Multiple Rate Changes on the Parallelogram

Why Multiple Changes Require Care

When several rate changes occur during an experience period, the earned premium does not all reflect the rate level in force at the period’s end. Policies earn over time, and renewal-effective changes reach the portfolio gradually. The parallelogram method estimates how much of the experience-period earned exposure reflects each change, then combines those pieces into an average earned rate level.

The key is to track each change’s effect through subsequent rate levels. A later change may raise or lower the rate. Either way, exposure affected by that later change is still exposure to which every earlier change in the sequence has already applied.

Definition: The average earned rate level is the earned-exposure-weighted average of the rate-level indices applying to exposure earned during the experience period. A rate-level index expresses a rate level relative to a chosen base level, commonly 1.000.

Model and Earned-Exposure Weights

Use the standard parallelogram assumptions: policies have annual terms; policy starts, including renewals, are distributed uniformly over time; exposure is earned pro rata over each policy term; and a rate change applies to policies starting or renewing on or after its effective date. Normalize the experience period to run from time 0 to time 1. A policy starting at time $s$ contributes the portion of its annual exposure that falls inside this period.

For a change at time $t$, where $0\le t\le1$, let $C(t)$ be the fraction of total earned exposure in the experience period that is at the rate level introduced by that change or at any subsequent rate level. This definition follows the sequence of changes, not the numerical size of the rate levels. If a later change decreases the rate, exposure at that later level is still included in $C(t)$.

C(t)=\frac{(1-t)^2}{2}

The denominator implicit in this fraction is total earned exposure. For a steady stream of annual policies, a policy starting at $s$ contributes $s+1$ units of exposure to the period when $-1\le s\le0$, and $1-s$ units when $0\le s\le1$. Total exposure, in units proportional to the policy-start rate, is therefore:

\int_{-1}^{0}(s+1)\,ds+\int_{0}^{1}(1-s)\,ds =\frac12+\frac12=1.

Policies starting at or after $t$ contribute to the rate level introduced by the change or a later one. Within this experience period their total earned exposure is:

\int_t^1(1-s)\,ds =\left[s-\frac{s^2}{2}\right]_t^1 =\frac{(1-t)^2}{2}.

Dividing this area by total exposure, which equals 1 in the chosen units, gives $C(t)$. This also explains why a change effective at the start of the period has weight $C(0)=0.5$, not 1: policies already in force at time 0 continue at their prior level until renewal. At the end of the period, $C(1)=0$, because no exposure is earned after the change within the period.

Combining Sequential Rate Changes

Suppose the initial rate-level index is $L_0$. There are $n$ changes during the experience period, with change $j$ occurring at time $t_j$, where $t_1<t_2<\cdots<t_n$. Let $L_j$ be the rate-level index after change $j$. The incremental change at step $j$ is $L_j-L_{j-1}$, which may be positive or negative.

Theorem: Under the parallelogram assumptions, the average earned rate level is $$ \bar L=L_0+\sum_{j=1}^{n}(L_j-L_{j-1})C(t_j). $$ Here $C(t_j)$ includes exposure at the level introduced at $t_j$ and at every later level, regardless of whether a later rate change increases or decreases the numerical index.

To see why this works, start with the initial level $L_0$ applied to all exposure. Change $j$ adds $L_j-L_{j-1}$ to the index for the fraction $C(t_j)$ of exposure affected by that change and subsequent levels. Its contribution to the average is therefore $(L_j-L_{j-1})C(t_j)$. Adding these contributions gives the formula.

Equivalently, the exposure shares at each successive level can be found by subtraction. The initial level has share $1-C(t_1)$; an intermediate level $j$ has share $C(t_j)-C(t_{j+1})$; and the final level has share $C(t_n)$. These shares are nonnegative and sum to 1 because $C(t)$ decreases as the change date moves later. The rate indices themselves need not increase.

Worked Example: An Increase Followed by a Decrease

A hypothetical insurer has an initial rate-level index of 1.000. A renewal-effective change increases rates by 8% at time $t_1=0.20$. A second change, effective at $t_2=0.60$, decreases the then-current rates by 5%. Calculate the average earned rate level for the experience period using the parallelogram method.

Step 1: Find the sequential rate levels. After the first change, $L_1=1.000(1.08)=1.080$. The second change is a 5% decrease from that level, so $L_2=1.080(0.95)=1.026$. Its incremental change is $L_2-L_1=1.026-1.080=-0.054$.

Step 2: Calculate the exposure weights. $C(0.20)=\frac12(1-0.20)^2=\frac12(0.80)^2=0.320$. $C(0.60)=\frac12(1-0.60)^2=\frac12(0.40)^2=0.080$.

Step 3: Apply the incremental-change formula. The first change adds $0.080(0.320)=0.02560$ to the average. The second adds $-0.054(0.080)=-0.00432$. Thus $\bar L=1.000+0.02560-0.00432=\mathbf{1.02128}$.

Check by allocating exposure to levels. The initial level has share $1-0.320=0.680$; the intermediate level has share $0.320-0.080=0.240$; and the final level has share $0.080$. The weighted average is $0.680(1.000)+0.240(1.080)+0.080(1.026)=0.680+0.25920+0.08208=1.02128$, matching the result.

Method and trap avoided. This uses cumulative change weights and then checks them by level allocation. Do not exclude the final exposure share from the first change merely because 1.026 is numerically below 1.080: the second change is applied after the first, so that exposure remains affected by the first change as well.

