Continuous Losses and Coverage Payment Densities

Why Payment Distributions Can Have Both Densities and Atoms

A continuous ground-up loss does not necessarily produce a payment distribution that is entirely continuous. An ordinary deductible creates a range of losses for which the insurer pays zero. A payment limit creates another range for which the insurer pays its maximum amount. Each range is collapsed to a single payment value, producing a point mass, even though the loss itself has a density. Between those ranges, payment changes with loss, and its distribution has a continuous density.

The distinction matters whenever a problem asks for a payment density, a probability at a particular payment, or a complete payment distribution. A density describes probability spread across intervals; it does not include probability concentrated at a point. For coverage with a deductible and a limit, omitting either boundary mass can make the probabilities appear not to add to one.

Definition: Let $X$ be a nonnegative ground-up loss with continuous distribution, density $f_X$, and distribution function $F_X$. Suppose the policy has an ordinary deductible $d\geq 0$, a payment limit $u>0$ measured after the deductible, and a coinsurance share $c$, where $0<c\leq 1$. The payment per loss is $$ Y^L=c\min\bigl((X-d)_+,u\bigr), \qquad (z)_+=\max(z,0). $$ Thus the maximum payment is $cu$, and the loss amount at which that cap is reached is $d+u$.

This tutorial uses a limit on the amount above the deductible before applying coinsurance. Under this convention, the payment increases at rate $c$ for losses between $d$ and $d+u$. If a problem instead defines its limit as a maximum insurer payment after coinsurance, the payment formula and the loss threshold at the cap must be adjusted to match that wording.

The Continuous Part and the Boundary Masses

For $X\leq d$, the payment is zero. For $d<X<d+u$, the payment is $c(X-d)$, which lies strictly between zero and $cu$. For $X\geq d+u$, the payment is capped at $cu$. These three regions of the ground-up loss determine the full payment distribution.

Theorem: Under the assumptions above, the payment distribution has a point mass at zero of size $F_X(d)$, a point mass at $cu$ of size $1-F_X(d+u)$, and a continuous density on the interior payment range: $$ f_{Y^L}(y)=\frac{1}{c}f_X\left(d+\frac{y}{c}\right), \qquad 0<y<cu. $$ The density is zero outside that interior range. There is no density contribution at the boundary points $0$ and $cu$; their probabilities are the point masses just stated.

Because $X$ is continuous, it assigns probability zero to the exact thresholds $d$ and $d+u$. It follows that $P(Y^L=0)=P(X\leq d)=F_X(d)$ and $P(Y^L=cu)=P(X\geq d+u)=1-F_X(d+u)$. These formulas do not require $X$ to have positive density at either threshold.

Why the Density Has This Form

On the interval where the payment is not capped and is positive, the transformation is one-to-one: $y=c(x-d)$, or, inversely, $x=d+y/c$. A small payment interval of width $\Delta y$ corresponds to a loss interval of width $\Delta x=\Delta y/c$. Since $c$ is positive, the density is scaled by the reciprocal factor $1/c$.

\begin{aligned} P(y<Y^L\leq y+\Delta y) &=P\left(d+\frac{y}{c}<X\leq d+\frac{y+\Delta y}{c}\right),\\[4pt] \frac{P(y<Y^L\leq y+\Delta y)}{\Delta y} &=\frac{F_X(d+(y+\Delta y)/c)-F_X(d+y/c)}{\Delta y},\\[4pt] \lim_{\Delta y\to 0} \frac{P(y<Y^L\leq y+\Delta y)}{\Delta y} &=\frac{1}{c}f_X\left(d+\frac{y}{c}\right). \end{aligned}

The factor $1/c$ is also necessary for probability to be preserved. For example, integrating the transformed density across the interior payment range gives $$ \int_0^{cu}\frac{1}{c}f_X\left(d+\frac{y}{c}\right)\,dy =\int_d^{d+u}f_X(x)\,dx =F_X(d+u)-F_X(d). $$ The substitution is $x=d+y/c$, so $dy=c\,dx$. The interior payment probability is exactly the probability that the ground-up loss lies between the deductible and the cap threshold.

