Why an Ordinary Deductible Creates a Zero Payment
A deductible changes the amount paid by an insurer, but it can also change the shape of the payment distribution. For an ordinary deductible, losses at or below the deductible produce no payment. Consequently, even when the original loss distribution is continuous, the insurer’s payment distribution has a point mass at zero. Calculating that probability correctly is important whenever a problem asks how often a claim generates no payment.
Let \(X\) be a nonnegative loss amount and let \(d\geq 0\) be an ordinary deductible. Assume the insurer pays the portion of a loss above \(d\), with no policy limit or coinsurance in this model. The payment per loss is \(Y^L\).
The inequality is inclusive: a loss exactly equal to the deductible produces a payment of \(d-d=0\). This matters particularly for discrete and mixed distributions, which can assign positive probability to the exact deductible amount. For a continuous distribution, the probability of any one exact loss amount is zero, but the same inclusive formula still applies.
Why the Formula Holds
The result follows directly from the payment rule. Since \(Y^L=\max(X-d,0)\), its value is zero when the expression inside the maximum is nonpositive; otherwise the payment is positive.
Taking probabilities of the equivalent events gives \(P(Y^L=0)=P(X\leq d)\). By the definition of a cumulative distribution function, \(P(X\leq d)=F_X(d)\). The same reasoning also identifies the complementary event:
The payment distribution has a jump at zero. For \(y<0\), \(P(Y^L\leq y)=0\), because a payment cannot be negative. At \(y=0\), the cumulative probability becomes \(P(Y^L\leq 0)=F_X(d)\). Thus the size of the jump at zero is precisely \(F_X(d)\). This remains true whether the original severity is continuous, discrete, or a mixture of the two.
Calculating the Atom for Different Severity Models
For a continuous severity distribution, evaluate its cumulative distribution function at the deductible. For a discrete distribution, add the probabilities of every loss value at or below the deductible. For a mixed distribution, account for each component that can produce a loss at or below the deductible. In every case, the condition is \(X\leq d\), not \(X<d\).
| Severity model | Zero-payment probability | Calculation detail |
|---|---|---|
| Continuous | \(F_X(d)\) | Evaluate the CDF at the deductible |
| Discrete | \(\sum_{x\leq d}P(X=x)\) | Include all probability values at or below \(d\) |
| Mixed | \(P(X\leq d)\) | Add relevant point masses and continuous probability below the threshold |
Worked Example: Exponential Severity and a Continuous Deductible Threshold
A hypothetical insurer models each claim amount \(X\) as exponential with rate \(\lambda=1/10{,}000\) per dollar. Thus the mean loss is \(\$10{,}000\), and \(F_X(x)=1-e^{-\lambda x}\) for \(x\geq 0\). The ordinary deductible is \(d=\$4{,}000\). Find the probability that the insurer makes no payment and check it using the positive-payment probability.
Method. Since the severity is continuous, evaluate its CDF at the deductible:
Using \(e^{-0.4}\approx 0.6703200\), the zero-payment probability is \(1-0.6703200=0.3296800\), or 0.3297 to four decimal places.
For the check, a positive payment requires \(X>4{,}000\), so \(P(Y^L>0)=e^{-0.4}\approx 0.6703\). The probabilities add to \(0.3296800+0.6703200=1.0000000\). The method is direct CDF evaluation; the common mistake it avoids is using the survival probability \(P(X>d)\) as the probability of zero payment rather than as its complement.
Worked Example: A Discrete Severity With a Loss at the Deductible
A hypothetical equipment policy has an ordinary deductible of \(\$1{,}500\). Its loss \(X\) takes the values \(\$0,\$500,\$1{,}500,\) and \(\$4{,}000\), with respective probabilities \(0.10, 0.25, 0.30,\) and \(0.35\). Find the probability of a zero payment, and identify the payment associated with each possible loss.
