Monotonicity Checks for Deductible and Limit Factors

Why Monotonicity Is a Useful Check

A deductible makes the insurer responsible for less of a covered loss, while a higher policy limit generally allows the insurer to pay more. These straightforward payment comparisons give actuaries a practical diagnostic: if a table purporting to represent the same severity basis shows expected costs moving in the opposite direction, the table merits investigation.

Monotonicity is a necessary consistency check, not a complete validation method. It does not establish that a factor table is correctly estimated, appropriately normalized, or suitable for a particular portfolio. The comparison must also hold other contract terms, exposure basis, severity distribution, and factor definitions constant.

Definition: Let $X$ be a nonnegative ground-up loss per claim, with $E[X]<\infty$. For a deductible $d\geq 0$, the insurer payment without a policy limit is $Y_d^L=(X-d)_+=\max(X-d,0)$. For a per-claim limit $u\geq 0$ and no deductible, the payment is $Y_u^P=\min(X,u)$. For a deductible and a limit that caps the insurer's payment above the deductible, define $Y_{d,u}=\min((X-d)_+,u)$. All comparisons below use these stated payment conventions.

The notation distinguishes the ground-up loss $X$ from the insurer's payment. In particular, $u$ in the combined payment function is the maximum insurer payment, not necessarily a stated cap on the ground-up loss. If a contract instead states a gross-loss cap $U$, its payment with deductible $d$ can be written $\min((X-d)_+,(U-d)_+)$, for $U\geq d$. The same basic ordering applies: raising the gross-loss cap cannot reduce payment, and raising the deductible cannot increase it.

Payment Inequalities and Expected Values

The simplest proof starts claim by claim. For $d_2\geq d_1$, every possible loss satisfies $(X-d_2)_+\leq(X-d_1)_+$. For $u_2\geq u_1$, every possible loss satisfies $\min(X,u_1)\leq\min(X,u_2)$. Taking expectations preserves each inequality. This reasoning does not require a particular severity distribution.

d_2\geq d_1 \implies E[(X-d_2)_+]\leq E[(X-d_1)_+], \qquad u_2\geq u_1 \implies E[\min(X,u_2)]\geq E[\min(X,u_1)].

For combined coverage, the same argument applies in each variable separately. Holding the limit fixed, increasing the deductible cannot increase payment. Holding the deductible fixed, increasing the maximum insurer payment cannot decrease payment:

d_2\geq d_1 \implies E[Y_{d_2,u}]\leq E[Y_{d_1,u}], \qquad u_2\geq u_1 \implies E[Y_{d,u_2}]\geq E[Y_{d,u_1}].

The inequalities need not be strict. For example, if no claim can exceed a particular deductible, raising the deductible further leaves every insurer payment at zero. More generally, the deductible expectation strictly decreases from $d_1$ to $d_2>d_1$ if $P(X>d_1)>0$; a limit expectation strictly increases from $u_1$ to $u_2>u_1$ if $P(X>u_1)>0$. A table with equal factors across adjacent levels can therefore be reasonable when the relevant range has no losses affected by the change.

Factor definitions: On a fixed severity and coverage basis, a deductible factor relative to no deductible is $D(d)=E[(X-d)_+]/E[X]$, provided $E[X]>0$. It is nonincreasing in $d$. An increased limits factor relative to a reference limit $u_0$ is $I(u)=E[\min(X,u)]/E[\min(X,u_0)]$, provided the denominator is positive. It is nondecreasing in $u$. A combined factor can be defined as $K(d,u)=E[Y_{d,u}]/E[Y_{d_0,u_0}]$, provided its fixed reference expectation is positive; it is nonincreasing in $d$ and nondecreasing in $u$.

The denominator in a normalized factor is fixed while the factor levels are compared. A factor table may use a reference other than the ones in these definitions, but the direction of change is unchanged as long as the denominator is the same positive constant throughout the comparison. Factors computed for different books of business, different reference contracts, or different severity assumptions cannot automatically be compared this way.

Why the Inequalities Hold: A Layer-Cake Derivation

The expected payment can also be understood as an area under the survival curve. For nonnegative $X$, the payment $(X-d)_+$ is the length of the interval from $d$ to $X$, when $X>d$, and zero otherwise. Equivalently, for each claim it is the integral of an indicator:

(X-d)_+=\int_d^\infty \mathbf{1}_{\{X>x\}}\,dx.

Taking expectations and interchanging the nonnegative integral and expectation gives the stop-loss identity. The probability inside the integral is the survival probability $S_X(x)=P(X>x)$. The capped payment has the analogous representation over the interval from zero to $u$:

E[(X-d)_+]=\int_d^\infty S_X(x)\,dx, \qquad E[\min(X,u)]=\int_0^u S_X(x)\,dx.

