Mixed Loss Distributions Under Coverage Modifications

Why Mixed Loss Distributions Need Care

A ground-up loss model may combine point masses with a continuous component. For example, a hypothetical coverage portfolio might assign positive probability to a small set of fixed repair costs while allowing other losses to vary continuously. An ordinary deductible and a payment limit then transform that model into the insurer’s payment distribution. The transformation can create new point masses at zero and at the maximum payment, even when the ground-up loss has no atoms at those values.

The important distinction is between the ground-up loss and the insurer payment. A loss at or below the deductible produces no payment. A loss at or above the deductible plus the maximum payment produces the maximum payment. A ground-up atom strictly between those thresholds remains an atom, translated by the deductible. Continuous losses between the thresholds remain continuous payments.

Definition: Let $X\geq 0$ be the ground-up loss, with distribution function $F_X(x)=P(X\leq x)$. Under an ordinary deductible $d\geq 0$ and a maximum insurer payment $u>0$, define the insurer payment per loss by $$ Y=\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ Thus the payment is zero when $X\leq d$, is $X-d$ when $d<X<d+u$, and is $u$ when $X\geq d+u$.

This definition treats the limit $u$ as the maximum payment, not as the maximum covered ground-up loss. Consequently, the ground-up threshold for reaching the limit is $d+u$. At the boundary $X=d$, the payment is zero; at $X=d+u$, the payment is $u$.

The Payment Distribution and Its Atoms

The CDF gives a compact description of the transformed distribution. Write $F_Y(y)=P(Y\leq y)$. For a payment strictly below the maximum, the event $Y\leq y$ is equivalent to $X\leq d+y$. At the maximum, every possible loss produces a payment no greater than $u$.

F_Y(y)= \begin{cases} 0, & y<0,\\ F_X(d+y), & 0\leq y<u,\\ 1, & y\geq u. \end{cases}

This CDF makes the endpoint probabilities visible. For any $a$, let $F_X(a-)=P(X<a)$, the left-hand limit of the ground-up CDF. Then

P(Y=0)=F_X(d), \qquad P(Y=u)=P(X\geq d+u)=1-F_X((d+u)-).

The second expression uses a left-hand limit: an atom exactly at $d+u$ reaches the payment limit and must be included. Likewise, the zero-payment probability includes a ground-up atom exactly at $d$. For an interior payment $0<y<u$,

$$ P(Y=y)=P(X=d+y)=F_X((d+y))-F_X((d+y)-). $$

In other words, the transformation moves any ground-up atom strictly between $d$ and $d+u$ to the payment value $x-d$. The continuous part of the ground-up distribution contributes probabilities throughout the corresponding interior payment range, but it does not create an interior point mass by itself.

Why the CDF Formula Holds

For $0\leq y<u$, the payment is at most $y$ precisely when the ground-up loss is at most $d+y$. To verify both directions, first suppose $X\leq d+y$. If $X\leq d$, then $Y=0\leq y$; otherwise $Y=X-d\leq y$. Conversely, if $Y\leq y<u$, the payment cannot have reached the limit, because reaching it would give $Y=u>y$. Therefore $Y=(X-d)_+$, and $Y\leq y$ implies $X\leq d+y$. Thus the two events are equal and their probabilities are equal:

\{Y\leq y\}=\{X\leq d+y\}, \qquad F_Y(y)=F_X(d+y), \qquad 0\leq y<u.

At $y=u$, the event $\{Y\leq u\}$ contains every outcome, so $F_Y(u)=1$. The jump in this CDF at zero has size $F_X(d)$. Its jump at $u$ has size $1-F_X((d+u)-)$, which includes all losses at or above the ground-up limit threshold.

Worked Distribution Calculations

Worked Example: Locating Zero, Interior, and Limit Masses

A hypothetical ground-up loss $X$ has probability $0.20$ at 0, probability $0.10$ at 6, and probability $0.70$ in a continuous uniform distribution on $[0,20]$. The continuous component therefore has total probability $0.70$, spread uniformly over that interval. The policy has an ordinary deductible $d=5$ dollars and a maximum payment $u=8$ dollars. Find the probabilities of zero payment, a payment of 1 dollar, and the maximum payment. Also find $P(Y\leq 3)$.

Method. The limit is reached at ground-up loss $d+u=5+8=13$. Calculate the two endpoint masses from the ground-up distribution, then translate any interior atom.

Zero payment. This occurs when $X\leq 5$. The atom at zero contributes $0.20$. The continuous component contributes $0.70(5/20)=0.175$. The atom at 6 is above the deductible. Hence

P(Y=0)=0.20+0.70\left(\frac{5}{20}\right)=0.20+0.175=0.375.

Payment of 1 dollar. An interior payment of 1 requires $X=d+1=6$. The ground-up atom at 6 has probability $0.10$, so $P(Y=1)=0.10$.

Maximum payment. This occurs for $X\geq 13$. There is no ground-up atom in this tail. The continuous component contributes $0.70(20-13)/20=0.70(7/20)=0.245$. Thus $P(Y=8)=0.245$.