Using Exposure Shares as a Check

The incremental-change formula is compact, while the level-allocation method is often easier to audit. The allocation method is especially useful when a problem gives several rate levels directly. Calculate $C(t_j)$ for each date, take differences between consecutive cumulative weights to get the shares, and multiply each share by its corresponding level. The shares should sum to 1. If they do not, there is likely an arithmetic or ordering error.

Worked Example: A Change at the Start of the Period

A hypothetical insurer starts at rate level 1.000. Rates increase by 10% at the beginning of the experience period. At time $t=0.50$, a second renewal-effective change reduces the resulting rate level by 4%. Find the average earned rate level.

Step 1: Establish the levels and increments. The first new level is $L_1=1.000(1.10)=1.100$, with increment $0.100$. The second level is $L_2=1.100(0.96)=1.056$, with increment $1.056-1.100=-0.044$.

Step 2: Find the weights. $C(0)=\frac12(1-0)^2=0.500$. At the second date, $C(0.50)=\frac12(0.50)^2=0.125$.

Step 3: Calculate the average. The first change contributes $0.100(0.500)=0.0500$, and the second contributes $-0.044(0.125)=-0.0055$. Therefore $\bar L=1.000+0.0500-0.0055=\mathbf{1.0445}$.

Check by level allocation. The initial level has share $1-0.500=0.500$, the intermediate level has share $0.500-0.125=0.375$, and the final level has share $0.125$. Then $0.500(1.000)+0.375(1.100)+0.125(1.056)=0.500+0.4125+0.1320=1.0445$.

Method and trap avoided. The beginning-of-period change has weight one-half under this annual-policy model. Treating it as fully earned immediately would incorrectly give it weight 1.000; renewal-effective changes are not instantly reflected in all policies in force.

Worked Example: Average Level and On-Level Factor

A hypothetical portfolio has initial rate level $L_0=1.000$. Rates increase by 12% at $t_1=0.25$, then decrease by 4% at $t_2=0.75$. Calculate the average earned rate level and the factor that brings that average to the final rate level.

Step 1: Calculate the levels. $L_1=1.000(1.12)=1.120$. The final level is $L_2=1.120(0.96)=1.0752$. The increments are $0.120$ and $1.0752-1.120=-0.0448$.

Step 2: Calculate the cumulative weights. $C(0.25)=\frac12(0.75)^2=0.28125$, and $C(0.75)=\frac12(0.25)^2=0.03125$.

Step 3: Find the average. The first change contributes $0.120(0.28125)=0.03375$. The second contributes $-0.0448(0.03125)=-0.00140$. Thus $\bar L=1.000+0.03375-0.00140=\mathbf{1.03235}$.

Step 4: Calculate the factor to the final level. Divide final level by average earned level: $1.0752/1.03235=1.041507\ldots$. Rounded to four decimal places, the factor is $\mathbf{1.0415}$. As a check, $1.03235(1.0415)=1.0751925$, which agrees with 1.0752 to the stated rounding.

Method and trap avoided. This calculation uses the ratio of the target final level to the average earned level. Do not add a factor based on the percentage changes alone: the changes affect different fractions of the earned exposure.

Common Mistakes and Model Limits

Comparing later levels numerically. A later level below an earlier one does not mean its exposure should be removed from the earlier change’s cumulative weight. The earlier change applied before the later one.
Applying the full rate change to all exposure. Each change is weighted by $C(t)$, not by 1, because renewal-effective changes phase in as policies renew.
Compounding the wrong amount. A percentage change is applied to the then-current level. For example, a 5% decrease after a 1.080 level gives $1.080(0.95)=1.026$, not 1.030.
Using level weights as if they were cumulative weights. $C(t_j)$ is the share at level $j$ or later. The share at exactly an intermediate level is $C(t_j)-C(t_{j+1})$.
Ignoring the assumptions. The formula for $C(t)$ depends on annual terms, uniform policy starts, pro-rata earning, and renewal-effective changes. If the policy term, writing pattern, earning basis, or application of the rate change differs materially, this parallelogram weight need not apply.

For ratemaking, the average earned rate level helps distinguish the rate level embedded in historical earned premium from the level desired for a prospective period. The calculation is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus. The essential discipline is to keep the time sequence and the earned-exposure weights separate: first determine how much exposure each change reaches, then apply the change in rate level.

Check Your Understanding

For each problem, use the annual-policy parallelogram assumptions stated above. Carry full precision through the calculations and round only where the question requests it.

A portfolio starts at level 1.000. Rates increase by 10% at $t=0.40$. Find the average earned rate level. (Answer: 1.0180)
A portfolio starts at 1.000, increases by 6% at $t=0.20$, and then increases by 3% at $t=0.70$, relative to the then-current level. Find the average earned rate level. (Answer: 1.020631)
A portfolio starts at 1.000, increases by 8% at $t=0.10$, and then decreases by 5% at $t=0.50$. Find the average earned rate level, retaining the effect of both sequential changes. (Answer: 1.025650)
A portfolio starts at 1.000 and increases by 10% at $t=0.20$. A second change at $t=0.60$ sets the average earned rate level for the period to 1.025. What must the second rate-level index be? (Answer: 1.0125)
A portfolio starts at 1.000, increases by 5% at $t=0.10$, and increases by 4% at $t=0.40$, relative to the then-current level. Find the factor to the final level from the average earned level, rounded to four decimal places. (Answer: 1.0625)

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