Deductible-and-Limit Contracts With Payment Atoms

Why the Boundary Payments Matter

A deductible and a per-claim limit transform the loss into the insurer’s payment. The transformation can create probability masses even when the original loss is continuous: every loss at or below the deductible produces zero payment, while every sufficiently large loss produces the maximum payment. Correctly locating these two groups is essential before calculating an expected payment or using the modified severity distribution in a broader model.

Let $X$ be a nonnegative loss, with distribution function $F_X(x)=P(X\leq x)$ and survival function $S_X(x)=P(X>x)$. Assume an ordinary deductible $d\geq0$ and a maximum insurer payment $u>0$. The limit $u$ is a cap on the insurer’s payment, not a cap on the loss amount before the deductible is applied.

Definition: The payment per loss is $$ Y^L=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ Thus $Y^L=0$ if $X\leq d$, $Y^L=X-d$ if $d<X<d+u$, and $Y^L=u$ if $X\geq d+u$. The loss threshold for a limit payment is $d+u$, because the insurer first absorbs the deductible and then pays up to $u$.

The Transformed Distribution

The payment distribution has a mass at zero, a transformed portion between zero and the limit, and a mass at the limit. The interior portion comes from losses between $d$ and $d+u$: a loss $x$ in that interval maps to payment $x-d$. The two boundary payments collect ranges of loss values into single payment values.

Theorem: For $0\leq y<u$, $$ P(Y^L\leq y)=F_X(d+y). $$ Consequently, $$ P(Y^L=0)=F_X(d)=P(X\leq d), \qquad P(Y^L=u)=P(X\geq d+u)=1-F_X((d+u)-). $$ For $0<y<u$, any probability mass at $y$ is $P(Y^L=y)=P(X=d+y)$. If $X$ has a density $f_X$, then the density of the interior payment values is $f_X(d+y)$ for $0<y<u$; the endpoint masses must still be included separately.

Here $F_X(c-)=P(X<c)$ denotes the left limit of the distribution function. This distinction matters for discrete or mixed losses: a loss exactly equal to $d+u$ already generates the limit payment, so the limit atom includes equality. Likewise, a loss exactly equal to $d$ generates zero payment and belongs in the zero atom.

For a continuous loss distribution, the boundary probabilities simplify to $F_X(d)$ and $S_X(d+u)$, because there is no probability mass at either threshold. The formula for the CDF still gives the full distribution: for $y<0$, the CDF is zero, and for $y\geq u$, it is one.

Worked Example: A Discrete Loss With Both Payment Atoms

A hypothetical repair policy covers a nonnegative loss $X$ with the following distribution. The deductible is $d=\$500$, and the maximum insurer payment is $u=\$700$. Find the complete payment distribution and its expected payment.

Loss $X$	Probability	Payment $Y^L=\min((X-500)_+,700)$
\$200	0.20	\$0
\$600	0.30	\$100
\$900	0.30	\$400
\$1,500	0.20	\$700

Method and boundary check. The limit-payment threshold is $d+u=\$500+\$700=\$1,200$. Losses at or below $\$500$ pay zero, so the $\$200$ loss produces $\$0$. The other losses are above the deductible: $\$600-\$500=\$100$, $\$900-\$500=\$400$, and $\$1,500-\$500=\$1,000$, capped at $\$700$. Only the $\$1,500$ loss reaches the cap.

Therefore $P(Y^L=0)=0.20$, $P(Y^L=100)=0.30$, $P(Y^L=400)=0.30$, and $P(Y^L=700)=0.20$. The probabilities sum to $0.20+0.30+0.30+0.20=1.00$.

The expected payment is $$ E[Y^L]=0(0.20)+100(0.30)+400(0.30)+700(0.20) =0+30+120+140=\$290.00. $$ The method is to transform each possible loss before weighting its payment by the loss probability. This avoids treating the limit as though it applied directly to $X$.

