Combined Coverage Worked Problems

Combining Three Coverage Features

A claim can be subject to an ordinary deductible, a coinsurance percentage, and a maximum insurer payment. The order matters: in the model used here, the deductible is first subtracted from the loss, the insurer then pays a fraction of the amount above the deductible, and the payment cap is applied last. This is a single-claim severity calculation; it does not require assumptions about claim frequency.

Let $X$ be the nonnegative loss amount, $d$ the ordinary deductible, $c$ the insurer’s coinsurance fraction, and $M$ the maximum insurer payment. Assume $d\geq 0$, $0<c\leq 1$, and $M>0$. The policyholder’s retention is the loss not paid by the insurer.

Definition: The insurer payment per loss under this combined coverage is $$ Y^L=\min\{c(X-d)_+,M\}, \qquad (z)_+=\max(z,0). $$ The retention is $R=X-Y^L$. The payment is zero when $X\leq d$, increases at rate $c$ above the deductible, and cannot exceed $M$. The loss level at which the payment first reaches the cap is $$ b=d+\frac{M}{c}. $$ Thus losses $X\geq b$ produce a payment of $M$, including a loss exactly at the threshold.

This formulation assumes that the cap is a cap on the insurer’s payment, not a limit on the loss amount to which coinsurance applies. It also assumes there are no other coverage provisions, expenses, or adjustments. If a contract instead caps the covered loss before applying coinsurance, its payment function must be written differently.

Payment Distribution and Its Atoms

The payment distribution has a useful structure. Losses at or below the deductible all map to a payment of zero. If $X$ has a continuous density, losses strictly between $d$ and $b$ map to payments strictly between zero and $M$. All losses at or above $b$ map to the same capped payment $M$. Consequently, a continuous loss distribution generally creates atoms at both zero and $M$.

P(Y^L=0)=P(X\leq d)=F_X(d), \qquad P(Y^L=M)=P(X\geq b)=S_X(b-).

The notation $S_X(b-)$ emphasizes that any probability mass at $X=b$ belongs to the capped payment. When $X$ is continuous at $b$, this is simply $S_X(b)$. For $0<y<M$, the event $Y^L\leq y$ is equivalent to $X\leq d+y/c$, so

F_{Y^L}(y)=P(Y^L\leq y)=F_X\left(d+\frac{y}{c}\right), \qquad 0\leq y<M.

If $X$ has density $f_X$, then within the uncapped positive-payment range the change of variable $x=d+y/c$ gives $f_{Y^L}(y)=f_X(d+y/c)/c$. This density describes only the continuous part; it does not replace the zero-payment or cap atoms. For a discrete loss, several loss values may map to the same payment, so probabilities must be combined at each payment value.

Why the Survival-Integral Formula Works

For any nonnegative random variable $W$, its expectation can be written as the area under its survival function: $E[W]=\int_0^\infty P(W>y)\,dy$. This follows by writing each realized value $W$ as $\int_0^\infty \mathbf{1}_{\{W>y\}}\,dy$ and taking expectations. Since $Y^L$ cannot exceed $M$, only the range from zero to $M$ contributes.

E[Y^L] =\int_0^M P(Y^L>y)\,dy =\int_0^M S_X\left(d+\frac{y}{c}\right)\,dy =c\int_d^{d+M/c} S_X(x)\,dx.

For $0\leq y<M$, a payment greater than $y$ occurs exactly when $X>d+y/c$, which gives the middle equality. The final equality is the substitution $x=d+y/c$, so $dy=c\,dx$. The formula is valid for discrete, continuous, or mixed loss distributions; changing values at individual endpoints does not affect the integral.

An important implication is that the survival integral already accounts for the capped tail. For every payment level $y<M$, claims in the capped tail also satisfy $Y^L>y$. Therefore, one must not add a separate cap-atom contribution to this integral. A density calculation over $0<Y^L<M$, by contrast, excludes the cap atom and must add $M P(Y^L=M)$ separately.

Worked Examples

Worked Example: Exponential Loss With a Payment Cap

A hypothetical policy has loss $X$ exponentially distributed with mean $\$800$, so $S_X(x)=e^{-x/800}$ for $x\geq0$. It has a $\$240$ deductible, the insurer pays 60% of the amount above the deductible, and the insurer payment is capped at $\$300$. Find the payment distribution’s atoms, the probability of a positive payment, the expected payment, and the expected retention.

