Worked Problems on Franchise Deductibles and Coinsurance

Franchise Deductibles and Coinsurance

Coverage terms determine not only how much an insurer pays, but also which losses produce a payment at all. A franchise deductible creates a threshold: losses at or below the deductible produce no insurer payment, while a loss above it is covered from its first dollar. Coinsurance then scales the covered payment by a stated percentage. The key is to translate these terms into one payment formula before calculating probabilities or expectations.

This tutorial considers nonnegative ground-up loss $X$, a franchise deductible $d$, and a coinsurance proportion $c$, with $0<c\leq 1$. Unless stated otherwise, assume there is no policy limit or other coverage modification. The examples are hypothetical.

Definition: Under a franchise deductible $d$, the insurer pays the full loss if $X>d$, and pays zero if $X\leq d$. With coinsurance proportion $c$, the insurer payment is $$ Y^L=cX\mathbf{1}_{\{X>d\}}, $$ where $\mathbf{1}_{\{X>d\}}$ equals 1 when $X>d$ and 0 otherwise. Thus a loss exactly equal to $d$ produces no payment. Here $Y^L$ denotes payment per loss.

This is different from an ordinary deductible, for which the insurer generally pays only the amount of loss above the deductible. With the same coinsurance proportion, an ordinary-deductible payment would be $c(X-d)_+$, where $(z)_+=\max(z,0)$. A franchise-deductible payment instead jumps from zero to $cX$ when the loss crosses the threshold. Do not subtract $d$ from a covered loss under a franchise deductible.

Thresholds and Conditional Probabilities

For a positive coinsurance proportion, a payment threshold can be translated directly into a loss threshold. A positive payment requires $X>d$. To have a payment greater than a specified amount $y\geq 0$, both the franchise condition and the payment condition must hold: $X>d$ and $cX>y$. Therefore, conditional on a positive payment, the probability of exceeding $y$ is

P(Y^L>y\mid Y^L>0) = \frac{P\bigl(X>\max(d,y/c)\bigr)}{P(X>d)}, \qquad P(X>d)>0.

The strict inequalities matter when $X$ has probability at individual loss amounts. For a question involving $Y^L\geq y$, use $X\geq y/c$ together with $X>d$, rather than automatically reusing a formula for a strict exceedance. For a continuous distribution, equality at a single threshold has probability zero; for a discrete or mixed distribution, it may have positive probability.

Expected payment follows from the same indicator rule. For any nonnegative loss distribution with a finite expectation,

E[Y^L]=cE\!\left[X\mathbf{1}_{\{X>d\}}\right].

If $X$ is continuous with density $f_X$, this becomes $c\int_d^\infty x f_X(x)\,dx$. If $X$ is discrete, sum $cxP(X=x)$ over the loss values strictly greater than $d$. The strict boundary is part of the coverage definition, not a detail to decide after calculating.

Why the Payment Formula Works

The indicator in the payment formula records whether the franchise condition is met. If $X\leq d$, then $\mathbf{1}_{\{X>d\}}=0$, so $Y^L=0$. If $X>d$, the indicator is 1, so the insurer pays $cX$. Taking expectations gives the expectation formula by multiplying each possible loss by its payment and averaging over the loss distribution:

E[Y^L] =E\!\left[cX\mathbf{1}_{\{X>d\}}\right] =cE\!\left[X\mathbf{1}_{\{X>d\}}\right].

Similarly, for $y\geq0$, the event $\{Y^L>y\}$ is exactly $\{X>d\}\cap\{cX>y\}$. Since $c>0$, the second condition is $X>y/c$; both conditions together require $X>\max(d,y/c)$. Dividing the probability of this event by $P(X>d)$ gives the conditional probability formula.

Worked Problems

Worked Example: Uniform Loss With a Franchise Threshold

A hypothetical equipment policy covers a loss $X\sim\operatorname{Uniform}(0,5000)$ dollars. It has a franchise deductible of $d=\$1{,}000$ and coinsurance of $c=0.80$, with no limit. Find (1) the probability of a positive payment, (2) the probability that the payment exceeds $\$1{,}600$, conditional on a positive payment, and (3) the expected insurer payment.

Method. First convert payment events into loss events using $Y^L=0.80X\mathbf{1}_{\{X>1000\}}$. For a uniform loss on $[0,5000]$, the probability of an interval is its length divided by 5,000.

