Distribution Under a Deductible and Policy Limit

Why a Deductible and a Limit Create a Mixed Distribution

A deductible and a policy limit change a loss into a payment in two distinct ways. Losses at or below the deductible produce no payment, while sufficiently large losses produce the maximum covered payment. Between those thresholds, each additional dollar of loss produces an additional dollar of payment. Consequently, the payment distribution has probability concentrated at zero, a continuous part, and probability concentrated at the cap.

This tutorial considers an ordinary deductible and a limit on the payment after the deductible. Assume $X$ is a nonnegative, absolutely continuous loss amount, $d\geq 0$ is the deductible, and $u>0$ is the maximum payment. Write $z^+=\max(z,0)$. The payment is $Y=\min((X-d)^+,u)$. The assumption that $X$ is absolutely continuous matters: it ensures there are no additional point masses at the deductible or at the loss threshold where the cap begins.

Definition: Under an ordinary deductible $d$ and payment limit $u$, the payment random variable is $Y=\min((X-d)^+,u)$. Its possible values lie between zero and $u$. Losses $X\leq d$ give payment zero; losses $d<X<d+u$ give payment $X-d$; and losses $X\geq d+u$ give payment $u$.

Distribution of the Payment

Let $F_X(x)=P(X\leq x)$, $S_X(x)=P(X>x)$, and $f_X(x)$ denote the loss distribution function, survival function, and density. Under the stated continuity assumption, the probability of zero payment is the probability that the loss does not exceed $d$. The cap starts to apply when the payment before the cap, $X-d$, reaches $u$. That occurs at loss $d+u$, not at loss $u$.

Theorem: For $X$ nonnegative and absolutely continuous, $d\geq0$, and $u>0$, the distribution of $Y=\min((X-d)^+,u)$ is $$ F_Y(y)= \begin{cases} 0, & y<0,\\ F_X(d+y), & 0\leq y<u,\\ 1, & y\geq u. \end{cases} $$ The probability masses at the endpoints are $$ P(Y=0)=F_X(d),\qquad P(Y=u)=S_X(d+u). $$ For $0<y<u$, the continuous density is $f_Y(y)=f_X(d+y)$. The continuous part has total probability $F_X(d+u)-F_X(d)$.

The CDF is worth reading carefully at its boundaries. At zero, $F_Y(0)=F_X(d)$, so its jump there is the probability of no payment. Just below $u$, the CDF approaches $F_X(d+u)$; at and above $u$, it is one. The jump at $u$ is therefore $1-F_X(d+u)=S_X(d+u)$, the probability of reaching the cap. Between zero and $u$, the CDF increases continuously as the loss threshold $d+y$ moves through the severity distribution.

Why the Formula Holds

The key is to translate a payment event into a loss event. For any $0\leq y<u$, a payment no greater than $y$ occurs exactly when the loss is no greater than $d+y$. Losses below the deductible produce zero, and losses above $d$ produce an uncapped payment $X-d$; since $y$ is below the cap, the cap does not change this event.

\{Y\leq y\}=\{X\leq d+y\},\quad 0\leq y<u \quad\Longrightarrow\quad F_Y(y)=P(X\leq d+y)=F_X(d+y).

At $y=u$, every possible payment is at most $u$, so $F_Y(u)=1$. The zero-payment event is $\{X\leq d\}$, and the cap-payment event is $\{X\geq d+u\}$. Absolute continuity lets us write the latter probability as $S_X(d+u)$ without a boundary atom. Differentiating the CDF expression on the open interval $0<y<u$ gives $f_Y(y)=f_X(d+y)$.

The continuous probability can also be checked by integration:

\int_0^u f_Y(y)\,dy =\int_0^u f_X(d+y)\,dy =F_X(d+u)-F_X(d).

The two endpoint masses and the continuous part therefore sum to $F_X(d)+[F_X(d+u)-F_X(d)]+S_X(d+u)=1$. This is a useful check whenever the distribution is calculated from a specific severity model.

Worked Example: Exponential Loss With a Deductible and Limit

A hypothetical equipment policy has deductible $d=\$200$ and maximum payment $u=\$500$. Suppose the loss $X$ is exponential with mean $\$1{,}000$, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. Find the masses at zero and at the cap, the probability in the continuous payment range, and $F_Y(250)$.

Method and thresholds. The zero-payment threshold is $X\leq200$. The cap threshold is $X\geq d+u=700$. For payments strictly between zero and 500, the payment density is $f_X(200+y)$.

Mass at zero. $P(Y=0)=F_X(200)=1-e^{-200/1000}=1-e^{-0.2}=0.1812692$, or $0.1813$ to four decimal places.

Mass at the cap. $P(Y=500)=S_X(700)=e^{-700/1000}=e^{-0.7}=0.4965853$, or $0.4966$.

Continuous probability. $P(0<Y<500)=F_X(700)-F_X(200)=e^{-0.2}-e^{-0.7}=0.8187308-0.4965853=0.3221455$, or $0.3221$.

Payment CDF and requested value. For $0<y<500$, $F_Y(y)=1-e^{-(200+y)/1000}$ and $f_Y(y)=0.001e^{-(200+y)/1000}$. Thus $F_Y(250)=1-e^{-450/1000}=1-e^{-0.45}=0.3623718$, or $0.3624$.

Check and trap avoided. The probabilities sum to $0.1812692+0.3221455+0.4965853=1.0000000$. A common error is to use $S_X(500)=e^{-0.5}$ as the cap mass. That would cap losses above $u$, but the contract caps the payment: the correct loss threshold is $d+u=700$.

