Why Bounds and Endpoints Matter
A loss elimination ratio (LER) describes the share of expected ground-up loss removed from the insurer’s payment by a coverage modification. For an ordinary deductible, the payment on a loss is reduced by the lesser of the loss and the deductible. Bounds on the LER provide a quick way to check a calculation, while the endpoint values clarify what happens when there is no deductible or when the deductible becomes very large.
The results here concern a nonnegative ground-up loss and an ordinary deductible only. A deductible is assumed to be a fixed, nonnegative amount; no policy limit, coinsurance, or inflation adjustment is included.
The denominator is the expected ground-up loss, not the expected insurer payment. The numerator is the expected amount eliminated by the deductible. A finite positive mean is necessary: if \(E[X]=0\), the ratio is undefined, and if the mean is infinite, this ratio does not have the stated finite-denominator interpretation.
Bounds from the Deductible Payment
For every possible loss \(X\geq 0\), the deductible eliminates a nonnegative amount, no more than the loss itself and no more than the deductible. Thus \(0\leq\min(X,d)\leq X\) and \(\min(X,d)\leq d\). Taking expectations gives bounds on the expected amount eliminated.
The two upper bounds have different sources. Since the amount eliminated cannot exceed the ground-up loss, its expectation cannot exceed \(E[X]\), giving \(\operatorname{LER}(d)\leq 1\). Since the amount eliminated cannot exceed \(d\) on any one loss, its expectation cannot exceed \(d\), giving \(\operatorname{LER}(d)\leq d/E[X]\). Both bounds apply, so the sharper combined bound is their minimum.
The lower bound is zero because the eliminated amount is nonnegative. For \(d>0\) and \(E[X]>0\), the actual LER is positive: positive mean implies \(P(X>0)>0\), and on the event \(X>0\), \(\min(X,d)>0\). However, there is no positive lower bound that works for every severity distribution with a given mean and deductible. For example, losses can be zero with high probability and very large with very low probability. The mean can stay fixed while the expected amount eliminated becomes arbitrarily small.
Why the Survival-Function Integral Holds
A useful way to calculate the numerator is to express it using the survival function \(S_X(x)=P(X>x)\). This applies to discrete, continuous, and mixed nonnegative loss distributions; no density is required.
To justify the formula, fix a realized loss \(X\). The length of the interval of values \(x\) between 0 and \(d\) for which \(x<X\) is exactly \(\min(X,d)\). Equivalently, apart from endpoint values that do not affect an integral, $$ \min(X,d)=\int_0^d \mathbf{1}_{\{X>x\}}\,dx. $$ Taking expectations and interchanging the expectation and integral for this nonnegative integrand gives $$ E[\min(X,d)] =\int_0^d E[\mathbf{1}_{\{X>x\}}]\,dx =\int_0^d P(X>x)\,dx =\int_0^d S_X(x)\,dx. $$
This representation is particularly helpful when the survival function is easier to work with than the density or probability mass function. It also makes clear that the numerator is an accumulated area under the survival curve up to the deductible.
When Can the Upper Bound Be Reached?
The pointwise inequalities also show when an upper bound is exact. If \(d<E[X]\), the tighter bound is \(d/E[X]\). Equality requires \(E[\min(X,d)]=d\), which holds exactly when \(\min(X,d)=d\) with probability 1, or \(X\geq d\) with probability 1. If \(d>E[X]\), the tighter bound is 1. Equality requires \(\min(X,d)=X\) with probability 1, or \(X\leq d\) with probability 1. When \(d=E[X]\), equality with the common bound 1 requires both conditions, so \(X=d\) with probability 1.
| Deductible relative to the mean | Upper bound | Condition for equality |
|---|---|---|
| \(d<E[X]\) | \(d/E[X]\) | \(P(X\geq d)=1\) |
| \(d=E[X]\) | 1 | \(P(X=d)=1\) |
| \(d>E[X]\) | 1 | \(P(X\leq d)=1\) |
Worked Example: A Discrete Loss and a Nonattained Bound
A hypothetical loss is \(0\) with probability \(0.40\), \(\$300\) with probability \(0.40\), and \(\$900\) with probability \(0.20\). An ordinary deductible is \(\$250\). Calculate the LER and the distribution-free upper bound, then check whether the bound is attained.