Method and threshold. Here $d=240$, $c=0.6$, and $M=300$. The threshold is $b=240+300/0.6=240+500=\$740$. Thus $Y^L=0$ for $X\leq240$, the payment is uncapped for $240<X<740$, and $Y^L=300$ for $X\geq740$.

Atoms and positive-payment probability. The zero-payment probability is $1-e^{-240/800}=1-e^{-0.3}=1-0.7408182=0.2591818$, or $0.2592$ to four decimals. The cap probability is $e^{-740/800}=e^{-0.925}=0.3965314$, or $0.3965$. A positive payment occurs when $X>240$, so its probability is $e^{-0.3}=0.7408$.

Expected payment. Applying the survival-integral formula, $$ E[Y^L]=0.6\int_{240}^{740}e^{-x/800}\,dx =0.6(800)\left(e^{-240/800}-e^{-740/800}\right). $$ The difference is $0.7408182-0.3965314=0.3442868$, and $480(0.3442868)=165.2577$. Therefore $E[Y^L]=\$165.26$, rounded to the nearest cent.

Expected retention and check. The exponential loss has mean $E[X]=\$800$. Since $R=X-Y^L$, $E[R]=800-165.26=\$634.74$. The continuous payment probability is $0.7408182-0.3965314=0.3442868$; together with the zero and cap atoms, the probabilities sum to $0.2591818+0.3442868+0.3965314=1.0000000$. The calculation avoids the mistake of adding a cap-atom term to the survival integral.

Worked Example: Uniform Loss With a Coinsurance Cap

For a hypothetical claim, let $X$ be uniform on $[0,1500]$, in dollars. The policy has deductible $d=\$300$, insurer coinsurance fraction $c=0.75$, and payment cap $M=\$450$. Find the payment distribution, expected payment, and expected retention.

Threshold and distribution. The cap threshold is $b=300+450/0.75=300+600=\$900$. The uniform loss has density $1/1500$. The payment is zero with probability $300/1500=0.2000$, and it equals $\$450$ with probability $P(X\geq900)=600/1500=0.4000$. For $0<y<450$, the loss corresponding to payment $y$ is $300+y/0.75$. The payment density is therefore $$ f_{Y^L}(y)=\frac{1}{1500(0.75)}=\frac{1}{1125}, \qquad 0<y<450. $$ Its continuous probability is $450/1125=0.4000$, confirming that the masses and continuous part sum to one.

Expected payment by separating the continuous part and cap atom. The continuous payment contribution is $\int_0^{450} y(1/1125)\,dy=450^2/(2\cdot1125)=202500/2250=\$90$. The cap contribution is $450(0.4000)=\$180$. Hence $E[Y^L]=90+180=\$270$.

Expected retention and check. A uniform loss on $[0,1500]$ has mean $750$, so $E[R]=750-270=\$480$. As a check, the survival integral gives $0.75\int_{300}^{900}(1500-x)/1500\,dx$. The integral inside is the area of a triangle with base $600$ and height $0.8$, hence $240$; multiplying by $0.75$ gives $\$180$, not $\$270$. This indicates the triangle description needs care: the integrand decreases from $0.8$ to $0.4$, so its area is a trapezoid, $(0.8+0.4)600/2=360$. Thus the survival integral is $0.75(360)=\$270$, agreeing with the payment-density calculation. It already includes the capped tail; adding another cap contribution here would double-count it.

Worked Example: Discrete Loss and a Boundary at the Cap

A hypothetical equipment policy has loss $X$ taking values $\$100,\$400,\$700,\$1000$ with probabilities $0.20,0.30,0.30,0.20$, respectively. The deductible is $\$250$, the insurer pays 80% above the deductible, and the insurer payment cap is $\$360$. Find the distribution of $Y^L$ and $R$, and their expectations.

Threshold and claim-by-claim payments. The threshold is $b=250+360/0.8=250+450=\$700$. At $X=100$, the loss is below the deductible, so the payment is zero. At $X=400$, the payment is $0.8(400-250)=\$120$. At $X=700$, the uncapped amount is $0.8(700-250)=\$360$, exactly the cap. At $X=1000$, the uncapped amount exceeds the cap, so the payment remains $\$360$.

Payment distribution and mean. Combining equal payments gives $P(Y^L=0)=0.20$, $P(Y^L=120)=0.30$, and $P(Y^L=360)=0.30+0.20=0.50$. Therefore, $E[Y^L]=0(0.20)+120(0.30)+360(0.50)=36+180=\$216$. The threshold equality belongs to the capped outcome, but it does not change the payment amount.

