Franchise Deductibles and Threshold Payments

What a Franchise Deductible Pays

A deductible can specify not only how large a loss must be before coverage applies, but also how much is paid after that threshold is crossed. Under a franchise deductible, a loss at or below the threshold receives no payment, while a loss strictly above it receives the full loss amount. The strict inequality matters whenever the loss distribution assigns positive probability to the threshold itself.

Let $X$ be a nonnegative loss severity and let $d\geq 0$ be a fixed franchise threshold. The payment is $Y=X\mathbf{1}(X>d)$, where the indicator equals 1 when $X>d$ and 0 otherwise. Thus, if $X=d$, then $Y=0$; if $X>d$, then $Y=X$. This tutorial focuses on the probability of no payment and on the expected payment.

Definition: For a nonnegative loss $X$ and a nonnegative franchise threshold $d$, the franchise-deductible payment is $Y=X\mathbf{1}(X>d)$. The insurer pays the entire loss only when it is strictly greater than $d$.

Write $F_X(x)=P(X\leq x)$ for the cumulative distribution function and $S_X(x)=P(X>x)=1-F_X(x)$ for the survival function. With this survival-function convention, the probability of exceeding the threshold is $S_X(d)$, not $P(X\geq d)$. That distinction preserves the strict trigger in the contract.

Probability of Zero Payment

Because losses are nonnegative, the payment is zero exactly when $X\leq d$. This includes losses equal to the threshold as well as losses below it. Consequently, the zero-payment probability and the positive-payment probability are complements.

P(Y=0)=P(X\leq d)=F_X(d), \qquad P(Y>0)=P(X>d)=S_X(d).

These equations apply to continuous, discrete, and mixed severity distributions. In a continuous model, $P(X=d)=0$, so using a weak or strict inequality at the threshold does not change the probability. In a discrete or mixed model, there may be positive probability at $d$, and that probability belongs in $P(Y=0)$.

Worked Example: A Threshold With an Atom at the Deductible

A hypothetical equipment policy has a franchise threshold of $5$ thousand dollars. Its loss $X$, measured in thousands of dollars, takes the following values: $0$ with probability $0.10$, $3$ with probability $0.15$, $5$ with probability $0.20$, $6$ with probability $0.25$, and $10$ with probability $0.30$. Find the probability of zero payment and the expected payment.

Method. Apply the strict trigger $X>5$ to each possible loss. The probabilities sum to $0.10+0.15+0.20+0.25+0.30=1.00$, so the distribution is valid.

Losses of $0$, $3$, and $5$ do not trigger payment. Therefore,

$$ P(Y=0)=0.10+0.15+0.20=0.45. $$

The payments for losses $6$ and $10$ are the full loss amounts. Hence,

$$ E[Y]=0.25(6)+0.30(10)=1.50+3.00=4.50 $$

in thousands of dollars, or $\$4{,}500.00$. As a check, the positive-payment probability is $0.25+0.30=0.55$, which equals $1-0.45$. The atom of probability $0.20$ at $X=5$ is correctly included among zero payments. A common error is to treat $X\geq5$ as the trigger and count that atom as a payment.

Expected Payment From the Survival Function

For a nonnegative random variable $Z$, the tail-integral identity is $E[Z]=\int_0^\infty P(Z>y)\,dy$. It holds as an extended expectation: the integral and expectation may be infinite. It does not require the expected payment to be finite.

To apply the identity, examine the survival probability of the franchise payment $Y$. If $0\leq y<d$, then $Y>y$ precisely when the loss crosses the franchise threshold, $X>d$. If $y\geq d$, then $Y>y$ precisely when $X>y$: a loss exceeding $y$ already exceeds $d$, and the full loss is paid. Integrating these two parts gives the expected-payment formula.

\begin{aligned} E[Y] &=\int_0^\infty P(Y>y)\,dy\\ &=\int_0^d S_X(d)\,dy+\int_d^\infty S_X(y)\,dy\\ &=dS_X(d)+\int_d^\infty S_X(y)\,dy. \end{aligned}

The value of the integrand at a single endpoint does not affect the integral, so atoms at $d$ do not require an endpoint adjustment. The first term accounts for the payment exceeding every level below $d$ when the threshold is crossed; the second accumulates the probability that payment exceeds each level at or above $d$. For finite $d$, the term $dS_X(d)$ is finite, so $E[Y]$ is finite exactly when $\int_d^\infty S_X(y)\,dy$ is finite. If that tail integral diverges, the expected payment is infinite.

Theorem: For $X\geq0$ and finite $d\geq0$, the franchise payment $Y=X\mathbf{1}(X>d)$ satisfies $P(Y=0)=F_X(d)$ and $E[Y]=dS_X(d)+\int_d^\infty S_X(y)\,dy$. The expectation formula is valid even when its value is infinity.

When $P(X>d)>0$, another useful representation is $E[Y]=P(X>d)E[X\mid X>d]$, interpreted as an extended expectation if necessary. It follows by conditioning on whether the threshold is crossed: payment is zero otherwise, and equals $X$ on the crossing event. This representation is convenient when the conditional severity above the threshold is known; the survival-integral form is useful when the survival function is easier to integrate.

Worked Example: Exponential Losses

A hypothetical claim severity $X$, measured in thousands of dollars, has an exponential distribution with rate $\lambda=0.2$ per thousand dollars. Thus $S_X(x)=e^{-0.2x}$ for $x\geq0$. For a franchise threshold $d=5$ thousand dollars, calculate the probability of zero payment and the expected payment.

