Conditional Payment Probabilities for Combined Coverage

Introduction

A claim payment is not the same quantity as the underlying loss. An ordinary deductible removes an initial layer of loss, a payment limit caps the insurer’s amount, and coinsurance scales the amount payable. A question about the probability of a payment above a threshold must therefore be translated back into a question about the loss that produced it.

The key technique is to invert the payment rule on the part of the loss range where payment increases. At the endpoints, take care: the payment has a probability mass at zero when losses do not exceed the deductible, and a cap can create a probability mass at the maximum payment even when the underlying loss is continuous.

Payment Rule and Its Thresholds

Definition: Let $X$ be a nonnegative ground-up loss. Suppose the contract has an ordinary deductible $d$, a payment limit $u$ applied after the deductible but before coinsurance, and a coinsurance fraction $c$, where $d\geq 0$, $u>0$, and $0<c\leq 1$. The insurer’s payment is $$ Y=c\min\bigl((X-d)_+,u\bigr),\qquad (z)_+=\max(z,0). $$ Thus $Y=0$ when $X\leq d$, $Y=c(X-d)$ when $d<X<d+u$, and $Y=cu$ when $X\geq d+u$.

This convention places the payment limit on the amount after the deductible and before coinsurance. Consequently, the maximum payment is $cu$, not $u$. A contract that defines its limit as the maximum amount payable after coinsurance has a different payment rule and requires different thresholds.

Write $F_X(x)=P(X\leq x)$ and $S_X(x)=P(X>x)$. If $0\leq y<cu$, reversing the increasing part of the payment rule gives

Y>y\quad\Longleftrightarrow\quad X>d+\frac{y}{c}.

This equivalence includes $y=0$: a positive payment occurs exactly when $X>d$. Therefore, provided $P(X>d)>0$,

P(Y>y\mid Y>0) =\frac{S_X(d+y/c)}{S_X(d)},\qquad 0\leq y<cu.

The strict inequality in $S_X(x)=P(X>x)$ matches the strict payment event $Y>y$. For a payment interval with $0\leq a<b<cu$, the event $a<Y\leq b$ corresponds to $d+a/c<X\leq d+b/c$. If the interval is conditioned on a positive payment, its probability is

P(a<Y\leq b\mid Y>0) =\frac{F_X(d+b/c)-F_X(d+a/c)}{S_X(d)}.

At the cap, use the separate equivalence $Y=cu$ if and only if $X\geq d+u$. Thus the cap probability is $P(X\geq d+u)$; it is not generally $S_X(d+u)$ when $X$ can have a point mass at $d+u$. If $X$ is continuous, the distinction between $>$ and $\geq$ at a fixed threshold does not affect the probability.

Why the Threshold Conversion Works

Below the cap, a payment threshold can only be reached on the linear section $Y=c(X-d)$. Solving $c(X-d)>y$ gives $X>d+y/c$. Losses in the zero-payment region cannot satisfy the event, and once the loss reaches $d+u$, the payment is already at its maximum. This establishes the threshold equivalence without requiring a particular severity distribution.

Check the payment threshold.
For a strict exceedance formula, first verify $0\leq y<cu$. At or above the cap, $P(Y>y)=0$.

Invert the payment rule.
For a threshold below the cap, calculate the corresponding loss threshold $d+y/c$.

Apply the conditioning event.
Translate both the event and the condition into loss events, intersect them if necessary, and divide by the probability of the conditioning event.

For example, conditioning on a loss above $r$ gives, when $P(X>r)>0$ and $0\leq y<cu$,

P(Y>y\mid X>r) =\frac{S_X\!\left(\max\left(r,d+\frac{y}{c}\right)\right)}{S_X(r)}.

Both events require $X$ to exceed their respective thresholds, so their intersection requires exceeding the larger one. This method also makes clear why a conditioning event must be translated rather than treated as though it were automatically equivalent to a positive payment.

Worked Applications

Worked Example: Exceeding a Payment Threshold Given a Positive Payment

A hypothetical contract has deductible $d=\$200$, payment limit $u=\$600$ before coinsurance, and coinsurance fraction $c=0.80$. The ground-up loss has the continuous distribution with survival function $S_X(x)=(1-x/2000)^2$ for $0\leq x\leq \$2{,}000$, with $S_X(x)=1$ below zero and $S_X(x)=0$ at or above $\$2{,}000$. Find the probability that the payment exceeds $\$240$, given that it is positive.

Method. Use the conditional survival formula. The maximum payment is $cu=0.80(\$600)=\$480$, so $\$240$ is below the cap. A positive payment requires $X>\$200$. A payment greater than $\$240$ requires $X>\$200+\$240/0.80=\$500$.

Evaluate the survival probabilities: $S_X(200)=(1-200/2000)^2=(0.90)^2=0.8100$, and $S_X(500)=(1-500/2000)^2=(0.75)^2=0.5625$.

Therefore, $P(Y>240\mid Y>0)=0.5625/0.8100=0.694444\ldots$, or 0.6944 to four decimal places.

