Payment Random Variables Under Policy Terms

One Claim, Three Amounts

A severity model describes the size of a claim before the contract changes what the insurer pays. To analyze coverage correctly, keep three quantities distinct: the ground-up loss, the insurer payment, and the policyholder retention. The last two depend on the ground-up loss and the policy wording.

This tutorial treats a single claim and the payment rules for that claim. All examples are hypothetical. There is no discounting or mortality assumption: the amounts are evaluated at the claim date, with no interest adjustment. Unless stated otherwise, amounts are in dollars and a deductible is applied to the ground-up loss.

Definition: Let $X$ be the nonnegative ground-up loss for one claim, before deductibles, limits, coinsurance, or other coverage modifications. Let $Y^L$ be the insurer's payment per loss under the stated policy terms. If the insurer payment and policyholder retention divide the entire ground-up loss, define the policyholder retention as $R=X-Y^L$. Thus $X=Y^L+R$. This identity assumes there are no other payers, recoveries, expenses, or amounts outside the loss being modeled.

The superscript $L$ indicates a payment amount for a loss, including a value of zero when the policy makes no payment. In some frequency-severity settings, $Y^P$ denotes payment per payment: the payment amount conditional on a positive insurer payment. In precise terms, when $P(Y^L>0)>0$, $Y^P$ has the conditional distribution of $Y^L$ given $Y^L>0$. It is not the same random variable as $Y^L$, because it excludes zero-payment losses.

Writing Policy Terms as Transformations

A policy rule is a function applied to $X$. For a number $z$, write $z^+=\max(z,0)$. Under an ordinary deductible $d$, the insurer pays the amount of loss above the deductible, but never a negative amount:

$$ Y^L=(X-d)^+. $$

The policyholder retains the part of the loss up to the deductible. For this deductible-only contract,

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

A payment cap $u$ limits the insurer's payment to at most $u$. When the contract has both an ordinary deductible and a cap on the insurer payment, the transformation is

Y^L=\min\bigl((X-d)^+,u\bigr), \qquad R=X-Y^L.

Here $u$ is specifically a cap on the payment, not a cap on the covered loss. Policy wording must determine which kind of limit is intended. The displayed formula applies only when the limit caps the insurer payment after applying the deductible.

A coinsurance rule can also be expressed as a transformation. For example, if the insurer pays a proportion $c$, where $0\leq c\leq1$, of the loss above an ordinary deductible, and there is no payment cap, then

Y^L=c(X-d)^+, \qquad R=X-c(X-d)^+.

The contract order matters: this formula applies a deductible first and then coinsurance to the excess. Other contract wording can specify a different order, so the transformation should always be built from the stated terms rather than assumed.

Worked Example: A Deductible Applied to a Discrete Severity

A hypothetical equipment policy has an ordinary deductible of $500 and no payment cap. One claim amount $X$ has the discrete severity distribution

P(X=0)=0.10,\quad P(X=300)=0.20,\quad P(X=900)=0.40,\quad P(X=1{,}800)=0.30.

Find the insurer payment per loss and policyholder retention for each possible claim amount. The loss distribution is discrete with the four stated support values and probabilities; there is no interest or mortality assumption.

Method. Apply $Y^L=(X-500)^+$ to each possible value of $X$, then calculate $R=X-Y^L$. The deductible is subtracted from the claim only when the claim exceeds $500.

Ground-up loss $X$	Insurer payment $Y^L=(X-500)^+$	Retention $R=X-Y^L$
$0	$0	$0
$300	$0	$300
$900	$400	$500
$1,800	$1,300	$500

For example, at $X=$900$, the payment is $(900-500)^+=$400$, and retention is $900-400=$500$. At $X=$300$, the payment is $(300-500)^+=0$, so the policyholder retains the full $300 loss.

Check. In every row, payment plus retention equals the ground-up loss. The probabilities also sum to $0.10+0.20+0.40+0.30=1.00$, so the stated distribution is complete. A common mistake is to treat the deductible as an amount always retained in full: when $X=$300$, retention is $300, not $500.

Why Payment and Retention Reconcile

For the deductible-and-payment-cap contract, the relationship between the policy terms and retention can be verified directly. This gives a useful check on piecewise calculations.

Claim does not exceed the deductible.
If $0\leq X\leq d$, then $(X-d)^+=0$, so $Y^L=0$ and $R=X$.

Claim exceeds the deductible but not the payment cap threshold.
If $d<X\leq d+u$, then $0<X-d\leq u$, so $Y^L=X-d$ and $R=X-(X-d)=d$.

Claim is large enough to reach the payment cap.
If $X>d+u$, then $X-d>u$, so $Y^L=u$ and $R=X-u$.

Combining the cases gives the retention function

R= \begin{cases} X, & 0\leq X\leq d,\\ d, & d<X\leq d+u,\\ X-u, & X>d+u. \end{cases}

The boundary values are consistent: at $X=d$, retention is $d$, and at $X=d+u$, the second and third expressions both give $d$. This derivation follows from subtracting the insurer's payment from the original loss in each region. It also shows why $R=\min(X,d)$ is not the retention formula once a payment cap can leave the policyholder responsible for loss above the cap.

Worked Example: Deductible and Payment Cap Together

A hypothetical repair policy has an ordinary deductible $d=$1,000$ and caps the insurer's payment at $u=$3,000$. The ground-up loss $X$ follows a uniform distribution on $[0,$10,000]$, written $X\sim\operatorname{Uniform}(0,10{,}000)$. For each specified claim amount, calculate the insurer payment and retention. The uniform distribution has lower endpoint $0 and upper endpoint $10,000; no probability calculation is required for the specified claims.