Find the payment boundaries.
The payment is zero through loss $d$, increases between $d$ and $d+u$, and reaches its maximum $cu$ at and above $d+u$.

Calculate the boundary probabilities.
Use $F_X(d)$ for the zero-payment mass and $1-F_X(d+u)$ for the maximum-payment mass.

Transform the density in the interior.
For $0<y<cu$, substitute $x=d+y/c$ into $f_X(x)$ and multiply by $1/c$.

Check total probability.
The two point masses and the integral of the interior density must sum to one.

Worked Applications

Worked Example: Uniform Loss With a Deductible, Limit, and Coinsurance

A hypothetical policy covers a ground-up loss $X\sim\text{Uniform}(0,100)$ dollars. It has an ordinary deductible of $d=\$20$, a post-deductible payment limit of $u=\$30$, and coinsurance $c=0.8$. Find the payment’s boundary masses and interior density, and verify that the probabilities total one.

Method. Identify the ground-up loss regions before transforming the density. For a uniform loss on $[0,100]$, the density is $f_X(x)=1/100=0.01$ for $0<x<100$, and $F_X(x)=x/100$ on that range.

The maximum payment is $cu=0.8(30)=\$24$. It is reached when $X\geq d+u=20+30=\$50$.

Zero-payment mass: $P(Y^L=0)=P(X\leq20)=20/100=0.20$.

Maximum-payment mass: $P(Y^L=24)=P(X\geq50)=(100-50)/100=0.50$.

Interior density: For $0<y<24$, the corresponding loss is $x=20+y/0.8=20+1.25y$. Therefore, $$ f_{Y^L}(y)=\frac{1}{0.8}(0.01)=0.0125,\qquad 0<y<24. $$ The density is zero outside the interior interval, apart from the separate point masses at $0$ and $24$.

The continuous probability is $0.0125(24)=0.30$. The total is $0.20+0.30+0.50=1.00$, as required. Common mistake avoided: writing $0.01$ as the payment density without the factor $1/c$ would give only $0.24$ of interior probability, instead of the correct $0.30$.

Worked Example: Increasing Density and a Large Cap Mass

A hypothetical loss has density $f_X(x)=x/5000$ for $0<x<100$ dollars and zero otherwise. A policy has $d=\$20$, $u=\$40$, and $c=0.5$. Find the point masses and interior payment density, then calculate the density at a payment of $\$10$ and check total probability.

Method. First integrate the ground-up density to obtain its distribution function on the support: $$ F_X(x)=\int_0^x\frac{t}{5000}\,dt=\frac{x^2}{10000},\qquad 0\leq x\leq100. $$ The maximum payment is $cu=0.5(40)=\$20$, reached at ground-up loss $d+u=20+40=\$60$.

The zero-payment mass is $F_X(20)=20^2/10000=400/10000=0.04$. The maximum-payment mass is $1-F_X(60)=1-60^2/10000=1-3600/10000=0.64$.

For $0<y<20$, the loss corresponding to payment $y$ is $x=20+y/0.5=20+2y$. The interior density is therefore $$ f_{Y^L}(y)=\frac{1}{0.5}\frac{20+2y}{5000} =\frac{20+2y}{2500},\qquad 0<y<20. $$ At $y=\$10$, this density is $(20+2(10))/2500=40/2500=0.016$ per dollar of payment.

The continuous probability is the ground-up probability between 20 and 60: $F_X(60)-F_X(20)=(3600-400)/10000=0.32$. The total is $0.04+0.32+0.64=1.00$. Common mistake avoided: the cap threshold is $\$60$, not $\$40$; the limit is applied to the loss above the deductible.

Worked Example: A Gap in the Ground-Up Density

A hypothetical ground-up loss is a 50–50 mixture of a uniform distribution on $[0,40]$ dollars and a uniform distribution on $[60,100]$ dollars. Thus $f_X(x)=0.0125$ on each of those two intervals and is zero elsewhere. A policy has $d=\$10$, $u=\$50$, and $c=0.6$. Determine the boundary masses and the full interior density, including any interval on which that density is zero.

