Identify Atoms in Deductible-Limit Payment Distributions

Why Boundary Masses Matter

A deductible and a policy limit change more than the size of a claim payment. They can also make the payment distribution have probability concentrated at particular values. Losses at or below the deductible all produce a zero payment, while sufficiently large losses all produce the maximum payment. These are atoms: individual payment values with positive probability.

This distinction matters whenever a problem asks for the probability of no payment, a payment equal to the limit, or a payment in a specified interior range. The endpoint details are especially important when the ground-up loss $X$ is discrete or mixed, because positive probability may occur exactly at the deductible or at the loss amount that reaches the limit.

Definition: Suppose $X$ is a nonnegative ground-up loss, $d\geq 0$ is a deductible, and $u>0$ is the maximum amount the insurer pays per loss. Under this contract convention, the payment per loss is $$ Y^L=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ Thus, the insurer pays nothing when $X\leq d$, pays $X-d$ when $d<X<d+u$, and pays $u$ when $X\geq d+u$.

Here $u$ is a limit on the insurer’s payment, not a limit on the ground-up loss. A contract whose stated limit instead caps the covered loss before subtracting the deductible has a different payment formula. Always translate the contract wording into a payment rule before calculating probabilities.

The Payment Distribution and Its Atoms

Write $F_X(x)=P(X\leq x)$, and let $F_X(x-)=P(X<x)$ denote the left-hand limit of the distribution function. This left-limit notation is needed to handle a possible atom exactly at a threshold.

Theorem: For the payment $Y^L=\min((X-d)_+,u)$, the boundary probabilities are $$ P(Y^L=0)=F_X(d) $$ and $$ P(Y^L=u)=P(X\geq d+u)=1-F_X((d+u)-). $$ For any interior payment amount $y$ with $0<y<u$, $$ P(Y^L=y)=P(X=d+y). $$ For $0\leq a<b<u$, $$ P(a<Y^L\leq b)=P(d+a<X\leq d+b). $$

The formula for the mass at zero includes losses exactly equal to $d$: the insurer pays zero at $X=d$. The formula for the mass at the limit includes losses exactly equal to $d+u$: at that point, $X-d=u$, so the insurer pays $u$. For a loss distribution with no atom at the threshold, the choice of strict or non-strict inequality does not affect the probability; with a discrete or mixed distribution, it can.

The full cumulative distribution function of the payment can also be written directly. Since the payment is nonnegative, $P(Y^L\leq y)=0$ for $y<0$. For $0\leq y<u$, the event $Y^L\leq y$ is equivalent to $X\leq d+y$, so

P(Y^L\leq y)=F_X(d+y),\qquad 0\leq y<u.

At $y=u$, the payment is always at most $u$, so $P(Y^L\leq u)=1$; it remains 1 for larger $y$. In a continuous model with density $f_X$, the payment has density $f_X(d+y)$ for $0<y<u$, in addition to the boundary probabilities at zero and $u$. In a discrete or mixed model, interior payment atoms occur wherever $X$ has positive probability at a value strictly between $d$ and $d+u$.

Why the Threshold Formulas Hold

The key is to translate an event about the payment back into an event about the loss. For zero payment, the positive-part operation gives $(X-d)_+=0$ exactly when $X\leq d$. For a payment at the limit, the capped amount equals $u$ exactly when the uncapped amount is at least $u$. These event equivalences establish the two endpoint formulas, including their boundary inequalities.

For an interior value $y$, the cap has not been reached and the deductible has been exceeded, so the payment equals $X-d$. The following equivalence also justifies the interior interval formula:

a<Y^L\leq b \quad\Longleftrightarrow\quad d+a<X\leq d+b, \qquad 0\leq a<b<u.

Indeed, if $X\leq d$, then $Y^L=0$, which does not satisfy $a<Y^L$ for $a\geq0$. If $d<X<d+u$, then $Y^L=X-d$, and the payment inequalities are equivalent to the displayed loss inequalities. If $X\geq d+u$, then $Y^L=u>b$, so neither event holds. This proves the equivalence across all three parts of the payment rule.

