Comparing Deductible and Limit Alternatives

Why Compare Coverage Designs on the Same Severity Model?

A deductible determines where insurer payments begin; a limit determines how far they can grow. When one design has a lower deductible but also a lower limit, neither feature alone settles which design has the larger expected payment. The comparison must account for the entire ground-up loss distribution, including how much probability lies in the ranges where the designs differ.

The objective here is to compare expected insurer payments per claim for alternative contracts applied to the same ground-up loss $X$. This is not a comparison of premiums, expenses, or aggregate portfolio costs. Unless stated otherwise, payments are undiscounted, so no interest rate or mortality assumption applies.

Definition: For a nonnegative ground-up loss $X$, a deductible $d$, and a limit $u$ on the insurer's payment, define the payment as $$ Y(d,u)=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ The payment starts above $d$, increases dollar for dollar until it reaches $u$, and then remains at $u$. Thus the layer attaches at $d$ and exhausts at $d+u$.

The limit convention matters. In this tutorial, $u$ is the maximum insurer payment, not the maximum ground-up loss to which coverage applies. A design with $d=10$ and $u=30$ can pay its full $30$ when the ground-up loss is $40$ or greater.

A Formula for Expected Payment

Define the stop-loss expectation at retention $a\geq 0$ by $L(a)=E[(X-a)_+]$. For any nonnegative severity distribution for which the expectations exist, the expected payment under the design $(d,u)$ is

$$ E[Y(d,u)]=L(d)-L(d+u). $$

This identity is useful because it turns a deductible-and-limit payment into a difference of two stop-loss expectations. It holds for continuous, discrete, and mixed severity distributions; probability atoms at the deductible or exhaustion point do not require a special correction.

Justification: For every possible value of $X$, the payment can be written as $$ \min\bigl((X-d)_+,u\bigr)=(X-d)_+-(X-d-u)_+. $$ If $X\leq d$, both terms on the right are zero. If $d<X<d+u$, the second term is zero and the first is $X-d$. If $X\geq d+u$, subtracting the two positive parts gives $u$. Taking expectations of this pointwise identity gives $E[Y(d,u)]=E[(X-d)_+]-E[(X-d-u)_+]=L(d)-L(d+u)$.

For two designs $A=(d_A,u_A)$ and $B=(d_B,u_B)$ applied to the same $X$, compare

\Delta=E[Y(d_A,u_A)]-E[Y(d_B,u_B)] =\bigl[L(d_A)-L(d_A+u_A)\bigr] -\bigl[L(d_B)-L(d_B+u_B)\bigr].

If $\Delta>0$, design A has the greater expected payment; if $\Delta<0$, design B does. If the designs trade a lower deductible for a lower payment limit, the payment functions can cross, and the sign of $\Delta$ can depend on the severity distribution.

When the Ranking Does Not Depend on the Distribution

A useful preliminary check is to compare the payment functions directly. If $d_A\leq d_B$ and $u_A\geq u_B$, design A has both the lower deductible and the higher payment cap. For every fixed loss $x\geq 0$, it follows that

\min\bigl((x-d_A)_+,u_A\bigr) \geq \min\bigl((x-d_B)_+,u_B\bigr).

Taking expectations preserves this inequality, so A cannot have a smaller expected payment under any shared severity distribution. This is a pointwise dominance result, not an assumption about the shape of $X$. When one design has a lower deductible but also a lower cap, this simple dominance check does not decide the ranking: calculate the expectations under the specified distribution.

Worked Example: Comparing Two Layers Under a Uniform Severity Model

A hypothetical property policy has ground-up loss $X\sim\operatorname{Uniform}(0,100)$, in dollars. Compare design A, with deductible $d_A=10$ and insurer payment limit $u_A=30$, against design B, with $d_B=20$ and $u_B=50$. The claims are undiscounted; no interest or mortality assumption is involved. Which design has the greater expected payment?

