Evaluating Multi-Year Rate Plans Against Ruin Constraints

Why Multi-Year Rate Plans Need Path-Based Tests

A rate plan can look adequate when judged by its average premium or its projected ending surplus, yet still allow the insurer’s surplus to fall below zero during the projection. Evaluating a plan over several years therefore requires two related calculations: project the cash flows and track surplus at each year-end; then, when claims are stochastic, calculate the probability that the surplus crosses below zero at any point in the horizon.

The calculations below use an annual time step. Premium is collected during each year and aggregate claims are paid by that year’s end. There is no investment income, so the effective annual interest rate is $i=0\%$. No mortality model applies; the examples concern non-life insurance claims. A negative surplus at a year-end is ruin. A surplus of exactly zero is not ruin under this convention. The model does not capture a shortfall that occurs and is resolved within a year.

Definition: Let $U_0$ be initial surplus, $P_t$ the premium income in year $t$, and $C_t$ the aggregate claim outgo in that year. With no interest income, the year-end surplus is $$ U_t=U_{t-1}+P_t-C_t,\qquad t=1,\ldots,T. $$ Finite-horizon ruin occurs if the surplus is negative at any year-end: $$ \min_{1\le t\le T} U_t<0. $$ The finite-horizon ruin probability is $P(\min_{1\le t\le T} U_t<0)$.

A rate plan specifies premium rates over time. If $e_t$ is the number of covered policies and $r_t$ is the premium per policy in year $t$, then $P_t=e_t r_t$. A stated annual rate increase applies to the rate, not automatically to the total premium: changes in exposure $e_t$ must also be reflected. For a plan that raises the rate by $g$ each year after year 1, $r_t=r_1(1+g)^{t-1}$, provided the plan defines increases as compounded annual changes.

When claims are random, the event of interest is a first-passage event: ruin occurs on the first year the path drops below zero. A projected path using expected claims is useful for planning, but it is not a ruin probability. The distinction matters because different claim paths can have the same expected total and very different chances of an early shortfall.

Projecting Cash Flows Before Testing Ruin

Begin by writing the premium and claim amounts for every year, in consistent units. The expected surplus path follows by applying the recurrence year by year. For a stochastic model, apply that recurrence separately to each possible claim path or to each simulated path. A plan meets a stated ruin constraint $\alpha$ only if its finite-horizon ruin probability is no greater than $\alpha$.

Translate the rate plan into premium income.
Apply each year’s rate to that year’s exposure. Check whether rate changes are compounded and whether exposure also changes.

Construct the surplus path.
For each year, add premium income and subtract aggregate claims from the prior year’s surplus.

Identify ruin paths.
For each possible or simulated path, check every year-end surplus, not only the final one.

Compare with the constraint.
Calculate or estimate the ruin probability for each plan and compare it with the specified maximum.

Worked Example: Comparing Expected Surplus Paths

A hypothetical insurer begins with $5,000 of surplus and covers 100 policies in each of three years. The year-1 premium rate is $550 per policy. Expected aggregate claims are $54,000, $56,000, and $58,000 in years 1, 2, and 3. Compare a level-rate plan with a plan that increases the rate by 5% annually after year 1. Treat the given claim amounts as expected claims; no claim distribution is specified or needed for this expected-value projection. The effective annual interest rate is $i=0\%$, and no mortality assumption applies.

Method. Calculate each plan’s premium income, then use the annual surplus recurrence with expected claims in place of $C_t$. This produces an expected surplus path, not a stochastic ruin probability.

Under the level-rate plan, the premium remains $550 per policy. Thus premium income is $100(550)=\$55{,}000$ in every year. Starting from $U_0=\$5{,}000$:

Year 1: $U_1=5{,}000+55{,}000-54{,}000=\$6{,}000$.
Year 2: $U_2=6{,}000+55{,}000-56{,}000=\$5{,}000$.
Year 3: $U_3=5{,}000+55{,}000-58{,}000=\$2{,}000$.

Under the 5% annual increase plan, the rates are $550, $550(1.05)=\$577.50$, and $577.50(1.05)=\$606.375$ per policy. Premium income is therefore $55,000, $57,750, and $60,637.50. The surplus calculations are:

Year 1: $U_1=5{,}000+55{,}000-54{,}000=\$6{,}000$.
Year 2: $U_2=6{,}000+57{,}750-56{,}000=\$7{,}750$.
Year 3: $U_3=7{,}750+60{,}637.50-58{,}000=\$10{,}387.50$.