Method. Find the mean ground-up loss and the expected amount eliminated directly from the probability distribution. The loss is nonnegative and has a finite, positive mean, so the LER is defined.
First calculate the mean: $$ E[X]=0.40(0)+0.40(300)+0.20(900)=0+120+180=\$300. $$
The amounts eliminated at losses of \(\$0\), \(\$300\), and \(\$900\) are respectively \(\$0\), \(\$250\), and \(\$250\). Therefore, $$ E[\min(X,250)] =0.40(0)+0.40(250)+0.20(250) =0+100+50 =\$150. $$
Thus, $$ \operatorname{LER}(250)=\frac{150}{300}=0.5000. $$ The upper bound is $$ \min\left(1,\frac{250}{300}\right)=\min(1,0.8333\ldots)=0.8333 $$ to four decimal places. The LER is below the bound, as required. The bound is not attained because \(P(X\geq 250)<1\): the loss is zero with probability \(0.40\).
Check. The expected insurer payment is \(0.40(0)+0.40(50)+0.20(650)=\$150\). The eliminated amount plus insurer payment is \(\$150+\$150=\$300\), equal to the ground-up mean. A common mistake is to report \(250/300=0.8333\) as the LER; that is only the upper bound, not the expected fraction eliminated for this distribution.
Worked Example: An Exponential Loss at a Specified Deductible
A hypothetical ground-up loss \(X\) has an exponential distribution with mean \(\$800\), so its rate is \(\lambda=1/800\) per dollar and \(S_X(x)=e^{-x/800}\) for \(x\geq0\). The deductible is \(d=800\ln 2\) dollars (approximately \(\$554.52\)). Calculate the LER and compare it with the upper bound.
Method. Use the survival-function integral for the expected amount eliminated. This avoids treating the deductible itself as the amount eliminated on every claim.
The mean is \(\$800\). The expected amount eliminated is $$ E[\min(X,d)] =\int_0^{800\ln 2}e^{-x/800}\,dx =800\left(1-e^{-(800\ln 2)/800}\right) =800\left(1-e^{-\ln 2}\right) =800\left(1-\frac12\right) =\$400. $$ Therefore, $$ \operatorname{LER}(d)=\frac{400}{800}=0.5000. $$
The universal upper bound is $$ \min\left(1,\frac{800\ln 2}{800}\right) =\min(1,\ln 2) =\ln 2 \approx 0.6931. $$ The calculated LER, \(0.5000\), is below \(0.6931\). The insurer’s expected payment is \(E[X]-E[\min(X,d)]=\$800-\$400=\$400\), so the two expected amounts sum to the ground-up mean.
Check. The exponential survival function is positive below the deductible, and its area over that interval is \(\$400\). A common mistake is to use \(d/E[X]\) as the LER; here that ratio is \(\ln 2\), the upper bound, whereas the LER is \(0.5000\).
Worked Example: A Loss Distribution That Attains the Bound
A hypothetical loss is \(\$400\) with probability \(0.60\) and \(\$1{,}000\) with probability \(0.40\). The ordinary deductible is \(\$300\). Calculate the mean, LER, and upper bound, and explain why equality holds.
Method. Evaluate the amount eliminated at each possible loss. Since both possible losses exceed the deductible, the deductible removes its full amount in every case.
The ground-up mean is $$ E[X]=0.60(400)+0.40(1{,}000)=240+400=\$640. $$ The amount eliminated is \(\$300\) at either loss value, so $$ E[\min(X,300)]=0.60(300)+0.40(300)=180+120=\$300. $$ Hence, $$ \operatorname{LER}(300)=\frac{300}{640}=0.46875. $$ The upper bound is $$ \min\left(1,\frac{300}{640}\right)=0.46875. $$
The bound is attained because \(P(X\geq300)=1\), making \(\min(X,300)=300\) with probability 1. As a payment check, the insurer pays \(\$100\) on the \(\$400\) loss and \(\$700\) on the \(\$1{,}000\) loss. Its expected payment is \(0.60(100)+0.40(700)=\$340\); adding the eliminated amount gives \(\$340+\$300=\$640\).
