Why the Deductible Type Matters
An ordinary deductible and a franchise deductible can have the same threshold but produce very different insurer payments. An ordinary deductible leaves the insured responsible for the deductible amount on every covered loss above the threshold. A franchise deductible instead makes the insurer pay nothing unless the loss exceeds the threshold; once it does, the insurer pays the full loss. The difference can be substantial, particularly when losses often exceed the deductible by only a small amount.
The comparison is about the payment generated by one loss, not the premium or the total cost of a policy. Assume throughout that \(X\) is a nonnegative loss and \(d\geq 0\) is the deductible. Unless a scenario states otherwise, use the strict franchise condition \(X>d\): a loss exactly equal to \(d\) produces no payment under either design.
Compare the Payments at the Threshold
The payment functions differ only once the loss is strictly above the deductible. Below or at the deductible, both pay zero. Above it, the ordinary deductible subtracts \(d\), while the franchise deductible pays the entire loss.
| Loss amount | Ordinary deductible payment | Franchise deductible payment |
|---|---|---|
| \(X<d\) | 0 | 0 |
| \(X=d\) | 0 | 0 under the strict \(X>d\) convention |
| \(X>d\) | \(X-d\) | \(X\) |
For any particular loss above \(d\), the franchise payment exceeds the ordinary payment by exactly \(d\). At or below \(d\), the difference is zero. Thus the franchise design does not merely reduce the amount the insured pays on a large loss: it creates a sharp threshold between no insurer payment and full payment.
Worked Example: Comparing Payments With a Loss at the Threshold
A hypothetical contract has deductible \(d=\$800\). Consider a loss distribution with \(P(X=\$200)=0.20\), \(P(X=\$800)=0.30\), \(P(X=\$1{,}200)=0.30\), and \(P(X=\$2{,}000)=0.20\). Find the payment under each design for each loss amount and then calculate each expected payment. This is a single-claim severity calculation: interest and mortality are not applicable, and no discounting is used.
Method. Apply each payment function separately at every possible loss, then weight the payments by their probabilities.
| \(X\) | \(P(X)\) | Ordinary payment | Franchise payment | Ordinary weighted payment | Franchise weighted payment |
|---|---|---|---|---|---|
| \(\$200\) | 0.20 | \(\$0\) | \(\$0\) | \(\$0\) | \(\$0\) |
| \(\$800\) | 0.30 | \(\$0\) | \(\$0\) | \(\$0\) | \(\$0\) |
| \(\$1{,}200\) | 0.30 | \(\$1{,}200-\$800=\$400\) | \(\$1{,}200\) | \(\$400(0.30)=\$120\) | \(\$1{,}200(0.30)=\$360\) |
| \(\$2{,}000\) | 0.20 | \(\$2{,}000-\$800=\$1{,}200\) | \(\$2{,}000\) | \(\$1{,}200(0.20)=\$240\) | \(\$2{,}000(0.20)=\$400\) |
Therefore, \(E[Y_o^L]=\$0+\$0+\$120+\$240=\$360\), while \(E[Y_f^L]=\$0+\$0+\$360+\$400=\$760\). The expected-payment difference is \(\$760-\$360=\$400\). Losses strictly above the deductible have total probability \(0.30+0.20=0.50\), so \(dP(X>d)=\$800(0.50)=\$400\), which checks the difference. The common mistake avoided here is treating the loss equal to \(\$800\) as exceeding the threshold; under the stated strict convention, it triggers neither payment.
Expected Payments and Their Relationship
Let \(S_X(x)=P(X>x)\) be the survival function of the nonnegative loss \(X\). For any severity distribution, including a distribution with discrete probability masses, the ordinary-deductible expected payment is the expected excess above \(d\). The franchise payment can be expressed as that same excess plus \(d\) whenever the loss exceeds \(d\).
The first integral identity follows by writing the ordinary payment for a fixed loss \(X\) as the length of an interval: \((X-d)_+=\int_d^\infty \mathbf{1}(X>t)\,dt\). If \(X\leq d\), the interval contributes zero; if \(X>d\), it contributes one for values of \(t\) from \(d\) up to \(X\), giving length \(X-d\). Taking expectations gives the integral of \(P(X>t)=S_X(t)\). For the franchise identity, on the event \(X>d\), write \(X=(X-d)+d\); outside that event the franchise payment is zero. The excess component is exactly the ordinary payment, and the remaining component has expectation \(dP(X>d)\).
If \(X\) has a continuous distribution, \(P(X=d)=0\), so using \(P(X\geq d)\) instead of \(P(X>d)\) happens to give the same probability. For a discrete or mixed distribution, the two probabilities can differ. The contract's exact threshold wording therefore matters whenever losses can equal the deductible with positive probability.
Worked Example: Exponential Severity and the Expected-Payment Gap
Suppose a hypothetical single-claim loss has an exponential distribution with mean \(\theta=\$1{,}000\), so \(S_X(x)=e^{-x/1000}\) for \(x\geq0\). The deductible is \(d=\$500\). Calculate the expected payment under each design and verify their difference. Interest and mortality are not applicable; payments are evaluated at claim time without discounting.