Both expected paths remain positive, and the increasing-rate plan ends with $10,387.50 rather than $2,000. This comparison alone cannot establish that either plan meets a ruin-probability constraint: that requires a claim distribution and path-based calculation. The main trap is to call a positive expected surplus path “zero ruin probability.” It describes expected cash flows, not every possible claim outcome.

Calculating Ruin Probability From Claim Paths

For a finite set of possible annual claim amounts, exact enumeration can be used. Suppose annual claims are independent and each has a specified distribution. For each sequence of claims, calculate its probability, construct its surplus path, and mark whether any year-end surplus is negative. The ruin probability is the sum of the probabilities of all marked paths.

This follows directly from the definition of ruin. The set of all claim paths is partitioned into paths that do and do not produce ruin; these events are mutually exclusive. If annual claims are independent, the probability of a particular path is the product of its annual claim probabilities. Summing those path probabilities over precisely the ruin paths gives the probability of ruin. If annual claims are dependent, use their joint path probabilities instead of multiplying marginal probabilities.

P(\text{ruin by }T) =\sum_{(c_1,\ldots,c_T)} \mathbf{1}\!\left\{\min_{1\le t\le T} \left(U_0+\sum_{j=1}^{t}(P_j-c_j)\right)<0\right\} P(C_1=c_1,\ldots,C_T=c_T).

Worked Example: Testing Three Constant-Premium Plans

A hypothetical insurer has initial surplus $U_0=\$1{,}000$ and operates for three years. In each year, aggregate claims $C_t$ are independently $0 with probability 0.75 or $4,000 with probability 0.25. Compare constant annual premiums of $2,000, $3,000, and $4,000. The effective annual interest rate is $i=0\%$; no mortality assumption applies. Find each plan’s ruin probability and determine which meet a 5% maximum.

Method. The claim distribution in each year is a two-point distribution with $P(C_t=0)=0.75$ and $P(C_t=4{,}000)=0.25$. Enumerate the paths that first cross below zero. The common error to avoid is testing only year-3 surplus.

Premium of $2,000. A $4,000 claim reduces surplus by $2,000; a zero claim increases it by $2,000. A claim in year 1 gives $U_1=1{,}000+2{,}000-4{,}000=-\$1{,}000$, so ruin occurs with probability 0.25. If year 1 has no claim, $U_1=\$3{,}000$. A year-2 claim then gives $U_2=\$1{,}000$; if year 3 also has a claim, $U_3=-\$1{,}000$. This additional ruin path has probability $0.75(0.25)(0.25)=0.046875$. In all other paths after a claim-free first year, surplus stays nonnegative. Thus the total ruin probability is $0.25+0.046875=0.296875$, or 29.6875%.

Premium of $3,000. A $4,000 claim reduces surplus by $1,000; a zero claim increases it by $3,000. A claim in year 1 gives $U_1=\$0$, which is not ruin. If year 2 is also a $4,000 claim, then $U_2=-\$1{,}000$, so ruin occurs. This path has probability $0.25^2=0.0625$. If the first two years are not both claims, direct path checks show no later negative surplus: after a claim followed by a zero claim, the year-2 surplus is $3,000 and a year-3 claim leaves $2,000; after a zero claim followed by a claim, year-2 surplus is also $3,000. The ruin probability is therefore 6.25%.

Premium of $4,000. A $4,000 claim leaves surplus unchanged, while a zero claim increases it by $4,000. Surplus can never fall below its initial $1,000, so the ruin probability is 0%.

Only the $4,000 plan meets the 5% constraint. To see why terminal-only testing is wrong, under the $2,000 plan the path “claim, zero claim, zero claim” has probability $0.25(0.75)(0.75)=0.140625$. It produces $U_1=-\$1{,}000$, $U_2=\$1{,}000$, and $U_3=\$3{,}000$. The terminal surplus is positive, but ruin has already occurred in year 1.

Rate Plans With Different Increases by Year

A multi-year plan need not use the same premium in every year. When rates differ by year, surplus changes by different amounts under the same claim outcome. A premium increase may prevent a path from reaching ruin in later years, but it does not erase ruin that occurred earlier. Exact enumeration makes this timing explicit when the horizon and claim distribution are small.

Worked Example: Testing a Stepped Plan Against a Tight Limit

A hypothetical insurer has initial surplus $3,000 and a three-year horizon. Annual aggregate claims are independent: $C_t=\$6{,}000$ with probability 0.20 and $C_t=0$ with probability 0.80 in each year. Compare Plan A, with premiums of $4,000 each year, to Plan B, with premiums of $4,000, $5,000, and $6,000. The effective annual interest rate is $i=0\%$, and no mortality assumption applies. The maximum permitted ruin probability is 2%.