Check and common mistake. The result is below 1, as it must be because exceeding $\$240$ is a stricter condition than merely receiving a positive payment. Using $200+0.80(240)$ as the loss threshold would incorrectly multiply by coinsurance; the threshold must be divided by $c$.

Worked Example: A Conditional Payment Interval

Use the same continuous loss distribution as in the previous example. A hypothetical contract has $d=\$100$, $u=\$500$, and $c=0.50$. Find the probability that the payment is greater than $\$80$ and at most $\$200$, given a positive payment.

Method. The maximum payment is $cu=0.50(\$500)=\$250$, so both interval thresholds are below the cap. Inverting the lower and upper payment bounds gives $X>100+80/0.50=\$260$ and $X\leq100+200/0.50=\$500$. The event is therefore $\$260<X\leq\$500$, and the conditioning event is $X>\$100$.

Evaluate the needed values: $S_X(260)=(1-260/2000)^2=(0.87)^2=0.7569$; $F_X(500)=1-S_X(500)=1-0.5625=0.4375$; and $F_X(260)=1-S_X(260)=1-0.7569=0.2431$. Thus the probability of the interval event is $F_X(500)-F_X(260)=0.4375-0.2431=0.1944$. The probability of a positive payment is $S_X(100)=(0.95)^2=0.9025$.

The required conditional probability is $0.1944/0.9025=0.215401\ldots$, or 0.2154 to four decimal places.

Check and common mistake. The loss interval has positive probability and lies wholly above the deductible. Subtracting $S_X(500)$ from $S_X(260)$ also gives $0.7569-0.5625=0.1944$. Be precise about the interval: using $F_X(500)-F_X(260)$ represents $\$260<X\leq\$500$, as required.

Worked Example: Reaching the Cap Given a Payment Exceedance

A hypothetical contract has $d=\$150$, $u=\$400$, and $c=0.75$. The ground-up loss has the continuous survival function $S_X(x)=(1-x/2000)^2$ on $0\leq x\leq \$2{,}000$, with the same endpoint conventions as above. Find the probability that the payment is at its cap, given that it exceeds $\$180$.

Method. The cap is $cu=0.75(\$400)=\$300$. Reaching it requires $X\geq d+u=\$550$. Exceeding $\$180$, which is below the cap, requires $X>\$150+\$180/0.75=\$390$. Since every loss at or above $\$550$ also exceeds $\$390$, the conditional probability is $S_X(550)/S_X(390)$.

The numerator is $S_X(550)=(1-550/2000)^2=(0.725)^2=0.525625$. The denominator is $S_X(390)=(1-390/2000)^2=(0.805)^2=0.648025$. Hence the probability is $0.525625/0.648025=0.811118\ldots$, or 0.8111 to four decimal places.

Check and common mistake. The answer is less than 1 because not every loss producing a payment above $\$180$ reaches the cap. Dividing by $P(X>150)$, the probability of a positive payment, would answer a different conditional question.

Common Exam Trap: Applying the Contract in the Wrong Order

The threshold $d+y/c$ follows from this contract’s specified payment rule. The deductible is removed first, the payment is limited next, and coinsurance then scales that limited amount. Do not replace this rule with a different order of operations.

Common mistake: With $d=\$100$, $u=\$500$, and $c=0.50$, someone might treat $u$ as the maximum payment after coinsurance and conclude that the cap is $\$500$. Under the stated convention, the cap is $cu=0.50(\$500)=\$250$. Thus $P(Y>\$250)=0$, not a positive tail probability. The threshold formula $d+y/c$ is valid only for $y<\$250$.

Also distinguish conditioning on $Y>0$ from conditioning on $X>r$. The former excludes all losses at or below the deductible; the latter may include losses that produce no payment. Translate each condition before forming the ratio.

Key takeaway: For a payment threshold below the cap, convert the threshold $y$ into the loss threshold $d+y/c$. Translate the conditioning event separately, intersect the resulting loss events, and divide by the probability of the condition. Treat the zero payment and maximum payment boundaries explicitly.

Check Your Understanding

Assume the continuous loss distribution has $S_X(x)=(1-x/2000)^2$ for $0\leq x\leq \$2{,}000$, as defined above, unless a question states otherwise. Give probabilities to four decimal places.

A contract has $d=\$100$, $u=\$500$, and $c=0.60$. Find $P(Y>\$120\mid Y>0)$. (Answer: 0.8006)
A contract has $d=\$100$, $u=\$700$, and $c=0.50$. Find $P(\$50<Y\leq\$150\mid Y>0)$. (Answer: 0.1884)
A contract has $d=\$200$, $u=\$400$, and $c=0.75$. Find the probability that the payment reaches its cap, given $Y>\$200$. (Answer: 0.8336)
A contract has $d=\$100$, $u=\$600$, and $c=0.80$. Find $P(Y>\$120\mid X>\$200)$. (Answer: 0.9452)
A contract has $d=\$100$, $u=\$600$, and $c=0.50$. Given $X>\$150$, find the probability that $\$80<Y\leq\$240$. (Answer: 0.2955)

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