Payment Quantiles From Ground-Up Loss Quantiles

Why Payment Quantiles Need Care

A ground-up loss quantile describes the loss before insurance terms are applied. An insurer-payment quantile describes the amount the insurer pays after those terms are applied. The second quantity is not always found by simply subtracting a deductible from the first: a deductible can create a point mass at zero, a payment limit can create a point mass at the cap, and a franchise deductible has a different threshold rule from an ordinary deductible.

The reliable method is to identify the payment as a function of the ground-up loss, then evaluate that function at the relevant ground-up quantile when its conditions allow. For a franchise deductible, use the threshold probability explicitly. In either case, keep the quantile convention and any strict or inclusive threshold condition visible.

Definition: For a loss $X$ with distribution function $F_X(x)=P(X\le x)$, define its $p$-quantile, for $0<p<1$, by $$ Q_X(p)=\inf\{x:F_X(x)\ge p\}. $$ For an insurer payment $Y$, define $Q_Y(p)$ in the same way using $F_Y(y)=P(Y\le y)$. This generalized-inverse convention applies whether the distribution is continuous, discrete, or mixed. In particular, a quantile can lie at an atom; interpolation between observed loss values is not part of the definition.

Ordinary Deductibles, Limits, and Coinsurance

Let $X\ge 0$ be the ground-up loss. Under an ordinary deductible $d\ge 0$, the insurer pays nothing on the first $d$ dollars of each loss. With a maximum insurer payment $u\ge 0$ above the deductible, the payment per loss is

Y^L=\min\bigl((X-d)_+,u\bigr),\qquad (x-d)_+=\max(x-d,0).

If coinsurance means the insurer pays a proportion $c$, where $0<c\le 1$, of this covered amount, then

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

These are continuous, nondecreasing functions of $X$. For a continuous, nondecreasing payment function $g$, the quantile transformation is

Q_{g(X)}(p)=g\bigl(Q_X(p)\bigr),\qquad 0<p<1.

The statement allows flat portions of $g$: those portions are exactly where different ground-up losses produce the same payment. In the deductible-limit case, losses at or below $d$ all pay zero, while sufficiently large losses all pay the capped amount. Thus the transformed distribution can have atoms even when $X$ is continuous.

Justification: The generalized inverse $Q_X(p)$ is nondecreasing in $p$. If $U$ is uniform on $(0,1)$, then $Q_X(U)$ has the distribution of $X$. Since $g$ is nondecreasing and continuous, $g(Q_X(U))$ is nondecreasing in $U$, and its generalized inverse at $p$ is $g(Q_X(p))$. But $g(Q_X(U))$ has the distribution of $g(X)$, so its quantile is $Q_{g(X)}(p)$. This proves the transformation, including when $g$ has flat portions.

Worked Example: A Discrete Loss With a Deductible, Limit, and Coinsurance

A hypothetical policy has ground-up loss $X$ with the following distribution: $P(X=0)=0.10$, $P(X=100)=0.20$, $P(X=300)=0.40$, and $P(X=700)=0.30$. The ordinary deductible is $d=\$150$, the insurer-payment limit above the deductible is $u=\$250$, and the insurer pays $c=0.80$ of the covered amount. Find $Q_{Y^L}(0.20)$, $Q_{Y^L}(0.50)$, and $Q_{Y^L}(0.75)$.

Method. Apply $g(x)=0.80\min((x-150)_+,250)$ to each ground-up loss. The resulting payments are:

At $X=0$, $Y^L=0.80\min(0,250)=\$0$. At $X=100$, $Y^L=0.80\min(0,250)=\$0$. At $X=300$, $Y^L=0.80\min(150,250)=\$120$. At $X=700$, $Y^L=0.80\min(550,250)=\$200$.

The payment distribution therefore has mass $0.10+0.20=0.30$ at zero, mass $0.40$ at $\$120$, and mass $0.30$ at $\$200$. Its cumulative probabilities at those payment values are $0.30$, $0.70$, and $1.00$, respectively.