Method. The payment cap is $cu=0.6(50)=\$30$, reached at $d+u=10+50=\$60$. The probability of a zero payment is the probability that $X\leq10$. Only the first mixture component contributes, with density $0.0125$ over a length of 10, so $P(Y^L=0)=10(0.0125)=0.125$.

All losses in the second component are at least 60, so they produce the capped payment. Its mixture probability is $0.50$, giving $P(Y^L=30)=0.50$.

For interior payments, $x=10+y/0.6$. The first component’s density applies when $10<x<40$, corresponding to $0<y<0.6(40-10)=18$. Thus $f_{Y^L}(y)=0.0125/0.6=0.0208333$ per dollar for $0<y<18$, rounded here to seven decimal places. For $18<y<30$, the corresponding losses satisfy $40<x<60$, where the ground-up density is zero, so the payment density is zero.

The continuous probability is $0.0208333\ldots(18)=0.375$, equivalently $30(0.0125)=0.375$. Adding the masses gives $0.125+0.375+0.50=1.00$. Common mistake avoided: treating the payment density as positive over all of $(0,30)$ would ignore the gap in the ground-up loss support.

Reading the Result Correctly

A full distribution description should state both the density on its continuous range and the probabilities at its atoms. Writing only $f_{Y^L}(y)$ can be incomplete: integrating that density gives the probability of an interior payment, not necessarily one. In the first example, for instance, the density integrates to $0.30$; the remaining probability is concentrated at zero and at the maximum payment.

For deductible-only coverage with no finite limit, the same reasoning gives a zero-payment mass $F_X(d)$ and an interior density $f_X(d+y/c)/c$ for $y>0$, wherever the transformed loss is in the support. There is no cap-created upper atom in that case. If there is no deductible but there is a finite limit, the zero-payment mass is $F_X(0)$, which is zero when losses are strictly positive, while the cap can still create a mass at the maximum payment.

The stated density formula relies on the continuous, one-to-one part of the payment transformation. It does not describe the probability at the flat parts of the transformation, and it should not be applied at the endpoints as if they were ordinary density values. For a loss distribution that already has discrete probability, additional point masses can arise from that probability; the continuous-density result alone would not describe those discrete contributions.

Key takeaway: On the uncapped positive-payment interval, the coverage transformation scales the ground-up density by $1/c$ and shifts its argument by the deductible. Separately calculate the probability mapped to zero and the probability mapped to the cap, then verify that all parts sum to one.

Check Your Understanding

For each problem, identify boundary point masses and the continuous payment density. Check that the probability components sum to one.

A hypothetical loss is uniform on $[0,80]$ dollars. Coverage has $d=\$16$, $u=\$24$, and $c=0.75$. Find the maximum payment, both boundary masses, and the interior density. (Answer: maximum payment $\$18$; mass at zero $0.2000$; mass at $\$18$ $0.5000$; density $1/60$ per dollar on $0<y<18$; continuous probability $0.3000$.)
A hypothetical loss has density $f_X(x)=x/7200$ on $0<x<120$. Coverage has $d=\$30$, $u=\$30$, and $c=0.4$. Give the maximum payment, the two boundary masses, and the interior density. (Answer: maximum payment $\$12$; mass at zero $0.0625$; mass at $\$12$ $0.7500$; density $(30+2.5y)/2880$ per dollar on $0<y<12$; continuous probability $0.1875$.)
A hypothetical loss is uniform on $[0,100]$ dollars. Coverage has deductible $d=\$25$, coinsurance $c=0.6$, and no finite limit. Find the atom and the density of positive payments. (Answer: mass at zero $0.2500$; no upper atom; density $1/60$ per dollar for $0<y<45$; continuous probability $0.7500$.)
A hypothetical loss is uniform on $[0,50]$ dollars. Coverage has $d=\$10$, $u=\$20$, and $c=0.5$. Find the maximum payment, both boundary masses, and the interior density, then verify the total probability. (Answer: maximum payment $\$10$; mass at zero $0.2000$; mass at $\$10$ $0.4000$; density $0.0400$ per dollar on $0<y<10$; continuous probability $0.4000$; total probability $1.0000$.)

What You'll Learn