Deductible Payments With Boundary Atoms

Why the Deductible Boundary Matters

A deductible does more than reduce a loss by a fixed amount. It determines whether a loss produces any insurer payment, and the precise contract wording matters when a loss can equal the deductible with positive probability. That can happen in a discrete loss model or a mixed model containing both continuous and discrete outcomes.

This tutorial studies a single nonnegative loss $X$ and a deductible $d\geq 0$. There is no policy limit, coinsurance, or discounting. Payments are measured at the same date as the loss. The key distinction is that an ordinary deductible subtracts $d$ from a covered loss, whereas a franchise deductible makes the insurer pay the full loss once the threshold is crossed.

Definition: For an ordinary deductible, the insurer payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. For a strict franchise deductible, the insurer pays the full loss only when it exceeds the deductible: $Y^L=X\mathbf{1}_{\{X>d\}}$. Thus a loss exactly equal to $d$ produces zero payment under either definition.

Unless a contract explicitly says otherwise, the franchise convention used here is strict: payment is made when $X>d$. Some contracts instead specify payment when $X\geq d$. That alternative is treated separately, because the two conventions differ if $P(X=d)>0$.

Payment Distributions

Write $F_X(x)=P(X\leq x)$ for the loss CDF. A point mass, or atom, at $a$ has probability $P(X=a)=F_X(a)-F_X(a^-)$, where $F_X(a^-)$ is the limit of the CDF from below. The payment distribution must include any probability concentrated at zero, not just the positive-payment portion.

Theorem: For a nonnegative loss and $d\geq0$, the ordinary-deductible payment has $P(Y^L=0)=F_X(d)$. Its CDF is zero for $y<0$, and for $y\geq0$, $F_{Y^L}(y)=F_X(d+y)$.

To see the zero-payment probability, $Y^L=0$ exactly when $X\leq d$. For $y\geq0$, the event $Y^L\leq y$ consists of every loss $X\leq d$, which pays zero, together with losses $d<X\leq d+y$, whose payment is at most $y$. Their union is $X\leq d+y$. This proves the CDF formula, including at $y=0$.

If the loss has an atom at $d$, that entire atom is included in $F_X(d)$ and hence in the zero-payment probability. An atom at a loss amount $x>d$ becomes an atom in the payment distribution at $x-d$. The CDF formula automatically preserves both facts.

Theorem: Under the strict franchise deductible, $P(Y^L=0)=F_X(d)$. Its CDF is zero for $y<0$, is $F_X(d)$ for $0\leq y\leq d$, and is $F_X(y)$ for $y>d$.

For $0\leq y\leq d$, a positive franchise payment must be greater than $d$, so it cannot be at most $y$. Only losses at or below $d$, which pay zero, contribute. If $y>d$, the event $Y^L\leq y$ includes all losses at or below $d$, plus losses between $d$ and $y$; together these are exactly $X\leq y$. In particular, an atom at $X=d$ contributes to the zero-payment mass under the strict convention.

Expectations From Tail Probabilities

For a nonnegative random variable $W$ with finite expectation, the tail-integral identity is $E[W]=\int_0^\infty P(W>y)\,dy$. It follows by writing each realized value $W$ as the area $\int_0^\infty \mathbf{1}_{\{W>y\}}\,dy$ and then taking expectations. Applying it to deductible payments gives formulas that work for continuous, discrete, and mixed loss models.

E[(X-d)_+] =\int_0^\infty P(X>d+y)\,dy =\int_d^\infty P(X>t)\,dt =\int_d^\infty S_X(t)\,dt.

The substitution is $t=d+y$. The strict tail $P(X>t)$ is used consistently. A single endpoint does not change the value of the integral, even when $X$ has an atom at $d$; that atom creates no ordinary-deductible payment.

For a strict franchise deductible, write the payment as $X\mathbf{1}_{\{X>d\}}=d\mathbf{1}_{\{X>d\}}+(X-d)_+$. The identity holds in both cases: if $X\leq d$, both sides are zero; if $X>d$, both sides equal $X$. Therefore:

E[X\mathbf{1}_{\{X>d\}}] =dP(X>d)+\int_d^\infty S_X(t)\,dt.

