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Exam ALTAM · Modified Reserve Bases · Tutorial 589 of 999

Profit Testing With Modified Reserves

Construct and check profit vectors and signatures when reserves are on a modified basis.

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What You'll Learn

  • Define a profit vector conditional on a policy being in force at each duration
  • Include modified reserves at the start and end of each policy year
  • Weight conditional profits by survival or persistence probabilities
  • Distinguish an undiscounted profit signature from its present value
  • Reconcile discounted profits with the present value of policy cash flows
  • Identify reserve-release and decrement-weighting errors

Why Profit Testing Uses Modified Reserves

A profit test tracks how a policy’s expected financial result emerges over time. It combines premiums and expenses, benefits on death or other exits, investment accumulation, and the release or establishment of reserves. If the reserve basis is modified, the profit vector must use those modified reserves at the relevant policy durations—not reserves from a different basis.

The profit vector records profit for each duration, conditional on a policy being in force at the start of that duration. The profit signature weights each entry by the probability that the policy has reached that duration in force. Keeping these two quantities distinct is essential: a conditional profit in a later year is not contributed by policies that have already exited.

Definition: In a discrete-time profit test, the profit vector is the sequence of expected profits for each policy year, conditional on being in force at the start of that year. The profit signature is the sequence of those profits weighted by the probability of being in force at the start of each year. Unless otherwise stated, its total is the sum of the weighted profits. A discounted profit signature additionally discounts each weighted profit to the chosen valuation date.

Set the Cash-Flow Timing and Reserve Convention

For the formulas below, consider an annual model. At the start of policy year \(t+1\), a policy in force has reserve \(V_t\), receives premium \(P_t\), and incurs expense \(E_t\). These opening amounts accumulate at effective annual interest rate \(i\). At year-end, a benefit may be paid on a decrement, and a reserve \(V_{t+1}\) is held only for policies that survive or persist into the next duration. The reserve values \(V_t\) are on the modified reserve basis being tested.

Let \(p_t\) be the probability of persisting through the year, and let \(q_{t,j}\) be the probability of decrement \(j\) during that year. For mutually exclusive decrement causes, the probabilities satisfy \(p_t+\sum_j q_{t,j}=1\). Let \(B_{t,j}\) be the year-end benefit on decrement \(j\). These are probabilities conditional on being in force at the year’s start. The conditional profit for the year is

$$ \Pi_t=(V_t+P_t-E_t)(1+i)-\sum_j q_{t,j}B_{t,j}-p_tV_{t+1}. $$

The final term is the expected year-end reserve held for persisting policies. It is not a reserve payable to every policyholder: only the \(p_t\) fraction continuing in force receives that reserve allocation in the profit calculation. If a contract ends at year-end, its terminal reserve is normally zero, though the formula allows a nonzero \(V_{t+1}\) when a continuing policy remains.

The formula assumes the stated premiums and expenses occur at the start of the year, benefits and the next reserve are measured at year-end, and the same annual interest rate is applied to opening funds. If cash flows occur at different times, they must be accumulated or discounted to a common time before applying the profit calculation. Continuous-time cash flows require a different timing treatment.

Constructing the Profit Signature

Let \({}_tp_x\) denote the probability that a policy issued at age \(x\) remains in force at duration \(t\), under the experience transition assumptions used in the profit test. Then the signature contribution at duration \(t\) is \({}_tp_x\Pi_t\). The first contribution has weight \({}_0p_x=1\). In a model with several decrement causes, persistence is the probability of avoiding all exits over the duration, not merely the probability of avoiding one selected cause.

1
Identify the modified reserve path.
Write the opening reserve and the reserve held for persisting policies at each year-end. Use a consistent reserve basis and duration convention.
2
Calculate conditional annual profit.
Accumulate the opening reserve and net cash inflow, subtract expected decrement benefits, and subtract the reserve held for persisting policies.
3
Weight by in-force probability.
Multiply each conditional profit by the probability the policy is in force at the start of that year.
4
Sum or discount the contributions.
For an undiscounted total signature, add the weighted profits. For a present-value comparison, discount each year-end profit to issue.

Worked Example: A Two-Year Contract With a Modified Reserve

Method. Calculate each year’s profit conditional on being in force, then weight the second-year profit by first-year persistence. This hypothetical contract has annual effective interest \(i=0\%\). At issue, the net premium less expense is $20. During year 1, the death probability is 0.25, with a death benefit of $80; the survival probability is 0.75. The modified reserve at duration 1 is $5. For a policy still in force at duration 1, the net premium less expense for year 2 is $10. During year 2, death probability is 0.20 with a benefit of $80, and maturity probability is 0.80 with a benefit of $40. The terminal reserve is zero. The modified reserve at issue is zero.