Worked Examples

Worked Example: Exponential Losses and a Payment Interval

A hypothetical warranty has a deductible of $d=\$200$ and a maximum payment of $u=\$500$ per loss. Assume the ground-up loss has an exponential distribution with mean $\$1{,}000$, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. Calculate the probability of zero payment, the probability of payment at the limit, and the probability that the payment is greater than $\$100$ but at most $\$300$.

Method. Apply the threshold formulas and map the interior payment interval to its corresponding loss interval. Because the exponential distribution is continuous, endpoints of the interval have probability zero.

Zero payment. The payment is zero when $X\leq200$:

P(Y^L=0)=F_X(200)=1-e^{-200/1000} =1-e^{-0.2} =1-0.8187308 =0.1812692.

Payment at the limit. The payment reaches $\$500$ when $X\geq d+u=700$:

P(Y^L=500)=P(X\geq700)=e^{-700/1000} =e^{-0.7} =0.4965853.

Interior interval. The event $100<Y^L\leq300$ corresponds to $300<X\leq500$:

P(100<Y^L\leq300) =P(300<X\leq500) =e^{-0.3}-e^{-0.5} =0.7408182-0.6065307 =0.1342875.

To check the boundary calculations, the entire interior-payment probability is $P(200<X<700)=e^{-0.2}-e^{-0.7}=0.3221455$. The two boundary probabilities and that interior probability sum to $0.1812692+0.4965853+0.3221455=1.0000000$, up to rounding. To four decimal places, the requested probabilities are $0.1813$, $0.4966$, and $0.1343$. This method avoids treating the payment limit as a threshold at $X=u$; the correct threshold is $X=d+u$.

Worked Example: A Discrete Loss at Both Contract Boundaries

A hypothetical equipment policy has $d=\$200$ and maximum payment $u=\$300$. The loss distribution is: $X=0$ with probability $0.10$, $X=150$ with probability $0.10$, $X=200$ with probability $0.15$, $X=260$ with probability $0.20$, $X=450$ with probability $0.15$, $X=500$ with probability $0.10$, and $X=720$ with probability $0.20$. Find the payment distribution and $P(0<Y^L\leq250)$.

Method. Apply the payment rule to each loss value, retaining the probability whenever multiple losses produce the same payment.

For $X=0,150,$ or $200$, the loss is at or below the deductible, so the payment is zero. The total probability at zero is $0.10+0.10+0.15=0.35$. For $X=260$, the payment is $260-200=60$, with probability $0.20$. For $X=450$, it is $450-200=250$, with probability $0.15$.

The cap threshold is $d+u=500$. A loss of exactly $500$ produces payment $300$, as does any larger loss. Therefore, the probability at the limit is $0.10+0.20=0.30$. The complete distribution is:

P(Y^L=0)=0.35,\quad P(Y^L=60)=0.20,\quad P(Y^L=250)=0.15,\quad P(Y^L=300)=0.30.

The event $0<Y^L\leq250$ includes the payments of $60$ and $250$, giving $0.20+0.15=0.35$. The probabilities in the full distribution sum to $0.35+0.20+0.15+0.30=1.00$. The important endpoint check is that the loss of exactly $500$ contributes to the limit payment, not to an interior payment.

Worked Example: A Mixed Loss Distribution

A hypothetical property contract has $d=\$100$ and a maximum payment $u=\$300$. With probability $0.10$, the loss is exactly $\$100$; with probability $0.20$, it is exactly $\$400$; otherwise, with probability $0.70$, it is uniformly distributed between $\$0$ and $\$800$. Find the boundary masses and $P(100<Y^L\leq250)$.

Method. Calculate each boundary probability from both the discrete components and the continuous component, then map the requested interval to the loss scale.

