LER for Discrete and Mixed Severities

Why Discrete and Mixed Severities Need Care

A deductible eliminates part of each ground-up loss, but how much it eliminates depends on the severity distribution. With discrete severity, losses occur at particular amounts; with a mixed severity, some probability may be concentrated at exact loss amounts while the rest is spread continuously. In either case, the loss elimination ratio (LER) is a ratio of expected dollars eliminated to expected ground-up dollars. It is not the probability that a loss is below the deductible.

The calculation is especially sensitive to the denominator and to the distinction between a point mass and a continuous component. The method below works directly from probability-weighted losses and applies to an ordinary per-loss deductible, with no policy limit or coinsurance.

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 amount eliminated from the ground-up loss is $X-Y^L=\min(X,d)$. The loss elimination ratio is $$ \operatorname{LER}(d)=\frac{E[\min(X,d)]}{E[X]}. $$

Discrete Severities

Suppose $X$ takes the distinct nonnegative values $x_1,\ldots,x_n$, with $P(X=x_i)=p_i$, where $\sum_{i=1}^{n}p_i=1$. Each outcome contributes its probability multiplied by its ground-up loss to $E[X]$, and its probability multiplied by the smaller of the loss and deductible to the expected eliminated amount.

E[X]=\sum_{i=1}^{n}p_i x_i,\qquad E[\min(X,d)]=\sum_{i=1}^{n}p_i\min(x_i,d),\qquad \operatorname{LER}(d)=\frac{\sum_{i=1}^{n}p_i\min(x_i,d)}{\sum_{i=1}^{n}p_i x_i}.

This sums over all possible losses, including zero losses if they have positive probability. A loss exactly equal to $d$ contributes $d$ to the eliminated amount and zero to the insurer payment. The probability of a loss below the deductible, by itself, is not enough to calculate LER: losses below $d$ can differ substantially in size.

Worked Example: A Four-Point Severity Distribution

A hypothetical coverage has ground-up loss $X$ with the following distribution, in dollars: $P(X=0)=0.10$, $P(X=1{,}000)=0.30$, $P(X=3{,}000)=0.40$, and $P(X=8{,}000)=0.20$. The policy has an ordinary per-loss deductible of $d=\$2{,}000$, with no limit. Calculate the LER and check it using expected insurer payments.

Method. Weight each severity by its probability to find the ground-up mean; then weight each severity capped at the deductible to find expected eliminated loss.

First, the probabilities sum to $0.10+0.30+0.40+0.20=1.00$. The ground-up expected loss is

E[X]=0.10(0)+0.30(1{,}000)+0.40(3{,}000)+0.20(8{,}000) =0+300+1{,}200+1{,}600=\$3{,}100.

The eliminated amount at each possible severity is $\min(x,2{,}000)$. Therefore,

E[\min(X,2{,}000)] =0.10(0)+0.30(1{,}000)+0.40(2{,}000)+0.20(2{,}000) =0+300+800+400=\$1{,}500.

Thus,

\operatorname{LER}(2{,}000)=\frac{1{,}500}{3{,}100}=0.483870\ldots\approx 0.4839.

As a check, expected insurer payment is $0.10(0)+0.30(0)+0.40(1{,}000)+0.20(6{,}000)=\$1{,}600$. Adding the expected eliminated amount gives $\$1{,}500+\$1{,}600=\$3{,}100=E[X]$. The LER is $0.4839$, or 48.39% to two decimal places. A common mistake is to divide by expected insurer payment, producing $1{,}500/1{,}600$; the denominator must be the ground-up expected loss.

Mixed Distributions: Point Masses and a Continuous Part

A mixed severity distribution assigns positive probability to one or more exact values and also has a continuous component. To calculate expectations, keep the point-mass contributions separate from the continuous contribution. One convenient notation is to let $x_j$ be the atom locations, with $P(X=x_j)=p_j$, and let $g(x)$ be the subdensity of the continuous component. A subdensity integrates to the probability assigned to that component; it does not need to integrate to one.

E[X]=\sum_j p_jx_j+\int_0^\infty xg(x)\,dx, \qquad E[\min(X,d)]=\sum_jp_j\min(x_j,d)+\int_0^\infty\min(x,d)g(x)\,dx.

