Introduction
An ordinary deductible changes not only the amount paid on a loss but also the shape of its probability distribution. Losses at or below the deductible produce no payment, so the payment distribution has a probability mass at zero. Losses above the deductible are shifted downward by the deductible. Describing both features correctly is essential when calculating probabilities for insurance payments.
Let \(X\) be a nonnegative loss and let \(d\geq 0\) be a fixed ordinary deductible. The insurer's payment per loss is \(Y=(X-d)^+\), where \(z^+=\max(z,0)\). This tutorial derives the CDF and survival function of \(Y\), then distinguishes the unconditional distribution of payment from the distribution conditional on a positive payment.
Deriving the Payment Distribution
The key is to translate an event about \(Y\) into an event about \(X\). Since \(Y\) cannot be negative, its CDF is zero below zero. For any \(y\geq 0\), the event \(Y\leq y\) occurs exactly when \(X\leq d+y\): a loss at or below \(d\) gives zero payment, and a loss above \(d\) gives payment no greater than \(y\) precisely when \(X-d\leq y\).
For \(y\geq 0\), \(Y\leq y\) is equivalent to \(X\leq d+y\). For \(y<0\), the event is impossible because \(Y\geq 0\).
Use the CDF of \(X\) on the translated event. At \(y=0\), this gives the probability that the loss is at or below the deductible.
For \(y\geq 0\), \(Y>y\) is equivalent to \(X>d+y\), so its probability is the loss survival function evaluated at \(d+y\).
The boundary at zero matters. The zero-payment event is \(X\leq d\), not \(X<d\), because a loss exactly equal to the deductible produces no payment. The positive-payment event is \(X>d\). These statements remain true whether \(X\) is continuous, discrete, or a mixture of the two.
The formula also handles atoms in the loss distribution. Any probability at loss amounts at or below \(d\) is combined into the mass at payment zero. An atom at a loss amount \(x>d\) becomes an atom at payment \(x-d\), with the same probability. No continuity assumption is needed for the CDF and survival formulas.
The Distribution Given a Positive Payment
Sometimes the question concerns payment size only after a claim has exceeded the deductible. This is a conditional distribution, not the unconditional distribution of \(Y\). It is defined only if \(S_X(d)=P(X>d)>0\). For \(y\geq 0\), the event \(0<Y\leq y\) is equivalent to \(d<X\leq d+y\). Therefore:
The denominator removes the probability of no payment. It is therefore incorrect to use the unconditional CDF \(F_Y(y)\) as the CDF conditional on a positive payment. For a continuous loss distribution with density \(f_X\), the unconditional payment distribution has mass \(F_X(d)\) at zero and, for \(y>0\), density \(f_Y(y)=f_X(d+y)\) wherever that density exists. Conditional on positive payment, the density is \(f_X(d+y)/S_X(d)\). For a discrete or mixed loss distribution, use the event and probability formulas rather than assuming a density.
Worked Examples
Worked Example: Exponential Loss With a Deductible
A hypothetical insurer models a loss as \(X\sim\operatorname{Exp}(\lambda=0.5)\), with rate \(0.5\) per dollar, and applies a deductible of \(d=2\) dollars. Find the probability of no payment, the payment CDF and survival probability at \(y=1\), and the conditional probability that a positive payment is at most \(1\) dollar. Give probabilities to four decimal places.
Method. For an exponential loss with rate \(0.5\), \(F_X(x)=1-e^{-0.5x}\) and \(S_X(x)=e^{-0.5x}\) for \(x\geq0\).
No payment. The zero-payment probability includes losses equal to the deductible: \(P(Y=0)=F_X(2)=1-e^{-0.5(2)}=1-e^{-1}=0.6321\). Thus \(P(Y>0)=e^{-1}=0.3679\).
Unconditional distribution at \(y=1\). Since \(d+y=3\), \(F_Y(1)=F_X(3)=1-e^{-0.5(3)}=1-e^{-1.5}=0.7769\). The survival probability is \(S_Y(1)=S_X(3)=e^{-1.5}=0.2231\). These add to \(1.0000\), as the CDF and strict survival probabilities should.
Conditional distribution. The probability of a positive payment no greater than \(1\) dollar is \(F_X(3)-F_X(2)=e^{-1}-e^{-1.5}=0.1447\). Dividing by \(P(Y>0)=e^{-1}=0.3679\) gives \(P(Y\leq1\mid Y>0)=0.1447/0.3679=0.3935\), with the ratio evaluated before rounding.
Check and trap avoided. The conditional probability \(0.3935\) is smaller than the unconditional CDF \(0.7769\), because the latter includes the \(0.6321\) probability mass at zero. Using \(F_Y(1)\) as the conditional answer would include no-payment outcomes.
Worked Example: A Discrete Loss With an Atom at the Deductible
A hypothetical equipment policy has loss distribution \(P(X=0)=0.10\), \(P(X=100)=0.20\), \(P(X=200)=0.20\), \(P(X=350)=0.30\), and \(P(X=500)=0.20\), with amounts in dollars. The deductible is \(d=200\). Find the payment distribution, \(F_Y(150)\), \(S_Y(150)\), and the conditional CDF of payment at \(150\) given positive payment.
Method. Apply \(Y=(X-200)^+\) to each loss value. Losses \(0,100,\) and \(200\) all produce payment zero. The losses \(350\) and \(500\) produce payments \(150\) and \(300\), respectively.
Payment probabilities. The probability at zero is \(0.10+0.20+0.20=0.50\). The probability at payment \(150\) is \(0.30\), and the probability at payment \(300\) is \(0.20\). The probabilities sum to \(0.50+0.30+0.20=1.00\).
