_X(x)=0$ for $x<0$, $F_X(x)=x/80$ for $0\leq x\leq40$, $F_X(x)=1/2+(x-40)/120$ for $40<x\leq100$, and $F_X(x)=1$ for $x>100$. Severity is then increased by a factor of 1.10. A current-dollar ordinary deductible is $66,000. Find the LER. Assume the same proportional factor applies to every loss.

Method. Convert the current deductible to base-year dollars before using the given CDF. Then calculate the limited expected value and the mean in that base. This avoids applying a current-dollar deductible to a historical-dollar distribution.

The deductible in base-year thousands is $66/1.10=60$. The survival function is $1-x/80$ from 0 to 40 and $(100-x)/120$ from 40 to 100. Thus the ground-up mean in base-year thousands is

\begin{aligned} E[X] &=\int_0^{40}\left(1-\frac{x}{80}\right)dx +\int_{40}^{100}\frac{100-x}{120}\,dx\\ &=30+\frac{60(0.5)}{2} =30+15=45. \end{aligned}

For the limited expected value at 60, the first integral is 30. On the interval from 40 to 60, the survival function falls linearly from $0.5$ to $(100-60)/120=1/3$. The area is the interval length times the average endpoint height:

\begin{aligned} \operatorname{LEV}_X(60) &=30+20\left(\frac{0.5+1/3}{2}\right)\\ &=30+\frac{25}{3} =\frac{115}{3} \approx38.3333. \end{aligned}

The base-year LER is $(115/3)/45=23/27\approx0.8519$, or 85.19%. In current dollars, the mean is $1.10(45)=49.5$ thousand and the limited expected value is $1.10(115/3)=253/6\approx42.1667$ thousand. Their ratio is still $23/27$. Scaling both loss and deductible by the same factor scales both expected amounts equally, so it does not change the ratio. The common mistake avoided is comparing a $66,000 deductible directly with a base-year severity in thousands.

Reading the Ratio Correctly

The numerator is the expected amount eliminated per ground-up loss, and the denominator is the expected ground-up loss per loss. Consequently, the LER is loss-weighted: it describes the share of expected dollars removed, not the share of claims that fall below the deductible. A small number of large losses can contribute substantially to the denominator even if many individual claims are small.

A useful check is $0\leq\operatorname{LEV}_X(d)\leq E[X]$, since $0\leq\min(X,d)\leq X$. When $E[X]>0$, the LER must therefore lie between 0 and 1. If the calculation gives a value outside that range, check the integration limits, the mean, and the units. If $d=0$, the limited expected value and LER are both zero. If $E[X]=0$, the ratio is undefined, which is why a positive ground-up mean is assumed here.

One especially common error is to put the insurer’s retained payment $(X-d)_+$ in the numerator. That computes expected payment, not expected loss eliminated. For every nonnegative loss,

X=(X-d)_+ + \min(X,d).

The two terms are respectively the retained payment and the eliminated amount. The LER uses the second term divided by $E[X]$. Keeping this distinction explicit makes the calculation easier to check and prevents confusing the eliminated share with the retained share.

Check Your Understanding

For each hypothetical severity, calculate the requested quantity using the stated units and ordinary deductible.

A severity has survival function $S_X(x)=(1-x/100)^2$ for $0\leq x\leq100$, and zero for $x>100$. Find the LER at $d=50$. (Answer: $\operatorname{LEV}_X(50)=175/6$, $E[X]=100/3$, and $\operatorname{LER}=7/8=0.8750$.)
A severity has CDF $F_X(x)=0$ below 0, $x/120$ for $0\leq x\leq60$, $1/2+(x-60)/80$ for $60<x\leq100$, and 1 above 100. Find the LER at $d=80$. (Answer: $\operatorname{LEV}_X(80)=52.5$, $E[X]=55$, and $\operatorname{LER}=21/22\approx0.9545$.)
A severity has survival function $1-x/80$ for $0\leq x\leq40$, $(100-x)/120$ for $40<x\leq100$, and zero above 100. Find the LER at $d=70$. (Answer: $\operatorname{LEV}_X(70)=41.25$, $E[X]=45$, and $\operatorname{LER}=11/12\approx0.9167$.)
A base-year severity has $E[X]=50$ thousand and $\operatorname{LEV}_X(40)=32$ thousand. Losses and a $55,000 deductible are both increased by a factor of 1.10. Find the current-dollar LER. (Answer: current mean $=55$ thousand, current limited expected value $=44$ thousand, and LER $=0.8000$.)
A severity has CDF $F_X(x)=x^2/10{,}000$ for $0\leq x\leq100$, with the usual zero and one tails. Find the LER for $d=25$. (Answer: $\operatorname{LEV}_X(25)=1175/48\approx24.4792$, $E[X]=200/3$, and $\operatorname{LER}=47/128\approx0.3672$.)

What You'll Learn