Franchise Deductibles and Their Payment Distributions

What a Franchise Deductible Pays

A deductible term changes the insurer’s payment for each covered loss. Under a franchise deductible, the threshold determines whether there is a payment at all: if the loss does not exceed the deductible, the payment is zero; if it does exceed the deductible, the full loss is paid. That strict threshold rule affects both the probability of a zero payment and the shape of the payment distribution.

Let $X$ be a nonnegative loss, and let $d\geq 0$ be the franchise deductible. In this tutorial, $Y^L$ denotes the insurer’s payment per loss. The loss model is specified per claim; no interest or mortality assumption is involved.

Definition: The payment under a franchise deductible $d$ is $$ Y^L = \begin{cases} 0, & X\leq d,\\ X, & X>d. \end{cases} $$ Equivalently, $Y^L=X\mathbf{1}_{\{X>d\}}$. Equality at the threshold is not enough to trigger payment.

The threshold does not reduce the size of a covered payment. It decides whether the loss is covered under this payment rule: once $X>d$, the payment equals $X$, not $X-d$. A consequence is that the possible payments have a gap: apart from zero, a payment must be greater than $d$.

Distribution of the Payment

The payment has a probability mass at zero. Because $Y^L=0$ exactly when $X\leq d$,

P(Y^L=0)=P(X\leq d)=F_X(d).

For positive payment values, the loss is passed through unchanged, but only when it exceeds $d$. The cumulative distribution function follows by considering the events that make $Y^L\leq y$.

Result: For a nonnegative loss $X$ and $d\geq 0$, the payment distribution is $$ F_{Y^L}(y)= \begin{cases} 0, & y<0,\\ F_X(d), & 0\leq y<d,\\ F_X(y), & y\geq d. \end{cases} $$ This formula permits discrete losses and atoms in the loss distribution. For a continuous loss with density $f_X$, there is an atom of size $F_X(d)$ at zero, no payment values strictly between zero and $d$, and density $f_X(y)$ for $y>d$.

To see the CDF result, first take $0\leq y<d$. A positive payment cannot be at most $y$, because every positive payment exceeds $d$. Thus the only way $Y^L\leq y$ is for $Y^L=0$, which occurs when $X\leq d$. If $y\geq d$, the event $Y^L\leq y$ consists of losses $X\leq d$, which produce zero, together with losses $d<X\leq y$, which are paid in full. Their union is simply $X\leq y$, giving $F_X(y)$.

If $X$ has a discrete distribution, positive payment atoms are inherited from losses above $d$, while all loss probability at or below $d$ is combined into the payment atom at zero. In particular, any probability mass at exactly $d$ contributes to the zero payment probability. For continuous models, the formula simplifies using $F_X(y)=1-S_X(y)$, where $S_X(y)=P(X>y)$.

Worked Example: Payment CDF for a Mixture of Exponentials

A hypothetical equipment policy has a franchise deductible of $500. Its loss $X$ is a two-component mixture: with probability 0.60, $X$ is exponential with mean $800; with probability 0.40, $X$ is exponential with mean $2,000. Find the probability of zero payment, the probability that payment is at most $1,000, and the expected payment. There is no interest or mortality component. Report probabilities to four decimals and money to the nearest cent.

Method. Calculate each loss CDF and expected franchise payment within each component, then weight by the component probabilities. For an exponential loss with mean $\theta$, $F_X(x)=1-e^{-x/\theta}$, and the expected franchise payment is $(d+\theta)e^{-d/\theta}$.

For the first component, with $\theta_1=800$, $$ F_1(500)=1-e^{-500/800}=1-e^{-0.625}\approx 0.4647. $$ For the second, with $\theta_2=2{,}000$, $$ F_2(500)=1-e^{-500/2000}=1-e^{-0.25}\approx 0.2212. $$ Therefore, $$ P(Y^L=0)=0.60(0.4647)+0.40(0.2212)\approx 0.3673. $$