Unconditional CDF and survival. Payment is at most \(150\) for the zero-payment outcomes and the outcomes paying \(150\), so \(F_Y(150)=0.50+0.30=0.80\). Strict survival means payment is greater than \(150\), not greater than or equal to it. Only payment \(300\) qualifies, so \(S_Y(150)=0.20\). The values sum to \(1.00\).
Conditional CDF. Positive payment has probability \(0.30+0.20=0.50\). Among these outcomes, payment is at most \(150\) with probability \(0.30\). Hence \(P(Y\leq150\mid Y>0)=0.30/0.50=0.60\).
Check and trap avoided. The loss \(X=200\) equals the deductible and contributes to the zero-payment mass. Also, \(S_Y(150)=P(Y>150)\) excludes payment exactly \(150\); treating survival as \(P(Y\geq150)\) would incorrectly give \(0.50\).
Worked Example: Uniform Loss and the Positive-Payment Distribution
For a hypothetical product, suppose \(X\sim\operatorname{Uniform}(0,10)\) dollars and the ordinary deductible is \(d=4\) dollars. Find the payment CDF for all \(y\), the zero-payment probability, \(P(1<Y\leq4)\), and the conditional CDF at \(y=3\) given a positive payment.
Method and loss distribution. For \(0\leq x\leq10\), \(F_X(x)=x/10\); below zero it is zero, and above \(10\) it is one. Thus \(F_X(4)=0.4\), so \(P(Y=0)=0.4\) and \(P(Y>0)=0.6\).
Payment CDF. For \(0\leq y\leq6\), \(d+y=4+y\) lies between zero and ten, giving \(F_Y(y)=F_X(4+y)=(4+y)/10\). For \(y>6\), \(F_Y(y)=1\); for \(y<0\), \(F_Y(y)=0\). In particular \(F_Y(0)=0.4\), \(F_Y(3)=0.7\), and \(F_Y(6)=1\).
Payment interval probability. The event \(1<Y\leq4\) corresponds to \(5<X\leq8\). Therefore its probability is \(F_X(8)-F_X(5)=0.8-0.5=0.3\).
Conditional CDF. At \(y=3\), \(P(Y\leq3\mid Y>0)=[F_X(7)-F_X(4)]/S_X(4)=(0.7-0.4)/0.6=0.5\).
Check and trap avoided. The conditional probability is \(0.5\), not the unconditional CDF \(0.7\). The zero-payment mass is \(0.4\); among positive payments, the amount is spread over the interval from zero to six dollars.
Common Boundary and Interpretation Errors
A reliable way to avoid mistakes is to write the event in terms of the original loss before evaluating a probability. In particular, use the strict or non-strict inequality requested rather than relying on a remembered formula.
- Putting the deductible boundary on the wrong side. Since \(X=d\) gives \(Y=0\), use \(P(Y=0)=P(X\leq d)=F_X(d)\). The probability of positive payment is \(P(X>d)=S_X(d)\).
- Forgetting the mass at zero. The unconditional CDF at a positive payment threshold includes the entire zero-payment mass. A CDF conditional on positive payment must exclude it and divide by \(S_X(d)\).
- Confusing survival with a non-strict tail. Here \(S_Y(y)=P(Y>y)\). At an atom, \(P(Y>y)\) can differ from \(P(Y\geq y)\), as the discrete example shows.
- Assuming continuity without being told. The CDF and survival formulas do not require a continuous loss distribution. Density formulas do require an appropriate density and do not describe atoms.
These distribution functions are useful whenever a problem asks for payment probabilities, payment quantiles, or the distribution conditional on a claim exceeding the deductible. They also provide a direct check on later calculations involving payment moments: the no-payment probability and positive-payment distribution must together account for all outcomes. Those later calculations should start from the correct payment distribution rather than treating the deductible as a simple rescaling of the loss.
Check Your Understanding
For each problem, identify the deductible boundary carefully and distinguish unconditional from conditional probabilities where requested.
- A hypothetical loss has \(X\sim\operatorname{Exp}(\lambda=0.2)\), with rate \(0.2\) per dollar, and deductible \(d=3\) dollars. Find \(P(Y=0)\) and \(S_Y(2)\), to four decimal places. (Answer: \(P(Y=0)=0.4512\), \(S_Y(2)=0.3679\))
- A loss takes values \(100,250,400\) dollars with probabilities \(0.25,0.35,0.40\), respectively. If \(d=250\), find the payment distribution and \(F_Y(150)\). (Answer: \(P(Y=0)=0.60\), \(P(Y=150)=0.40\); \(F_Y(150)=1.00\))
- Suppose \(X\sim\operatorname{Uniform}(0,20)\) dollars and \(d=5\). Find \(P(Y=0)\) and \(P(2<Y\leq7)\). (Answer: \(P(Y=0)=0.2500\), \(P(2<Y\leq7)=0.2500\))
- A hypothetical loss has \(X\sim\operatorname{Exp}(\lambda=0.4)\), with rate \(0.4\) per dollar, and deductible \(d=1\) dollar. Find \(P(Y\leq2\mid Y>0)\), to four decimal places. (Answer: \(0.5507\))
- A discrete loss has \(P(X=0)=0.10\), \(P(X=100)=0.30\), and \(P(X=300)=0.60\), with deductible \(d=100\). Find \(P(Y=0)\), \(F_Y(200)\), and \(S_Y(200)\). (Answer: \(P(Y=0)=0.40\), \(F_Y(200)=1.00\), \(S_Y(200)=0.00\))