For zero payment, the point mass at $\$100$ is included because it equals the deductible. The uniform component also produces zero payment when its loss is at most $\$100$, which has conditional probability $100/800=0.125$. Thus,

P(Y^L=0)=0.10+0.70\left(\frac{100}{800}\right) =0.10+0.0875 =0.1875.

The limit threshold is $d+u=\$400$. The point mass at $\$400$ is therefore included at the limit. For the uniform component, the probability of a loss at least $\$400$ is $(800-400)/800=0.5$. Consequently,

$$ P(Y^L=300)=0.20+0.70(0.5)=0.55. $$

For the requested interior interval, $100<Y^L\leq250$ corresponds to $200<X\leq350$. Neither point mass lies in that loss interval. The uniform component assigns it probability $(350-200)/800=0.1875$, so

P(100<Y^L\leq250)=0.70\left(\frac{150}{800}\right) =0.13125.

As a check, the remaining interior-payment probability is $1-0.1875-0.55=0.2625$. The requested interval probability $0.13125$ is no larger than this total. Rounded to four decimal places, the answers are $0.1875$, $0.5500$, and $0.1313$. The mixture’s atoms must be handled separately from its continuous component; a density calculation alone would omit both point masses.

Common Exam Trap: Misplacing the Cap Boundary

A frequent error is to treat the payment limit $u$ as the ground-up loss threshold. Under the convention used here, the insurer first subtracts the deductible and then caps its payment. The cap is reached at $X=d+u$, not at $X=u$. In the discrete example above, incorrectly using $X\geq300$ as the cap event would count the losses of $450$, $500$, and $720$, producing $0.45$ instead of the correct $0.30$.

A second endpoint error is to calculate the cap probability as $1-F_X(d+u)$. This excludes losses exactly equal to $d+u$. In the same discrete example, $1-F_X(500)=0.20$, because it counts only $X>500$. The correct probability is $P(X\geq500)=1-F_X(500-)=0.30$, including the $0.10$ mass at $X=500$.

Key takeaway: Under $Y^L=\min((X-d)_+,u)$, the zero-payment atom is $P(X\leq d)$, the limit-payment atom is $P(X\geq d+u)$, and interior payment events are translated by adding $d$ to the payment bounds. Check equality at both thresholds whenever $X$ may have atoms.

Check Your Understanding

For each problem, use the stated convention that $u$ is the maximum insurer payment per loss.

A hypothetical loss is exponential with mean $\$500$. A contract has $d=\$100$ and $u=\$200$. Find $P(Y^L=0)$ and $P(Y^L=200)$. (Answer: $0.1813$ and $0.5488$)
A discrete loss equals $\$50,\$200,\$300,\$500,$ or $\$700$ with respective probabilities $0.10,0.20,0.25,0.15,$ and $0.30$. For $d=\$200$ and $u=\$300$, find $P(Y^L=0)$, $P(Y^L=100)$, and $P(Y^L=300)$. (Answer: $0.3000,\ 0.2500,\ 0.4500$)
A loss is uniform on $[0,1000]$, with $d=\$200$ and $u=\$300$. Find $P(Y^L=0)$, $P(Y^L=300)$, and $P(50<Y^L\leq200)$. (Answer: $0.2000,\ 0.5000,\ 0.1500$)
A loss equals $\$250$ with probability $0.30$; otherwise it is uniform on $[0,1000]$, with probability $0.70$. For $d=\$250$ and $u=\$400$, find $P(Y^L=0)$, $P(Y^L=400)$, and $P(100<Y^L\leq300)$. (Answer: $0.4750,\ 0.2450,\ 0.1400$)
A loss equals $\$100$ with probability $0.15$, $\$350$ with probability $0.25$, $\$600$ with probability $0.20$, and $\$900$ with probability $0.40$. For $d=\$100$ and $u=\$500$, list the distinct payment values and their probabilities. (Answer: $P(Y^L=0)=0.15,\ P(Y^L=250)=0.25,\ P(Y^L=500)=0.60$)

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