Since $1,000 is at least the deductible, $F_{Y^L}(1000)=F_X(1000)$. The component CDFs there are $$ F_1(1000)=1-e^{-1.25}\approx 0.7135,\qquad F_2(1000)=1-e^{-0.5}\approx 0.3935. $$ Thus, $$ P(Y^L\leq 1000)=0.60(0.7135)+0.40(0.3935)\approx 0.5855. $$

For the expected payment, the first component contributes $$ (500+800)e^{-500/800}=1300e^{-0.625}\approx 695.84, $$ and the second contributes $$ (500+2000)e^{-500/2000}=2500e^{-0.25}\approx 1947.00. $$ Weighting the component expectations gives $$ E[Y^L]=0.60(695.84)+0.40(1947.00)\approx \$1{,}196.30. $$ The results are consistent: about 36.73% of losses produce no payment, and the payment CDF at $1,000 includes both the zero-payment probability and paid losses up to $1,000.

Expected Payment

For any nonnegative loss distribution, the expected payment can be written as an integral over losses strictly above the deductible:

E[Y^L]=E[X\mathbf{1}_{\{X>d\}}] =\int_{(d,\infty)}x\,dF_X(x).

If the loss is continuous with density $f_X$, this becomes

E[Y^L]=\int_d^\infty x f_X(x)\,dx.

A useful equivalent expression uses the survival function. It also makes clear how both the trigger probability and the size of losses above the threshold affect the expected payment.

Result: For a nonnegative loss with finite expected payment, $$ E[Y^L]=dS_X(d)+\int_d^\infty S_X(t)\,dt, $$ where $S_X(t)=P(X>t)$. This form applies to continuous and discrete loss distributions.

To justify the formula, for every possible loss value $X$, $$ X\mathbf{1}_{\{X>d\}} =d\mathbf{1}_{\{X>d\}}+(X-d)_+, $$ where $(X-d)_+=\max(X-d,0)$. If $X\leq d$, both sides are zero; if $X>d$, the right side is $d+(X-d)=X$. Taking expectations gives $$ E[Y^L]=dP(X>d)+E[(X-d)_+]. $$ For each fixed $X$, the excess above the threshold can be written as $$ (X-d)_+=\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt. $$ Taking expectations inside this nonnegative integral yields $$ E[(X-d)_+]=\int_d^\infty P(X>t)\,dt =\int_d^\infty S_X(t)\,dt. $$ Since $P(X>d)=S_X(d)$, the stated formula follows. This reasoning handles a possible atom at $d$: that atom is not paid, and it contributes neither to $P(X>d)$ nor to the excess above $d$.

Worked Example: Uniform Loss With a Franchise Threshold

A hypothetical repair contract has franchise deductible $d=\$500$. Its loss $X$ is uniformly distributed from $0 to $2,000, with density $f_X(x)=1/2000$ on that interval. Find the zero-payment probability, $P(Y^L\leq \$1{,}000)$, and the expected payment. No interest or mortality assumption applies. Report probabilities to four decimals and money to the nearest cent.

Method. Use the payment CDF for the probabilities and integrate $x f_X(x)$ over losses greater than $500 for the mean.

The loss CDF at the deductible is $$ F_X(500)=\frac{500}{2000}=0.25, $$ so $P(Y^L=0)=0.2500$.

Because $1,000 exceeds the deductible, $$ P(Y^L\leq 1000)=F_X(1000)=\frac{1000}{2000}=0.5000. $$ This includes the zero payments as well as payments from losses between $500 and $1,000.

The expected payment is $$ E[Y^L]=\int_{500}^{2000}x\left(\frac{1}{2000}\right)\,dx =\frac{1}{2000}\left[\frac{x^2}{2}\right]_{500}^{2000} =\frac{4{,}000{,}000-250{,}000}{4{,}000} =\$937.50. $$ As a check, the probability of a loss above $500$ is $1-0.25=0.75$, and the positive payments range from just above $500 to $2,000; an expected payment of $937.50 is consistent with that range and trigger probability.

Discrete Losses and the Strict Threshold

A discrete model makes the strict inequality especially visible. When a possible loss equals $d$, it produces no payment. When a possible loss is greater than $d$, it is paid in full. The payment distribution is therefore found by collecting all probability at or below $d$ at zero and retaining the original loss probabilities above $d$.

