Calculating Payments Under A Per-Loss Deductible

Introduction

A deductible determines how much of a covered loss the insured pays before the insurer contributes. With a per-loss deductible, that amount applies separately to each loss. The key calculation is therefore not just how large the total losses are, but how many distinct losses the policy treats separately.

This tutorial develops the insurer’s payment rule as a function of the ground-up loss and applies it to multiple losses. The examples use hypothetical fixed dollar amounts; no interest rate, mortality assumption, or probability distribution is involved.

The Per-Loss Deductible Rule

Definition: A per-loss deductible of $d$ dollars is the amount the insured bears first for each covered loss. If $X$ is the ground-up amount of one covered loss, the insurer’s payment for that loss is $Y^L$, where $Y^L=\max(X-d,0)$. Here, $X\geq 0$ and $d\geq 0$.

The ground-up loss is the covered loss amount before applying the deductible. If the ground-up loss is less than or equal to the deductible, the insurer pays nothing. If the loss exceeds the deductible, the insurer pays the excess over the deductible. Equivalently, the rule is

Y^L = \begin{cases} 0, & 0 \leq X \leq d,\\ X-d, & X>d. \end{cases}

The amount retained by the insured from this loss is $\min(X,d)$: it is the whole loss when $X\leq d$, and it is $d$ when $X>d$. Thus, for every loss,

X=Y^L+\min(X,d).

This identity is a useful check: the insurer’s payment plus the insured’s retained portion must equal the ground-up loss. A deductible is not a payment made by the insurer, and it does not make the covered loss disappear; it allocates the first $d$ dollars of each loss to the insured.

Why the Payment Formula Holds

The formula follows directly from the two possible relationships between $X$ and $d$. If $X\leq d$, the insured’s first-dollar responsibility covers the entire loss, so the insurer’s payment is zero. If $X>d$, the insured bears $d$ dollars and the insurer pays the remaining $X-d$ dollars. A payment cannot be negative, which is why the two cases can be written together as a maximum.

Y^L=\max(X-d,0).

For separate losses $X_1,\ldots,X_n$, each subject to the same per-loss deductible $d$, calculate the payment for each loss first. The total insurer payment is the sum of those individual payments:

\sum_{j=1}^{n}Y_j^L = \sum_{j=1}^{n}\max(X_j-d,0).

The deductible is applied once to each loss, not once to the sum of all losses. If the losses have different deductibles, use the deductible specified for each loss in its own payment calculation.

Applying the Rule to Individual Losses

A reliable calculation keeps the loss amount, deductible, insurer payment, and insured retention separate. For each loss, compare $X$ with $d$, calculate $Y^L$, and check that the payment does not exceed the loss. If there are multiple losses, repeat this process for every one before adding payments.

Worked Example: Several Losses With One Per-Loss Deductible

A hypothetical property policy has a per-loss deductible of $d=\$1{,}500$. Three separate covered losses have ground-up amounts of $\$900$, $\$2{,}400$, and $\$6{,}200$. Find the insurer’s payment and the insured’s retained amount for each loss, then find the totals. The amounts are fixed; no probability distribution, interest rate, or mortality assumption applies.

Method. Apply $Y^L=\max(X-d,0)$ separately to each loss, then add the resulting payments.

For the $\$900$ loss,

Y_1^L=\max(900-1{,}500,0)=\max(-600,0)=\$0.

The insured retains the whole $\$900$, since $\min(900,1{,}500)=\$900$.

For the $\$2{,}400$ loss,

Y_2^L=\max(2{,}400-1{,}500,0)=\max(900,0)=\$900.

The insured retains $\min(2{,}400,1{,}500)=\$1{,}500$.

For the $\$6{,}200$ loss,

Y_3^L=\max(6{,}200-1{,}500,0)=\$4{,}700.

The insured retains $\min(6{,}200,1{,}500)=\$1{,}500$. The total insurer payment is $\$0+\$900+\$4{,}700=\$5{,}600$. The total insured retention is $\$900+\$1{,}500+\$1{,}500=\$3{,}900$. The total ground-up loss is $\$900+\$2{,}400+\$6{,}200=\$9{,}500$, and $\$5{,}600+\$3{,}900=\$9{,}500$, as required.

Common mistake avoided: Subtracting one deductible from the combined loss would give $\max(9{,}500-1{,}500,0)=\$8{,}000$. That calculation applies one deductible to three losses. Under the stated per-loss term, the correct total is $\$5{,}600$.

Worked Example: One Loss With Multiple Cost Components

A hypothetical covered incident produces two repair bills, $\$1{,}350$ and $\$2{,}050$. The policy treats both bills as components of one covered loss and has a per-loss deductible of $d=\$1{,}000$. Find the insurer’s payment and the insured’s retention. The bills are fixed amounts, with no probability distribution, interest rate, or mortality assumption.

