Ordinary Deductibles and Excess-Loss Payments

Introduction

An ordinary deductible determines how much of a covered loss the insurer pays, not whether the loss itself occurred. For a loss below the deductible, the insurer pays nothing. For a loss above it, the insurer pays the amount by which the loss exceeds the deductible. The essential calculation is simple, but it is easy to confuse the ground-up loss with the insurer’s payment or to subtract the deductible when doing so would produce a negative amount.

This tutorial defines the ordinary-deductible payment for one loss and develops its basic interpretation. It assumes a covered loss with no policy limit, coinsurance, or other adjustment to the payment. The focus is the transformation from a ground-up loss to the payment for that loss, rather than probabilities or moments of the transformed amount.

Definition: Let $X$ be a nonnegative ground-up loss, measured before applying the insurance contract, and let $d\geq 0$ be an ordinary deductible. The insurer’s payment for the loss is $$ Y=(X-d)^+=\max(X-d,0). $$ In loss-model notation, this payment per loss may also be written $Y^L$. The positive-part notation $z^+$ means $\max(z,0)$.

How the Payment Rule Works

The deductible is applied to each covered loss under the ordinary per-loss arrangement considered here. If $X\leq d$, the payment is zero. If $X>d$, the payment is $X-d$. The equality case matters: a loss exactly equal to the deductible produces no insurer payment.

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

The policyholder’s retained amount is the part of the ground-up loss not paid by the insurer. For this contract, it is $X-Y$, so

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

This identity gives a useful check. When $X\leq d$, the policyholder retains the full loss, $X$. When $X>d$, the policyholder retains exactly $d$, while the insurer pays the excess. In either case, the two amounts add back to the original loss:

$$ X=Y+(X-Y). $$

The deductible is not itself an additional payment by the insurer, nor is it subtracted from every loss without regard to the result. The positive-part operation enforces the contract’s zero-payment floor. This rule describes an ordinary deductible; it should not be replaced by a different payment rule unless the contract specifies different terms.

Why the Positive-Part Formula Is Correct

The formula follows directly by considering the two possible positions of the loss relative to the deductible. If $X\leq d$, then $X-d\leq 0$, and the maximum of $X-d$ and zero is zero. If $X>d$, then $X-d>0$, and the maximum is $X-d$. Thus the single expression $\max(X-d,0)$ implements both branches of the contract without allowing a negative insurer payment.

The same reasoning verifies the retained amount. If $X\leq d$, then $Y=0$ and $X-Y=X$, which is the smaller of $X$ and $d$. If $X>d$, then $Y=X-d$, giving $X-Y=d$, again the smaller of $X$ and $d$. Therefore the retention is $\min(X,d)$ in both cases.

Identify the ground-up loss.
Use the covered loss amount before any deductible is applied, and ensure it is expressed in the same monetary units as the deductible.

Compare the loss with the deductible.
If $X\leq d$, the payment is zero. If $X>d$, continue by subtracting $d$ from $X$.

Check the allocation.
Calculate the retained amount as $X-Y$, or directly as $\min(X,d)$, and verify that payment plus retention equals $X$.

Worked Applications

Worked Example: Applying a Deductible to Several Claims

A hypothetical insurer covers four separate claims under a contract with a per-loss ordinary deductible of $1,800. The realized ground-up losses are $700, $1,800, $2,500, and $6,200. Calculate the insurer payment and policyholder retention for each claim, then check the totals. Each loss is treated as a fixed realized amount; no probability distribution is assumed. Interest and mortality assumptions are not relevant to this single-loss severity calculation.

Method. Apply $Y=(X-d)^+$ separately to each claim, then calculate retention as $X-Y$. A common mistake avoided here is to treat a claim equal to the deductible as producing a positive payment.

For the $700 loss, $X-d=700-1{,}800=-1{,}100$, so the payment is $\max(-1{,}100,0)=\$0$. Retention is $700-0=\$700$.

For the $1,800 loss, $X-d=1{,}800-1{,}800=0$, so the payment is $\max(0,0)=\$0$. Retention is $1{,}800-0=\$1{,}800$.

For the $2,500 loss, $X-d=2{,}500-1{,}800=700$, so the payment is $\max(700,0)=\$700$. Retention is $2{,}500-700=\$1{,}800$.

For the $6,200 loss, $X-d=6{,}200-1{,}800=4{,}400$, so the payment is $\max(4{,}400,0)=\$4{,}400$. Retention is $6{,}200-4{,}400=\$1{,}800$.

The total ground-up loss is $700+1{,}800+2{,}500+6{,}200=\$11{,}200$. Total insurer payments are $0+0+700+4{,}400=\$5{,}100$, and total retentions are $700+1{,}800+1{,}800+1{,}800=\$6{,}100$. The check is $5{,}100+6{,}100=11{,}200$, matching the total ground-up loss.

Worked Example: Mapping Loss Scenarios to Payments

A hypothetical coverage has an ordinary deductible of $1,800. For illustration, the ground-up loss $X$ has a discrete severity model with possible values $0, $900, $2,800, and $6,400, with respective probabilities 0.15, 0.25, 0.35, and 0.25. The probabilities sum to $0.15+0.25+0.35+0.25=1.00$. Interest and mortality are not part of this model. Find the insurer payment and policyholder retention associated with each possible loss.