For $p=0.20$, the first payment value whose cumulative probability is at least $0.20$ is zero, so $Q_{Y^L}(0.20)=\$0$. For $p=0.50$, the cumulative probability first reaches $0.50$ at $\$120$, so $Q_{Y^L}(0.50)=\$120$. For $p=0.75$, it first reaches $0.75$ at $\$200$, so $Q_{Y^L}(0.75)=\$200$.

The ground-up quantiles provide the same answers: $Q_X(0.20)=100$, $Q_X(0.50)=300$, and $Q_X(0.75)=700$. Applying $g$ gives $0$, $\$120$, and $\$200$. This calculation also checks the payment-distribution result. The common mistake avoided here is to treat the quantile as though it had to be a loss value with positive payment; the deductible creates a zero-payment atom.

The cap affects quantiles at and above the ground-up loss level $d+u$. For $X\ge d+u$, the payment is $u$ before coinsurance, or $cu$ after coinsurance. The probability of reaching that payment is determined by the ground-up distribution above the cap threshold, not by whether the ground-up loss itself has an atom there.

Franchise Deductibles: Separate the Threshold Case

A franchise deductible has a different payment rule. With franchise threshold $d$, the insurer pays nothing when $X\le d$, but pays the loss in full when $X>d$. If there is also a payment limit $u$, the payment is $Y^L=0$ for $X\le d$ and $Y^L=\min(X,u)$ for $X>d$. The strict inequality matters: a loss exactly equal to $d$ receives no payment under this convention.

Write $r=F_X(d)=P(X\le d)$. For $0<p\le r$, the insurer-payment quantile is zero. For $p>r$, the ground-up quantile is above the threshold, and the payment quantile is $\min(Q_X(p),u)$ when a limit applies. If there is no limit, it is $Q_X(p)$. This split is essential: the franchise payment function jumps just above $d$, so the continuous-transform result should not be applied to it as though it were continuous.

Worked Example: A Franchise Threshold With an Atom at the Threshold

A hypothetical loss has $P(X=100)=0.20$, $P(X=200)=0.25$, $P(X=250)=0.25$, and $P(X=600)=0.30$. Coverage has a franchise deductible of $d=\$200$: the insurer pays zero for $X\le \$200$, and otherwise pays the loss, subject to a limit of $u=\$400$. Find the payment quantiles at $p=0.45$, $p=0.50$, and $p=0.80$.

The threshold probability is $r=P(X\le 200)=0.20+0.25=0.45$. Therefore, at $p=0.45$, the payment quantile is zero. In ground-up terms, $Q_X(0.45)=\$200$, which is exactly at the threshold and is not eligible for payment.

At $p=0.50>r$, $Q_X(0.50)=\$250$, since the cumulative ground-up probability reaches $0.70$ at $\$250$. This loss exceeds the franchise threshold and is below the limit, so $Q_{Y^L}(0.50)=\min(250,400)=\$250$.

At $p=0.80>r$, $Q_X(0.80)=\$600$, since the cumulative ground-up probability reaches $1.00$ at $\$600$. The limit reduces this payment to $Q_{Y^L}(0.80)=\min(600,400)=\$400$.

The payment distribution has probability $0.45$ at zero, probability $0.25$ at $\$250$, and probability $0.30$ at $\$400$. Its cumulative probabilities are $0.45$, $0.70$, and $1.00$, confirming all three quantiles. The key method is to compare $p$ with $F_X(d)$ before mapping a ground-up quantile. This avoids treating a loss equal to the franchise threshold as payable.

Continuous Losses Can Still Produce Payment Atoms

A continuous ground-up distribution does not guarantee a continuous payment distribution. Under an ordinary deductible, every loss in an interval from zero through $d$ maps to payment zero. Under a limit, every loss at or above $d+u$ maps to the capped amount. These intervals can create positive probabilities at the payment endpoints, even though no single ground-up loss value has positive probability.

Worked Example: Quantiles From a Uniform Loss Distribution

Suppose $X$ is uniform on $[0,\$1{,}000]$. A hypothetical policy has an ordinary deductible of $d=\$200$, a payment limit of $u=\$300$ above the deductible, and coinsurance $c=0.60$. Find the payment quantiles at $p=0.10$, $p=0.50$, and $p=0.95$, and identify the endpoint atom probabilities.