Method. First combine the components because the problem identifies them as one loss. Then apply the deductible once to that loss.

The ground-up amount of the single loss is

X=1{,}350+2{,}050=\$3{,}400.

The insurer’s payment is

Y^L=\max(3{,}400-1{,}000,0)=\$2{,}400.

The insured retains $\min(3{,}400,1{,}000)=\$1{,}000$. The check is $\$2{,}400+\$1{,}000=\$3{,}400$, which equals the ground-up loss.

Common mistake avoided: Subtracting the deductible separately from each repair bill would calculate $\max(1{,}350-1{,}000,0)+\max(2{,}050-1{,}000,0)=\$350+\$1{,}050=\$1{,}400$. That treats the two components as separate losses, contrary to the stated facts. The relevant unit is the loss identified by the policy and problem, not automatically each invoice or cost item.

Separating Losses From Their Combined Total

A per-loss deductible depends on the individual loss amounts, so the total ground-up amount alone is not generally enough to determine the insurer’s total payment. The same combined total can produce different payments if it is divided into different numbers or sizes of losses.

Worked Example: Two Losses From One Hypothetical Event

A hypothetical event results in two separate covered losses, $X_1=\$4{,}600$ and $X_2=\$3{,}100$. Each is subject to a per-loss deductible of $d=\$2{,}000$. Calculate the payment on each loss, the total insurer payment, and the total insured retention. The loss amounts are fixed; no probability distribution, interest rate, or mortality assumption applies.

Method. Treat the two stated losses separately, even though they arose from the same event. Apply the deductible to each, then sum the payments.

For the first loss,

Y_1^L=\max(4{,}600-2{,}000,0)=\$2{,}600.

The insured retains $\min(4{,}600,2{,}000)=\$2{,}000$.

For the second loss,

Y_2^L=\max(3{,}100-2{,}000,0)=\$1{,}100.

The insured retains $\min(3{,}100,2{,}000)=\$2{,}000$. The total insurer payment is $\$2{,}600+\$1{,}100=\$3{,}700$, and total insured retention is $\$2{,}000+\$2{,}000=\$4{,}000$. The combined ground-up loss is $\$4{,}600+\$3{,}100=\$7{,}700$, and $\$3{,}700+\$4{,}000=\$7{,}700$.

Common mistake avoided: Applying one deductible to the combined amount gives $\max(7{,}700-2{,}000,0)=\$5{,}700$, which is not the payment under the stated per-loss term. The correct total is $\$3{,}700$, because the deductible applies twice. Whether losses count separately is determined by the coverage terms and the facts; it cannot be inferred merely from the fact that they share an event.

Common Checks and Exam Traps

Check the loss amount against the deductible. When $X\leq d$, the payment is zero. At exactly $X=d$, the payment is still zero.
Apply the deductible at the stated unit. A per-loss deductible is applied to each loss. Do not replace it with one deduction from a combined total.
Distinguish separate losses from cost components. Several bills may make up one loss, while one event may produce multiple separate losses. Follow the stated classification.
Check the payment cannot exceed the loss. For $X\geq0$ and $d\geq0$, $0\leq Y^L\leq X$.
Reconcile the amounts. For each loss, insurer payment plus insured retention must equal ground-up loss.

These checks matter because the deductible’s unit changes the calculation even when the total loss amount is unchanged. A sound solution states what counts as one loss, applies the deductible at that level, and shows the payment for each loss before reporting a total.

Key takeaway: Under a per-loss deductible $d$, the insurer pays $\max(X-d,0)$ for each covered loss $X$. For multiple losses, calculate each payment separately and then add them; do not deduct $d$ just once from the combined loss amount.

Check Your Understanding

For each problem, use the stated per-loss deductible and show the individual payments before totaling them.

A single covered loss is $\$2{,}750$ and the deductible is $\$3{,}000$. What does the insurer pay? (Answer: $\$0$)
Three separate covered losses are $\$1{,}800$, $\$5{,}200$, and $\$10{,}000$, each subject to a $\$2{,}500$ per-loss deductible. What is the total insurer payment? (Answer: $\$10{,}200$)
A covered loss is $\$13{,}750$ with a $\$2{,}250$ deductible. Find both the insurer’s payment and the insured’s retention. (Answer: payment $\$11{,}500$; retention $\$2{,}250$)
One hypothetical event produces two separate covered losses of $\$4{,}600$ and $\$3{,}100$. Each has a $\$2{,}000$ per-loss deductible. What are the total insurer payment and total insured retention? (Answer: payment $\$3{,}700$; retention $\$4{,}000$)
A single covered loss consists of two repair items costing $\$700$ and $\$950$, plus a separate covered loss of $\$2{,}400$. Each loss has a $\$1{,}200$ deductible. What is the total insurer payment? (Answer: $\$1{,}650$)

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