Worked Example: Uniform Loss and a Payment Interval

A hypothetical cargo policy has $d=\$200$ and $u=\$400$. Its loss $X$ is uniform on $[0,1200]$, with density $1/1200$ on that interval. Find both endpoint masses, the continuous probability, the payment density, and $P(100<Y\leq300)$.

Method and thresholds. Zero payment occurs for $X\leq200$; the cap applies for $X\geq200+400=600$. The continuous payment range corresponds to $200<X<600$.

Endpoint masses. $P(Y=0)=200/1200=1/6=0.1667$. Also, $P(Y=400)=P(X\geq600)=(1200-600)/1200=1/2=0.5000$.

Continuous probability and density. $P(0<Y<400)=P(200<X<600)=(600-200)/1200=1/3=0.3333$. For $0<y<400$, $f_Y(y)=f_X(200+y)=1/1200$. The three components sum to $1/6+1/3+1/2=1$.

Payment interval. For $0<y<400$, $F_Y(y)=(200+y)/1200$. Therefore $F_Y(300)=500/1200=5/12$, and $F_Y(100)=300/1200=1/4$. Since there is no point mass at either interior endpoint, $P(100<Y\leq300)=5/12-1/4=1/6=0.1667$.

Check and trap avoided. The interval corresponds to losses $300<X\leq500$, a width of 200, so its probability is $200/1200=1/6$, matching the CDF calculation. Do not treat the limit $400$ as the loss threshold for the cap; the cap threshold is $600$.

Worked Example: Pareto Loss With a High Deductible

For a hypothetical specialty-property contract, let $X$ have a Lomax Pareto distribution with survival function $S_X(x)=(500/(500+x))^3$ for $x\geq0$. The deductible is $d=\$250$, and the payment limit is $u=\$750$. Find the endpoint masses, $P(0<Y<750)$, $F_Y(400)$, and $P(100<Y<500)$.

Method and thresholds. The cap activates at loss $d+u=1000$. The relevant CDF is $F_X(x)=1-(500/(500+x))^3$.

Mass at zero. $P(Y=0)=F_X(250)=1-(500/750)^3=1-(2/3)^3=19/27=0.7037$.

Mass at the cap. $P(Y=750)=S_X(1000)=(500/1500)^3=(1/3)^3=1/27=0.0370$.

Continuous probability. $P(0<Y<750)=1-19/27-1/27=7/27=0.2593$. Equivalently, it is $F_X(1000)-F_X(250)=(26/27)-(19/27)=7/27$.

CDF value. Since $400<750$, $F_Y(400)=F_X(250+400)=F_X(650)=1-(500/1150)^3=1-(10/23)^3=11167/12167=0.9178$.

Payment interval. The event $100<Y<500$ corresponds to $350<X<750$. Thus $P(100<Y<500)=F_X(750)-F_X(350)$. Here $F_X(750)=1-(2/5)^3=117/125$, and $F_X(350)=1-(10/17)^3=3913/4913$. Their difference is $117/125-3913/4913=85696/614125=0.1395$.

Check and trap avoided. The endpoint masses plus continuous probability are $19/27+7/27+1/27=1$. The cap probability uses survival at 1000, because that is the loss required to produce a payment of 750 after the 250 deductible.

Exam Checks and Common Mistakes

For a continuous severity, a reliable solution starts by translating payment boundaries into loss boundaries. A payment $y$ in the uncapped range corresponds to loss $d+y$; zero corresponds to losses through $d$; and the cap corresponds to losses from $d+u$ upward. The following checks catch many calculation errors:

Check the cap threshold. The cap is a payment limit, so the loss that reaches it is $d+u$.
Check the support. Payments cannot be negative or exceed $u$.
Check the jumps. The CDF jump at zero must be $F_X(d)$; the jump at $u$ must be $S_X(d+u)$.
Check the total probability. The two endpoint masses and the probability in the continuous interval must add to one.

A further boundary caution is that the formulas here rely on an absolutely continuous loss distribution. If $X$ has positive probability at particular loss amounts, those losses can create additional payment atoms. That requires accounting for the point probabilities as well as the continuous part; the density-only formula should not be used without the stated assumption.

The full distribution is useful whenever a question asks for payment probabilities, a CDF value, or a payment interval under both contract terms. In each case, first identify the relevant payment region, translate it into a region for $X$, and then use $F_X$ or $S_X$. This keeps the contract mechanics separate from the chosen severity family.

Check Your Understanding

For each hypothetical contract, identify the relevant payment probabilities or distribution value. Give probabilities to four decimal places where appropriate.

A loss is exponential with mean $\$2{,}000$, deductible $d=\$1{,}000$, and payment limit $u=\$2{,}000$. Find $P(Y=0)$ and $P(Y=u)$. (Answer: $0.3935,\ 0.2231$)
A loss is uniform on $[0,1000]$, with $d=\$100$ and $u=\$300$. Find the masses at zero and at the cap, and the probability in the continuous payment range. (Answer: $0.1000,\ 0.6000,\ 0.3000$)
A loss is exponential with mean $\$500$, with $d=\$100$ and $u=\$200$. Find $F_Y(100)$. (Answer: $1-e^{-0.4}=0.3297$)
A loss is uniform on $[0,1000]$, with $d=\$250$ and $u=\$250$. Find $P(0<Y\leq150)$. (Answer: $0.1500$)
A loss has survival function $S_X(x)=(1000/(1000+x))^2$ for $x\geq0$. If $d=\$500$ and $u=\$1000$, find $P(Y=0)$ and $P(Y=u)$. (Answer: $0.5556,\ 0.1600$)

What You'll Learn