Method. Use $Y^L=\min((X-1{,}000)^+,3{,}000)$ and $R=X-Y^L$. The cap applies to the insurer payment, so a claim above $4,000 can leave additional loss with the policyholder.

For $X=$700$: $(700-1{,}000)^+=0$, so $Y^L=\min(0,3{,}000)=$0$ and $R=700-0=$700$.
For $X=$2,400$: $(2{,}400-1{,}000)^+=$1,400$, so $Y^L=\min(1{,}400,3{,}000)=$1,400$ and $R=2{,}400-1{,}400=$1,000$.
For $X=$5,200$: $(5{,}200-1{,}000)^+=$4,200$, so $Y^L=\min(4{,}200,3{,}000)=$3,000$ and $R=5{,}200-3{,}000=$2,200$.
For $X=$9,000$: $(9{,}000-1{,}000)^+=$8,000$, so $Y^L=\min(8{,}000,3{,}000)=$3,000$ and $R=9{,}000-3{,}000=$6,000$.

Check. The cap threshold is $d+u=$4,000$. The payment reaches $3,000 at and above that claim amount, as seen for $5,200 and $9,000. In each case $Y^L+R=X$. The common mistake is to report retention as just the $1,000 deductible for a $9,000 claim; the cap means the insurer pays only $3,000, leaving $6,000 retained.

Worked Example: Coinsurance After a Deductible

A hypothetical mobile-device contract has a $1,000 ordinary deductible. After the deductible, the insurer pays 80% of the remaining loss, with no payment cap. The ground-up loss has the discrete distribution

P(X=500)=0.15,\quad P(X=1{,}000)=0.25,\quad P(X=2{,}500)=0.40,\quad P(X=7{,}000)=0.20.

Calculate payment and retention at each support value. The loss model is discrete with the stated parameters; there is no interest or mortality assumption. The exact contract order is deductible first, then an 80% insurer share.

Method. Set $c=0.80$ and use $Y^L=0.80(X-1{,}000)^+$, followed by $R=X-Y^L$.

For $X=$500$, the amount above the deductible is $(500-1{,}000)^+=0$. Thus $Y^L=0.80(0)=$0$ and $R=$500$.
For $X=$1,000$, the amount above the deductible is $(1{,}000-1{,}000)^+=0$. Thus $Y^L=$0$ and $R=$1,000$.
For $X=$2,500$, the amount above the deductible is $2{,}500-1{,}000=$1,500$. Thus $Y^L=0.80(1{,}500)=$1,200$ and $R=2{,}500-1{,}200=$1,300$.
For $X=$7,000$, the amount above the deductible is $7{,}000-1{,}000=$6,000$. Thus $Y^L=0.80(6{,}000)=$4,800$ and $R=7{,}000-4{,}800=$2,200$.

Check. Payment and retention sum to $500, $1,000, $2,500, and $7,000 in the respective rows. At $X=$2,500$, a common error is to apply 80% to the whole loss and then subtract the deductible: $0.80(2{,}500)-1{,}000=$1,000$. That is not this contract's rule. Applying the deductible first gives the correct payment of $1,200.

Common Interpretation Traps

Confusing ground-up loss with insurer payment. $X$ is the original claim amount. A deductible, coinsurance percentage, or payment cap changes $Y^L$, not the definition of $X$.
Forgetting the zero-payment case. Under an ordinary deductible, a loss at or below $d$ has $Y^L=0$. The insurer payment cannot be negative.
Using the wrong limit. The formula $\min((X-d)^+,u)$ treats $u$ as a cap on insurer payment. If contract wording instead limits the covered loss, the transformation must be derived from that wording; the same formula should not be assumed.
Applying policy terms in the wrong order. A deductible followed by coinsurance gives $c(X-d)^+$. A different order describes a different contract and can produce a different payment.
Omitting retained loss above a cap. With a deductible and payment cap, retention can exceed the deductible. Always check $R=X-Y^L$.
Mixing per-loss and per-payment amounts. $Y^L$ includes the possibility of no payment. $Y^P$ describes payment amounts conditional on a positive payment, when such payments are possible.

Key takeaway: Start with the ground-up loss $X$, translate each policy term into a function, and obtain retention from $R=X-Y^L$ when payment and retention divide the full loss. Check both the contract order and the identity $X=Y^L+R$.

Check Your Understanding

For each problem, identify the payment transformation and use it to determine the requested values. Amounts are hypothetical and in dollars.

A policy has a $600 ordinary deductible and no cap. For a ground-up loss of $1,450, find insurer payment and retention. (Answer: Payment $850; retention $600.)
A contract has a $900 deductible and a $2,500 cap on insurer payment. A claim is $4,200. Find payment and retention. (Answer: Payment $2,500; retention $1,700.)
A policy pays 70% of the amount above a $500 deductible, with no cap. For a $2,000 loss, calculate payment and retention. (Answer: Payment $1,050; retention $950.)
A claim has a discrete severity distribution $P(X=400)=0.25$, $P(X=1{,}200)=0.50$, and $P(X=3{,}000)=0.25$. Under an ordinary $800 deductible with no cap, give the payment and retention at each support value. (Answer: At $400, payment $0 and retention $400; at $1,200, payment $400 and retention $800; at $3,000, payment $2,200 and retention $800.)
A hypothetical policy applies a $1,000 deductible, then pays 90% of the amount above it, subject to a $3,000 payment cap. For a $6,000 claim, calculate payment and retention, and state whether the cap binds. (Answer: Uncapped share $=0.90(6{,}000-1{,}000)=$4{,}500$; payment $3,000; retention $3,000; the cap binds.)

What You'll Learn