Conditional Probabilities Under Deductibles

Why Condition on a Positive Payment?

A deductible can eliminate some losses from the insurer’s payments. A probability conditional on a positive payment therefore describes the distribution among claims that pass the deductible, rather than among all losses. The distinction matters: the condition changes the probability denominator, and the deductible type determines how a loss threshold translates into a payment threshold.

This tutorial considers an individual nonnegative loss $X$ and a deductible $d\ge 0$. We use strict exceedance throughout: a loss must be greater than $d$ to produce a positive payment. This convention is important for discrete loss models, which can assign positive probability to exactly $d$. Results use the survival function $S_X(t)=P(X>t)$, so the strict inequality is explicit. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Under an ordinary deductible, the payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. Under a franchise deductible, the insurer pays the full loss if it exceeds the deductible and otherwise pays nothing: $Y^F=X\mathbf{1}_{\{X>d\}}$. In either case, the positive-payment event is $\{X>d\}$, and $P(Y>0)=S_X(d)$. Conditional probabilities given a positive payment are defined only if $S_X(d)>0$.

The payment amounts differ once the loss exceeds $d$. With an ordinary deductible, a loss of $d+4$ produces a payment of $4$; with a franchise deductible, it produces a payment of $d+4$. Thus the conditional loss probability $P(X>x\mid Y>0)$ is the same for either contract, but the conditional payment probability $P(Y>y\mid Y>0)$ generally is not.

Conditional Loss Probabilities

Conditioning on a positive payment is equivalent to conditioning on $X>d$, for either deductible. For a threshold $x\ge d$, the event $X>x$ is contained in $X>d$, so the intersection in the conditional probability is just $X>x$. If $x<d$, every loss that passes the deductible is already greater than $x$.

P(X>x\mid Y>0)=P(X>x\mid X>d)= \begin{cases} 1, & x<d,\\[4pt] \dfrac{S_X(x)}{S_X(d)}, & x\ge d, \end{cases} \qquad S_X(d)>0.

At $x=d$, the ratio equals one, as it should: conditional on $X>d$, the loss exceeds $d$ with certainty. At $x>d$, the survival-function ratio accounts for both the losses that exceed $x$ and the reduced population of losses that passed the deductible. No continuity assumption is needed; using $S_X(t)=P(X>t)$ makes the result valid even when $X$ has atoms.

Conditional Payment Probabilities

For nonnegative payment thresholds $y$, the payment event can be translated back into an event about $X$. Under an ordinary deductible, a positive payment greater than $y$ requires the loss to exceed $d+y$. Under a franchise deductible, every loss above $d$ is paid in full, so a payment greater than $y$ requires the loss to exceed $y$, as well as to pass the deductible.

Theorem: For $y\ge0$ and $S_X(d)>0$, $$ P(Y^L>y\mid Y^L>0)=\frac{S_X(d+y)}{S_X(d)}. $$ For a franchise deductible, $$ P(Y^F>y\mid Y^F>0)= \begin{cases} 1, & 0\le y\le d,\\[4pt] \dfrac{S_X(y)}{S_X(d)}, & y>d. \end{cases} $$

At $y=d$, every positive franchise payment is strictly greater than $d$, since it comes from a loss strictly greater than $d$. The ratio $S_X(d)/S_X(d)$ would also give one at this boundary. For an ordinary deductible, $P(Y^L>0\mid Y^L>0)=1$, and the formula at $y=0$ gives exactly that result.

Identify the conditioning event.
For either deductible under the stated strict-threshold convention, $Y>0$ means $X>d$. Check that $S_X(d)>0$.

Translate the event of interest.
For a loss threshold, retain $X>x$. For a payment threshold, use $X>d+y$ for an ordinary deductible and $X>y$ for a franchise deductible, while also retaining $X>d$.

Divide by the positive-payment probability.
Use $P(X>d)=S_X(d)$ as the denominator. If the translated event is automatic among positive payments, the conditional probability is one.