Increasing $d$ removes a nonnegative segment from the first integral. Increasing $u$ adds a nonnegative segment to the second. This proves the expected-payment directions directly and also explains why the results do not depend on a density or on the absence of severity atoms.

Worked Examples

Worked Example: Checking Deductible Factors

A hypothetical small commercial portfolio has ground-up claim severity $X$ with the following distribution: $P(X=0)=0.20$, $P(X=1{,}000)=0.50$, and $P(X=3{,}000)=0.30$, with amounts in dollars. Calculate the expected insurer payment and deductible factor for deductibles of $0, $500, and $1,500. Check the factor ordering.

Method and basis. Use $Y_d^L=(X-d)_+$, and normalize each expected payment by $E[X]$, the expected payment at a zero deductible. The claim probabilities sum to $0.20+0.50+0.30=1.00$.

Step 1: Calculate the denominator. The expected ground-up loss is $0.20(0)+0.50(1{,}000)+0.30(3{,}000)=500+900=\$1{,}400$.

Step 2: Calculate payments at each deductible. At $d=0$, the expected payment is $0.50(1{,}000)+0.30(3{,}000)=\$1{,}400$. At $d=500$, it is $0.50(1{,}000-500)+0.30(3{,}000-500)=250+750=\$1{,}000$. At $d=1{,}500$, it is $0.30(3{,}000-1{,}500)=\$450$; the $1,000 loss is below the deductible and produces no payment.

Step 3: Normalize and check. The factors are $1{,}400/1{,}400=1.0000$, $1{,}000/1{,}400=0.7143$, and $450/1{,}400=0.3214$, rounded to four decimals. Thus the factors are nonincreasing as the deductible rises. The mistake this check guards against is treating a higher deductible as if it raised insurer payment.

Check: Each payment is no greater than the corresponding ground-up loss, and the expected payments satisfy $\$1{,}400\geq \$1{,}000\geq \$450$.

Worked Example: Checking Increased Limits Factors

For a hypothetical equipment policy, the ground-up loss distribution is $P(X=0)=0.10$, $P(X=800)=0.40$, $P(X=2{,}000)=0.30$, and $P(X=5{,}000)=0.20$. There is no deductible. Calculate the limited expected losses at limits of $1,000, $2,500, and $6,000, then calculate increased limits factors relative to the $1,000 limit.

Method and basis. The insurer pays $Y_u^P=\min(X,u)$. The probability check is $0.10+0.40+0.30+0.20=1.00$. The reference expected payment is at $u_0=\$1{,}000$.

Step 1: Limit of $1,000. The expected payment is $0.40(800)+0.30(1{,}000)+0.20(1{,}000)=320+300+200=\$820$. This is the denominator for the factors.

Step 2: Limit of $2,500. The expected payment is $0.40(800)+0.30(2{,}000)+0.20(2{,}500)=320+600+500=\$1{,}420$. The factor is $1{,}420/820=1.7317$, rounded to four decimals.

Step 3: Limit of $6,000. This limit exceeds every possible loss, so the expected payment is $0.40(800)+0.30(2{,}000)+0.20(5{,}000)=320+600+1{,}000=\$1{,}920$. The factor is $1{,}920/820=2.3415$, rounded to four decimals.

Check: The expected payments, and their factors, increase with the limit: $\$820\leq \$1{,}420\leq \$1{,}920$ and $1.0000\leq1.7317\leq2.3415$. A factor below one at the higher limit would contradict this payment basis. The factors are relative to the $1,000 reference, not to unlimited losses.

Worked Example: Diagnosing a Combined Factor Table

A hypothetical liability book has $P(X=0)=0.20$, $P(X=1{,}000)=0.30$, $P(X=3{,}000)=0.30$, and $P(X=5{,}000)=0.20$. Its factor basis is $Y_{d,u}=\min((X-d)_+,u)$, with the factor normalized by the expected payment at $d=\$500,\ u=\$1{,}000$. Calculate the four expected payments for $d\in\{\$500,\$1{,}500\}$ and $u\in\{\$1{,}000,\$2{,}000\}$. Then assess a submitted table that reports $1.00,\ 1.77,\ 0.77,\ 0.70$, respectively.

Method and basis. Keep the severity distribution, payment convention, and denominator fixed. The probabilities sum to $1.00$. Calculate payments claim by claim before taking expectations; in particular, the limit caps insurer payment above the deductible.

Step 1: Establish the reference expectation. At $d=500,\ u=1{,}000$, losses of $1,000, $3,000, and $5,000 produce payments of $500, $1,000, and $1,000. The expectation is $0.30(500)+0.30(1{,}000)+0.20(1{,}000)=150+300+200=\$650$.

Step 2: Raise the limit to $2,000, holding $d=500$. The three nonzero payments are $500, $2,000, and $2,000. The expectation is $0.30(500)+0.30(2{,}000)+0.20(2{,}000)=150+600+400=\$1{,}150$.