Method. First check that the probabilities define a distribution: \(0.10+0.25+0.30+0.35=1.00\). A zero payment occurs for every loss satisfying \(X\leq \$1{,}500\). The relevant outcomes are \(\$0,\$500,\) and \(\$1{,}500\), so
The corresponding payments are \(\max(0-1{,}500,0)=\$0\), \(\max(500-1{,}500,0)=\$0\), \(\max(1{,}500-1{,}500,0)=\$0\), and \(\max(4{,}000-1{,}500,0)=\$2{,}500\). Thus \(P(Y^L=0)=0.65\), and \(P(Y^L=\$2{,}500)=0.35\). These payment probabilities sum to \(1.00\).
The method is to sum the probabilities of all discrete losses at or below the deductible. The common mistake it avoids is excluding the \(\$1{,}500\) outcome: equality still gives a zero payment, so its probability \(0.30\) must be included.
Worked Example: A Mixed Severity With a Point Mass at the Deductible
For a hypothetical class of claims, the loss \(X\) has a mixed distribution: with probability \(0.30\), \(X=\$2{,}000\); with probability \(0.70\), \(X\) follows an exponential distribution with mean \(\$4{,}000\). The deductible is \(d=\$2{,}000\). Find the probability of a zero payment, including the contribution from the point mass.
Method. Separate the two mixture components. The point-mass component always produces a loss equal to the deductible, so all of its \(0.30\) probability produces zero payment. In the exponential component, the rate is \(\lambda=1/4{,}000\) per dollar, and the probability of a loss at or below \(\$2{,}000\) is
Weight this conditional component probability by its mixture probability and add the point mass:
The zero-payment probability is 0.5754 to four decimal places. The complementary positive-payment probability is \(0.70e^{-0.5}\approx 0.4245715\), and the two probabilities sum to \(1.0000000\). The method is to combine mixture components using their weights; the common mistake it avoids is applying only the exponential CDF and omitting the \(0.30\) point mass at the deductible.
Common Exam Trap: Strict Versus Inclusive Inequalities
The deductible condition is determined by the payment function, not by a convention about how a threshold is usually written. For an ordinary deductible, \(X=d\) gives \(Y^L=0\), so the zero-payment event is \(X\leq d\). The positive-payment event is \(X>d\). These events are complements, including when the loss distribution has a jump at \(d\).
In the discrete example, an incorrect strict-inequality calculation would use only \(P(X<\$1{,}500)=0.10+0.25=0.35\). That is not the zero-payment probability: it leaves out the \(0.30\) probability at \(X=\$1{,}500\), even though that loss produces no insurer payment. The correction is
For a continuous severity, \(P(X=d)=0\), so \(P(X<d)=P(X\leq d)\). That equality can make the strict-inequality error invisible in a continuous-only calculation. It does not justify using \(X<d\) as the general rule: discrete and mixed models may have positive probability at the threshold.
Check Your Understanding
For each hypothetical severity model, calculate the probability that an ordinary deductible produces zero payment. Give probabilities to four decimal places.
- \(X\) is uniform on \([\$0,\$12{,}000]\), and \(d=\$3{,}000\). What is \(P(Y^L=0)\)? (Answer: 0.2500)
- \(X\) takes values \(\$0,\$800,\$1{,}600,\) and \(\$5{,}000\) with probabilities \(0.08,0.22,0.40,\) and \(0.30\), respectively. If \(d=\$1{,}600\), what is \(P(Y^L=0)\)? (Answer: 0.7000)
- \(X\) is exponential with mean \(\$5{,}000\), and \(d=\$2{,}500\). What is \(P(Y^L=0)\)? (Answer: 0.3935)
- With probability \(0.20\), \(X=\$0\); with probability \(0.80\), \(X\) is uniform on \([\$0,\$10{,}000]\). If \(d=\$2{,}500\), what is \(P(Y^L=0)\)? (Answer: 0.4000)
- With probability \(0.15\), \(X=\$3{,}000\); otherwise \(X\) is exponential with mean \(\$6{,}000\). For \(d=\$3{,}000\), what is \(P(Y^L=0)\)? (Answer: 0.4844)