The first term reflects the full deductible amount paid by the insurer on every loss that crosses the threshold. The integral accounts for the amount above the deductible on those losses. If $P(X=d)>0$, the atom at $d$ is excluded: it neither triggers the strict franchise payment nor contributes to the expectation.

Identify the contract rule.
Write the payment as $(X-d)_+$ for an ordinary deductible, or $X\mathbf{1}_{\{X>d\}}$ for a strict franchise deductible.

Check the boundary.
Determine whether a loss exactly equal to $d$ pays zero or triggers payment. Find $P(X=d)$ if the model may have an atom there.

Find the zero-payment mass.
Under either convention defined here, it is $P(X\leq d)=F_X(d)$. Then use the relevant CDF or expectation formula.

Worked Examples

Worked Example: A Discrete Loss With Mass at the Deductible

A hypothetical equipment policy has a loss $X$ with this distribution: $P(X=0)=0.10$, $P(X=500)=0.20$, $P(X=1{,}000)=0.40$, and $P(X=2{,}000)=0.30$. The deductible is $d=\$1{,}000$. Find the zero-payment probability and expected insurer payment under an ordinary deductible and a strict franchise deductible.

Method. List the payment at each possible loss, including the loss exactly at the threshold. No interest or mortality assumption is relevant; this is a single-loss distribution.

For the ordinary deductible, the payments at $X=0,500,1{,}000,2{,}000$ are respectively $\$0,\$0,\$0,\$1{,}000$. The zero-payment probability is $0.10+0.20+0.40=0.70$. The expectation is $0.10(0)+0.20(0)+0.40(0)+0.30(1{,}000)=\$300$.

For the strict franchise deductible, losses $0,500,$ and $1{,}000$ do not exceed the deductible, so their payments are zero. A loss of $\$2{,}000$ triggers payment of the full $\$2{,}000$. The zero-payment probability remains $0.70$, and the expected payment is $0.30(2{,}000)=\$600$.

Check. The probabilities sum to $1.00$, and the only positive payment in either model occurs with probability $0.30$. The franchise payment is larger because it does not subtract the deductible. The common mistake avoided here is treating the loss equal to $\$1{,}000$ as triggering a strict franchise payment.

Worked Example: Exponential Loss Payments

Suppose a hypothetical single claim has an exponential loss $X$ with mean $\$2{,}000$, so its rate is $\lambda=1/2{,}000$ per dollar. The deductible is $\$500$. Find the zero-payment probability and expected payment for each deductible type. Assume there is no atom at any loss amount.

For an exponential loss, $S_X(t)=e^{-t/2{,}000}$ for $t\geq0$, and $F_X(t)=1-e^{-t/2{,}000}$. Thus the zero-payment probability for either contract is $F_X(500)=1-e^{-500/2{,}000}=1-e^{-0.25}=0.2212$, to four decimals.

For the ordinary deductible, $E[Y^L]=\int_{500}^\infty e^{-t/2{,}000}\,dt =2{,}000e^{-500/2{,}000} =2{,}000e^{-0.25}$. Using $e^{-0.25}=0.7788008$, the expectation is $\$1{,}557.60$, rounded to the nearest cent.

For the strict franchise deductible, $E[Y^L]=500P(X>500)+\int_{500}^\infty S_X(t)\,dt$. The first term is $500e^{-0.25}=\$389.40$; the integral is $\$1{,}557.60$. The total is $\$1{,}947.00$.

Check. The franchise expectation exceeds the ordinary expectation by $\$389.40$, exactly $dP(X>d)$. The common mistake avoided is using the ordinary payment expectation for a franchise contract, even though the latter pays the full loss whenever the threshold is exceeded.

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

A hypothetical loss equals $\$1{,}000$ with probability $0.20$. With probability $0.80$, it follows an exponential distribution with mean $\$1{,}000$. The deductible is $d=\$1{,}000$. Find the zero-payment probability and expected payment under the ordinary and strict franchise conventions.

The model is a mixture: the discrete outcome is exactly the deductible, and the continuous component has survival function $e^{-t/1{,}000}$. For the continuous component, the probability of a loss at or below $\$1{,}000$ is $1-e^{-1}=0.6321$. Including the atom, the overall zero-payment probability is $0.20+0.80(1-e^{-1})=1-0.80e^{-1}=0.7057$, to four decimals. The atom produces no payment under either convention as defined here.