Step 3: Raise the deductible to $1,500, holding $u=1,000$. The $1,000 loss pays zero; the $3,000 and $5,000 losses each pay $1,000. The expectation is $0.30(1{,}000)+0.20(1{,}000)=\$500$.

Step 4: Use $d=1,500,\ u=2,000$. The $3,000 and $5,000 losses each pay $1,500. The expectation is $0.30(1{,}500)+0.20(1{,}500)=\$750$.

Step 5: Normalize and diagnose. Dividing each expectation by $650 gives $1.0000,\ 1.7692,\ 0.7692,\ 1.1538$, rounded to four decimals. The submitted value $0.70$ at $d=1,500,\ u=2,000$ violates monotonicity: at the same deductible, increasing the limit from $1,000 to $2,000 must not reduce the factor from $0.77$ to $0.70$. On the stated data and basis, the calculated value is $1.1538$. The monotonicity test identifies an inconsistency; it does not by itself determine whether the cause is a calculation error, a transcription error, or a changed basis.

Check: At each deductible the payment rises with the limit ($\$650$ to $\$1{,}150$, and $\$500$ to $\$750$). At each limit it falls with the deductible ($\$650$ to $\$500$, and $\$1{,}150$ to $\$750$).

How to Screen a Factor Table

A reliable screen compares like with like. For a deductible series, hold the limit, exposure definition, severity basis, and normalization constant. For a limit series, hold the deductible and the corresponding basis constant. For a two-way table, check each row and column in the appropriate direction:

Confirm the payment convention.
Determine whether a stated limit caps ground-up loss or caps insurer payment above a deductible. Apply one convention consistently.

Confirm the comparison basis.
Check that the severity assumptions, covered population, and fixed factor denominator are the same across the levels being compared.

Check the directions.
Expected payment and deductible factors should not rise with a higher deductible; expected payment and increased limits factors should not fall with a higher limit.

Investigate, do not overinterpret.
A violation flags a problem to investigate. Monotone values can still be inaccurate, poorly estimated, or inappropriate for the intended rating basis.

A common exam trap is to compare two entries that change both deductible and limit, then attribute the entire movement to one of them. The inequalities are separate-variable statements: raise the deductible while holding the limit fixed, or raise the limit while holding the deductible fixed. If both change, the effects can oppose one another, so the direction of the net change cannot be inferred from monotonicity alone.

Key takeaway: Payment functions establish the expected direction before any severity model is fitted: increasing a deductible cannot increase insurer payment, and increasing a limit cannot decrease it. Use those directions as screening rules only when the contract definition, severity basis, and normalization are held constant.

Check Your Understanding

For each problem, use the specified payment convention and compare factors only on the stated common basis.

A hypothetical severity distribution has $P(X=0)=0.25$, $P(X=1{,}000)=0.50$, and $P(X=4{,}000)=0.25$. Calculate $E[X]$, then the expected payments and factors relative to no deductible at $d=500$ and $d=2{,}000$. (Answer: $E[X]=\$1{,}500$; payments are $\$1{,}125$ and $\$500$; factors are $0.7500$ and $0.3333$, respectively.)
A hypothetical policy has $P(X=500)=0.40$, $P(X=1{,}500)=0.40$, and $P(X=4{,}000)=0.20$, with no deductible. Find expected capped losses at limits of $1,000 and $3,000 and the increased limits factor relative to $1,000. (Answer: expected payments are $\$800$ and $\$1{,}400$; the factor at $3,000 is $1.7500$.)
For a hypothetical book, $P(X=0)=0.20$, $P(X=2{,}000)=0.50$, and $P(X=6{,}000)=0.30$. Using $Y_{d,u}=\min((X-d)_+,u)$, calculate expected payments at $(d,u)=(1{,}000,1{,}500),(1{,}000,3{,}000),(2{,}000,1{,}500),(2{,}000,3{,}000)$. (Answer: $\$950,\ \$1{,}400,\ \$450,\ \$900$, respectively.)
A limit factor table on one fixed basis gives factors $1.000,\ 1.120,\ 0.980$ at three successively higher limits, with the deductible unchanged. Identify the monotonicity issue and state the minimum conclusion justified by it. (Answer: The last factor falls from $1.120$ to $0.980$, violating nondecreasing limit-factor monotonicity; the table requires investigation but the violation alone does not identify its cause.)
A combined-factor table has factors $1.00$ at $(d_1,u_1)$, $1.10$ at $(d_2,u_1)$, and $1.30$ at $(d_1,u_2)$, where $d_2>d_1$ and $u_2>u_1$. Which comparisons violate monotonicity, assuming one fixed positive denominator and otherwise identical basis? (Answer: The deductible comparison violates monotonicity because the factor rises from $1.00$ to $1.10$ as the deductible increases; the limit comparison is consistent because it rises from $1.00$ to $1.30$ as the limit increases.)

What You'll Learn