Worked Examples

Worked Example: Exponential Losses Under Both Deductibles

A hypothetical insurer models a claim amount by an exponential distribution with rate $\lambda=0.1$ per dollar, so $S_X(t)=e^{-0.1t}$ for $t\ge0$. The deductible is $d=\$5$. Calculate (a) $P(X>10\mid Y>0)$, (b) the probability that an ordinary-deductible payment exceeds $\$3$, given a positive payment, and (c) the probability that a franchise-deductible payment exceeds $\$8$, given a positive payment. Give probabilities to four decimals. Method: translate each event to a loss threshold and divide its survival probability by $S_X(5)$. The main trap is using an unconditional tail probability as the answer.

First, the positive-payment probability for either contract is $S_X(5)=e^{-0.1(5)}=e^{-0.5}=0.6065307$.

(a) Conditional loss tail. Since $10\ge5$, $$ P(X>10\mid Y>0)=\frac{S_X(10)}{S_X(5)} =\frac{e^{-1}}{e^{-0.5}} =e^{-0.5}=0.6065307. $$ Rounded to four decimals, the answer is $0.6065$.

(b) Ordinary payment. The event $Y^L>3$ is $X-5>3$, or $X>8$. Therefore $$ P(Y^L>3\mid Y^L>0)=\frac{S_X(8)}{S_X(5)} =\frac{e^{-0.8}}{e^{-0.5}} =e^{-0.3}=0.7408182. $$ The answer is $0.7408$.

(c) Franchise payment. Here a positive payment equals the full loss. Since $8>d$, the event $Y^F>8$, conditional on a positive payment, has probability $$ P(Y^F>8\mid Y^F>0)=\frac{S_X(8)}{S_X(5)} =\frac{e^{-0.8}}{e^{-0.5}} =e^{-0.3}=0.7408182. $$ The answer is $0.7408$. As a check, a franchise payment greater than $8$ requires a loss greater than $8$, whereas an ordinary payment greater than $3$ requires a loss greater than $5+3=8$; the matching probabilities are expected.

Worked Example: Discrete Losses and a Boundary Atom

A hypothetical warranty loss $X$ takes values $\$2,\$6,\$9,\$14$ with probabilities $0.15, 0.25, 0.35, 0.25$, respectively. The deductible is $d=\$6$, and the contract pays only when $X>d$. Calculate (a) $P(X>9\mid Y>0)$, (b) $P(Y^L>3\mid Y^L>0)$, and (c) $P(Y^F>8\mid Y^F>0)$. Give probabilities to four decimals. Method: list the outcomes in each strict-threshold event. The trap is treating a loss equal to the deductible as a positive payment.

The loss probabilities sum to $0.15+0.25+0.35+0.25=1$. Because the loss of $\$6$ does not exceed the deductible, positive payment occurs only for losses of $\$9$ and $\$14$. Thus $$ P(Y>0)=P(X>6)=0.35+0.25=0.60. $$

(a) Conditional loss tail. Of the positive-payment outcomes, only $X=14$ exceeds $9$. Hence $$ P(X>9\mid Y>0)=\frac{0.25}{0.60}=0.4166667, $$ which rounds to $0.4167$.

(b) Ordinary payment. Payments at $X=9$ and $X=14$ are $\$3$ and $\$8$, respectively. The strict event $Y^L>3$ excludes the $\$3$ payment, so $$ P(Y^L>3\mid Y^L>0)=\frac{P(X=14)}{0.60} =\frac{0.25}{0.60}=0.4167. $$

(c) Franchise payment. The payments are $\$9$ and $\$14$. Only the $\$14$ payment exceeds $\$8$, so $$ P(Y^F>8\mid Y^F>0)=\frac{0.25}{0.60}=0.4167. $$ Check: the ordinary event $Y^L>3$ and franchise event $Y^F>8$ both select just the loss of $\$14$. The mass at $X=6$ contributes to neither the numerator nor the positive-payment denominator.

Worked Example: Pareto Losses and Threshold Shifts

A hypothetical equipment portfolio uses a Pareto loss model with survival function $S_X(t)=(20/(20+t))^2$ for $t\ge0$. The deductible is $d=\$10$. Find (a) the probability that the loss exceeds $\$30$, given a positive payment, (b) the probability that an ordinary-deductible payment exceeds $\$10$, given a positive payment, and (c) the probability that a franchise payment exceeds $\$20$, given a positive payment. Give probabilities to four decimals. Method: substitute each correctly translated threshold into the survival ratio. The trap is forgetting the deductible shift for the ordinary payment.