For $d\geq0$, the continuous contribution to eliminated loss can be split at the deductible:

\int_0^\infty\min(x,d)g(x)\,dx =\int_0^d xg(x)\,dx+d\int_d^\infty g(x)\,dx.

Below the deductible, the eliminated amount equals the loss $x$; above it, the eliminated amount is $d$. If $X$ has an atom exactly at $d$, its contribution is handled in the sum over atoms: the eliminated amount is $d$. The continuous integral is unaffected by whether its endpoint includes $d$, since a single point has zero probability under a continuous component.

Derivation: For each realized loss $x\geq0$, the insurer pays $(x-d)_+$. If $x\leq d$, then $x-(x-d)_+=x-0=x=\min(x,d)$. If $x>d$, then $x-(x-d)_+=x-(x-d)=d=\min(x,d)$. Thus $X=(X-d)_++\min(X,d)$ for every outcome. Taking expectations gives $E[X]=E[(X-d)_+]+E[\min(X,d)]$, which justifies both the eliminated-loss numerator and the insurer-payment check.

Worked Example: An Atom at Zero and a Uniform Component

A hypothetical ground-up loss has probability $0.20$ of being zero, probability $0.30$ of being exactly $\$4{,}000$, and probability $0.50$ of coming from a uniform distribution on $[0,\$10{,}000]$. Conditional on being in that continuous component, every loss in the interval is equally likely. An ordinary deductible is $d=\$3{,}000$, with no limit. Calculate the LER.

Method. Treat the uniform part as a conditional distribution with weight $0.50$. Its conditional mean is $(0+10{,}000)/2=\$5{,}000$; multiply that mean by $0.50$ when calculating the overall expectation.

The overall ground-up mean is

E[X]=0.20(0)+0.30(4{,}000)+0.50(5{,}000) =0+1{,}200+2{,}500=\$3{,}700.

For the uniform component, the conditional expected eliminated loss is the area-weighted average of $x$ from 0 to 3,000 and the deductible amount 3,000 from 3,000 to 10,000:

E[\min(U,3{,}000)] =\frac{1}{10{,}000}\int_0^{3{,}000}x\,dx +\frac{1}{10{,}000}\int_{3{,}000}^{10{,}000}3{,}000\,dx =\frac{4{,}500{,}000}{10{,}000} +\frac{21{,}000{,}000}{10{,}000} =\$2{,}550.

The continuous component has overall probability $0.50$, so its contribution to expected eliminated loss is $0.50(2{,}550)=\$1{,}275$. The atom at zero contributes $0.20(0)=\$0$, and the atom at $\$4{,}000$ contributes $0.30(3{,}000)=\$900$. Hence,

E[\min(X,3{,}000)]=0+900+1{,}275=\$2{,}175,\qquad \operatorname{LER}(3{,}000)=\frac{2{,}175}{3{,}700}=0.587837\ldots\approx0.5878.

The insurer-payment check is $\$3{,}700-\$2{,}175=\$1{,}525$. Directly, the $\$4{,}000$ atom contributes $0.30(1{,}000)=\$300$. Conditional on the uniform component, expected payment is

\frac{1}{10{,}000}\int_{3{,}000}^{10{,}000}(x-3{,}000)\,dx =\frac{24{,}500{,}000}{10{,}000}=\$2{,}450,

so its overall contribution is $0.50(2{,}450)=\$1{,}225$, and total expected insurer payment is $\$300+\$1{,}225=\$1{,}525$. The check agrees. The reported LER is $0.5878$, or 58.78% to two decimal places.

A Second Mixed Model and a Boundary Check

The same method applies when the continuous component is uniform on an interval that does not start at zero, and when a point mass lies exactly at the deductible. The atom at the deductible is not an insurer payment: the payment is $(d-d)_+=0$. Its entire loss is eliminated.

Worked Example: A Deductible-Boundary Atom

A hypothetical severity has probability $0.25$ of being exactly $\$2{,}000$, probability $0.25$ of being exactly $\$6{,}000$, and probability $0.50$ of coming from a uniform distribution on $[\$1{,}000,\$5{,}000]$. The deductible is $d=\$2{,}000$. Find the LER and verify the result using expected insurer payments.

Method. Calculate each atom’s contribution directly. For the uniform component, integrate the loss from $\$1{,}000$ to $\$2{,}000$, then use the deductible for losses above $\$2{,}000$.

