LER from Deductible-Truncated Loss Data

Why Truncated Claims Need a Ground-Up Model

An ordinary deductible prevents the insurer from paying the first $c$ dollars of each loss. A database containing only claims with positive insurance payments therefore omits losses that do not exceed $c$. But a database of reported claims may include those losses, depending on its reporting and recording rules. The distinction matters: positive-payment observations are conditional on exceeding the deductible, and that conditional distribution does not generally reveal the severity distribution of all losses.

The loss elimination ratio (LER) measures the expected amount eliminated by a deductible as a proportion of expected ground-up loss. Calculating it from truncated observations requires care. The missing portion below the observation threshold must be supplied by an explicit model or other information; it cannot simply be treated as zero.

Definition: Let $X\geq 0$ be ground-up loss, with $0<E[X]<\infty$. Under an ordinary deductible $d$, the insurer payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$, and the amount eliminated is $\min(X,d)$. The ground-up loss elimination ratio is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]} =1-\frac{E[(X-d)_+]}{E[X]}. $$

The expectation in the denominator is over all ground-up losses, including losses below the deductible. If observations are restricted to $X>c$, they describe the conditional distribution of loss given that it exceeds $c$. They do not directly give $E[X]$, or the expected eliminated amount across all losses.

What Truncation Does—and Does Not—Tell Us

Suppose a database records ground-up amounts only when $X>c$. For a continuous severity distribution, the observed density is the conditional density $f_X(x\mid X>c)=f_X(x)/S_X(c)$ for $x>c$, where $S_X(c)=P(X>c)$. Equivalently, the conditional survival function above the threshold is $S_X(x)/S_X(c)$. The data describe the shape of the tail relative to its value at $c$; they do not reveal $S_X(c)$ or the distribution below $c$ without further assumptions.

If the database contains only payment amounts, a positive payment under deductible $c$ is $X-c$, not $X$. When $X>c$, the ground-up amount can be recovered by adding $c$ to the recorded payment. That adjustment recovers the observed ground-up severity for positive-payment claims, but it does not restore losses that were never included in the database.

For discrete losses, the threshold condition must match the contract and data definition. For example, a positive payment under an ordinary deductible occurs when $X>c$; losses equal to $c$ produce zero payment. With a continuous model, the probability of exact equality is zero, but that is not necessarily true for a discrete model.

Key result: For a nonnegative loss with finite mean, $$ E[\min(X,d)]=\int_0^d S_X(x)\,dx, \qquad E[(X-d)_+]=\int_d^\infty S_X(x)\,dx. $$ Thus LER depends on the ground-up survival function, including values below an observation threshold. Tail-only data do not generally determine the ground-up LER unless a model or additional data supply the missing information.

The integral identity follows by writing a nonnegative random variable $Z$ as the area under its indicator: $Z=\int_0^\infty \mathbf{1}(Z>t)\,dt$. Taking expectations gives $E[Z]=\int_0^\infty P(Z>t)\,dt$. For $Z=\min(X,d)$, the event $Z>t$ is $X>t$ for $0\leq t<d$, and is impossible for $t\geq d$. For $Z=(X-d)_+$, the event $Z>t$ is $X>d+t$. Integrating these probabilities gives the two identities.

Identify the observation rule.
Determine whether the data contain all reported losses, only losses above a threshold, or positive payment amounts. If amounts are payments, distinguish those from ground-up severities.

Specify the ground-up model.
State the model and its parameters, including how they are known or supplied. Do not substitute the conditional sample mean for the ground-up mean.

Calculate both expectations.
Find $E[X]$ and $E[\min(X,d)]$, or find $E[(X-d)_+]$ and use the equivalent LER formula.

Form the ratio and check it.
Divide eliminated expected loss by ground-up expected loss. The result must be between 0 and 1 for a nonnegative deductible and positive finite mean.

Worked Calculations Under Stated Models

In each example, the severity model is explicitly supplied. This is the information that permits the ground-up calculation despite truncation. In an estimation problem, the likelihood must also reflect the observation rule: for continuous observations known only to exceed $c$, each observed severity contributes a density proportional to $f_X(x)/S_X(c)$, rather than the untruncated density $f_X(x)$.

