Loss Elimination Ratios for Franchise Deductibles

Why the Franchise Rule Needs Its Own LER

A deductible usually reduces the insurer's payment on a covered loss. Under an ordinary deductible, the insurer pays only the amount above the deductible. A franchise deductible works differently: the insurer pays nothing when a loss does not exceed the threshold, but pays the full loss when it does. Because the payment rules differ, their loss elimination ratios (LERs) differ too.

This distinction matters especially for discrete or mixed severities. If a loss can equal the deductible with positive probability, the contract's rule at equality changes which party bears that loss. In this tutorial, the franchise contract pays only when the loss is strictly greater than the deductible; a loss equal to the deductible is not paid by the insurer. The convention must be stated rather than assumed.

Definition: Let $X$ be a nonnegative ground-up loss with $0<E[X]<\infty$, and let $d\geq 0$ be a franchise deductible. Under the strict-threshold convention, the insurer payment per loss is $Y^L=X\mathbf{1}_{\{X>d\}}$. The loss eliminated is $X-Y^L=X\mathbf{1}_{\{X\leq d\}}$. The franchise loss elimination ratio is $$ \operatorname{LER}_{\mathrm{fr}}(d) =\frac{E[X\mathbf{1}_{\{X\leq d\}}]}{E[X]}. $$

The numerator is the expected ground-up loss borne below or at the threshold. The denominator is the expected ground-up loss before applying the contract. Thus, the franchise LER is a loss-weighted proportion, not the probability that a loss is below the deductible. A small loss and a large loss do not contribute equally to the ratio.

For comparison, an ordinary deductible pays $Y^L=(X-d)_+$, where $(z)_+=\max(z,0)$. It eliminates $\min(X,d)$, so its LER is $E[\min(X,d)]/E[X]$. For any $X$, $\min(X,d)\geq X\mathbf{1}_{\{X\leq d\}}$; therefore the ordinary-deductible LER is at least as large as the franchise LER for the same $X$ and $d$. The difference comes from losses above $d$: the ordinary deductible eliminates $d$ of each such loss, whereas the franchise deductible eliminates none of it.

Calculating the Eliminated Expected Loss

For a severity distribution with density $f_X$, the lower partial expectation can be calculated directly as an integral. For a discrete distribution, sum each loss amount at or below $d$, weighted by its probability. A survival-function formula provides a convenient approach that also works when the severity has atoms.

Result: Write $S_X(t)=P(X>t)$. For nonnegative $X$ with finite mean, $$ E[X\mathbf{1}_{\{X\leq d\}}] =\int_0^d S_X(t)\,dt-dS_X(d), \qquad E[X]=\int_0^\infty S_X(t)\,dt. $$ Consequently, $$ \operatorname{LER}_{\mathrm{fr}}(d) =\frac{\int_0^d S_X(t)\,dt-dS_X(d)} {\int_0^\infty S_X(t)\,dt}. $$

The term $S_X(d)$ is $P(X>d)$, not $P(X\geq d)$. This is important when there is probability mass exactly at $d$: under the stated contract convention, that mass is included in the eliminated loss.

Why the Survival-Function Formula Holds

For any nonnegative loss, the capped amount $\min(X,d)$ can be represented as the length of the interval from zero up to the smaller of $X$ and $d$. Taking expectations gives the usual tail-integral identity. The capped amount includes $d$ for every loss strictly above $d$, but the franchise lower partial expectation does not include any of those losses. Subtracting that contribution gives the result.

\begin{aligned} E[\min(X,d)] &=\int_0^d P(X>t)\,dt =\int_0^d S_X(t)\,dt,\\ \min(X,d) &=X\mathbf{1}_{\{X\leq d\}}+d\mathbf{1}_{\{X>d\}},\\ E[X\mathbf{1}_{\{X\leq d\}}] &=E[\min(X,d)]-dP(X>d)\\ &=\int_0^d S_X(t)\,dt-dS_X(d). \end{aligned}

For a continuous severity with density $f_X$, the same numerator is $\int_0^d x f_X(x)\,dx$. If $P(X=d)>0$, this density integral alone may omit the atom at $d$; use a sum or the survival formula, or explicitly add $dP(X=d)$. A contract that instead pays at equality would use $X\mathbf{1}_{\{X\geq d\}}$ as its payment rule, so the treatment of an atom at $d$ would change.

