Loss Elimination Ratios for Deductible Contracts

Why Measure Loss Elimination?

An ordinary deductible assigns the first part of each covered loss to the insured. The expected insurer payment tells us what remains for the insurer to pay, but it does not by itself describe how much of the expected ground-up loss the deductible removes. The loss elimination ratio (LER) expresses that removed amount as a proportion of expected ground-up loss.

The ratio is useful for comparing deductible effects relative to the size of the underlying losses. A deductible of a fixed dollar amount can eliminate a large share of expected loss for one severity distribution and a much smaller share for another. The ratio depends on the loss distribution as well as on the deductible.

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

The ratio is defined only when the expected ground-up loss is finite and positive. It is between zero and one: the eliminated amount is nonnegative and cannot exceed the original loss. If $d=0$, nothing is eliminated and the ratio is zero. As the deductible increases, the eliminated amount for each possible loss cannot decrease, so the LER cannot decrease.

Relating the Ratio to Expected Insurer Payment

The key relationship follows directly from the payment definition. For every realized loss $X$, the amount retained by the insured and the insurer payment add back to the ground-up loss:

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

Taking expectations gives $E[X]=E[\min(X,d)]+E[Y^L]$. Rearranging and dividing by $E[X]$ yields the equivalent formula below. This identity applies to discrete, continuous, and mixed loss distributions, as long as the expected loss is finite and positive.

Theorem: For an ordinary deductible $d$, $$ \operatorname{LER}(d)=1-\frac{E[Y^L]}{E[X]}, \qquad E[Y^L]=E[X]\bigl(1-\operatorname{LER}(d)\bigr). $$ Thus, the LER is the proportion of expected ground-up loss removed, while $1-\operatorname{LER}(d)$ is the proportion remaining as expected insurer payment.

The ratio is a ratio of expectations, not the expected value of a ratio calculated separately for each loss. In particular, dividing the deductible by expected loss does not generally give the LER: losses below the deductible are only partly eliminated, and losses above it have only the first $d$ dollars eliminated.

Survival-Function Formulas

For a nonnegative loss, write $S_X(t)=P(X>t)$. The survival function gives a convenient way to calculate both the expected loss eliminated and expected insurer payment. The formulas do not require a density and therefore also apply when the distribution has point masses.

Result: For $d\geq 0$ and $E[X]<\infty$, $$ E[\min(X,d)]=\int_0^d S_X(t)\,dt, \qquad E[(X-d)_+]=\int_d^\infty S_X(t)\,dt, \qquad E[X]=\int_0^\infty S_X(t)\,dt. $$ Consequently, $$ \operatorname{LER}(d)= \frac{\int_0^d S_X(t)\,dt}{\int_0^\infty S_X(t)\,dt}. $$

To justify these formulas, observe that for a fixed loss $X$, 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: the indicator is one at thresholds $t$ below the loss and zero above it. Similarly, $(X-d)_+$ is the length of the portion of the loss above $d$.

\min(X,d)=\int_0^d \mathbf{1}_{\{X>t\}}\,dt, \qquad (X-d)_+=\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt.

Taking expectations inside these integrals gives the survival-function expressions because $E[\mathbf{1}_{\{X>t\}}]=P(X>t)=S_X(t)$. This interchange is valid for nonnegative integrands. Whether the loss distribution has an atom at a particular threshold does not change an integral over thresholds: a single endpoint has zero length.

Identify the target.
Find the expected ground-up loss $E[X]$, or obtain it from the survival function.

Calculate the amount eliminated.
Use $E[\min(X,d)]$, the survival integral from zero to $d$, or the identity $E[X]-E[Y^L]$.

Form the ratio and check it.
Divide the expected eliminated loss by $E[X]$. Confirm the result is between zero and one and that eliminated loss plus expected payment equals expected ground-up loss.

Worked Calculations

Worked Example: Uniform Losses With a Fixed Deductible

A hypothetical coverage has ground-up loss $X$ uniformly distributed from $0 to $1,200. The ordinary deductible is $300. Find the expected ground-up loss, the expected loss eliminated, the expected insurer payment, and the LER. Report dollar amounts to the cent and the ratio to four decimal places.

Method. Use the uniform density $f_X(x)=1/1{,}200$ for $0\leq x\leq1{,}200$. First, the mean of a uniform distribution on this interval is

E[X]=\frac{0+1{,}200}{2}=600.

The eliminated amount is $X$ when $X\leq300$ and $300 when $X>300$. Therefore,

\begin{aligned} E[\min(X,300)] &=\int_0^{300}\frac{x}{1{,}200}\,dx +300P(X>300)\\ &=\frac{300^2}{2(1{,}200)} +300\left(\frac{1{,}200-300}{1{,}200}\right)\\ &=37.5+225\\ &=262.5. \end{aligned}

For the insurer payment, losses from zero to $300 produce no payment. For losses above $300, the payment is $x-300$. Hence,

\begin{aligned} E[Y^L] &=\int_{300}^{1{,}200}\frac{x-300}{1{,}200}\,dx\\ &=\frac{(1{,}200-300)^2}{2(1{,}200)}\\ &=\frac{900^2}{2{,}400}\\ &=337.5. \end{aligned}

The ratio is $262.5/600=0.4375$. The results are therefore an expected eliminated loss of $262.50, an expected insurer payment of $337.50, and an LER of 0.4375. The check $262.50+337.50=$600.00$ agrees with $E[X]$.

Common mistake avoided: Do not count the deductible as fully eliminated only for losses exceeding it. That would give $300P(X>300)=\$225$ and omit the losses below $300, which are also eliminated in full. The correct calculation includes both $E[X\mathbf{1}_{\{X\leq300\}}]=\$37.50$ and $300P(X>300)=\$225$.