For the ordinary deductible, only the exponential component can produce a positive payment. Its conditional expected payment is $\int_{1{,}000}^\infty e^{-t/1{,}000}\,dt=1{,}000e^{-1}$. Weighting by its mixture probability gives $0.80(1{,}000e^{-1})=\$294.30$, rounded to the nearest cent.

For the strict franchise deductible, the conditional expectation from the exponential component is $dP(X>d)+\int_d^\infty S_X(t)\,dt =1{,}000e^{-1}+1{,}000e^{-1} =\$2{,}000e^{-1}$. After weighting by $0.80$, the total is $0.80(2{,}000e^{-1})=\$588.61$. The atom at $\$1{,}000$ contributes zero, since the trigger is $X>d$, not $X\geq d$.

Check. The expected franchise payment exceeds the ordinary payment by $\$294.30$, which equals $dP(X>d)=1{,}000(0.80e^{-1})$. The common mistake avoided is counting the $0.20$ atom at the deductible as a franchise payment.

When the Contract Uses an Inclusive Threshold

If the policy says that the insurer pays the full loss when $X\geq d$, use $Y^L=X\mathbf{1}_{\{X\geq d\}}$, not the strict-franchise formula. The zero-payment probability becomes $P(X<d)=F_X(d^-)$. The expectation is the strict-franchise expectation plus the payment on the boundary atom:

E[X\mathbf{1}_{\{X\geq d\}}] =dP(X>d)+\int_d^\infty S_X(t)\,dt+dP(X=d).

The added term is $dP(X=d)$, because every boundary loss now receives a payment of $d$. If $P(X=d)=0$, strict and inclusive franchise conventions give the same results. This distinction is most important in discrete or mixed models; in a continuous model, the probability of any one exact value is zero.

Key takeaway: First translate the contract wording into a payment random variable. Then identify the probability of no payment and handle the event $X=d$ explicitly. Ordinary and strict franchise payments share the same zero-payment probability $F_X(d)$, but their positive payments—and therefore their expectations—are different.

Check Your Understanding

For each problem, state the threshold convention before calculating.

A loss is $\$200,\$800,$ or $\$1{,}200$ with probabilities $0.25,0.35,$ and $0.40$, respectively. With an $\$800$ deductible, find the zero-payment probability and expected payment for an ordinary deductible and a strict franchise deductible. (Answer: Both zero-payment probabilities are $0.6000$; expected payments are $\$160$ ordinary and $\$480$ strict franchise.)
A loss is exponential with mean $\$4{,}000$, and the deductible is $\$1{,}000$. Find the ordinary-deductible zero-payment probability and expected payment. Give the probability to four decimals and payment to the nearest cent. (Answer: $0.2212$; $\$3{,}115.20$.)
With probability $0.30$, a loss is exactly $\$500$; otherwise it is uniform on $[0,\$1{,}000]$. For a $\$500$ deductible, find the zero-payment probability and expected payment under each convention. (Answer: Both zero-payment probabilities are $0.6500$; expected payments are $\$87.50$ ordinary and $\$262.50$ strict franchise.)
A loss is $\$1{,}000,\$1{,}500,$ or $\$2{,}000$ with probabilities $0.15,0.25,$ and $0.60$. A franchise contract pays the full loss when $X\geq \$1{,}500$. Find the zero-payment probability and expected payment, and compare with a strict threshold. (Answer: Inclusive zero-payment probability $0.1500$ and expected payment $\$1{,}575$; strict zero-payment probability $0.4000$ and expected payment $\$1{,}200$.)
A loss has $P(X=900)=0.10$, $P(X=1{,}000)=0.20$, and $P(X=1{,}500)=0.70$. For deductible $d=\$1{,}000$, find the CDF of the ordinary payment at $y=0$ and $y=\$500$, and the strict franchise payment at those same values. (Answer: Ordinary CDF values are $0.3000$ at $0$ and $1.0000$ at $\$500$; strict franchise CDF values are $0.3000$ at $0$ and $0.3000$ at $\$500$.)

What You'll Learn