Worked Calculations

Worked Example: Uniform Loss With a Franchise Threshold

A hypothetical equipment policy has a ground-up loss $X$ uniformly distributed from \$0 to \$1,200. Its franchise deductible is $d=\$300$, with payment only for losses strictly greater than \$300. Calculate the expected loss eliminated, the franchise LER, and the insurer's expected payment. Then compare with an ordinary deductible of \$300.

Method. Use the density $f_X(x)=1/1200$ for $0\leq x\leq1200$. There is no probability mass at the threshold, so integrating through $d$ gives the lower partial expectation directly.

The ground-up expected loss is $$ E[X]=\frac{0+1200}{2}=\$600.00. $$ The expected loss eliminated by the franchise deductible is $$ E[X\mathbf{1}_{\{X\leq300\}}] =\int_0^{300}x\frac{1}{1200}\,dx =\frac{300^2}{2(1200)} =\$37.50. $$ Thus, $$ \operatorname{LER}_{\mathrm{fr}}(300) =\frac{37.50}{600} =0.0625. $$ The insurer's expected payment is the ground-up mean less the eliminated amount: $$ E[Y^L]=600.00-37.50=\$562.50. $$ As a check, the insurer pays 0 for losses up to \$300 and the full loss above \$300; the expected payment is therefore $\int_{300}^{1200}x(1/1200)\,dx=\$562.50$, consistent with the subtraction.

With an ordinary deductible, the eliminated amount is $\min(X,300)$. Its expectation is $$ E[\min(X,300)] =\int_0^{300}\left(1-\frac{t}{1200}\right)\,dt =300-\frac{300^2}{2400} =\$262.50, $$ so the ordinary-deductible LER is $262.50/600=0.4375$. The common mistake is to use this ordinary-deductible numerator for the franchise contract. That would count \$300 eliminated on every loss above the threshold, even though the franchise insurer pays the full loss in that range.

Worked Example: Exponential Severity and a Comparison of Payment Rules

A hypothetical repair-loss amount $X$ follows an exponential distribution with mean $\theta=\$1,000$, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. A franchise deductible is $d=\$500$, and payment requires $X>500$. Calculate its LER and expected insurer payment, then calculate the ordinary-deductible LER at the same threshold.

Method. Apply the survival-function expression for the franchise lower partial expectation. The mean is $\theta=\$1,000$, and $S_X(500)=e^{-0.5}$.

First, $$ \int_0^{500}S_X(t)\,dt =\int_0^{500}e^{-t/1000}\,dt =1000(1-e^{-0.5}) =\$393.469340. $$ Also, $500S_X(500)=500e^{-0.5}=\$303.265330$. Therefore, $$ E[X\mathbf{1}_{\{X\leq500\}}] =393.469340-303.265330 =\$90.204010. $$ The franchise LER is $90.204010/1000$. Rounded to four decimal places, it is 0.0902. The expected insurer payment is $$ E[Y^L]=1000.00-90.204010=\$909.80, $$ rounded to the nearest cent.

For the ordinary deductible, the eliminated expectation is $E[\min(X,500)]=\int_0^{500}e^{-t/1000}\,dt=\$393.469340$. Its LER is $393.469340/1000$, or 0.3935 to four decimal places. As a check, the franchise insurer's expected payment is also $E[X\mathbf{1}_{\{X>500\}}]=(500+1000)e^{-0.5}=\$909.80$.

The trap is to confuse the probability of crossing the threshold, $P(X>500)=e^{-0.5}$, with the fraction of expected loss eliminated. The probability of payment is about 0.6065, but the franchise LER is about 0.0902: they measure different quantities.

Worked Example: A Discrete Loss With an Atom at the Deductible

A hypothetical cargo loss $X$ is \$200 with probability 0.20, \$500 with probability 0.30, and \$900 with probability 0.50. A franchise deductible is $d=\$500$, and the insurer pays only if $X>500$. Find the expected eliminated loss and LER. Compare with an ordinary deductible of \$500.

