Uniform Losses Under Deductibles and Limits

Why Uniform Losses Make Coverage Effects Visible

A uniform ground-up loss model assigns equal probability to every loss amount in a fixed interval. When an ordinary deductible removes small losses and a payment limit caps large covered losses, the insurer’s payment is no longer simply uniform: there may be a point mass at zero, a continuous portion, and a point mass at the cap. The uniform model makes each part especially transparent and provides a useful way to check coverage calculations.

Assume throughout that the ground-up loss $X$ has distribution $X\sim\operatorname{Uniform}(0,b)$, where $b>0$, so its density is $f_X(x)=1/b$ for $0<x<b$. Let $d\geq 0$ be an ordinary deductible and $u>0$ the maximum insurer payment after the deductible. Let $c$, with $0\leq c\leq 1$, be the insurer’s coinsurance share, applied to the deductible-adjusted, limited amount. Under this stated order of operations, the payment per loss is

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

Thus the deductible is applied to the ground-up loss, the limit caps the amount above the deductible, and coinsurance scales the resulting payment. A contract that applies coinsurance at a different stage must be modeled using its actual terms. The analysis below assumes $c>0$, unless a special case is stated.

Payment Distribution Under a Deductible and Limit

First suppose $0\leq d<b$. Write $L=b-d$ for the largest possible loss amount above the deductible. A claim produces no payment when $X\leq d$, so the probability of zero payment is $d/b$. Losses above the deductible generate an uncapped amount $X-d$, up to the limit $u$. Whether any positive probability reaches the cap depends on comparing $u$ with $L$.

Result: For $X\sim\operatorname{Uniform}(0,b)$, $0\leq d<b$, $u>0$, and $0<c\leq1$, let $L=b-d$. The payment has a point mass $d/b$ at zero. If $u<L$, it also has a point mass $(L-u)/b$ at $cu$, and its density is $1/(bc)$ on $0<y<cu$. If $u\geq L$, there is no positive cap mass; the density is $1/(bc)$ on $0<y<cL$. In either case the density is zero outside the stated interval.

At $u=L$, the limit equals the largest possible amount above the deductible. The loss distribution is continuous, so the single endpoint $X=b$ has probability zero: there is no point mass at the cap in this case. Likewise, if $u>L$, the cap cannot be reached. If $d\geq b$, every possible loss is at or below the deductible and $Y^L=0$ with probability one. If $c=0$, every payment is zero regardless of the other terms.

For clarity, when $u<L$, the distribution function includes both atoms. For $y\geq0$, it can be written

F_{Y^L}(y)= \begin{cases} \dfrac{d}{b}+\dfrac{y}{bc},&0\leq y<cu,\\[6pt] 1,&y\geq cu. \end{cases}

The jump at zero is $d/b$, and the jump at $cu$ is $(L-u)/b$. Between those jumps, the probability increases linearly. If $u\geq L$, the same linear expression holds for $0\leq y<cL$, and the distribution reaches one at $cL$ without a jump there.

Expected Payment and Its Derivation

The expected payment can be found directly by averaging the payment over the uniform loss interval. The part of the interval between $d$ and $\min(b,d+u)$ produces a payment that increases linearly with the loss; if the limit binds, the remaining interval produces the constant payment $cu$.

E[Y^L] = \frac{c}{b} \left[ \int_d^{\min(b,d+u)}(x-d)\,dx + u\max(b-d-u,0) \right].

For a short derivation, condition on the value $X=x$. The payment is zero for $0\leq x\leq d$, equals $c(x-d)$ for $d<x<d+u$, and equals $cu$ for $x\geq d+u$. Since the density is $1/b$, the expectation is the integral of this payment multiplied by $1/b$ over the loss interval. The first integral accounts for the increasing-payment region; the second term is the cap payment times the length of the capped region. Dividing that length by $b$ converts it to its probability.

Evaluating the integral gives a convenient piecewise formula:

E[Y^L]= \begin{cases} \dfrac{c}{b}\left(uL-\dfrac{u^2}{2}\right),&0<u<L,\\[8pt] \dfrac{cL^2}{2b},&u\geq L. \end{cases}

When the limit binds, the uncapped portion contributes $u^2/(2b)$ before coinsurance and the capped portion contributes $u(L-u)/b$. Their sum is $(uL-u^2/2)/b$. When the limit does not bind, all losses above the deductible are paid in proportion to $x-d$, and the integral from $d$ to $b$ is $L^2/2$. Coinsurance multiplies either expected amount by $c$.