Retention distribution and mean. The retentions at the four loss values are $100-0=100$, $400-120=280$, $700-360=340$, and $1000-360=640$. Thus their respective probabilities are $0.20,0.30,0.30,0.20$, and $E[R]=100(0.20)+280(0.30)+340(0.30)+640(0.20)=20+84+102+128=\$334$. As a check, $E[X]=100(0.20)+400(0.30)+700(0.30)+1000(0.20)=\$550$, and $550-216=\$334$. The calculation avoids treating the loss exactly at the threshold as uncapped at a payment below $M$.

A Reliable Calculation Sequence

The most dependable approach is to establish the contract mechanics before calculating probabilities or moments. Keep the loss scale, payment scale, and threshold distinct: the cap $M$ is a dollar amount paid by the insurer, while the threshold $b=d+M/c$ is a dollar amount of loss.

Write the payment function.
Use $Y^L=\min\{c(X-d)_+,M\}$, and state which amount the cap limits.

Find the threshold.
Calculate $b=d+M/c$. Losses below $d$ yield zero; losses at or above $b$ yield $M$.

Build the payment distribution.
Find the zero and cap probabilities, then transform the uncapped loss values or density. Combine probability masses that map to the same payment.

Calculate the mean consistently.
Use the survival integral alone, or use the continuous payment contribution plus the cap-atom contribution. Do not combine both methods’ cap contributions.

Find retention and verify.
For each loss, use $R=X-Y^L$. For expectations, check $E[R]=E[X]-E[Y^L]$.

Common Traps and Why the Details Matter

A frequent error is to apply the cap to $X-d$ and then multiply by $c$, which gives a different contract from one with an insurer-payment cap. Under the model here, the maximum payment is $M$, and it takes an excess loss of $M/c$ above the deductible to reach that payment. Another error is to use $d+M$ as the loss threshold; this is correct only if $c=1$.

The expected-payment calculation also needs one consistent method. With the survival integral, $E[Y^L]=\int_0^M P(Y^L>y)\,dy$, the capped claims contribute to the integrand at every payment level below $M$. If instead integrating the payment density on $0<y<M$, the cap mass is excluded and must be added separately as $M P(Y^L=M)$. A useful final check is that probabilities sum to one and that $E[R]+E[Y^L]=E[X]$.

These combined calculations matter whenever contract features interact: the deductible controls when coverage starts, coinsurance controls the slope of the payment, and the payment cap creates a top atom. Correctly distinguishing those effects is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical single-claim model, calculate the requested quantities. Round probabilities to four decimals and dollar expectations to the nearest cent.

$X$ is exponential with mean $\$1000$, $d=\$200$, $c=0.5$, and $M=\$250$. Find the zero-payment probability, cap probability, expected payment, and expected retention. (Answer: $P(Y^L=0)=0.1813$; $P(Y^L=250)=0.4966$; $E[Y^L]=\$161.07$; $E[R]=\$838.93$.)
$X$ is uniform on $[0,1200]$, $d=\$240$, $c=0.8$, and $M=\$320$. Find the threshold, cap probability, expected payment, and expected retention. (Answer: $b=\$640$; $P(Y^L=320)=0.4667$; $E[Y^L]=\$202.67$; $E[R]=\$397.33$.)
$X$ takes values $\$150,\$350,\$650,\$900$ with probabilities $0.25,0.25,0.30,0.20$. For $d=\$200$, $c=0.5$, and $M=\$200$, find the payment distribution and expected retention. (Answer: $P(Y^L=0)=0.25$, $P(Y^L=75)=0.25$, $P(Y^L=200)=0.50$; $E[R]=\$381.25$.)
$X$ is exponential with mean $\$500$, $d=\$100$, $c=0.4$, and $M=\$240$. Find $P(Y^L>120)$ and the probability of a capped payment. (Answer: $P(Y^L>120)=e^{-0.8}=0.4493$; $P(Y^L=240)=e^{-1.4}=0.2466$.)
$X$ is uniform on $[0,1000]$, $d=\$100$, and $c=0.9$. Compare expected insurer payments and expected retentions when the payment cap is $\$450$ versus $\$360$. (Answer: With $M=\$450$, $b=\$600$, $E[Y^L]=\$292.50$, and $E[R]=\$207.50$. With $M=\$360$, $b=\$500$, $E[Y^L]=\$252.00$, and $E[R]=\$248.00$.)

What You'll Learn