Method. Sum the loss amounts by probability, applying the strict payment rule to each possible outcome. In particular, the \$500 outcome is not paid by the franchise insurer because it equals, rather than exceeds, the deductible.

The ground-up expected loss is $$ E[X]=200(0.20)+500(0.30)+900(0.50) =40+150+450 =\$640.00. $$ The franchise contract eliminates the \$200 and \$500 losses, so $$ E[X\mathbf{1}_{\{X\leq500\}}] =200(0.20)+500(0.30) =40+150 =\$190.00. $$ The LER is $190/640=0.296875$, which is 0.2969 to four decimal places. The insurer pays only on the \$900 outcome, giving expected payment $900(0.50)=\$450.00$. The check $190+450=640$ reconciles eliminated and insured expected loss to the ground-up mean.

Under an ordinary deductible, the \$200 loss is eliminated in full, while \$500 is eliminated from each loss of \$500 or \$900. Thus, $$ E[\min(X,500)] =200(0.20)+500(0.30)+500(0.50) =40+150+250 =\$440.00, $$ and the ordinary-deductible LER is $440/640=0.6875$. The common error is to treat the \$500 atom as if it were above the threshold. Under this problem's strict rule, it is not paid; under an equality-inclusive payment rule, the result would differ.

Common Checks and Exam Traps

A reliable calculation begins by writing the payment rule, not by recalling a formula from an ordinary deductible problem. Then identify which losses are eliminated, calculate their expected dollar amount, and divide by the ground-up mean. These steps keep the contract interpretation separate from the severity calculation.

Write the contract rule.
For the convention used here, $Y^L=X\mathbf{1}_{\{X>d\}}$. Confirm whether equality is paid if the problem specifies a different convention.

Find the eliminated amount.
For the strict-threshold rule it is $X\mathbf{1}_{\{X\leq d\}}$, not $\min(X,d)$.

Divide by the ground-up mean.
Compute $E[X\mathbf{1}_{\{X\leq d\}}]/E[X]$, using consistent loss units in numerator and denominator.

Reconcile expected amounts.
Check that expected eliminated loss plus expected insurer payment equals $E[X]$.

The most frequent errors are using the ordinary-deductible expression $E[\min(X,d)]$, using a probability such as $P(X\leq d)$ in place of an expected dollar amount, and overlooking a point mass at $d$. For a continuous distribution, a single point has probability zero, so the equality convention does not affect the LER. For a discrete or mixed distribution, it can matter materially.

Key takeaway: Under a strict franchise deductible, losses at or below $d$ are eliminated in full and losses above $d$ are not reduced. Therefore, $\operatorname{LER}_{\mathrm{fr}}(d)=E[X\mathbf{1}_{\{X\leq d\}}]/E[X]$. This is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, apply the stated franchise payment rule and distinguish expected loss eliminated from the probability of falling below the threshold.

A hypothetical component loss is uniform on $[0,\$800]$, with a strict franchise deductible of \$200. Find the expected loss eliminated and the LER. (Answer: expected loss eliminated \$25.00; LER 0.0625.)
A hypothetical claim severity is exponential with mean \$600, and a franchise deductible of \$300 pays only for losses strictly above \$300. Find the LER to four decimal places. (Answer: 0.0902.)
A discrete loss is \$100 with probability 0.25, \$400 with probability 0.25, and \$700 with probability 0.50. A franchise deductible of \$400 pays only when the loss is strictly greater than \$400. Find the expected ground-up loss and franchise LER. (Answer: expected ground-up loss \$475.00; LER 0.2632.)
A hypothetical machine loss is \$0 with probability 0.20, \$300 with probability 0.30, and \$1,000 with probability 0.50. The franchise deductible is \$300 with strict-threshold payment. Find the expected eliminated loss and LER. (Answer: expected eliminated loss \$90.00; LER 0.1525.)
For $X$ uniform on $[0,\$1,000]$, find the franchise deductible $d$ that gives an LER of 0.1600, assuming $0\leq d\leq \$1,000$. (Answer: $d=\$400.00$.)

What You'll Learn