Worked Example: Exponential Losses Above a Deductible

A hypothetical portfolio has ground-up losses $X\sim\operatorname{Exponential}(\lambda)$, with $\lambda=0.001$ per dollar. The database contains only positive-payment claims under an ordinary deductible of $c=\$500$, and records their ground-up severities. Calculate the ground-up LER for a deductible of $d=\$500$. Also check the calculation using the expected insurer payment.

Method and observation interpretation. The exponential model is specified for all ground-up losses, not just those in the database. Its mean is $E[X]=1/\lambda=\$1{,}000$. The observed claims are conditional on $X>\$500$. Under the exponential model, the memoryless property gives $E[X\mid X>500]=500+1/\lambda=\$1{,}500$, but this conditional mean is not the ground-up mean.

For an exponential severity, $S_X(x)=e^{-\lambda x}$. The expected insurer payment is $E[(X-d)_+]=\int_d^\infty e^{-\lambda x}\,dx=e^{-\lambda d}/\lambda$. At $d=\$500$, $\lambda d=0.001(500)=0.5$, so $E[(X-500)_+]=e^{-0.5}/0.001=\$606.5307$.

The expected amount eliminated is $E[\min(X,500)]=E[X]-E[(X-500)_+]=1{,}000-606.5307=\$393.4693$. Therefore, $\operatorname{LER}(500)=393.4693/1{,}000=0.3934693$, or 0.3935 to four decimals.

Check. The equivalent formula is $1-E[(X-500)_+]/E[X]=1-606.5307/1{,}000=0.3934693$, matching the direct calculation. The common mistake is to use the observed conditional mean, $\$1{,}500$, as $E[X]$. Doing so would use the wrong denominator because it excludes all losses at or below $\$500$.

Worked Example: A Discrete Severity With Omitted Small Losses

A hypothetical model assigns ground-up loss probabilities $P(X=200)=0.20$, $P(X=600)=0.50$, and $P(X=1{,}200)=0.30$, with amounts in dollars. A database includes only positive-payment claims under an ordinary deductible of $\$500$. The observed tail frequencies are 62.5% at $\$600$ and 37.5% at $\$1{,}200$. Calculate the ground-up LER for the $\$500$ deductible.

Method. First verify what the truncated frequencies mean. The probability of exceeding $\$500$ is $0.50+0.30=0.80$. Conditional on exceeding the deductible, the probabilities are $0.50/0.80=0.625$ at $\$600$ and $0.30/0.80=0.375$ at $\$1{,}200$, as observed. These conditional frequencies alone do not show that 80% of all losses exceed the deductible; that probability comes from the stated ground-up model.

The ground-up mean is $E[X]=0.20(200)+0.50(600)+0.30(1{,}200)=40+300+360=\$700$.

The deductible eliminates $\min(X,500)$ for each loss. Its expected amount is $E[\min(X,500)]=0.20(200)+0.50(500)+0.30(500)=40+250+150=\$440$. Thus $\operatorname{LER}(500)=440/700=0.628571$, or 0.6286 to four decimals.

Check. The expected insurer payment is $0.20(0)+0.50(600-500)+0.30(1{,}200-500)=0+50+210=\$260$. Since $440+260=700$, the eliminated and paid amounts sum to the ground-up mean. A common error is to calculate a ratio using only the conditional tail frequencies; that ignores the omitted $\$200$ losses and changes the weighting of the entire portfolio.

Worked Example: Pareto-Type Losses With a Known Ground-Up Model

Suppose a hypothetical portfolio has a Lomax (Pareto Type II) ground-up severity with survival function $S_X(x)=(\theta/(\theta+x))^\alpha$, where $x\geq0$, $\theta=\$1{,}000$, and $\alpha=2$. A positive-payment database under an ordinary deductible of $c=\$500$ contains only severities above that threshold. Calculate the ground-up LER for a deductible of $d=\$1{,}000$.

Method. Use the supplied ground-up survival function to calculate both the mean and the expected eliminated amount. The model has $\alpha>1$, so its mean exists and equals $E[X]=\theta/(\alpha-1)=\$1{,}000$.

