Deriving LER from Ground-Up and Insurer Losses

Why Compare Ground-Up Loss with Insurer Payment?

A deductible reduces the amount an insurer pays on each covered loss. To measure the reduction, compare the expected loss before the deductible with the expected payment after it. The loss elimination ratio (LER) expresses the expected amount removed by the coverage modification as a proportion of expected ground-up loss.

The key is to define the payment on an individual loss first, then take its expectation. The deductible applies to each loss, not to the average loss. This distinction matters for discrete severities, distributions with probability mass at the deductible, and any calculation where the payment function is nonlinear.

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

This definition concerns expected loss amounts, not the probability that a claim is below the deductible. It uses a per-loss severity model: $X$ describes the covered loss before the deductible, and $Y^L$ is the corresponding insurer payment. The ground-up loss and payment must be measured in the same currency and on the same price basis.

The Ordinary-Deductible Identity

For an ordinary deductible, the insured bears the portion of a loss up to $d$, while the insurer pays the excess above $d$. Thus the eliminated amount on one loss is never more than the loss itself and never more than the deductible.

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

To see why, consider the two possible ranges for a particular loss $x$. If $x\leq d$, the insurer pays zero, so the eliminated amount is $x$. If $x>d$, the insurer pays $x-d$, so the eliminated amount is $d$. In both cases the eliminated amount is $\min(x,d)$. Taking expectations gives the first form of the identity; rearranging gives the form that directly connects the LER to expected insurer payment.

Theorem: Under an ordinary per-loss deductible, if $0<E[X]<\infty$, then $$ \operatorname{LER}(d) =\frac{E[X-Y^L]}{E[X]} =\frac{E[\min(X,d)]}{E[X]} =1-\frac{E[Y^L]}{E[X]}. $$ Equivalently, $$ E[Y^L]=E[X]\bigl(1-\operatorname{LER}(d)\bigr). $$

The result does not require a continuous severity distribution. It holds for discrete, continuous, or mixed distributions because the pointwise payment identity holds for every nonnegative loss value. In particular, a loss exactly equal to the deductible has zero insurer payment and contributes its full amount to the eliminated loss.

Specify the loss and coverage.
Define the ground-up loss $X$, the ordinary deductible $d$, and the per-loss payment $Y^L=(X-d)_+$.

Find the expected ground-up loss.
Calculate $E[X]$ from the loss model. This is the denominator of the LER.

Find the expected insurer payment.
Apply the deductible to each possible loss, then average the resulting payments to obtain $E[Y^L]$.

Calculate and check the ratio.
Use $1-E[Y^L]/E[X]$, or calculate expected eliminated loss directly. The two answers must agree.

Worked Applications

Worked Example: Calculating LER from a Discrete Severity Distribution

A hypothetical insurer covers individual losses $X$ under an ordinary deductible of $d=\$1{,}000$. The loss distribution is $P(X=\$0)=0.10$, $P(X=\$1{,}000)=0.30$, $P(X=\$2{,}500)=0.40$, and $P(X=\$5{,}000)=0.20$. Find the expected ground-up loss, expected insurer payment, and LER. No interest or mortality model is involved.

Method. Apply the ordinary-deductible payment function to each possible loss, calculate the two expectations, and use the identity. The probabilities sum to $0.10+0.30+0.40+0.20=1.00$.