These results also provide a useful consistency check: the probabilities of the zero mass, continuous part, and cap mass must sum to one. In the binding-limit case, the continuous probability is $u/b$, so $d/b+u/b+(L-u)/b=1$, because $d+L=b$.

Worked Example: A Binding Limit With Coinsurance

A hypothetical insurer covers a ground-up loss $X\sim\operatorname{Uniform}(0,1000)$ dollars. The policy has a deductible of $d=\$200$, a maximum payment after the deductible of $u=\$300$, and an insurer coinsurance share of $c=0.80$. Find the payment distribution and the expected payment per loss. Give probabilities to four decimals and the expectation to the nearest cent.

Method. Compare the limit with the maximum amount above the deductible. Here $b=1000$, $L=b-d=1000-200=800$, and $u=300<800$, so the limit binds.

The probability of no payment is $d/b=200/1000=0.2000$. The probability of reaching the cap is $(L-u)/b=(800-300)/1000=0.5000$. The probability of the continuous portion is $u/b=300/1000=0.3000$. These probabilities sum to $0.2000+0.5000+0.3000=1.0000$.

The maximum payment is $cu=0.80(300)=\$240$. Coinsurance scales the continuous payment interval from $0$ to $300$ dollars to an interval from $0$ to $240$ dollars. Its density is $1/(bc)=1/(1000(0.80))=1/800$ per dollar on $0<y<240$. Thus there is a mass of $0.2000$ at zero, density $1/800$ on that interval, and a mass of $0.5000$ at $\$240$.

The expected payment is \[ E[Y^L]=\frac{0.80}{1000}\left(300(800)-\frac{300^2}{2}\right) =\frac{0.80}{1000}(240000-45000) =\frac{0.80}{1000}(195000) =\$156.00. \] As a check, the uncapped expected payment is $\$195.00$, and multiplying by $0.80$ gives $\$156.00$. Common mistake avoided: treating the limit as a maximum ground-up loss would use the wrong cap threshold. The limit applies to the amount above the deductible here.

Worked Example: A Limit That Cannot Be Reached

For a hypothetical portfolio loss, suppose $X\sim\operatorname{Uniform}(0,600)$ dollars, with deductible $d=\$150$, payment limit $u=\$500$, and coinsurance share $c=0.60$. Find the payment distribution, $P(Y^L\leq \$90)$, and the expected payment. Give probabilities to four decimals and the expectation to the nearest cent.

Method. The largest amount above the deductible is $L=600-150=450$. Since $u=500\geq450$, the limit does not bind. There is a zero-payment probability of $150/600=0.2500$, no cap mass, and a maximum payment of $cL=0.60(450)=\$270$.

The continuous density on $0<y<270$ is $1/(bc)=1/(600(0.60))=1/360$ per dollar. For $0\leq y<270$, the distribution function is $0.2500+y/360$. Therefore, $P(Y^L\leq\$90)=0.2500+90/360=0.2500+0.2500=0.5000$.

The expected payment is \[ E[Y^L]=\frac{0.60(450^2)}{2(600)} =\frac{0.60(202500)}{1200} =\frac{121500}{1200} =\$101.25. \] The distribution has total probability $0.2500+270/360=1.0000$, as required. Common mistake avoided: assuming every policy limit creates a point mass at its cap. Since $u>L$, no loss in this model reaches the limit.

Worked Example: Checking the Capped Mass and Mean

Suppose a hypothetical loss has $X\sim\operatorname{Uniform}(0,500)$ dollars, with deductible $d=\$100$, payment limit $u=\$200$, and no coinsurance reduction ($c=1$). Find the point masses, continuous density, and expected payment. Give probabilities to four decimals and the expectation to the nearest cent.

Method. Here $L=500-100=400$, so $u=200<400$ and the limit binds. The zero-payment mass is $100/500=0.2000$. The capped mass is $(L-u)/b=(400-200)/500=0.4000$. The continuous portion has probability $u/b=200/500=0.4000$, so the masses and continuous probability total $1.0000$.