(1) Positive payment. A payment is positive exactly when $X>1{,}000$. Thus

P(Y^L>0)=P(X>1000)=\frac{5000-1000}{5000} =\frac{4000}{5000}=0.8000.

(2) Payment exceeds $\$1{,}600$, given a positive payment. When covered, the payment is $0.80X$. The strict payment condition is $0.80X>1600$, or $X>2000$. Since this already implies $X>1000$,

P(Y^L>1600\mid Y^L>0) =\frac{P(X>2000)}{P(X>1000)} =\frac{(5000-2000)/5000}{(5000-1000)/5000} =\frac{3000}{4000}=0.7500.

(3) Expected payment. The uniform density is $1/5000$ on $[0,5000]$. Integrate only above the franchise threshold:

\begin{aligned} E[Y^L] &=0.80\int_{1000}^{5000}x\frac{1}{5000}\,dx\\ &=\frac{0.80}{5000}\left[\frac{x^2}{2}\right]_{1000}^{5000}\\ &=\frac{0.80}{5000}\cdot\frac{25{,}000{,}000-1{,}000{,}000}{2}\\ &=\frac{0.80}{5000}\cdot12{,}000{,}000 =1920. \end{aligned}

The answers are $0.8000$, $0.7500$, and $\$1{,}920.00$. The expectation is also consistent with the covered portion: the conditional mean loss above 1,000 is $(1000+5000)/2=3000$, so $0.80\times 3000\times0.80=\$1{,}920$. The common mistake avoided here is treating the franchise as an ordinary deductible and subtracting $\$1{,}000$ from each covered loss.

Worked Example: Exponential Loss and Conditional Payment Exceedance

For a hypothetical repair claim, suppose $X$ has an exponential distribution with mean $\$2{,}000$, so its rate is $\lambda=1/2000=0.0005$ per dollar and $P(X>x)=e^{-0.0005x}$ for $x\geq0$. The contract has a franchise deductible $d=\$1{,}000$ and coinsurance $c=0.75$. Find the probability of a positive payment, the probability that payment exceeds $\$1{,}500$ given a positive payment, and the expected payment.

Method. Use the exponential survival function for threshold probabilities and integrate $cxf_X(x)$ over losses above the deductible for the expectation.

Positive payment. The payment is positive if $X>1000$:

P(Y^L>0)=e^{-0.0005(1000)} =e^{-0.5}=0.6065307\approx0.6065.

Conditional payment exceedance. The condition $0.75X>1500$ requires $X>2000$. Therefore,

\begin{aligned} P(Y^L>1500\mid Y^L>0) &=\frac{P(X>2000)}{P(X>1000)}\\ &=\frac{e^{-0.0005(2000)}}{e^{-0.0005(1000)}}\\ &=e^{-0.5}=0.6065307\approx0.6065. \end{aligned}

Expected payment. The density is $f_X(x)=0.0005e^{-0.0005x}$. Evaluate the integral using integration by parts:

\begin{aligned} E[Y^L] &=0.75\int_{1000}^{\infty}x(0.0005)e^{-0.0005x}\,dx\\ &=0.75e^{-0.0005(1000)}\left(1000+\frac{1}{0.0005}\right)\\ &=0.75e^{-0.5}(1000+2000)\\ &=2250e^{-0.5}\\ &=1364.693984\ldots \end{aligned}

Rounded to the nearest cent, the expected payment is $\$1{,}364.69$. The three results are $0.6065$, $0.6065$, and $\$1{,}364.69$. The equality of the first two probabilities is specific to these thresholds and this exponential model: the threshold for the payment event is exactly $\$1{,}000$ above the deductible.

Worked Example: Discrete Losses and a Mass at the Deductible

A hypothetical small-business policy has a franchise deductible of $d=\$1{,}000$ and coinsurance $c=0.50$. Its loss distribution is

\begin{array}{c|rrrrrr} x&0&800&1000&1500&2500&4000\\ \hline P(X=x)&0.10&0.20&0.15&0.25&0.20&0.10 \end{array}

Find the probability of a positive payment, the probability that payment exceeds $\$1{,}000$ given a positive payment, and the expected payment.

Method. Apply the strict condition $X>1000$ to the listed loss amounts. In particular, the probability mass at exactly $\$1{,}000$ produces no payment.

