Zero Payment Probability Under a Deductible

Why the Boundary Matters

An ordinary deductible leaves the policyholder responsible for the first $d$ dollars of a covered loss. The insurer pays only the amount above that deductible. To find the probability of no insurer payment, the key question is therefore whether the loss is at or below the deductible. The boundary is important: a loss exactly equal to $d$ produces no payment, even if the loss distribution assigns positive probability to that exact value.

This is immediate for a continuous severity model with no point masses, but it requires care for discrete and mixed models. In particular, a CDF evaluated at $d$ includes losses equal to $d$, whereas a left-limit CDF does not. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a nonnegative ground-up loss and let $d\geq 0$ be an ordinary deductible. The insurer's payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. A zero payment occurs exactly when $X\leq d$. Thus, for $F_X(d)=P(X\leq d)$, the probability of zero payment is $P(Y^L=0)=F_X(d)$, and the probability of a positive payment is $P(Y^L>0)=P(X>d)=1-F_X(d)$.

The notation $F_X(d)$ means the probability that the loss is no greater than $d$, including any probability mass at $d$. When a survival function is defined by $S_X(d)=P(X>d)$, the same result can be written $P(Y^L=0)=1-S_X(d)$. Always check the stated survival-function convention: some settings use a tail probability that includes equality, which would require adjusting the complement.

Deriving the Zero-Payment Event

The result follows directly from the payment rule. Since $Y^L=\max(X-d,0)$, the payment is zero if and only if the expression $X-d$ is nonpositive. Rearranging gives $X\leq d$. The equality case is included because a loss equal to the deductible leaves nothing above the deductible for the insurer to pay.

\begin{aligned} Y^L=0 &\iff \max(X-d,0)=0\\ &\iff X-d\leq 0\\ &\iff X\leq d,\\[4pt] P(Y^L=0) &=P(X\leq d)=F_X(d). \end{aligned}

No assumption of continuity is needed for this event argument. Continuity matters only for shortcuts used to evaluate $F_X(d)$. If $X$ has a density and no atom at $d$, then $P(X=d)=0$, so $P(X\leq d)=P(X<d)$. If there is an atom at $d$, those probabilities differ by $P(X=d)$.

Theorem: For any nonnegative loss distribution and any ordinary deductible $d\geq0$, including a distribution with discrete masses or a mixture of discrete and continuous components, $P(Y^L=0)=F_X(d)$. In particular, an atom at the deductible contributes to the zero-payment probability, not to the positive-payment probability.

Worked Calculations

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

A hypothetical policy has an ordinary deductible of $d=\$1{,}000$. Its ground-up loss $X$ is generated as follows: with probability $0.10$, $X=\$0$; with probability $0.20$, $X=\$1{,}000$; and with probability $0.70$, $X$ follows an exponential distribution with mean $\$2{,}000$. Find the probability of no insurer payment. No interest or mortality assumption is involved.

Method. Use $P(Y^L=0)=P(X\leq d)$, adding the contributions from each mixture component. The exponential component has rate $\lambda=1/2{,}000$ per dollar, so its CDF at $\$1{,}000$ is $1-e^{-1000/2000}$.

The point mass at zero is at or below the deductible, so it contributes $0.10$. The point mass at $\$1{,}000$ is exactly at the deductible, so it contributes $0.20$. The exponential component contributes

0.70\left(1-e^{-1000/2000}\right) =0.70(1-e^{-0.5}) \approx 0.70(0.393469) =0.275428.

Therefore,

P(Y^L=0)=0.10+0.20+0.275428=0.575428\approx 0.5754.

To check, the probability of a positive payment is the exponential component's remaining tail, $0.70e^{-0.5}\approx0.4246$. The two probabilities sum to $1.0000$. The common mistake avoided here is omitting the $0.20$ atom at the deductible: equality still produces zero payment.

Worked Example: A Discrete Loss Exactly at the Deductible

A hypothetical equipment policy has a deductible of $d=\$900$. The loss distribution is $P(X=\$0)=0.40$, $P(X=\$900)=0.30$, $P(X=\$1{,}500)=0.20$, and $P(X=\$3{,}000)=0.10$. Find the probability of no payment and the probability of a positive payment. No interest or mortality assumption is involved.

Method. Add the probabilities for all loss values $x\leq \$900$. The eligible outcomes are $\$0$ and $\$900$:

P(Y^L=0)=P(X=\$0)+P(X=\$900)=0.40+0.30=0.70.

The remaining outcomes exceed the deductible, so

P(Y^L>0)=P(X=\$1{,}500)+P(X=\$3{,}000)=0.20+0.10=0.30.

The check $0.70+0.30=1.00$ confirms the partition. A frequent error is to use $P(X<\$900)=0.40$ as the zero-payment probability. That excludes the $\$900$ outcome even though the insurer pays $\max(900-900,0)=\$0$.