CDF at 3 dollars. Since $3<u$, $P(Y\leq3)=F_X(5+3)=F_X(8)$. This includes the atoms at 0 and 6 and the continuous probability through 8:

P(Y\leq3)=0.20+0.10+0.70\left(\frac{8}{20}\right) =0.30+0.28=0.58.

Check. The continuous losses strictly between 5 and 13 contribute $0.70(8/20)=0.28$, while the atom at 6 contributes $0.10$ to the interior payment range. The total is $0.375+0.28+0.10+0.245=1.000$. The endpoint and interior probabilities account for all outcomes.

Common mistake avoided: The atom at 6 is not part of the zero-payment mass, and it is not absorbed into the limit mass. Its payment is the translated interior value $6-5=1$.

Worked Example: An Atom Exactly at the Limit Threshold

A hypothetical loss has an atom of $0.15$ at 0, an atom of $0.20$ at 4, an atom of $0.25$ at 9, and a continuous component of total probability $0.40$ that is uniform on $[0,12]$. The policy has deductible $d=4$ dollars and maximum payment $u=5$ dollars. Find $P(Y=0)$, $P(Y=5)$, and $P(Y\leq3)$.

Method. The ground-up loss threshold for the maximum payment is $d+u=9$. In particular, the atom at 9 is included in the maximum payment.

Zero payment. The losses at 0 and 4 are both at or below the deductible. The continuous probability on $[0,4]$ is $0.40(4/12)=0.1333\ldots$. Therefore

P(Y=0)=0.15+0.20+0.40\left(\frac{4}{12}\right) =0.35+0.1333\ldots=0.4833\ldots.

Maximum payment. The atom at 9 contributes $0.25$, since $X=9$ gives $Y=9-4=5$. The continuous component on $[9,12]$ contributes $0.40(3/12)=0.10$. Thus $P(Y=5)=0.25+0.10=0.35$.

CDF at 3 dollars. Since $3<5$, use $P(Y\leq3)=F_X(4+3)=F_X(7)$. The atoms at 0 and 4 contribute $0.15+0.20$, and the continuous component through 7 contributes $0.40(7/12)=0.2333\ldots$:

P(Y\leq3)=0.15+0.20+0.40\left(\frac{7}{12}\right) =0.35+0.2333\ldots=0.5833\ldots.

Check. The interior continuous mass between ground-up losses 4 and 9 is $0.40(5/12)=0.1667\ldots$. The total probability is $0.4833\ldots+0.1667\ldots+0.35=1.0000$.

Common mistake shown and corrected: Using $P(X>9)$ for the maximum-payment mass would give only $0.10$ and omit the atom of $0.25$ at 9. The correct event is $X\geq9$, so the atom is included and the answer is $0.35$.

Separating Atoms from the Continuous Component

For a mixed distribution, it is often clearest to list the payment masses and the continuous range separately. Suppose the ground-up distribution has atoms at several specified losses and a continuous component. Each atom at or below $d$ joins the zero-payment mass. Each atom strictly between $d$ and $d+u$ becomes an interior payment atom. Each atom at or above $d+u$ joins the maximum-payment mass. The continuous component is handled over the same three ground-up regions.

Ground-up loss $X$	Payment $Y$	Effect on payment distribution
$X\leq d$	0	Contributes to the atom at zero
$d<X<d+u$	$X-d$	Interior losses translate by $d$; ground-up atoms remain atoms
$X\geq d+u$	$u$	Contributes to the atom at the maximum payment

The strict inequalities in the middle row matter: losses exactly at either boundary belong to an endpoint payment. These conventions also make the CDF useful for interval probabilities. For example, when $0\leq a<b<u$, the probability $P(a<Y\leq b)$ is $F_X(d+b)-F_X(d+a)$. The subtraction follows directly from the CDF, so it automatically handles any atom at the upper endpoint $d+b$ and excludes an atom at the lower endpoint $d+a$.

Worked Example: Full Mixed Payment Distribution and Mean

A hypothetical loss has probability $0.10$ at 2 dollars, probability $0.20$ at 7 dollars, probability $0.15$ at 14 dollars, and a continuous component of total probability $0.55$ that is uniform on $[0,20]$. Coverage has deductible $d=5$ dollars and maximum payment $u=6$ dollars. Find the masses at zero, at the interior payment of 2 dollars, and at the maximum payment; find the total continuous probability of an interior payment; and calculate $E[Y]$.

Method. The zero-payment region is $X\leq5$, the interior region is $5<X<11$, and the maximum-payment region is $X\geq11$, because $d+u=11$.

Mass at zero. The atom at 2 contributes $0.10$. The continuous part from 0 to 5 contributes $0.55(5/20)=0.1375$. Thus $P(Y=0)=0.2375$.

Interior atom. The atom at 7 gives payment $7-5=2$, so $P(Y=2)=0.20$. The atom at 14 is above 11 and therefore contributes to the maximum-payment mass, not to an interior payment.