The denominator is $$ S_X(10)=\left(\frac{20}{20+10}\right)^2 =\left(\frac{2}{3}\right)^2=\frac{4}{9}. $$

(a) Conditional loss tail. Since $S_X(30)=(20/50)^2=4/25$, $$ P(X>30\mid Y>0)=\frac{4/25}{4/9} =\frac{9}{25}=0.36. $$ The answer is $0.3600$.

(b) Ordinary payment. A payment greater than $\$10$ requires $X>10+10=20$. Since $S_X(20)=(20/40)^2=1/4$, $$ P(Y^L>10\mid Y^L>0)=\frac{1/4}{4/9} =\frac{9}{16}=0.5625. $$

(c) Franchise payment. A franchise payment greater than $\$20$ requires $X>20$, so $$ P(Y^F>20\mid Y^F>0)=\frac{S_X(20)}{S_X(10)} =\frac{1/4}{4/9}=\frac{9}{16}=0.5625. $$ The check is that these last two questions translate to the same loss event, $X>20$; their conditional probabilities must therefore agree.

Why the Denominator and Deductible Type Matter

The conditional-probability identity $P(A\mid B)=P(A\cap B)/P(B)$, for $P(B)>0$, explains the survival ratios. For example, when $x\ge d$, the events $\{X>x\}$ and $\{X>d\}$ intersect in $\{X>x\}$. Therefore

P(X>x\mid Y>0) =\frac{P(X>x,\ X>d)}{P(X>d)} =\frac{P(X>x)}{P(X>d)} =\frac{S_X(x)}{S_X(d)}.

The payment results follow by the same reasoning. Under an ordinary deductible, $Y^L>y$ is equivalent to $X>d+y$ for $y\ge0$. Under a franchise deductible, $Y^F>y$, together with $Y^F>0$, is equivalent to $X>\max(d,y)$. These event equivalences produce the respective formulas and show why a deductible shift belongs in the ordinary-payment numerator, but not in the franchise-payment threshold.

A frequent exam error is to report $S_X(d+y)$ itself as the ordinary conditional payment probability. In the exponential example, for $d=5$ and $y=3$, that incorrect calculation gives $S_X(8)=e^{-0.8}=0.4493$. It is an unconditional probability. The required conditional probability is $$ \frac{S_X(8)}{S_X(5)} =\frac{0.4493}{0.6065} \approx 0.7408. $$ Another error is to apply the ordinary shift $d+y$ to a franchise payment. A franchise payment of $y$ dollars is the full loss, not the loss in excess of the deductible.

Key takeaway: First condition on the event $X>d$, then translate the requested threshold into a loss threshold. Use $S_X(d)$ as the denominator whenever it is positive. An ordinary payment threshold $y$ becomes the loss threshold $d+y$; a franchise payment threshold $y$ remains $y$, subject to the requirement that the loss exceed $d$.

Check Your Understanding

Use strict exceedance and give probabilities to four decimals. Each problem is hypothetical.

A loss is exponential with rate $0.2$ per dollar and has an ordinary deductible of $\$3$. Find $P(X>8\mid Y^L>0)$. (Answer: $0.3679$)
A loss is exponential with rate $0.1$ per dollar and has a franchise deductible of $\$4$. Find $P(Y^F>7\mid Y^F>0)$. (Answer: $0.7408$)
A loss is $\$1,\$5,\$9,$ or $\$15$, with respective probabilities $0.10,0.20,0.40,0.30$. Under an ordinary deductible of $\$6$, find $P(Y^L>4\mid Y^L>0)$. (Answer: $0.4286$)
A loss has survival function $S_X(t)=(20/(20+t))^3$, $t\ge0$. Under an ordinary deductible of $\$5$, find $P(Y^L>10\mid Y^L>0)$. (Answer: $0.3644$)
A loss is uniform on $[0,20]$, and a franchise deductible is $\$8$. Find $P(X>12\mid Y^F>0)$ and $P(Y^F>12\mid Y^F>0)$. (Answer: $0.6667$ for each)

What You'll Learn