Common mistake avoided. The ratio is not automatically 1 just because the deductible is applied to every claim. The numerator is the expected amount eliminated, \(\$300\), while the denominator is the full expected loss, \(\$640\).
Zero Deductible and the Large-Deductible Limit
At \(d=0\), \(\min(X,0)=0\) for every loss. Therefore the expected amount eliminated is zero and, when \(0<E[X]<\infty\), the LER is exactly zero. This is consistent with both upper bounds.
As the deductible grows, \(\min(X,d)\) increases toward \(X\) for each realized finite loss. The monotone convergence theorem gives \(E[\min(X,d)]\to E[X]\) as \(d\to\infty\). Because the mean is finite and positive, dividing by \(E[X]\) yields the limiting LER:
For any finite deductible, the LER need not equal 1: losses above that deductible still produce an insurer payment. It equals 1 at a finite \(d\) only if \(X\leq d\) with probability 1. The limit of 1 does not require that all losses be bounded; finite expected loss is enough. If the mean is infinite, the ratio in this tutorial is not defined, and this limiting conclusion does not apply in the stated form.
Common Exam Traps
- Reporting the bound as the LER. The quantity \(d/E[X]\) is an upper bound, not generally the LER. The numerator must be calculated or obtained from the survival integral.
- Using the insurer payment in the denominator. The denominator is \(E[X]\), the ground-up mean. The insurer’s expected payment is \(E[(X-d)_+]\), not the denominator of the LER.
- Ignoring zero or small losses. A deductible eliminates less than \(d\) whenever \(X<d\), including when \(X=0\). This can make the bound loose.
- Applying the large-deductible limit at a finite amount. The limit is 1 as \(d\) tends to infinity. At a particular finite deductible, an LER of 1 requires \(X\leq d\) almost surely.
- Using the result without its assumptions. The loss must be nonnegative and have a finite, positive mean for the stated ratio and bounds to apply.
Check Your Understanding
For each hypothetical loss model, identify the mean, calculate the requested LER or bound, and check that the result is consistent with the stated inequalities.
- A loss is \(0\) with probability \(0.25\), \(\$200\) with probability \(0.50\), and \(\$800\) with probability \(0.25\). For \(d=\$100\), calculate the mean, LER, and upper bound. (Answer: \(E[X]=\$300\); LER \(=0.2500\); upper bound \(=0.3333\).)
- A loss is exponential with mean \(\$500\), and \(d=500\ln 2\) dollars. Find the expected amount eliminated, LER, and upper bound. (Answer: expected amount eliminated \(=\$250.00\); LER \(=0.5000\); upper bound \(=\ln 2\approx0.6931\).)
- A loss is \(\$400\) with probability \(0.50\) and \(\$1{,}000\) with probability \(0.50\). For a \(\$700\) deductible, calculate the LER and upper bound, and state whether equality holds. (Answer: \(E[X]=\$700\); LER \(=0.7857\); upper bound \(=1.0000\); equality does not hold.)
- A loss is \(\$50\) with probability \(0.40\) and \(\$150\) with probability \(0.60\). For a \(\$200\) deductible, calculate the LER and explain why it reaches the upper bound. (Answer: \(E[X]=\$110\); LER \(=1.0000\); upper bound \(=1.0000\), attained because \(X\leq200\) with probability 1.)
- A nonnegative loss has finite positive mean \(\$2{,}400\). State the LER at a zero deductible and its limit as the deductible tends to infinity. (Answer: \(\operatorname{LER}(0)=0\); \(\lim_{d\to\infty}\operatorname{LER}(d)=1\).)