Method. Use $E[Y(d,u)]=L(d)-L(d+u)$. For $X\sim\operatorname{Uniform}(0,100)$, the density is $1/100$ on $[0,100]$. For $0\leq a\leq100$, $$ L(a)=E[(X-a)_+] =\frac{1}{100}\int_a^{100}(x-a)\,dx =\frac{(100-a)^2}{200}. $$

Design A. Its attachment is $10$, and its exhaustion point is $10+30=40$. Therefore, $$ L(10)=\frac{90^2}{200}=40.5, \qquad L(40)=\frac{60^2}{200}=18, $$ so $E[Y_A]=40.5-18=22.5$ dollars.

Design B. Its attachment is $20$, and its exhaustion point is $20+50=70$. Therefore, $$ L(20)=\frac{80^2}{200}=32, \qquad L(70)=\frac{30^2}{200}=4.5, $$ so $E[Y_B]=32-4.5=27.5$ dollars.

Conclusion and check. Design B pays $27.5-22.5=5.00$ dollars more in expectation per claim. The result is plausible: B has a higher attachment, but it continues paying through a substantially higher loss range. The calculation uses the same uniform severity model for both alternatives and rounds to cents only at the final comparison.

Common mistake avoided. The exhaustion point for design B is $70$, not its payment limit $50$. Using $L(20)-L(50)$ would incorrectly treat $50$ as a ground-up loss threshold.

Crossing Designs: The Severity Distribution Can Decide

When one design begins paying earlier but the other can pay more at high losses, their payment functions generally cross. A distribution with many moderate claims may favor the lower deductible; a distribution with enough large claims may favor the higher cap. These are tendencies rather than universal rules: the actual ranking follows from the expectation under the specified model.

Worked Example: Lower Deductible Versus Higher Cap Under an Exponential Model

For a hypothetical equipment policy, suppose ground-up loss $X$ is exponential with mean $100$ dollars, so its rate parameter is $\lambda=0.01$ per dollar and $S_X(x)=e^{-0.01x}$ for $x\geq0$. Compare A, with $d_A=0$, $u_A=40$, and B, with $d_B=20$, $u_B=100$. Payments are undiscounted; interest and mortality are not applicable.

For an exponential loss with this rate, direct integration gives $$ L(a)=\int_a^\infty (x-a)(0.01)e^{-0.01x}\,dx =100e^{-0.01a}. $$ Thus the expected payment is $L(d)-L(d+u)$.

Design A. The attachment is $0$ and the exhaustion point is $40$: $$ E[Y_A]=100e^0-100e^{-0.4} =100(1-0.670320)=32.967995\ldots. $$ Hence $E[Y_A]=\$32.97$ to the nearest cent.

Design B. The attachment is $20$ and the exhaustion point is $120$: $$ E[Y_B]=100e^{-0.2}-100e^{-1.2} =100(0.818731-0.301194) =51.7537\ldots. $$ Hence $E[Y_B]=\$51.75$ to the nearest cent.

Conclusion and check. B's expected payment exceeds A's by about $51.7537-32.9680=\$18.79$. A starts paying at zero, but its cap is only $40$; B's larger payment range more than offsets its higher deductible under this severity model. Each expected payment is below its own payment cap, as it must be.

Common mistake avoided. Do not compare deductibles or limits in isolation. The formula uses both the attachment and exhaustion point for each design.

Worked Example: Discrete Losses Can Reverse the Ranking

A hypothetical small-business policy has ground-up severity $X$ with the following distribution, in dollars:

Loss $x$	0	25	60	120
Probability	0.20	0.30	0.30	0.20

Compare design A, $d_A=0,u_A=40$, and design B, $d_B=20,u_B=60$. The probabilities sum to $0.20+0.30+0.30+0.20=1.00$. There is no discounting, so interest and mortality assumptions do not apply.