Worked Example: A Discrete Loss With an Atom at the Deductible

A hypothetical cargo policy has franchise deductible $d=\$600$. A covered loss $X$ has the following distribution: $200 with probability 0.10, $600 with probability 0.20, $1,000 with probability 0.40, and $1,500 with probability 0.30. Find the payment distribution, $P(Y^L\leq \$1{,}000)$, and the expected payment. No interest or mortality assumption applies. Report probabilities to four decimals and money to the nearest cent.

Method. Apply the condition $X>d$ to each loss value; in particular, the loss equal to $600 does not trigger payment.

The $200 loss and the $600 loss both produce zero. Thus $$ P(Y^L=0)=0.10+0.20=0.30. $$ The $1,000 loss is paid in full with probability 0.40, and the $1,500 loss is paid in full with probability 0.30. The payment distribution is therefore $$ P(Y^L=0)=0.30,\qquad P(Y^L=1000)=0.40,\qquad P(Y^L=1500)=0.30. $$

The event $Y^L\leq 1000$ includes zero payment and the $1,000 payment, so $$ P(Y^L\leq 1000)=0.30+0.40=0.7000. $$ The expected payment is $$ E[Y^L]=0(0.30)+1000(0.40)+1500(0.30) =400+450=\$850.00. $$ The probabilities sum to $0.30+0.40+0.30=1.00$, and the $600 loss has correctly been included in the zero-payment mass.

Common Mistake: Subtracting the Deductible

A frequent error is to calculate a franchise payment as $(X-d)_+$. That expression pays only the amount above the threshold. It is not the franchise payment defined here, which is $X\mathbf{1}_{\{X>d\}}$. The threshold determines whether the full loss is paid, not the size of an excess payment.

For example, let $X$ be exponential with mean $1,000 and let $d=\$400$. Using the excess-above-deductible expression would give $$ E[(X-400)_+]=1000e^{-400/1000}\approx \$670.32. $$ The franchise payment instead has expectation $$ E[Y^L]=(400+1000)e^{-400/1000} =1400e^{-0.4}\approx \$938.45. $$ The difference is $400P(X>400)$, the deductible amount paid on every loss that crosses the threshold. The CDF gives a separate check: $P(Y^L=0)=F_X(400)=1-e^{-0.4}\approx 0.3297$, so the crossing probability is $e^{-0.4}$. Multiplying that crossing probability by $400 gives the difference between the two expectations.

Key takeaway: Under a franchise deductible, $Y^L=0$ for $X\leq d$ and $Y^L=X$ for $X>d$. The payment CDF has mass $F_X(d)$ at zero, and its expected value can be calculated either by integrating $x$ over losses above $d$ or by using $dS_X(d)+\int_d^\infty S_X(t)\,dt$. These distribution and expectation techniques are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical loss model, find the requested values using the strict trigger $X>d$. Give probabilities to four decimals and dollar amounts to the nearest cent.

A loss is exponential with mean $500 and franchise deductible $200. Find $P(Y^L=0)$ and $E[Y^L]$. (Answer: $P(Y^L=0)=0.3297$; $E[Y^L]=\$469.22$.)
A loss is uniform on $[0,3000]$, and $d=\$750$. Find the zero-payment probability and expected payment. (Answer: $0.2500$; $\$1{,}406.25$.)
A loss is $100, $400, $900, or $1,600 with respective probabilities 0.15, 0.25, 0.40, and 0.20. With $d=\$400$, find $P(Y^L=0)$, $E[Y^L]$, and $P(Y^L\leq \$900)$. (Answer: $0.4000$; $\$680.00$; $0.8000$.)
A loss is an exponential mixture: with probability 0.60 its mean is $800, and with probability 0.40 its mean is $2,000. If $d=\$500$, find $P(Y^L=0)$ and $E[Y^L]$. (Answer: $0.3673$; $\$1{,}196.30$.)
A loss is exponential with mean $1,200 and $d=\$600$. Find the probability of zero payment, the probability that the payment exceeds $1,000, and the expected payment. (Answer: $0.3935$; $0.4346$; $\$1{,}091.76$.)

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