The density is $1/b=1/500$ per dollar on $0<y<200$, and the cap mass $0.4000$ is at $\$200$. The expected payment can be checked by adding the continuous contribution and the cap contribution: \[ E[Y^L] =\int_0^{200} y\frac{1}{500}\,dy+200(0.4000) =\frac{200^2}{2(500)}+80 =40+80 =\$120.00. \]

The same answer follows from the piecewise formula: \[ E[Y^L]=\frac{1}{500}\left(200(400)-\frac{200^2}{2}\right) =\frac{80000-20000}{500} =\$120.00. \] Both checks agree. Common mistake avoided: the continuous part contributes $\int_0^{200}y/500\,dy=\$40$, not $\$80$; the additional $\$80$ comes from the capped claims.

Common Exam Traps

Putting the deductible in the wrong place. With a payment limit $u$ applied above the deductible, the cap begins at ground-up loss $X=d+u$, not at $X=u$.
Forgetting the atom at zero. A deductible creates a probability $d/b$ of no payment when $0\leq d<b$. The continuous density alone does not describe the full payment distribution.
Adding a cap mass when the cap is unreachable. A positive cap mass occurs only if $u<L=b-d$. At $u=L$, the cap is reached only at the endpoint, which has probability zero under a continuous uniform distribution.
Scaling the wrong quantity for coinsurance. Under the contract defined here, the final payment and its expectation are multiplied by $c$; the ground-up loss distribution and deductible are not scaled.
Using the wrong density after scaling. If $Z$ has density $1/b$ on a continuous interval and $Y^L=cZ$, the transformed density is $1/(bc)$, not $1/b$. The interval is scaled by $c$, so the density must change to preserve probability.

Uniform losses are useful for isolating the mechanics of coverage modifications: the deductible determines the zero-payment probability, the limit determines whether a capped mass appears, and coinsurance changes the payment scale. The same reasoning is useful when checking more elaborate severity calculations, but the formulas depend on the assumed uniform ground-up distribution and the stated order of contract terms. These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume the stated uniform ground-up loss model and the payment convention defined above. Use dollars throughout, give probabilities to four decimals, and give expected payments to the nearest cent.

A hypothetical loss has $X\sim\operatorname{Uniform}(0,800)$, deductible $d=160$, payment limit $u=240$, and coinsurance share $c=0.75$. Find the zero-payment mass, capped-payment mass and location, continuous density and interval, and $E[Y^L]$. (Answer: zero mass $0.2000$; capped mass $0.5000$ at $\$180$; density $1/600$ on $0<y<180$; $E[Y^L]=\$117.00$.)
For $X\sim\operatorname{Uniform}(0,300)$, let $d=60$, $u=400$, and $c=0.50$. Find the zero-payment probability, maximum payment, continuous density, and expected payment. (Answer: zero mass $0.2000$; maximum payment $\$120$, with no cap mass; density $1/150$ on $0<y<120$; $E[Y^L]=\$48.00$.)
For $X\sim\operatorname{Uniform}(0,1000)$, let $d=250$, $u=500$, and $c=0.40$. Find the two point masses, the continuous density and interval, and the expected payment. (Answer: zero mass $0.2500$; capped mass $0.2500$ at $\$200$; density $1/400$ on $0<y<200$; $E[Y^L]=\$100.00$.)
For $X\sim\operatorname{Uniform}(0,500)$, let $d=100$, $u=400$, and $c=0.80$. Determine whether a cap mass exists, and find the zero mass, continuous density and interval, and expected payment. (Answer: zero mass $0.2000$; no cap mass; density $1/400$ on $0<y<320$; $E[Y^L]=\$128.00$.)
A hypothetical insurer compares two contracts for $X\sim\operatorname{Uniform}(0,1000)$. Contract A has $d=100$, $u=300$, $c=0.50$; Contract B has $d=200$, $u=500$, $c=0.50$. Find the expected payment under each contract and the difference. (Answer: Contract A $\$112.50$; Contract B $\$137.50$; Contract B has the higher expected payment by $\$25.00$.)

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