For a uniform distribution on this interval, $Q_X(p)=1000p$. The payment function is $g(x)=0.60\min((x-200)_+,300)$. At $p=0.10$, $Q_X(0.10)=\$100$, so $g(100)=\$0$. Thus $Q_{Y^L}(0.10)=\$0$.

At $p=0.50$, $Q_X(0.50)=\$500$. The amount above the deductible is $\$500-\$200=\$300$, which equals the payment limit. Hence $g(500)=0.60(300)=\$180$, and $Q_{Y^L}(0.50)=\$180$.

At $p=0.95$, $Q_X(0.95)=\$950$, which is above the cap threshold $d+u=\$500$. Therefore $g(950)=0.60(300)=\$180$, and $Q_{Y^L}(0.95)=\$180$.

The payment is zero for $X\le \$200$, an event of probability $200/1000=0.20$. It is capped at $\$180$ for $X\ge \$500$, an event of probability $(1000-500)/1000=0.50$. Thus the payment distribution has an atom of $0.20$ at zero and an atom of $0.50$ at $\$180$. This verifies why the median payment is already at the cap: the cumulative probability at the cap includes the entire capped tail.

Common Quantile Traps

Using the wrong deductible rule. An ordinary deductible pays $(X-d)_+$; a franchise deductible pays zero through its threshold and then pays the loss itself. They do not have the same payment quantiles.
Ignoring the atom. If $F_Y(0)=0.30$, then every probability level $p\le 0.30$ has payment quantile zero. Do not interpolate between zero and the next positive payment.
Applying coinsurance at the wrong stage. If coinsurance applies to the insurer's capped covered amount, calculate $\min((X-d)_+,u)$ first and multiply by $c$ second. Multiplying the loss by $c$ before applying the deductible generally changes the contract.
Misreading the limit. In this tutorial, $u$ is the maximum payment above an ordinary deductible before coinsurance. The cap threshold in ground-up loss is $d+u$, not $u$.
Using a strict threshold as inclusive. Under the stated franchise convention, $X=d$ produces zero payment. Check whether the loss is strictly above the threshold before assigning a positive payment.

Key takeaway: Define the payment variable first. For ordinary deductible-limit coverage with coinsurance, apply its continuous nondecreasing payment function to the ground-up quantile. For a franchise deductible, first compare the quantile level $p$ with $F_X(d)$, then apply the payment rule on the eligible side of the threshold. These steps preserve the atoms and jumps that determine the insurer-payment quantiles.

Check Your Understanding

For each problem, use the generalized-inverse quantile convention and state the payment quantile requested.

A loss is $50$ with probability $0.20$, $150$ with probability $0.30$, and $400$ with probability $0.50$. An ordinary deductible is $100$, and the insurer-payment limit above it is $200$. Find the payment quantile at $p=0.40$. (Answer: $\$50$)
A loss is $200$ with probability $0.25$, $300$ with probability $0.25$, $450$ with probability $0.30$, and $700$ with probability $0.20$. A franchise deductible pays zero for $X\le300$; above that threshold it pays the loss, subject to a limit of $500$. Find the payment quantile at $p=0.50$. (Answer: $\$0$)
$X$ is uniform on $[0,\$600]$. An ordinary deductible is $\$100$, the limit above the deductible is $\$250$, and coinsurance is $0.80$. Find the payment quantile at $p=0.75$. (Answer: $\$200$)
A loss is $100$ with probability $0.10$, $200$ with probability $0.20$, $500$ with probability $0.40$, and $900$ with probability $0.30$. The insurer pays $0.50\min((X-200)_+,300)$. Find the payment quantile at $p=0.60$. (Answer: $\$150$)
$X$ is uniform on $[0,\$800]$. A franchise deductible pays zero when $X\le \$200$, and above $\$200$ pays $\min(X,\$500)$. Find the payment quantile at $p=0.25$ and at $p=0.70$. (Answer: $\$0$ at $p=0.25$; $\$500$ at $p=0.70$)

The quantile transformations in this tutorial are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

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