Method. For each plan, calculate the surplus change under a high-claim year and a zero-claim year, then identify which ordered claim paths cross below zero. The common mistake is to count the number of high-claim years without considering their order.

Under Plan A, a high-claim year changes surplus by $4{,}000-6{,}000=-\$2{,}000$, while a zero-claim year changes it by $+\$4{,}000$. Starting from $3,000, a high claim in year 1 leaves $1,000. A second consecutive high claim leaves $-\$1{,}000$, so the path with high claims in both years 1 and 2 causes ruin. Its probability is $0.20(0.20)=0.04$. If year 1 is a zero-claim year, surplus becomes $7,000; even high claims in years 2 and 3 leave $3,000 at the end. If year 1 is a high-claim year and year 2 is a zero-claim year, surplus becomes $5,000, so year 3 cannot cause ruin. Therefore Plan A’s ruin probability is 4%.

Under Plan B, the year-specific changes for a high claim are $-\$2{,}000$, $-\$1{,}000$, and $0. The changes for a zero claim are $+\$4{,}000$, $+\$5{,}000$, and $+\$6{,}000$. The lowest possible path is three consecutive high claims: surplus is $1,000 after year 1, $0 after year 2, and $0 after year 3. Since zero is not negative, this path is not ruin. Any zero-claim year raises the path above this lowest path. Thus Plan B’s ruin probability is 0%.

Plan A fails the 2% constraint because $0.04>0.02$; Plan B meets it because $0\le 0.02$. The zero result is exact for this specified model and horizon, not a claim that real-world ruin risk is absent. A different claim distribution, time step, or cash-flow timing could change the result.

Common Traps and Practical Interpretation

Checking only the final year. Ruin is a path event. A later surplus recovery does not undo an earlier negative year-end surplus.
Treating zero as negative. Under the convention used here, $U_t=0$ is solvent. If a problem defines ruin differently, follow that definition consistently.
Applying a rate change to the wrong quantity. A rate increase changes $r_t$; total premium is $e_t r_t$. A changing number of policies affects premium income separately.
Calling an expected projection a probability. Expected claims produce a useful central projection, but they do not describe the frequency of adverse claim paths.
Ignoring the horizon and time step. A three-year annual test checks three year-end balances. It does not automatically detect intra-year shortfalls or imply an infinite-horizon ruin probability.

For larger or continuous claim distributions, exact enumeration may be impractical. Claim-path simulation can estimate the same finite-horizon event by applying each year’s premium and simulated claims to the surplus recurrence and recording whether any year-end value is negative. The estimated ruin probability is the proportion of simulated paths with at least one negative surplus. As with any simulation estimate, its precision depends on the number of runs, particularly when ruin is rare.

Key takeaway: Compare multi-year rate plans by projecting annual cash flows and testing the entire surplus path. A plan satisfies a ruin constraint only when its finite-horizon probability of at least one negative surplus is within the stated limit.

Check Your Understanding

For each problem, use the stated annual time step and count ruin only when year-end surplus is negative.

A hypothetical insurer starts with $12,000. Premiums are $8,000, $9,000, and $10,000 over three years; expected claims are $7,000, $12,000, and $8,000. With $i=0\%$, calculate the three expected year-end surpluses and the minimum of the path. (Answer: $13,000, $10,000, and $12,000; minimum $10,000.)
Initial surplus is $2,000 and annual premium is $3,000 for two years. Independent annual aggregate claims are $0 with probability 0.80 or $5,000 with probability 0.20. With $i=0\%$, find the two-year ruin probability. (Answer: 0.0400.)
Initial surplus is $1,000; premiums are $2,000, $3,000, and $4,000. Independent annual claims are $0 with probability 0.70 or $5,000 with probability 0.30. With $i=0\%$, find the three-year ruin probability. (Answer: 0.3000.)
A two-year insurer starts with $4,000. Independent annual claims are $0 or $8,000, with probability 0.75 and 0.25, respectively. Plan L collects $4,000 each year; Plan H collects $5,000 in year 1 and $7,000 in year 2. With $i=0\%$, which plan meets a 5% ruin limit? (Answer: Plan L has ruin probability 0.0625; Plan H has 0.0000, so only Plan H meets the limit.)
A hypothetical insurer has 120 policies in each of three years, an initial surplus of $2,000, and a year-1 rate of $400 per policy. Rates rise by 6% each year after year 1. Expected aggregate claims are $46,000, $48,000, and $51,000. With $i=0\%$, calculate the expected year-end surplus path. (Answer: $4,000, $6,880, and $9,812.80.)

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