Empirical Loss Elimination Ratios

Why Estimate a Loss Elimination Ratio?

A deductible reduces the amount an insurer pays on each covered loss. The loss elimination ratio summarizes the reduction as a proportion of ground-up losses: it compares the average amount removed by the deductible with the average loss before the deductible. With complete ground-up claim amounts, both averages can be estimated directly from the claims.

This estimate can also be used to calculate the insurer’s expected payment per claim. The calculation depends on having the full loss amount for each observed claim, not just the amount paid after the deductible. It assumes an ordinary deductible and, unless stated otherwise, no policy limit or coinsurance.

Definition: Let $X$ be a nonnegative ground-up loss and let $d\geq 0$ be an ordinary deductible. The amount eliminated by the deductible is $\min(X,d)$, and the insurer’s payment is $(X-d)_+=\max(X-d,0)$. The loss elimination ratio (LER) is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]}, $$ provided $0<E[X]<\infty$. For $n$ complete ground-up claims $x_1,\ldots,x_n$, the empirical LER is $$ \widehat{\operatorname{LER}}(d)= \frac{\sum_{j=1}^{n}\min(x_j,d)}{\sum_{j=1}^{n}x_j}, $$ provided the observed ground-up total is positive.

Computing the Empirical Ratio

Each claim contributes its full amount to the ground-up total and at most $d$ to the eliminated total. A loss below the deductible is eliminated in full; a loss above it contributes exactly $d$ as eliminated. The empirical LER is the ratio of these two totals. Equivalently, it is the ratio of the empirical mean eliminated amount to the empirical mean ground-up loss, since both means divide their totals by the same number of claims.

The data must be on a consistent ground-up basis: the recorded $x_j$ must be the claim amount before applying the deductible. If records contain only insurer payments, losses below the deductible may be absent and losses above it may not reveal their full amounts. Such records do not provide the complete ground-up sample required for this direct calculation.

Worked Example: From Complete Claims to Payment per Policy-Year

A hypothetical portfolio has 200 policy-years and four complete ground-up claims of $500, $1,000, $1,500, and $3,000. Estimate the LER for a $1,000 ordinary deductible, the insurer’s mean payment per claim, and its payment per policy-year. Assume no policy limit or coinsurance.

Method and ground-up total. Sum the observed claims: $$ 500+1{,}000+1{,}500+3{,}000=6{,}000. $$ The empirical ground-up mean is $6{,}000/4=\$1{,}500$ per claim.

Eliminated amounts. Apply $\min(x_j,1{,}000)$ separately to each complete claim: $$ \min(500,1{,}000)=500,\quad \min(1{,}000,1{,}000)=1{,}000,\quad \min(1{,}500,1{,}000)=1{,}000,\quad \min(3{,}000,1{,}000)=1{,}000. $$ Their sum is $500+1{,}000+1{,}000+1{,}000=\$3{,}500$. Thus $$ \widehat{\operatorname{LER}}(1{,}000)=\frac{3{,}500}{6{,}000}=0.5833 $$ to four decimal places.

Insurer payment per claim. The insurer payments are $0$, $0$, $500$, and $2{,}000$ dollars, totaling $2,500. The empirical mean payment is $2{,}500/4=\$625$ per claim.

Payment per policy-year. The observed claim frequency is $4/200=0.02$ claims per policy-year. Multiplying by the mean payment per claim gives $0.02\times \$625=\$12.50$ per policy-year. Directly, the total insurer payment is $\$2{,}500$, and $\$2{,}500/200=\$12.50$ per policy-year, which checks the result.

Common mistake avoided. Do not average the individual fractions $\min(x_j,d)/x_j$. Those fractions are $1$, $1$, $2/3$, and $1/3$, whose average is $0.75$, not the empirical LER. The requested ratio compares total eliminated dollars with total ground-up dollars, so larger claims appropriately contribute more to its denominator.

Why the LER Gives the Insurer’s Payment

The connection between the LER and the insurer’s payment follows from an identity for each individual loss. If $X\leq d$, the eliminated amount is $X$ and the insurer pays zero. If $X>d$, the eliminated amount is $d$ and the insurer pays $X-d$. In either case, the two amounts sum to the ground-up loss.

Result: Under an ordinary deductible, with finite mean ground-up loss, $$ X=\min(X,d)+(X-d)_+. $$ Taking expectations gives $E[(X-d)_+]=E[X]-E[\min(X,d)]$. Therefore, if $E[X]>0$, $$ E[(X-d)_+]=E[X]\bigl(1-\operatorname{LER}(d)\bigr). $$ For complete observations, the corresponding empirical mean insurer payment is the empirical ground-up mean multiplied by $1-\widehat{\operatorname{LER}}(d)$.

This is a short derivation rather than an extra modeling assumption: the identity holds claim by claim. It also provides a useful check. The observed total insurer payments must equal the ground-up total less the eliminated total. For positive ground-up losses, the LER lies between zero and one. Increasing the deductible cannot decrease any individual eliminated amount, so the empirical LER cannot decrease as the deductible increases.

Worked Example: A Different Deductible and an Exposure Estimate

A hypothetical insurer observes five complete ground-up claims of $200, $500, $900, $1,300, and $2,100 over 250 policy-years. For an $800 ordinary deductible, calculate the empirical LER, the mean insurer payment per claim, and the payment per policy-year. Assume no limit or coinsurance.

Method. First total the ground-up losses: $$ 200+500+900+1{,}300+2{,}100=5{,}000. $$ The empirical mean ground-up severity is $5{,}000/5=\$1{,}000$.