Positive payment. The covered loss values are $1500,2500,4000$, so

$$ P(Y^L>0)=0.25+0.20+0.10=0.55. $$

Payment exceeds $\$1{,}000$, given a positive payment. For covered losses, payment is $0.50X$. A payment greater than 1,000 requires $X>2000$, so only losses of 2,500 and 4,000 qualify:

P(Y^L>1000\mid Y^L>0) =\frac{0.20+0.10}{0.55} =\frac{0.30}{0.55} =0.5455.

Expected payment. Multiply each covered payment by its probability:

\begin{aligned} E[Y^L] &=(0.50)(1500)(0.25)+(0.50)(2500)(0.20) +(0.50)(4000)(0.10)\\ &=187.50+250.00+200.00\\ &=637.50. \end{aligned}

The answers are $0.55$, $0.5455$, and $\$637.50$. The mass at the deductible is excluded, while the mass at $\$1{,}500$ is included. Including the deductible mass would incorrectly raise both the positive-payment probability and the expected payment.

Common Exam Trap: Mixing Up the Thresholds

A frequent error is to use the deductible itself as the ground-up loss threshold for a payment amount, even after coinsurance has scaled the payment. For example, in the first worked example, the incorrect calculation for a payment above $\$1{,}600$ might start with $X>1600$, giving a conditional probability of $(5000-1600)/(5000-1000)=3400/4000=0.8500$. This is wrong because a covered loss of $\$1{,}600$ produces only a $\$1{,}280$ payment at 80% coinsurance.

The correction is to solve $cX>y$ for $X$, then impose the franchise condition too. Here $0.80X>1600$ means $X>2000$, and the conditional probability is $3000/4000=0.7500$. Always distinguish the insurer payment $Y^L$ from the ground-up loss $X$.

Practical Checks

Before finalizing a calculation, verify that the payment rule matches the contract and that every threshold is expressed in loss units before using the loss distribution. Useful checks include:

A franchise payment must be zero for $X\leq d$, including $X=d$ under the convention used here.
For $c>0$, a positive-payment probability is $P(X>d)$, independent of the numerical value of $c$.
A payment threshold $y$ corresponds to the loss threshold $y/c$; compare it with $d$ before calculating.
The expected insurer payment cannot exceed $cE[X]$, since the insurer pays at most $cX$ on any loss.
For a discrete or mixed distribution, check whether an equality event has positive probability before replacing $>$ with $\geq$, or vice versa.

These checks are useful because a small change in coverage wording can alter both the payment amount and the event being measured. Keeping the indicator condition visible is often the simplest way to avoid errors.

Key takeaway: For a franchise deductible with coinsurance and no limit, write $Y^L=cX\mathbf{1}_{\{X>d\}}$ first. Then derive payment events from this rule and calculate expectations using only losses strictly above the deductible. This approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical contract, identify the loss threshold implied by the payment condition before calculating. Answers are given for checking; monetary expectations are rounded to the nearest cent.

Let $X\sim\operatorname{Uniform}(0,6000)$, with franchise deductible $d=\$1{,}500$, coinsurance $c=0.80$, and no limit. Find the positive-payment probability and expected payment. (Answer: $0.7500$; $\$2{,}250.00$)
Let $X$ be exponential with mean $\$4{,}000$, $d=\$1{,}000$, and $c=0.50$. Find the probability of a positive payment and the expected payment. (Answer: $0.7788$; $\$1{,}947.00$)
A discrete loss has $P(X=500)=0.20$, $P(X=1000)=0.10$, $P(X=2000)=0.40$, and $P(X=4000)=0.30$. With $d=\$1{,}000$ and $c=0.75$, find $P(Y^L>2250\mid Y^L>0)$. (Answer: $0.4286$)
Let $X\sim\operatorname{Uniform}(0,5000)$, with $d=\$1{,}000$ and $c=0.60$. Find the probability that payment exceeds $\$1{,}800$, given a positive payment, and find the expected payment. (Answer: $0.5000$; $\$1{,}440.00$)
A loss takes values $\$1{,}000,\$1{,}500,\$3{,}000$ with probabilities $0.25,0.25,0.50$, respectively. For a franchise deductible $d=\$1{,}000$ and coinsurance $c=0.40$, find $P(Y^L>600\mid Y^L>0)$ and $E[Y^L]$. (Answer: $0.6667$; $\$750.00$)

What You'll Learn