Expected Payment From the Survival Function

The expectation can be calculated without first finding a density for the transformed payment. Since $Y^L$ is bounded between zero and $u$, its expectation is the area under its survival function. At a payment level $y$ below the cap, $Y^L>y$ exactly when $X>d+y$. This gives a useful formula that also works when $X$ is discrete or mixed.

Write the payment as accumulated indicators.
For any value $0\leq Y^L\leq u$, $Y^L=\int_0^u \mathbf{1}(Y^L>y)\,dy$, where the indicator is one when its condition holds and zero otherwise.

Take expectations.
Integrating the indicator probabilities gives $E[Y^L]=\int_0^u P(Y^L>y)\,dy$.

Translate each payment threshold back to a loss threshold.
For $0\leq y<u$, $Y^L>y$ if and only if $X>d+y$. Thus $P(Y^L>y)=S_X(d+y)$.

Change the integration variable.
Setting $x=d+y$ yields $E[Y^L]=\int_d^{d+u}S_X(x)\,dx$.

Result: For a nonnegative loss with finite survival probabilities, an ordinary deductible $d\geq0$, and a payment limit $u>0$, $$ E[Y^L]=\int_d^{d+u}S_X(x)\,dx. $$ The formula integrates the loss survival function only over the interval where an additional increase in loss can increase the insurer’s payment. Losses above $d+u$ still generate a payment, but that payment is already capped.

Worked Example: Uniform Loss With a Deductible and Payment Cap

A hypothetical equipment policy has loss $X$ uniformly distributed on $[0,\$1,200]$, deductible $d=\$300$, and maximum insurer payment $u=\$500$. Find the probabilities of zero payment and limit payment, then calculate the expected payment. Assume the stated uniform distribution applies over the entire interval and has density $1/1200$ per dollar.

Zero-payment probability. A zero payment occurs for $X\leq300$. The interval has length $\$300$ out of a total length of $\$1,200$, so $$ P(Y^L=0)=\frac{300}{1200}=0.2500. $$

Limit-payment probability. The cap is reached when $X\geq d+u=300+500=800$. The interval from $\$800$ to $\$1,200$ has length $\$400$, so $$ P(Y^L=500)=\frac{1200-800}{1200}=\frac{400}{1200}=0.3333. $$

Expected payment by survival integration. On the support of $X$, $S_X(x)=(1200-x)/1200$. Therefore, $$ E[Y^L]=\int_{300}^{800}\frac{1200-x}{1200}\,dx =\left[x-\frac{x^2}{2400}\right]_{300}^{800}. $$ At $x=800$, the expression is $800-640000/2400=533.3333$. At $x=300$, it is $300-90000/2400=262.5000$. Their difference is $533.3333-262.5000=270.8333$, so the expected payment is $\$270.83$, rounded to the nearest cent.

Independent check. The payment graph over the loss interval has a triangle from $X=300$ to $X=800$, with area $\frac12(500)(500)=125,000$, and a rectangle from $X=800$ to $X=1,200$, with area $(400)(500)=200,000$. Dividing their total area, $325,000$, by the support length $1,200$ gives $270.8333$ dollars. This confirms the result.

A Continuous Severity Can Still Produce Atoms

The zero and cap atoms are not limited to discrete losses. A continuous distribution assigns no probability to an individual loss value, but the payment transformation maps whole intervals of losses to the same boundary payment. A continuous severity can therefore produce positive probability at both zero and $u$.

Worked Example: Exponential Loss and the Two Atoms

A hypothetical claim amount $X$ has an exponential distribution with mean $\$1,000$, so its rate is $\lambda=1/1000$ per dollar and $S_X(x)=e^{-x/1000}$ for $x\geq0$. The contract has deductible $d=\$200$ and maximum payment $u=\$600$. Find both boundary probabilities and the expected payment.