Method. Use the survival function at the threshold for the trigger probability, then use the derived tail-integral formula. The exponential model has a finite mean, and the integral below converges.

First, the probability of zero payment is

P(Y=0)=1-S_X(5)=1-e^{-0.2(5)}=1-e^{-1} =1-0.367879=0.632121.

To find the expected payment, evaluate the threshold term and the tail integral:

\begin{aligned} E[Y] &=5e^{-1}+\int_5^\infty e^{-0.2y}\,dy\\ &=5e^{-1}+\left[-5e^{-0.2y}\right]_5^\infty\\ &=5e^{-1}+5e^{-1}\\ &=10e^{-1}=3.678794 \end{aligned}

in thousands of dollars. The zero-payment probability is $0.6321$ to four decimal places. The expected payment is $\$3{,}678.79$, rounded to the nearest cent. As a check, $S_X(5)=0.367879$, so positive payments occur with probability $0.3679$; the computed mean is positive and less than the unconditional exponential mean $1/\lambda=5$ thousand dollars, as expected after zeroing out losses that do not cross the threshold. A common mistake is to subtract the threshold from each loss above it; that would calculate a different payment rule, not this franchise payment.

Heavy Tails and Infinite Expected Payments

A positive probability of crossing the franchise threshold does not guarantee a finite expected payment. The threshold changes which losses are paid, but it does not cap the amount paid on a triggered claim. Therefore, sufficiently heavy tails can still produce an infinite expected payment.

Worked Example: A Pareto Severity With Infinite Expected Payment

A hypothetical loss $X$, in thousands of dollars, has a Pareto Type I distribution with scale $x_m=2$ and shape $\alpha=1$. Its survival function is $S_X(x)=1$ for $0\leq x<2$, and $S_X(x)=(2/x)^\alpha=2/x$ for $x\geq2$. A franchise threshold is $d=4$. Find the zero-payment probability and determine whether the expected payment is finite.

Method. Evaluate the survival probability at $4$, then test the tail integral in the expected-payment formula. The Pareto shape and scale are stated explicitly; here the tail decreases proportionally to $1/x$.

S_X(4)=\frac{2}{4}=0.5, \qquad P(Y=0)=1-S_X(4)=0.5.

For the expected payment, the threshold term is $4(0.5)=2$. The remaining integral is

\begin{aligned} \int_4^\infty S_X(y)\,dy &=\int_4^\infty \frac{2}{y}\,dy\\ &=\lim_{b\to\infty}2(\ln b-\ln 4)\\ &=\infty. \end{aligned}

Thus $E[Y]=2+\infty=\infty$. The zero-payment probability is $0.5000$, but the expected payment is infinite. This is consistent with the tail-integral identity for nonnegative variables; finiteness is not a condition for using that identity. The common mistake is to assume that removing all losses at or below a positive threshold must make the expected payment finite. It does not remove the arbitrarily large losses in this Pareto tail.

Exam Checks and Practical Interpretation

For a franchise-deductible calculation, first identify the exact contractual trigger, then determine the payment on a triggered claim. Under the rule taught here, crossing the threshold produces payment of the full loss, not merely the amount above the threshold. Next, use $F_X(d)$ for the probability of zero payment and choose an expected-value method suited to the distribution: sum payments for a discrete model, or integrate the survival function for a continuous or mixed model.

Write the payment rule.
Use $Y=X\mathbf{1}(X>d)$, retaining the strict inequality.

Handle the threshold correctly.
Losses equal to $d$ produce zero payment, so $P(Y=0)=F_X(d)$.

Calculate the expected amount.
Use a probability-weighted sum or $dS_X(d)+\int_d^\infty S_X(y)\,dy$.

Check whether the tail is integrable.
The expected payment may be infinite; the tail-integral formula remains valid in that case.

The key modeling question is whether the contract pays the full loss after the trigger or only a portion of it. Applying a formula for a different payment rule can give a plausible-looking but incorrect answer. In particular, a threshold condition and a deduction from the loss amount are separate features: the formula must reflect the contract actually described. These distinctions and calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, apply the strict franchise trigger and state the requested probability or expected payment. Monetary severities are in thousands of dollars unless otherwise stated.

A loss takes values $0,4,8$ with probabilities $0.20,0.30,0.50$, respectively. If $d=4$, find $P(Y=0)$ and $E[Y]$. (Answer: $P(Y=0)=0.50$, $E[Y]=4.00$ thousand dollars.)
A loss is exponential with rate $\lambda=0.25$ per thousand dollars and $d=4$. Find $P(Y=0)$ and $E[Y]$. (Answer: $P(Y=0)=1-e^{-1}=0.6321$, $E[Y]=8e^{-1}=2.9430$ thousand dollars.)
A loss is uniform on $[0,12]$, and $d=3$. Find the probability of zero payment and the expected payment. (Answer: $P(Y=0)=0.2500$, $E[Y]=5.625$ thousand dollars.)
A loss equals $2,6,9$ with probabilities $0.25,0.35,0.40$, respectively. For $d=6$, find the zero-payment probability and expected payment. (Answer: $P(Y=0)=0.60$, $E[Y]=3.60$ thousand dollars.)
A Pareto Type I loss has scale $x_m=3$, shape $\alpha=2$, and threshold $d=6$. Its survival function above the scale is $S_X(x)=(3/x)^2$. Find $P(Y=0)$ and $E[Y]$. (Answer: $P(Y=0)=0.7500$, $E[Y]=3.00$ thousand dollars.)

What You'll Learn