Worked Example: A Zero Deductible With a Point Mass at Zero

A hypothetical small-claims portfolio has $d=\$0$. With probability $0.25$, the loss is exactly zero; with probability $0.75$, the loss is uniformly distributed from $\$0$ to $\$1{,}200$. Find the probability of no insurer payment. No interest or mortality assumption is involved.

Method. A zero deductible gives zero payment exactly when $X\leq0$. Because losses are nonnegative, this event is $X=0$. The point mass at zero contributes $0.25$. The uniform component is continuous, so its probability of being exactly zero is zero:

$$ P(Y^L=0)=P(X=0)=0.25+0.75(0)=0.2500. $$

Thus the probability of a positive payment is $1-0.25=0.75$, and the two probabilities sum to one. The common mistake is to conclude that a zero deductible means the zero-payment probability must be zero. That would be true only if the loss distribution had no probability mass at zero.

Boundary Cases and Calculation Choices

The general formula handles the main boundary cases without a special adjustment:

At the deductible: if $P(X=d)>0$, that mass is included in $F_X(d)$, because $X=d$ gives $Y^L=0$.
At zero deductible: $P(Y^L=0)=P(X=0)$ for nonnegative losses. This may be positive or zero, depending on the model.
Above the deductible: losses strictly greater than $d$ produce positive payment, so the relevant event is $X>d$.
Continuous loss at the boundary: if $X$ has no atom at $d$, then $P(X=d)=0$. In that case, using $P(X<d)$ gives the same answer as $P(X\leq d)$, but this shortcut is not valid in general.

For a distribution with point masses, calculate $F_X(d)$ by adding all masses at or below $d$ and the probability contributed by any continuous components over values at or below $d$. For a fully discrete model, sum the probabilities of all outcomes $x\leq d$. For a continuous model with a known CDF, evaluate that CDF at $d$. The method changes with the model, but the event being calculated does not.

Identify the payment rule.
For an ordinary deductible, use $Y^L=(X-d)_+$, with $d\geq0$.

Translate zero payment into a loss event.
From the rule, $Y^L=0$ if and only if $X\leq d$.

Evaluate the event using the full distribution.
Compute $F_X(d)=P(X\leq d)$, including any atom at $d$.

Check the complement.
Verify that the positive-payment probability $P(X>d)$ adds to the zero-payment probability to equal one.

Common Exam Trap: Using the Left Limit

For a distribution that may have an atom at $d$, the left limit $F_X(d-)=P(X<d)$ is not the required probability of zero payment. The relationship is

$$ F_X(d)=F_X(d-)+P(X=d). $$

For example, if $F_X(d-)=0.42$ and $P(X=d)=0.13$, then the probability of no payment is $0.42+0.13=0.55$, not $0.42$. The error comes from treating “at or below” as “strictly below.” A second version of the same error is to count an atom at $d$ as a positive insurer payment; under an ordinary deductible, positive payment requires $X>d$, not $X\geq d$.

Do not confuse a deductible with a franchise deductible. The payment rule used here is the ordinary deductible rule: the insurer pays only the amount in excess of $d$. A different contract rule can create a different zero-payment event, so the payment definition must be checked before calculating.

Key takeaway: For an ordinary deductible, calculate the zero-payment probability as $F_X(d)=P(X\leq d)$. Include every loss at the deductible, and use $P(X<d)$ only when the model establishes that there is no probability mass at $d$.

Check Your Understanding

For each hypothetical loss model, find the requested probability. Treat each deductible as an ordinary deductible. No interest or mortality assumptions are involved.

A loss is uniformly distributed from $\$0$ to $\$500$, and the deductible is $\$125$. Find the probability of no payment. (Answer: $0.2500$)
A loss is $\$0$ with probability $0.08$, $\$600$ with probability $0.22$, and otherwise follows an exponential distribution with mean $\$600$; the deductible is $\$600$. Find the probability of no payment. (Answer: $0.7425$)
In a hypothetical sample of 20 losses, 13 observations are at or below $\$250$, including observations equal to $\$250$. Using the empirical distribution, estimate the zero-payment probability for a $\$250$ deductible. (Answer: $0.6500$)
A loss is zero with probability $0.17$; otherwise it has a continuous distribution strictly above zero. Find the zero-payment probability for a zero deductible. (Answer: $0.1700$)
A loss is $\$0$, $\$500$, or $\$800$ with probabilities $0.10$, $0.20$, and $0.10$, respectively; otherwise it is uniform from $\$0$ to $\$1{,}200$. Find the zero-payment probabilities for deductibles of $\$500$ and $\$800$. (Answer: $0.5500$ for $\$500$; $0.8000$ for $\$800$)

What You'll Learn