For the target deductible, $E[\min(X,1{,}000)]=\int_0^{1{,}000}(1{,}000/(1{,}000+x))^2\,dx$. An antiderivative is $-1{,}000^2/(1{,}000+x)$. Evaluating at the endpoints gives $-1{,}000^2/2{,}000-(-1{,}000^2/1{,}000)=-500+1{,}000=\$500$.

Therefore, $\operatorname{LER}(1{,}000)=500/1{,}000=\mathbf{0.5000}$. As a check, the expected payment is the remaining tail integral: $\int_{1{,}000}^{\infty}(1{,}000/(1{,}000+x))^2\,dx=\$500$. The eliminated and paid expectations sum to $\$1{,}000$.

The observation threshold is $\$500$, but the target deductible is $\$1{,}000$; they play different roles. The threshold describes which claims were observed, while $d$ specifies the coverage whose LER is required. The common mistake is to calculate only from the tail above $\$500$, omitting the modeled ground-up losses below it.

Interpreting Results and Avoiding Identification Errors

A stated severity model makes the calculation possible, but it does not turn truncated observations into direct observations of the missing losses. Any estimate of ground-up LER then depends on how well the model represents the unobserved portion. If the model is fitted from truncated data, parameter identifiability and model fit must be assessed under the truncation mechanism; a conditional sample average is not a substitute for that work.

For example, the conditional distribution above $c$ may determine a parameter in a particular parametric family, but that depends on the family and the observation process. Other models can share the same conditional tail while assigning different probabilities and severities below $c$. Those models can have different ground-up means and therefore different LERs. The ratio is not identified from the tail alone in general.

What the data contain	What can be used directly	What still needs care
All recorded ground-up losses, including losses at or below the deductible	Ground-up observations, subject to the recording rules	Check whether any other reporting or sampling restrictions apply
Ground-up severities only for claims above $c$	The conditional tail distribution given $X>c$	The probability of exceeding $c$ and the severity below $c$ may be missing
Positive insurer payment amounts above $c$	For each positive payment, recover its ground-up loss by adding $c$	Losses with zero payment remain absent if the database includes only positive payments

A practical check is to compare the expected eliminated amount and expected insurer payment with the ground-up mean. Under an ordinary deductible, $\min(X,d)+(X-d)_+=X$ for every $X\geq0$, so their expectations must sum to $E[X]$. For a finite positive mean, the LER must lie in $[0,1]$. A result outside that range signals a likely error in the model, units, threshold, or calculation.

Key takeaway: A positive-payment database under an ordinary deductible is a truncated view of ground-up severity, not the full loss distribution. Calculate ground-up LER only after stating how the missing ground-up severity is supplied; then use the full-model expectations, not tail-only averages.

Check Your Understanding

For each problem, distinguish the observation threshold from the target deductible, and use the stated ground-up model when calculating LER.

Ground-up losses are exponential with mean $\$2{,}000$. A database records only positive-payment claims above a $\$400$ ordinary deductible. Find the ground-up LER at a target deductible of $\$800$. (Answer: $1-e^{-0.4}=0.3297$)
A discrete model has $P(X=100)=0.25$, $P(X=500)=0.45$, and $P(X=900)=0.30$. Only positive-payment claims above $\$400$ are observed, and the target deductible is $\$400$. Calculate the ground-up LER. (Answer: $0.6250$)
Ground-up loss is uniform on $[0,\$3{,}000]$. A database includes only severities above $\$600$. Find the ground-up LER at a target deductible of $\$1{,}200$. (Answer: $0.6400$)
A Lomax severity has survival function $S_X(x)=(600/(600+x))^3$, for $x\geq0$. Data are observed only above $\$300$. Find the ground-up LER for a $\$600$ deductible. (Answer: $0.7500$)
Under a supplied exponential model, positive payment amounts above an ordinary deductible of $\$250$ have mean $\$750$. Use the exponential memoryless property to find the ground-up LER for a target deductible of $\$500$. (Answer: $1-e^{-2/3}=0.4866$)

What You'll Learn