Worked Problems on Deductible-Limit Transformations

Why the Payment Distribution Has Boundary Atoms

A deductible removes the part of a loss that the policyholder must absorb, while a policy limit caps the insurer’s payment. Combining the two changes the loss distribution: there can be a probability mass at zero payment and another at the maximum payment. The upper mass depends on the loss exceeding the deductible plus the payment limit, not merely the payment limit.

Let $X$ be a nonnegative ground-up loss, let $d\geq 0$ be an ordinary deductible, and let $u>0$ be the insurer’s maximum payment per loss. The payment is measured in the same monetary units as $X$, $d$, and $u$. This tutorial concerns this ordinary deductible-limit transformation; it does not apply the limit to the ground-up loss before subtracting the deductible.

Definition: The insurer payment per loss under an ordinary deductible $d$ and a limit $u$ on payment is $$ Y^L=\min\bigl((X-d)_+,u\bigr),\qquad (z)_+=\max(z,0). $$ Equivalently, $Y^L=0$ when $X\leq d$, $Y^L=X-d$ when $d<X<d+u$, and $Y^L=u$ when $X\geq d+u$. Thus, the ground-up loss threshold for a capped payment is $d+u$.

These boundary conventions include equality: a loss exactly equal to $d$ produces zero payment, and a loss exactly equal to $d+u$ produces the full payment $u$. Those distinctions matter when $X$ is discrete or has atoms. For a continuous loss, an exact point typically has probability zero, but the same conventions still define the transformation.

Distribution and Boundary Probabilities

Write $F_X(x)=P(X\leq x)$ for the ground-up loss distribution function. Since $Y^L$ cannot be negative and cannot exceed $u$, its distribution is concentrated on $[0,u]$. For $0\leq y<u$, the event that the payment is at most $y$ is exactly the event that the loss is at most $d+y$. At $y=u$, every possible payment is at most $u$.

F_{Y^L}(y)=P(Y^L\leq y)= \begin{cases} 0, & y<0,\\ F_X(d+y), & 0\leq y<u,\\ 1, & y\geq u. \end{cases}

In particular, the probability at zero is the probability that the loss does not exceed the deductible. The probability at the cap includes equality at the threshold $d+u$. For an interior payment $y$, where $0<y<u$, an atom occurs precisely when the ground-up loss has an atom at $d+y$.

P(Y^L=0)=F_X(d),\qquad P(Y^L=u)=P(X\geq d+u)=1-F_X\bigl((d+u)^-\bigr).

The left limit in the upper-atom formula is important for a discrete or mixed loss distribution: $1-F_X(d+u)$ would exclude any probability at exactly $d+u$. For a continuous distribution, there is no atom at the threshold, so the upper-atom probability is $S_X(d+u)$, where $S_X(x)=P(X>x)$.

If $X$ is continuous with density $f_X$, the payment has a density $f_{Y^L}(y)=f_X(d+y)$ for $0<y<u$, together with boundary probabilities $F_X(d)$ at zero and $S_X(d+u)$ at $u$. A density on the interior does not describe these two point masses; they must be reported separately.

Expected Payment: A Tail-Integral Derivation

For a nonnegative random variable bounded above by $u$, its expectation is the integral of its exceedance probabilities over that range. To see why, for each realized payment $Y^L$, its value is the length of the interval of levels $t$ between zero and $Y^L$. In indicator form, $Y^L=\int_0^u \mathbf{1}_{\{Y^L>t\}}\,dt$. Taking expectations gives the tail-integral formula. For $0\leq t<u$, a payment exceeds $t$ exactly when the loss exceeds $d+t$.

E[Y^L]=\int_0^u P(Y^L>t)\,dt =\int_0^u P(X>d+t)\,dt =\int_d^{d+u}P(X>x)\,dx.

The final expression is valid for discrete, continuous, and mixed loss distributions; a probability mass at an individual boundary does not change the value of an integral over $t$. For continuous $X$, the integrand is $S_X(x)$. This approach avoids having to calculate a separate expectation for every part of the payment distribution, although explicitly summing the payment values is often convenient for discrete losses.

Worked Example: A Discrete Loss With an Atom at the Cap Threshold

A hypothetical policy has deductible $d=\$5$ and maximum payment $u=\$10$. The loss $X$, in dollars, has the following distribution: $P(X=0)=0.10$, $P(X=5)=0.20$, $P(X=9)=0.15$, $P(X=15)=0.25$, and $P(X=22)=0.30$. Find the payment distribution and expected payment. The method is to transform each loss value; this directly preserves atoms and avoids overlooking equality at the cap threshold.