Ground-up expectation. $$ E[X]=0(0.10)+1{,}000(0.30)+2{,}500(0.40)+5{,}000(0.20) =0+300+1{,}000+1{,}000=\$2{,}300. $$

Payment at each loss value. The payments are $0$, $0$, $2{,}500-1{,}000=\$1{,}500$, and $5{,}000-1{,}000=\$4{,}000$, respectively. Therefore, $$ E[Y^L]=0(0.10)+0(0.30)+1{,}500(0.40)+4{,}000(0.20) =0+0+600+800=\$1{,}400. $$

LER and check. $$ \operatorname{LER}(1{,}000)=1-\frac{1{,}400}{2{,}300} =\frac{900}{2{,}300}\approx 0.3913. $$ Directly, the eliminated amounts are $0$, $\$1{,}000$, $\$1{,}000$, and $\$1{,}000$, giving $0(0.10)+1{,}000(0.30)+1{,}000(0.40)+1{,}000(0.20)=\$900$. Thus $\$2{,}300-\$1{,}400=\$900$, as required. The LER is $0.3913$, or 39.13%, rounded to four decimal places. A common mistake is to count the probability of losses at or below the deductible instead of measuring the expected amount eliminated; the direct check uses amounts, not probabilities alone.

Worked Example: Applying the Deductible Before Averaging

A hypothetical repair portfolio has per-claim severity $X$ with $P(X=\$0)=0.20$, $P(X=\$500)=0.30$, $P(X=\$2{,}000)=0.30$, and $P(X=\$5{,}000)=0.20$. Each claim has an ordinary deductible of $d=\$1{,}000$. Calculate the LER and assess the tempting shortcut of applying the deductible to the mean loss. No interest or mortality model is involved.

Ground-up mean. The probabilities sum to one, and $$ E[X]=0(0.20)+500(0.30)+2{,}000(0.30)+5{,}000(0.20) =0+150+600+1{,}000=\$1{,}750. $$

Expected payment per claim. The deductible is applied to each possible claim: payments are $0$, $0$, $\$1{,}000$, and $\$4{,}000$. Hence $$ E[Y^L]=0(0.20)+0(0.30)+1{,}000(0.30)+4{,}000(0.20) =0+0+300+800=\$1{,}100. $$ It follows that $$ \operatorname{LER}(1{,}000)=1-\frac{1{,}100}{1{,}750} =\frac{650}{1{,}750}\approx 0.3714. $$

Check and shortcut. The eliminated amounts are $0$, $\$500$, $\$1{,}000$, and $\$1{,}000$. Their expectation is $0+150+300+200=\$650$, matching $\$1{,}750-\$1{,}100$. The incorrect shortcut gives $(E[X]-d)_+=(1{,}750-1{,}000)=\$750$ as an alleged expected insurer payment. That is not $E[(X-d)_+]$: the deductible acts on each claim before averaging. The correct LER is $0.3714$, rounded to four decimal places.

Worked Example: Keeping an Inflation-Adjusted Loss on One Price Basis

A hypothetical severity model has base-year losses $X_0=\$1{,}000$, $\$3{,}000$, and $\$6{,}000$, with probabilities $0.20$, $0.50$, and $0.30$. Losses are trended by a deterministic factor of $1.10$, and the ordinary deductible is a fixed $\$2{,}500$ in the trended price basis. Find the trended expected ground-up loss, expected insurer payment, and LER. No interest or mortality model is involved.

Trend the possible losses first. The current-basis losses are $1.10(\$1{,}000)=\$1{,}100$, $1.10(\$3{,}000)=\$3{,}300$, and $1.10(\$6{,}000)=\$6{,}600$. The ground-up expectation is $$ E[X]=1{,}100(0.20)+3{,}300(0.50)+6{,}600(0.30) =220+1{,}650+1{,}980=\$3{,}850. $$

Apply the deductible in that same basis. The insurer payments are $0$, $\$3{,}300-\$2{,}500=\$800$, and $\$6{,}600-\$2{,}500=\$4{,}100$. Thus $$ E[Y^L]=0(0.20)+800(0.50)+4{,}100(0.30) =0+400+1{,}230=\$1{,}630. $$

Calculate and verify. $$ \operatorname{LER}=1-\frac{1{,}630}{3{,}850} =\frac{2{,}220}{3{,}850}\approx 0.5766. $$ The eliminated amounts are $\$1{,}100$, $\$2{,}500$, and $\$2{,}500$, so their expected value is $220+1{,}250+750=\$2{,}220$. This agrees with $\$3{,}850-\$1{,}630$. The LER is 0.5766, or 57.66%, rounded to four decimal places. The common mistake is to trend the loss but compare it with a deductible expressed in a different price basis; both must be on the same basis before calculating the ratio.