The eliminated amounts are $200$, $500$, $800$, $800$, and $800$, totaling $200+500+800+800+800=\$3{,}100$. Hence $$ \widehat{\operatorname{LER}}(800)=\frac{3{,}100}{5{,}000}=0.6200. $$

The retained insurer payments are $0$, $0$, $100$, $500$, and $1{,}300$, totaling $1,900. The mean payment is $1{,}900/5=\$380$ per claim. The identity gives the same answer: $\$1{,}000(1-0.6200)=\$380$.

There were five claims in 250 policy-years, so the observed frequency is $5/250=0.02$ claims per policy-year. The implied payment is $0.02\times \$380=\$7.60$ per policy-year. Directly, $\$1{,}900/250=\$7.60$, confirming the calculation.

Common mistake avoided. The deductible is applied to each claim before totals are formed. It is not a single $800 reduction from the combined portfolio loss: the eliminated amount is capped at $800 for each claim.

Comparing Deductibles on the Same Sample

When the complete claim amounts are available, the same ground-up total can be used to compare candidate deductibles. The denominator of the empirical LER stays fixed, while each claim’s eliminated amount changes with the deductible. The resulting comparison describes the claim sample; it does not by itself account for expenses, behavioral responses, or changes in the mix of insured risks.

Worked Example: Checking How the Ratio Changes with the Deductible

A hypothetical risk pool has five complete ground-up losses of $300, $700, $1,200, $2,800, and $5,000. Compare the empirical LER and mean insurer payment per claim at deductibles of $1,000 and $2,000. Assume no policy limit or coinsurance.

Common denominator. The total ground-up loss is $$ 300+700+1{,}200+2{,}800+5{,}000=10{,}000, $$ so the empirical ground-up mean is $10{,}000/5=\$2{,}000$.

$1,000 deductible. The eliminated amounts are $300$, $700$, $1{,}000$, $1{,}000$, and $1{,}000$, totaling $4,000. Thus the LER is $4{,}000/10{,}000=0.4000$. Insurer payments are $0$, $0$, $200$, $1{,}800$, and $4{,}000$, totaling $6,000, or $6{,}000/5=\$1{,}200$ per claim.

$2,000 deductible. The eliminated amounts are $300$, $700$, $1{,}200$, $2{,}000$, and $2{,}000$, totaling $6,200. Thus the LER is $6{,}200/10{,}000=0.6200$. Insurer payments are $0$, $0$, $0$, $800$, and $3{,}000$, totaling $3,800, or $3{,}800/5=\$760$ per claim.

The LER increases from 0.4000 to 0.6200 as the deductible rises, while the mean insurer payment decreases from $1,200 to $760. The check from the identity gives $\$2{,}000(1-0.4000)=\$1{,}200$ and $\$2{,}000(1-0.6200)=\$760$, respectively.

Common mistake avoided. The $2,000 deductible does not eliminate $2,000 from every claim. For losses of $300, $700, and $1,200, the eliminated amount is the loss itself, not the full deductible.

Conditions, Interpretation, and Common Traps

The direct empirical ratio is most useful when every observed claim has its full ground-up amount and the claims represent the severity experience of interest. It is a ratio of totals, not generally the average of claim-level ratios. If observations have explicitly assigned weights, use the intended weighted empirical distribution consistently in both numerator and denominator; do not introduce weights merely because claim amounts differ.

The estimate is undefined if the observed ground-up total is zero. If there are no claims in an exposure period, that period alone supplies no claim-severity sample from which to estimate an LER, although its claim frequency is zero for that observed period. The LER concerns the severity payment conditional on the claim amounts used; a payment estimate per policy-year also needs the relevant claim frequency.

Using payments as if they were ground-up losses. A payment after a deductible is not the original claim amount. Applying a deductible to it again does not estimate the ground-up LER.
Using the wrong denominator. The denominator is the sum of complete ground-up losses, not the number of claims or the sum of insurer payments.
Mixing per-claim and per-exposure quantities. A mean insurer payment per claim is not a payment per policy-year. To convert, multiply by claims per policy-year, using the matching exposure basis.
Ignoring coverage terms. The identity here is for an ordinary deductible without a limit or coinsurance. Additional payment modifications must be included in the claim payment calculation rather than silently treated as part of this formula.

Key takeaway: With complete ground-up claims, divide the sum of $\min(x_j,d)$ by the sum of $x_j$ to estimate the LER. The empirical mean insurer payment per claim is the ground-up mean multiplied by one minus that ratio; a per-exposure estimate additionally requires claim frequency.

Check Your Understanding

For each problem, assume an ordinary deductible and no policy limit or coinsurance unless the problem states otherwise.

A hypothetical sample contains complete claims of $400, $1,200, and $2,400. At a $1,000 deductible, find the empirical LER and mean insurer payment per claim. (Answer: LER $=0.6000$; mean payment $=\$533.33$ per claim.)
Six complete claims are $100, $300, $600, $900, $1,500, and $2,600. At a $700 deductible, find the LER and mean insurer payment per claim. (Answer: LER $=0.5167$; mean payment $=\$483.33$ per claim.)
Eight complete claims over 400 policy-years have a ground-up total of $12,000 and eliminated amounts totaling $3,000 at the selected deductible. Find the LER, mean insurer payment per claim, and insurer payment per policy-year. (Answer: LER $=0.2500$; mean payment $=\$1,125$ per claim; payment $=\$22.50$ per policy-year.)
Complete claims are $500, $1,000, and $2,500, with a $500 deductible. Find the LER and mean insurer payment per claim. (Answer: LER $=0.3750$; mean payment $=\$833.33$ per claim.)
For claims of $200, $800, $2,000, and $4,000, use a $1,000 deductible. Compare the ratio-of-totals LER with the average of the four individual fractions $\min(x_j,d)/x_j$. (Answer: ratio-of-totals LER $=0.4286$; average individual fraction $=0.6875$; use $0.4286$ as the empirical LER.)

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