Method. Use the survival-function formulas, with the exponential survival function and its mean parameter stated above.
First, the probability that the loss strictly exceeds the deductible is \(S_X(500)=e^{-500/1000}=e^{-0.5}=0.6065307\), or 0.6065 to four decimal places.
For an exponential severity with mean 1,000, integrate the tail above 500:
The franchise expected payment is the ordinary expected payment plus the deductible times the exceedance probability: \(E[Y_f^L]=\$606.5307+\$500(0.6065307)=\$606.5307+\$303.2653=\$909.7960\), or \(\$909.80\) to the nearest cent. The difference is \(\$909.7960-\$606.5307=\$303.2653\), which also equals \(\$500(0.6065307)\). For an individual loss of \(\$300\), both payments are zero; at \(\$500\), both are zero; at \(\$900\), the ordinary payment is \(\$400\) and the franchise payment is \(\$900\). The common mistake avoided is adding the deductible to every loss rather than adding it only on the event \(X>d\).
A Second Severity Shape
The comparison does not depend on an exponential model. The payment functions and expectation relationship apply to other severity distributions as well. The following uniform example shows that the expected-payment gap is still determined by the deductible and the probability of exceeding it.
Worked Example: Uniform Severity With a Different Tail Shape
A hypothetical claim amount is uniformly distributed from \(\$0\) to \(\$2{,}000\), with density \(f_X(x)=1/2000\) on that interval. The deductible is \(d=\$500\). Find both expected payments and check the result using the exceedance probability. This is an undiscounted, single-claim calculation; interest and mortality are not applicable.
Method. Integrate the relevant payment over the uniform density. The density is constant, and the loss exceeds \(\$500\) with probability \((2000-500)/2000=0.75\).
The ordinary payment is zero below the deductible and \(x-500\) above it:
For the franchise payment, the insurer pays the full loss when \(x>500\):
The difference is \(\$937.50-\$562.50=\$375.00\). The identity gives the same value: \(dP(X>d)=\$500(0.75)=\$375.00\). At a loss of \(\$300\), both payments are zero; at \(\$500\), both are zero; at \(\$1{,}300\), the ordinary payment is \(\$800\) and the franchise payment is \(\$1{,}300\). The common mistake avoided is calculating the franchise expectation as though the insurer paid only the excess above \(\$500\); that would calculate the ordinary payment instead.
Common Exam Trap: Using the Wrong Exceedance Probability
When the loss distribution has a point mass at the deductible, replacing \(P(X>d)\) by \(P(X\geq d)\) overstates the franchise payment. Consider the discrete example with \(d=\$800\): \(P(X=\$200)=0.20\), \(P(X=\$800)=0.30\), \(P(X=\$1{,}200)=0.30\), and \(P(X=\$2{,}000)=0.20\). The ordinary expected payment is \(\$360\), as calculated above.
The incorrect calculation uses \(P(X\geq800)=0.30+0.30+0.20=0.80\), giving a purported franchise expected payment of \(\$360+\$800(0.80)=\$1{,}000\). This is wrong because the \(\$800\) loss itself receives no franchise payment. The correct strict exceedance probability is \(P(X>800)=0.30+0.20=0.50\); thus \(E[Y_f^L]=\$360+\$800(0.50)=\$760\), consistent with direct probability weighting.
A reliable solution sequence is to write the threshold condition exactly, evaluate the payment at losses below, equal to, and above the deductible, and only then calculate the expectation. State whether the model has a probability mass at the threshold; a continuous model has no such mass, but a discrete model may.
Check Your Understanding
For each problem, use the strict franchise convention that losses equal to the deductible produce no payment. Amounts are hypothetical and undiscounted.
- A loss takes values \(\$0\), \(\$400\), \(\$1{,}000\), and \(\$1{,}600\) with probabilities 0.10, 0.20, 0.40, and 0.30, respectively. If \(d=\$600\), calculate both expected payments. (Answer: Ordinary \(\$460.00\); franchise \(\$880.00\).)
- A single-claim loss is exponential with mean \(\$800\), and \(d=\$400\). Find the exceedance probability and both expected payments. (Answer: \(P(X>d)=0.6065\); ordinary \(\$485.22\); franchise \(\$727.84\).)
- A loss is uniform on \([0,\$1{,}200]\), with deductible \(d=\$300\). Find both expected payments and verify the gap. (Answer: Ordinary \(\$337.50\); franchise \(\$562.50\); gap \(\$225.00\).)
- A loss takes values \(\$500\), \(\$1{,}000\), \(\$1{,}500\), and \(\$2{,}500\) with probabilities 0.40, 0.10, 0.30, and 0.20. For \(d=\$1{,}000\), calculate each expected payment, paying particular attention to the probability mass exactly at the deductible. (Answer: Ordinary \(\$450.00\); franchise \(\$950.00\).)
- A loss takes values \(\$250\), \(\$750\), and \(\$1{,}250\) with probabilities 0.20, 0.50, and 0.30. For \(d=\$750\), calculate both expected payments and the probability that creates the gap between them. (Answer: Ordinary \(\$150.00\); franchise \(\$375.00\); \(P(X>d)=0.3000\), so the gap is \(\$225.00\).)