Assumptions, Scope, and Exam Traps

The identity depends on the ordinary deductible payment function. For every loss $x\geq 0$, the insurer pays $(x-d)_+$. Under that contract, the eliminated portion is $\min(x,d)$, which is why the expectation can be expressed either as expected eliminated loss or as the difference between ground-up and insurer expected losses.

The identity does not apply unchanged if the coverage has a different payment function. For example, a franchise deductible may pay the full loss once the loss exceeds the threshold, rather than paying only the excess above it. A policy limit or coinsurance also changes the payment function. In those cases, define the actual payment per loss and compare its expectation with $E[X]$; do not assume the ordinary-deductible expression $Y^L=(X-d)_+$.

A further condition is $0<E[X]<\infty$. If $E[X]=0$, the ratio has a zero denominator and is undefined. If expected ground-up loss is infinite, this finite-expectation identity cannot be used as a ratio of finite expected amounts. For the ordinary deductible with a finite expected loss, the identity is valid whether the severity distribution is discrete, continuous, or mixed.

Key takeaway: For an ordinary per-loss deductible, expected ground-up loss equals expected insurer payment plus expected eliminated loss. Divide the eliminated amount by expected ground-up loss to obtain LER, or calculate it as one minus the expected insurer payment divided by expected ground-up loss.

Check Your Understanding

For each ordinary deductible problem, work per loss before taking expectations. Unless stated otherwise, no interest or mortality model is involved. Round ratios to four decimal places.

A hypothetical loss has values $\$0$, $\$1{,}000$, and $\$3{,}000$ with probabilities $0.25$, $0.25$, and $0.50$. With an ordinary deductible of $\$1{,}000$, find $E[X]$, $E[Y^L]$, and the LER. (Answer: $E[X]=\$1{,}750$, $E[Y^L]=\$1{,}000$, LER $=0.4286$.)
A hypothetical claim severity is $\$400$, $\$1{,}400$, or $\$3{,}400$, with probabilities $0.40$, $0.40$, and $0.20$. An ordinary deductible is $\$900$. Find the LER and verify it from expected eliminated loss. (Answer: $E[X]=\$1{,}400$, $E[Y^L]=\$700$, expected eliminated loss $=\$700$, LER $=0.5000$.)
A hypothetical severity has values $\$0$, $\$1{,}000$, and $\$3{,}000$, with probabilities $0.20$, $0.30$, and $0.50$. Under an ordinary $\$500$ deductible, calculate the expected payment and LER. (Answer: $E[X]=\$1{,}800$, $E[Y^L]=\$1{,}400$, LER $=0.2222$.)
Base-basis losses are $\$1{,}000$ and $\$4{,}000$, each with probability $0.50$. Trend both by a factor of $1.20$, then apply a fixed ordinary deductible of $\$2{,}500$ in the trended basis. Find the LER. (Answer: trended $E[X]=\$3{,}000$, $E[Y^L]=\$1{,}150$, expected eliminated loss $=\$1{,}850$, LER $=0.6167$.)
A loss is $\$500$ or $\$1{,}500$, each with probability $0.50$, and the threshold is $\$1{,}000$. Find the LER under an ordinary deductible, then explain why it changes if the contract instead uses a franchise deductible that pays the full loss only when it exceeds the threshold. (Answer: ordinary deductible $E[X]=\$1{,}000$, $E[Y^L]=\$250$, LER $=0.7500$; franchise expected payment $=\$750$, giving an eliminated-loss ratio of $0.2500$, so the ordinary-deductible identity does not apply to that payment function.)

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