Method. Evaluate the payment rule at each specified loss value. The common mistake avoided is to calculate a payment for every positive loss; a positive loss still produces zero payment if it does not exceed the deductible.

At $X=\$0$, $Y=\max(0-1{,}800,0)=\$0$, and retention is $0-0=\$0$.

At $X=\$900$, $Y=\max(900-1{,}800,0)=\max(-900,0)=\$0$, and retention is $900-0=\$900$.

At $X=\$2{,}800$, $Y=\max(2{,}800-1{,}800,0)=\$1{,}000$, and retention is $2{,}800-1{,}000=\$1{,}800$.

At $X=\$6{,}400$, $Y=\max(6{,}400-1{,}800,0)=\$4{,}600$, and retention is $6{,}400-4{,}600=\$1{,}800$.

The probability attached to each scenario does not change the payment formula; it describes how likely that ground-up loss is under the stated model. In every scenario, the calculated payment plus retention equals the listed loss: respectively, $0+0=0$, $0+900=900$, $1{,}000+1{,}800=2{,}800$, and $4{,}600+1{,}800=6{,}400$.

Worked Example: Recovering the Ground-Up Loss From a Payment

A hypothetical policy has an ordinary deductible of $1,250 and no payment limit or coinsurance. A covered claim produces an insurer payment of $3,750. Find the ground-up loss and the policyholder’s retention. Then determine the payment and retention for that same loss if the deductible were instead $2,000. Treat the claim amount as a fixed realized loss; no probability distribution, interest rate, or mortality assumption is needed.

Method. Because the observed payment is positive, the ground-up loss must exceed the deductible, and the payment equation is $Y=X-d$. The common mistake avoided is to report the payment itself as the ground-up loss.

Under the original deductible, solve $3{,}750=X-1{,}250$. Adding $1,250 to both sides gives $X=\$5{,}000$. The original retention is $\min(5{,}000,1{,}250)=\$1{,}250$. The allocation check is $3{,}750+1{,}250=\$5{,}000$.

Under the revised deductible of $2,000, the ground-up loss remains $5,000. The payment is $\max(5{,}000-2{,}000,0)=\$3{,}000$, and retention is $\min(5{,}000,2{,}000)=\$2{,}000$. The revised allocation check is $3{,}000+2{,}000=5{,}000$. Increasing the deductible by $750 reduces the payment by $750 for this loss, while the ground-up loss itself does not change.

Common Mistakes and Checks

The positive-part notation is not decorative: it is part of the payment definition. For example, suppose $X=\$1{,}300$ and $d=\$1{,}700$. The incorrect calculation $Y=X-d=1{,}300-1{,}700=-\$400$ gives a negative payment. The correction is

Y=(1{,}300-1{,}700)^+=(-400)^+=\$0.

The policyholder retains the full $1,300 loss. More generally, a reliable calculation should pass three checks: the payment is nonnegative; it is zero whenever $X\leq d$; and payment plus retention equals the ground-up loss. For a loss above the deductible, the retention should equal $d$; for a loss at or below it, retention should equal $X$.

Subtracting without taking the positive part. This can create a negative payment for losses below the deductible. Use $\max(X-d,0)$.
Paying the full loss above the threshold. When $X>d$, the insurer pays $X-d$, not $X$; the policyholder retains the deductible.
Misreading equality. At $X=d$, the excess is zero, so the payment is zero.
Mixing payment and retention. The payment is $(X-d)^+$; the retention is $\min(X,d)$. They are different amounts that sum to $X$.
Applying the rule to inconsistent units. Convert losses and deductibles to the same currency and scale before comparing or subtracting them.

Key takeaway: An ordinary deductible pays only the amount of a ground-up loss above $d$: $Y=(X-d)^+$. The insured retains $\min(X,d)$, and these two amounts must add to $X$.

Check Your Understanding

For each item, apply the ordinary-deductible payment rule and check the allocation between insurer and policyholder.

A covered loss is $650 and the deductible is $900. Find the payment and retention. (Answer: payment $0; retention $650.)
A covered loss is $2,350 and the deductible is $1,400. Find the payment and retention. (Answer: payment $950; retention $1,400.)
A hypothetical discrete severity model assigns probabilities 0.20, 0.50, and 0.30 to losses of $500, $2,000, and $5,500, respectively. With a $1,500 ordinary deductible, give the payment for each loss in the same order. (Answer: $0, $500, and $4,000.)
A claim has an insurer payment of $2,750 under a $750 ordinary deductible, with no limit or coinsurance. Recover the ground-up loss and retention. (Answer: ground-up loss $3,500; retention $750.)
A $4,800 covered loss is subject to a $1,100 deductible. A contract change raises the deductible to $1,600. Find the payment under each deductible and the decrease in payment. (Answer: payments $3,700 and $3,200, respectively; decrease $500.)

What You'll Learn