Design A. Applying $Y_A=\min(X,40)$ to each possible loss gives payments $0,25,40,40$. Therefore, $$ E[Y_A]=(0.20)(0)+(0.30)(25)+(0.30)(40)+(0.20)(40) =0+7.50+12.00+8.00 =\$27.50. $$

Design B. Applying $Y_B=\min((X-20)_+,60)$ gives payments $0,5,40,60$. Therefore, $$ E[Y_B]=(0.20)(0)+(0.30)(5)+(0.30)(40)+(0.20)(60) =0+1.50+12.00+12.00 =\$25.50. $$

Conclusion and check. A pays $\$2.00$ more in expectation. The result reverses the ranking in the exponential example: the answer depends on where this distribution places its probability, not just on the numerical deductible and limit values. Both computed expectations lie between zero and their respective payment limits.

Common mistake avoided. With discrete severities, calculate the payment at each possible loss, including losses exactly at or above a boundary. Here the loss of $120$ produces B's full payment limit of $60$, not $120-20=100$.

A Practical Comparison Method and Common Traps

For an exam-style comparison, first write down the payment function and confirm what the stated limit applies to. Then calculate each design separately under the same $F_X$, using a stop-loss formula when available or probability-weighted payments for a discrete model. Only after both expectations are calculated should you state which is larger.

Identify the limit convention.
For a payment limit $u$, the payment is $\min((X-d)_+,u)$, and the exhaustion point is $d+u$.

Check for pointwise dominance.
If one design has both a no-higher deductible and a no-lower cap, its payment is at least as large for every loss.

Calculate both expectations under the shared model.
Use $L(d)-L(d+u)$, or compute and weight the payments at each discrete severity value.

Compare and interpret.
Report the difference in expected payments and explain which feature of the distribution makes the result plausible.

A frequent error is to use $L(u)$ as the second stop-loss term. That would calculate $L(d)-L(u)$, but the correct term is $L(d+u)$ when $u$ is a payment limit. For example, with $d=10$ and $u=30$, the correct exhaustion point is $40$, so the expected payment is $L(10)-L(40)$. The incorrect $L(10)-L(30)$ truncates the layer too early.

Another trap is to assume that a lower deductible automatically produces the larger expected payment. It does not when the alternative has a sufficiently higher cap. Conversely, a larger cap alone does not guarantee a larger expectation if the deductible is also higher. When the payment functions cross, the severity distribution is essential to the comparison.

Key takeaway: Compare alternative deductible-and-payment-limit designs through their expected payments under the same ground-up severity model. The formula $E[Y(d,u)]=L(d)-L(d+u)$ accounts for both the attachment and the exhaustion point; distribution-free conclusions are justified only when one payment function dominates the other pointwise.

Check Your Understanding

For each problem, compare expected insurer payments per claim. Unless a problem says otherwise, payments are undiscounted and there is no interest or mortality assumption.

A hypothetical policy has $X\sim\operatorname{Uniform}(0,80)$ dollars. Compare A with $d=5,u=25$ and B with $d=15,u=40$. Which expected payment is larger, and by how much? (Answer: B is larger by $\$2.97$; $E[Y_A]=\$19.53$, $E[Y_B]=\$22.50$.)
Suppose a hypothetical severity is exponential with mean $\$50$. Compare A with $d=0,u=30$ and B with $d=10,u=50$. Which expected payment is larger? (Answer: B; $E[Y_A]=\$22.56$, $E[Y_B]=\$25.88$, to the nearest cent.)
A discrete ground-up loss is $\$0,\$10,\$40,\$90$ with respective probabilities $0.10,0.40,0.30,0.20$. Compare A with $d=0,u=30$ and B with $d=10,u=50$. (Answer: They tie; each expected payment is $\$19.00$.)
For any nonnegative severity distribution, compare A with $d_A=5,u_A=40$ and B with $d_B=10,u_B=30$. Which design has the greater or equal expected payment, and why? (Answer: A has an expected payment at least as large, because its payment is at least B's for every ground-up loss.)
A hypothetical loss is $\$20$ with probability $p$ and $\$100$ with probability $1-p$. Compare A with $d=0,u=30$ and B with $d=10,u=60$. For what values of $p$ does A have the greater expected payment? (Answer: A is greater when $p>0.75$; the designs tie at $p=0.75$, and B is greater when $p<0.75$.)

What You'll Learn