Maximum-payment mass. The atom at 14 contributes $0.15$. The continuous component from 11 to 20 contributes $0.55(9/20)=0.2475$. Hence $P(Y=6)=0.15+0.2475=0.3975$.

Continuous interior probability. Continuous losses between 5 and 11 produce interior payments. Their total probability is $0.55(6/20)=0.165$. As a check on the distribution, $0.2375+0.20+0.165+0.3975=1.0000$.

Expected payment. The atom at payment 2 contributes $2(0.20)=0.40$. The maximum-payment mass contributes $6(0.3975)=2.385$. For the continuous interior component, the uniform ground-up density including its mixture weight is $0.55/20=0.0275$ per dollar. Therefore its expected-payment contribution is

0.0275\int_5^{11}(x-5)\,dx =0.0275\left[\frac{(x-5)^2}{2}\right]_5^{11} =0.0275(18)=0.495.

Adding the contributions gives $E[Y]=0.40+2.385+0.495=3.28$ dollars.

Check. The atom at 2 dollars contributes no payment, as it lies below the deductible. The expected payment is between 0 and the maximum of 6 dollars, as required. The probability masses and continuous interior probability sum to 1.

Common mistake avoided: The atom at 7 is translated to a payment of 2, while the atom at 14 is capped at 6. Treating both as continuous losses would lose two distinct payment atoms and give an incorrect distribution.

Exam Traps and a Reliable Check

Using strict inequalities at a boundary. For zero payment use $X\leq d$, not $X<d$. For maximum payment use $X\geq d+u$, not $X>d+u$.
Confusing the payment limit with a ground-up loss limit. If $u$ is the maximum insurer payment after a deductible, the cap begins at ground-up loss $d+u$.
Dropping interior atoms. A ground-up atom $x$ with $d<x<d+u$ creates a payment atom at $x-d$; it does not become part of the continuous component.
Counting a threshold atom twice. An atom at $d$ belongs to zero payment, and an atom at $d+u$ belongs to maximum payment. Neither is an interior atom.

A useful final check is to partition all ground-up outcomes into $X\leq d$, $d<X<d+u$, and $X\geq d+u$. Their probabilities must sum to 1. Within the middle region, keep discrete atoms and continuous probability distinct when reporting the payment distribution. This partition is a dependable way to avoid losing or double-counting probability.

Key takeaway: Under an ordinary deductible $d$ and maximum payment $u$, the payment CDF below the limit is $F_Y(y)=F_X(d+y)$. The endpoint atoms are $P(Y=0)=F_X(d)$ and $P(Y=u)=1-F_X((d+u)-)$; interior ground-up atoms translate by subtracting $d$.

Check Your Understanding

For each problem, identify the relevant ground-up thresholds before calculating. Probabilities are rounded to four decimal places where needed.

A ground-up loss has probability $0.30$ at 0, probability $0.20$ at 5, and continuous probability $0.50$ uniformly distributed on $[0,10]$. With deductible $d=3$ and maximum payment $u=4$, find $P(Y=0)$, $P(Y=2)$, and $P(Y=4)$. (Answer: $0.4500,\ 0.2000,\ 0.1500$, respectively.)
A ground-up loss has atoms of $0.10$ at 2 and $0.30$ at 8, plus continuous probability $0.60$ uniformly distributed on $[0,12]$. With $d=4$ and $u=4$, find $P(Y=0)$, $P(Y=4)$, and $P(Y=2)$. (Answer: $0.3000,\ 0.5000,\ 0.0000$, respectively.)
A ground-up loss has probability $0.25$ at 6, probability $0.10$ at 15, and continuous probability $0.65$ uniformly distributed on $[0,20]$. With $d=5$ and $u=7$, find $P(Y=0)$, $P(Y=1)$, $P(Y=7)$, and the continuous probability of an interior payment. (Answer: $0.1625,\ 0.2500,\ 0.3600,\ 0.2275$, respectively.)
A ground-up loss has atoms of $0.20$ at 4 and $0.15$ at 10, with continuous probability $0.65$ uniformly distributed on $[0,16]$. Under $d=4$ and $u=6$, find $P(Y=0)$, $P(Y=6)$, and $P(Y\leq3)$. (Answer: $0.3625,\ 0.3938,\ 0.4844$, respectively.)
A ground-up loss has probability $0.20$ at 3, probability $0.25$ at 9, probability $0.10$ at 18, and continuous probability $0.45$ uniformly distributed on $[0,20]$. Under $d=6$ and $u=8$, find $P(Y=0)$, $P(Y=3)$, $P(Y=8)$, and $E[Y]$. (Answer: $0.3350,\ 0.2500,\ 0.2350,\ \$3.35$.)

Ground-up loss \(X\)	Payment \(Y\)	Effect on payment distribution
\(X\leq d\)	0	Contributes to the atom at zero
\(d<X<d+u\)	\(X-d\)	Interior losses translate by \(d\); ground-up atoms remain atoms
\(X\geq d+u\)	\(u\)	Contributes to the atom at the maximum payment

What You'll Learn