Step 1: Identify the boundaries. The zero-payment boundary is $d=\$5$. The full-payment threshold is $d+u=\$15$, and the maximum payment is $\$10$.

Step 2: Transform each possible loss. At $X=0$ and $X=5$, the payment is zero. At $X=9$, it is $9-5=\$4$. At $X=15$, it is $15-5=\$10$, exactly the cap. At $X=22$, the uncapped payment would be $22-5=\$17$, so the actual payment is $\$10$.

\begin{aligned} P(Y^L=0)&=0.10+0.20=0.30,\\ P(Y^L=4)&=0.15,\\ P(Y^L=10)&=0.25+0.30=0.55. \end{aligned}

Step 3: Calculate the expectation.

E[Y^L]=0(0.30)+4(0.15)+10(0.55)=0+0.60+5.50=\$6.10.

Check. The probabilities sum to $0.30+0.15+0.55=1.00$, and the expected payment lies between $\$0$ and the cap of $\$10$. Counting only $X>\$15$ as capped would incorrectly omit the $\$0.25$ mass at $X=\$15$.

Worked Example: Exponential Loss With a Limited Payment

Suppose a hypothetical claim loss has an exponential distribution with rate $\lambda=0.1$ per dollar, so $S_X(x)=e^{-0.1x}$ for $x\geq0$. A policy has $d=\$3$ and $u=\$7$. Find the zero-payment probability, the capped-payment probability, the payment density on its interior, and the expected payment. The method uses the survival function; it avoids treating the payment cap as a ground-up loss threshold.

Step 1: Find the boundaries. A payment is zero for $X\leq3$, and it reaches the cap of $\$7$ at the ground-up loss $d+u=3+7=\$10$.

Step 2: Calculate the boundary probabilities.

\begin{aligned} P(Y^L=0)&=P(X\leq3)=1-e^{-0.1(3)} =1-e^{-0.3}=0.2592,\\ P(Y^L=7)&=P(X\geq10)=e^{-0.1(10)} =e^{-1}=0.3679. \end{aligned}

The loss is continuous, so equality at either boundary has probability zero. The probability of an interior payment is therefore $1-0.2592-0.3679=0.3729$, with rounding applied only to the displayed probabilities.

Step 3: Describe the interior density. For $0<y<7$, payment $y$ corresponds to loss $3+y$. The exponential density at that loss is $0.1e^{-0.1(3+y)}$, so

f_{Y^L}(y)=0.1e^{-0.1(3+y)},\qquad 0<y<7.

This is the density on the open interval only; the probabilities at zero and seven are separate atoms.

Step 4: Calculate the expected payment. Using the tail integral,

\begin{aligned} E[Y^L] &=\int_3^{10}e^{-0.1x}\,dx\\ &=10\left(e^{-0.3}-e^{-1}\right)\\ &=10(0.740818-0.367879)\\ &=3.72939\ldots=\$3.73. \end{aligned}

Check. The answer is between zero and the $\$7$ maximum. The interior mass is $e^{-0.3}-e^{-1}=0.3729$, consistent with the sum of the two boundary probabilities and the total probability of one. The answer is rounded to the nearest cent.

Worked Example: Uniform Loss and Two Ways to Find the Mean

A hypothetical loss is uniform on $[0,\$20]$, with deductible $d=\$4$ and payment limit $u=\$8$. Find the payment distribution and its expected value. The method first identifies the boundary masses, then checks the mean using both the tail integral and the payment density.

Step 1: Identify the thresholds. Losses at most $\$4$ produce zero payment. The cap is reached at $d+u=4+8=\$12$. Since the loss is uniform, $F_X(x)=x/20$ for $0\leq x\leq20$.

Step 2: Calculate the boundary probabilities.

P(Y^L=0)=P(X\leq4)=\frac{4}{20}=0.20,\qquad P(Y^L=8)=P(X\geq12)=\frac{20-12}{20}=0.40.

On the interior $0<y<8$, the payment density is the ground-up density shifted by $d$, so $f_{Y^L}(y)=1/20=0.05$. The interior probability is $8(0.05)=0.40$; all three probabilities sum to $0.20+0.40+0.40=1.00$.