Worked Example: Exponential Severity Using the Survival Function

For a hypothetical portfolio, suppose each ground-up loss has an exponential distribution with mean $1,000, so $S_X(t)=e^{-t/1{,}000}$ for $t\geq0$. An ordinary deductible is $500. Calculate the LER and expected insurer payment. Report dollars to the cent and the ratio to four decimal places.

Method. The mean loss is $1,000. Apply the survival-function formulas to the eliminated and remaining layers:

\begin{aligned} E[\min(X,500)] &=\int_0^{500}e^{-t/1{,}000}\,dt\\ &=1{,}000(1-e^{-0.5})\\ &=393.4693\ldots,\\[4pt] E[Y^L] &=\int_{500}^\infty e^{-t/1{,}000}\,dt\\ &=1{,}000e^{-0.5}\\ &=606.5307\ldots. \end{aligned}

Thus the expected eliminated loss is $393.47, and the expected insurer payment is $606.53. The LER is $393.4693\ldots/1{,}000=0.3934693\ldots$, or 0.3935 to four decimal places. As a check, the unrounded expected amounts add to $1,000.

The deductible is half the mean, but that fact alone does not make the ratio one-half. The ratio depends on the distribution of losses around and above the deductible; here it is approximately 39.35%.

Worked Example: A Discrete Loss Distribution

A hypothetical device policy has the following ground-up loss distribution. The ordinary deductible is $400. Calculate the expected ground-up loss, expected loss eliminated, expected insurer payment, and LER. Report the ratio to four decimal places.

Loss $X$	Probability
$100	0.20
$500	0.50
$1,500	0.30

Method. Multiply each possible loss by its probability to obtain the expected ground-up loss:

$$ E[X]=100(0.20)+500(0.50)+1{,}500(0.30)=20+250+450=720. $$

For the three possible losses, the eliminated amounts $\min(X,400)$ are $100, $400, and $400. Their expectation is

E[\min(X,400)]=100(0.20)+400(0.50)+400(0.30)=20+200+120=340.

The corresponding insurer payments $(X-400)_+$ are $0, $100, and $1,100. Thus,

E[Y^L]=0(0.20)+100(0.50)+1{,}100(0.30)=0+50+330=380.

The LER is $340/720=0.472222\ldots$, or 0.4722. The eliminated and paid amounts add to $340+380=720$, as required. This example also shows that a discrete distribution is handled by averaging the contract amounts at each possible loss; no density is needed.

Common mistake avoided: At a loss of $500, the insurer pays $100, not $500 and not $400. The deductible is subtracted from each loss before computing the insurer payment.

Interpretation and Exam Traps

An LER of 0.4722 means that the deductible eliminates 47.22% of expected ground-up loss. It does not mean that 47.22% of claims are eliminated, nor does it mean that every loss is reduced by that percentage. The ratio summarizes expected dollars, so both the sizes and probabilities of losses matter.

Keep the denominator tied to the question. The denominator is $E[X]$, the expected loss before applying the deductible. If a problem supplies expected insurer payment instead, first use $E[\min(X,d)]=E[X]-E[Y^L]$, then divide the eliminated expectation by $E[X]$. Dividing by expected insurer payment would produce a different ratio, not the LER defined here.

Ignoring losses below the deductible: These losses are not paid by the insurer, but they are still part of ground-up loss and are fully eliminated by the deductible.
Using the deductible as the eliminated amount for every loss: If $X<d$, the amount eliminated is $X$, not $d$. The correct amount is $\min(X,d)$.
Confusing frequency with severity: The LER defined here compares expected loss amounts for the loss variable $X$. It is not the probability that a claim has no insurer payment.
Skipping the arithmetic check: For an ordinary deductible, expected eliminated loss plus expected insurer payment must equal $E[X]$.

This measure is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In a calculation, state the deductible payment rule, identify the ground-up mean, and keep the eliminated and paid amounts distinct until the final ratio is formed.

Key takeaway: For an ordinary deductible, the LER is $E[\min(X,d)]/E[X]$, equivalently $1-E[Y^L]/E[X]$. The identity $X=\min(X,d)+(X-d)_+$ supplies both the justification and a reliable check.

Check Your Understanding

For each hypothetical loss model, calculate the requested quantity. Answers are provided for checking.

A loss is uniform on $0 to $800, and the ordinary deductible is $200. Find the expected eliminated loss, expected insurer payment, and LER. (Answer: $175.00 eliminated; $225.00 insurer payment; LER $=0.4375$.)
A loss is exponential with mean $600, and the ordinary deductible is $300. Find the expected insurer payment and LER. Report dollars to the cent and the ratio to four decimal places. (Answer: expected insurer payment $=600e^{-0.5}=\$363.92$; LER $=1-e^{-0.5}=0.3935$.)
A loss is $100 with probability 0.10, $600 with probability 0.60, and $1,000 with probability 0.30. With an ordinary deductible of $500, find $E[X]$, expected eliminated loss, expected insurer payment, and the LER. (Answer: $E[X]=\$670$; eliminated $=\$460$; insurer payment $=\$210$; LER $=0.6866$.)
A severity model has expected ground-up loss $1,500. Under an ordinary deductible, the expected insurer payment is $975. Find the expected loss eliminated and the LER. (Answer: expected loss eliminated $=\$525$; LER $=525/1{,}500=0.3500$.)
A nonnegative loss has survival function $S_X(t)=(800/(800+t))^2$ for $t\geq0$. An ordinary deductible is $800. Use survival integrals to find the expected ground-up loss, expected eliminated loss, expected insurer payment, and LER. (Answer: $E[X]=\$800$; eliminated $=\$400$; insurer payment $=\$400$; LER $=0.5000$.)

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