The conditional mean of the uniform component is $(1{,}000+5{,}000)/2=\$3{,}000$. Therefore,

E[X]=0.25(2{,}000)+0.25(6{,}000)+0.50(3{,}000) =500+1{,}500+1{,}500=\$3{,}500.

The atom at $\$2{,}000$ contributes $0.25(2{,}000)=\$500$ to expected eliminated loss. The atom at $\$6{,}000$ contributes $0.25(2{,}000)=\$500$. For the conditional uniform component, whose interval width is $\$4{,}000$,

E[\min(U,2{,}000)] =\frac{1}{4{,}000}\int_{1{,}000}^{2{,}000}x\,dx +\frac{1}{4{,}000}\int_{2{,}000}^{5{,}000}2{,}000\,dx =\frac{1{,}500{,}000}{4{,}000}+\frac{6{,}000{,}000}{4{,}000} =\$1{,}875.

Weighting this conditional expectation by the component probability gives $0.50(1{,}875)=\$937.50$. Thus expected eliminated loss is $\$500+\$500+\$937.50=\$1{,}937.50$, and

\operatorname{LER}(2{,}000)=\frac{1{,}937.50}{3{,}500} =0.553571\ldots\approx0.5536.

For a check, the atom at $\$2{,}000$ has zero insurer payment; the atom at $\$6{,}000$ contributes $0.25(4{,}000)=\$1{,}000$. The uniform component’s conditional expected payment is

\frac{1}{4{,}000}\int_{2{,}000}^{5{,}000}(x-2{,}000)\,dx =\frac{4{,}500{,}000}{4{,}000}=\$1{,}125.

Its overall contribution is $0.50(1{,}125)=\$562.50$, so expected insurer payment is $\$1{,}562.50$. The sum $\$1{,}937.50+\$1{,}562.50=\$3{,}500$ matches the ground-up mean. The LER is $0.5536$, or 55.36% to two decimal places.

Common Mistakes and a Reliable Workflow

The ratio has a clear interpretation only when its numerator and denominator are calculated on the same ground-up severity distribution. In particular, do not replace the denominator with expected insurer payment, or use the probability of a loss below the deductible as the numerator. The former changes the ratio’s meaning; the latter counts claims rather than dollars.

Using the wrong denominator. LER divides expected eliminated loss by $E[X]$, not by $E[(X-d)_+]$.
Using a probability instead of an expected amount. For $X<d$, the eliminated amount is $X$, which can be less than $d$.
Forgetting component weights. A conditional mean within a component must be multiplied by that component’s probability to obtain its contribution to the overall mean.
Misreading the deductible boundary. At $X=d$, eliminated loss is $d$ and insurer payment is zero.
Normalizing a subdensity incorrectly. A density for a component conditional on that component occurring integrates to one; its weighted subdensity integrates to the component’s probability.

A dependable calculation proceeds in three stages: find $E[X]$ from all atoms and continuous components; find $E[\min(X,d)]$ using the same component weights; and check that expected eliminated loss plus expected insurer payment equals $E[X]$. This direct approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus and works without a closed-form LER formula for the severity distribution.

Check Your Understanding

For each hypothetical loss model, calculate the LER for the stated ordinary per-loss deductible. Give probabilities to four decimal places.

$X$ equals $\$0,\$2{,}000,\$5{,}000$ with probabilities $0.20,0.50,0.30$, respectively. The deductible is $\$2{,}500$. (Answer: $0.7000$)
$X$ is zero with probability $0.40$, equals $\$5{,}000$ with probability $0.20$, and otherwise comes from a uniform distribution on $[0,\$4{,}000]$, with probability $0.40$. The deductible is $\$1{,}000$. (Answer: $0.3056$)
$X$ equals $\$3{,}000$ with probability $0.25$, and otherwise comes from a uniform distribution on $[\$2{,}000,\$6{,}000]$, with probability $0.75$. The deductible is $\$4{,}000$. (Answer: $0.9000$)
$X$ equals $\$1{,}000,\$4{,}000,\$9{,}000$ with probabilities $0.20,0.50,0.30$, respectively. The deductible is $\$3{,}000$. (Answer: $0.5306$)
$X$ equals $\$2{,}000$ with probability $0.40$, and otherwise comes from a uniform distribution on $[0,\$6{,}000]$, with probability $0.60$. The deductible is $\$2{,}000$. (Answer: $0.6923$)

What You'll Learn