Zero-payment atom. The zero-payment event is $X\leq200$, so $$ P(Y^L=0)=1-e^{-200/1000}=1-e^{-0.2} =1-0.8187308=0.1813. $$

Limit-payment atom. The loss threshold for the cap is $d+u=200+600=800$, not $600$. Hence $$ P(Y^L=600)=P(X\geq800)=e^{-800/1000}=e^{-0.8}=0.4493. $$

Expected payment. Applying the survival integral, $$ E[Y^L]=\int_{200}^{800}e^{-x/1000}\,dx =1000\left(e^{-0.2}-e^{-0.8}\right). $$ Substituting $e^{-0.2}=0.8187308$ and $e^{-0.8}=0.4493290$ gives $1000(0.3694018)=\$369.40$, rounded to the nearest cent.

As a check, the probability of an interior payment is $1-0.1812692-0.4493290=0.3694018$, which is positive and leaves the three parts of the payment distribution summing to one. The interior probability need not equal the expected payment divided by the limit; payments in that interval vary in size.

Common Exam Trap: Applying the Limit Before the Deductible

A frequent error is to treat $u$ as the loss threshold at which the insurer begins paying the maximum. It is instead the maximum payment. The cap is reached at loss $d+u$. For example, in the discrete example, using $X\geq700$ as the cap event would incorrectly classify both the $\$900$ and $\$1,500$ losses as $\$700$ payments. That wrong calculation would give $$ 0(0.20)+100(0.30)+700(0.30)+700(0.20)=\$380.00, $$ rather than the correct $\$290.00$. The $\$900$ loss pays $\$900-\$500=\$400$; it does not reach the cap because $\$900<\$1,200$.

For a continuous uniform loss, another useful check is to compute the area under the payment curve. With support $[0,1200]$, deductible $300$, and payment cap $500$, the payment is zero up to $300$, increases linearly to $500$ by loss $800$, and stays at $500$ thereafter. The triangle-plus-rectangle area is $325,000$, and the mean is $325,000/1200=\$270.83$. A different numerical result for these assumptions fails this direct check.

Before calculating, write down both boundary conditions: zero payment when $X\leq d$, and limit payment when $X\geq d+u$. Then check that the probabilities of the zero, interior, and limit portions add to one. This is especially important when the loss distribution has atoms exactly at either threshold.

Check Your Understanding

For each hypothetical contract, identify the boundary probabilities and calculate any requested expected payment. Treat each stated limit as a maximum insurer payment.

A discrete loss is $\$100,\$400,\$700,\$1,000$, with probabilities $0.25,0.35,0.25,0.15$, respectively. The deductible is $\$250$ and the payment limit is $\$500$. Give the payment distribution and its mean. (Answer: $P(Y^L=0)=0.25$, $P(Y^L=150)=0.35$, $P(Y^L=450)=0.25$, $P(Y^L=500)=0.15$; $E[Y^L]=\$240.00$)
A loss is uniform on $[0,\$1,000]$. The deductible is $\$200$ and the payment limit is $\$300$. Find the zero-payment probability, limit-payment probability, and expected payment. (Answer: $0.2000$, $0.5000$, and $\$195.00$)
An exponential loss has mean $\$500$. The deductible is $\$100$ and the payment limit is $\$400$. Find both boundary probabilities and the expected payment, rounding money to the nearest cent. (Answer: $P(Y^L=0)=0.1813$, $P(Y^L=400)=0.3679$, and $E[Y^L]=\$225.43$)
A discrete loss equals $\$0,\$250,\$600,\$900$ with probabilities $0.20,0.30,0.25,0.25$, respectively. The deductible is $\$250$, and the payment limit is $\$400$. Give the payment distribution and mean, paying particular attention to losses exactly at a boundary. (Answer: $P(Y^L=0)=0.50$, $P(Y^L=350)=0.25$, $P(Y^L=400)=0.25$; $E[Y^L]=\$187.50$)
A loss is uniform on $[0,\$1,500]$, with deductible $\$300$ and payment limit $\$700$. Find the probability of zero payment, the probability of a limit payment, and the expected payment to the nearest cent. (Answer: $0.2000$, $0.3333$, and $\$396.67$)

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