Step 3: Find the expectation by integrating survival probabilities. For payment level $t$ between zero and eight, $P(Y^L>t)=P(X>4+t)=(16-t)/20$.

\begin{aligned} E[Y^L] &=\int_0^8\frac{16-t}{20}\,dt\\ &=\frac{1}{20}\left[16t-\frac{t^2}{2}\right]_0^8\\ &=\frac{1}{20}(128-32)=\$4.80. \end{aligned}

Step 4: Check by summing the continuous and capped contributions. The interior contribution is $\int_0^8 y(0.05)\,dy=0.05(8^2/2)=\$1.60$. The cap contributes $8(0.40)=\$3.20$. Their sum is $\$1.60+\$3.20=\$4.80$, agreeing with the tail-integral result. The zero-payment atom contributes nothing to the mean.

Common Exam Trap: Applying the Limit at the Wrong Threshold

A limit $u$ in this setting is the maximum insurer payment, not a maximum ground-up loss. The payment reaches $u$ when $X-d=u$, or $X=d+u$. For a continuous loss, the capped-payment probability is therefore $S_X(d+u)$, not $S_X(u)$.

For example, for a hypothetical continuous loss with $S_X(x)=(10/(10+x))^2$, deductible $d=10$, and payment limit $u=10$, the correct cap threshold is $20$. The cap probability is

S_X(20)=\left(\frac{10}{30}\right)^2=\frac{1}{9}=0.1111,

not $S_X(10)=(10/20)^2=0.25$. Likewise, the correct expected payment is

\int_{10}^{20}\left(\frac{10}{10+x}\right)^2dx =100\left(\frac{1}{20}-\frac{1}{30}\right) =\frac{5}{3} =\$1.67

to the nearest cent. Integrating from $d$ to $d+u$ follows directly from the payment definition. A reliable check is to state both the payment cap $u$ and the ground-up threshold $d+u$ before calculating.

Key takeaway: Under an ordinary deductible and a limit on insurer payment, $Y^L=\min((X-d)_+,u)$. The zero-payment mass is $F_X(d)$, the capped-payment mass is $P(X\geq d+u)$, and the expected payment is $\int_d^{d+u}P(X>x)\,dx$. Keep the ground-up threshold distinct from the maximum payment.

Check Your Understanding

For each hypothetical loss model, identify the relevant boundaries before calculating. Answers involving probabilities are given to four decimal places where needed; dollar amounts are rounded to the nearest cent.

A loss is exponential with rate $0.2$ per dollar. The deductible is $\$2$, and the payment limit is $\$3$. Find the probabilities of zero payment and full payment, and find the expected payment. (Answer: $P(Y^L=0)=0.3297$, $P(Y^L=3)=0.3679$, and $E[Y^L]=5(e^{-0.4}-e^{-1})=\$1.51$.)
A loss is uniform on $[0,\$30]$, with deductible $\$6$ and payment limit $\$9$. Find the zero-payment probability, capped-payment probability, and expected payment. (Answer: $P(Y^L=0)=0.2000$, $P(Y^L=9)=0.5000$, and $E[Y^L]=\$5.85$.)
A discrete loss has $P(X=2)=0.10$, $P(X=8)=0.25$, $P(X=11)=0.20$, $P(X=14)=0.15$, and $P(X=30)=0.30$. With deductible $d=8$ and limit $u=6$, find the payment distribution and mean. (Answer: $P(Y^L=0)=0.35$, $P(Y^L=3)=0.20$, $P(Y^L=6)=0.45$, and $E[Y^L]=\$3.30$.)
A loss has survival function $S_X(x)=(10/(10+x))^2$ for $x\geq0$. With deductible $d=5$ and payment limit $u=10$, calculate the zero-payment probability, capped-payment probability, and expected payment. (Answer: $P(Y^L=0)=0.5556$, $P(Y^L=10)=0.1600$, and $E[Y^L]=100(1/15-1/25)=\$2.67$.)
A loss equals $\$4$, $\$10$, or $\$17$, with respective probabilities $0.20$, $0.30$, and $0.50$. Under deductible $d=\$4$ and limit $u=\$8$, find the probability at each possible payment and the expected payment. (Answer: $P(Y^L=0)=0.20$, $P(Y^L=6)=0.30$, $P(Y^L=8)=0.50$, and $E[Y^L]=\$5.80$.)

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