Year 1. The opening reserve plus net cash inflow is \(0+20=20\). Expected death benefits are \(0.25(80)=20\). The expected reserve for survivors is \(0.75(5)=3.75\). Therefore,

$$ \Pi_0=20-0.25(80)-0.75(5)=20-20-3.75=-3.75. $$

Year 2. The opening reserve plus net cash inflow is \(5+10=15\). Expected benefits are \(0.20(80)+0.80(40)=16+32=48\), and the terminal reserve is zero. Thus,

$$ \Pi_1=15-48=-33. $$

The profit vector is \((-3.75,-33.00)\). The year-2 profit is conditional on reaching duration 1, so its signature weight is 0.75. The total signature is

$$ -3.75+0.75(-33)=-3.75-24.75=-28.50. $$

Check. At issue, expected net inflows are \(20+0.75(10)=27.50\). Expected benefits are \(0.25(80)+0.75[0.20(80)+0.80(40)]=20+36=56\). Their difference is \(27.50-56=-28.50\), matching the total signature. The trap is to subtract year-2 benefits without weighting them for first-year survival, or to subtract the full $5 reserve rather than the $3.75 expected reserve held for survivors.

Why Reserve Terms Reconcile Across Durations

A reserve is an accounting amount in this calculation, not an additional benefit paid to every policy. A reserve deducted at one year-end is carried forward as the opening reserve for the policies that persist. When profits are discounted consistently, these reserve terms cancel between adjacent years. This explains why changing a consistent reserve path can alter the timing of profit emergence without changing the present value of the underlying net policy cash flows.

To see this, let \(\pi_t\) be the probability of being in force at duration \(t\), with \(\pi_0=1\) and \(\pi_{t+1}=\pi_t p_t\). Multiply the annual profit formula by \(\pi_t v^{t+1}\), where \(v=1/(1+i)\). The opening-reserve term becomes \(\pi_tV_t v^t\), while the closing-reserve term becomes \(-\pi_{t+1}V_{t+1}v^{t+1}\). The latter cancels the next year’s opening-reserve term. Summing over years leaves the initial reserve plus the present value of premiums less expenses and benefits, with any terminal reserve treated according to the contract’s terminal condition.

$$ \sum_{t=0}^{n-1}\pi_t v^{t+1}\Pi_t = V_0+ \sum_{t=0}^{n-1}\pi_t v^t(P_t-E_t) - \sum_{t=0}^{n-1}\pi_t v^{t+1}\sum_j q_{t,j}B_{t,j} - \pi_n v^n V_n. $$

This identity follows because \(\pi_t p_t=\pi_{t+1}\), so each intermediate reserve appears once with a negative sign and once with a positive sign. It assumes the reserve held at a duration is the same amount whether regarded as the closing reserve from the preceding year or the opening reserve for the next year. If the basis or reserve definition changes between those points, an adjustment is needed; the identity as written does not omit such a change.

Worked Example: Interest, Benefits, and a Present-Value Check

Method. Use the year-end profit formula at \(i=5\%\), then compare the discounted, weighted profit signature with the present value of net policy cash flows. This hypothetical two-year endowment has modified reserves \(V_0=0\), \(V_1=$20\), and \(V_2=0\). At the start of each year in force, premium less expense is $25. In year 1, death probability is 0.10 with benefit $100, and survival probability is 0.90. In year 2, conditional on reaching it, death probability is 0.20 with benefit $100 and maturity probability is 0.80 with benefit $60.

Year 1 conditional profit. Opening funds accumulate to \((0+25)(1.05)=$26.25\). Expected death benefits are \(0.10(100)=$10\). The expected reserve for survivors is \(0.90(20)=$18\). Hence \(\Pi_0=26.25-10-18=-$1.75\).

Year 2 conditional profit. Opening funds accumulate to \((20+25)(1.05)=$47.25\). Expected benefits are \(0.20(100)+0.80(60)=20+48=$68\), with no terminal reserve. Thus \(\Pi_1=47.25-68=-$20.75\).

The undiscounted profit vector is \((-1.75,-20.75)\). Weighting the second year by the 0.90 probability of survival gives the total undiscounted signature \(-1.75+0.90(-20.75)=-$20.425\), or \(-$20.43\) to the nearest cent.

For the present-value check, the discounted signature is

$$ \frac{-1.75}{1.05}+\frac{0.90(-20.75)}{1.05^2} =-1.6667-16.9388=-18.6054. $$

The net premium and expense cash flows have present value \(25+0.90(25)/1.05=$46.4286\). Benefits have present value \(10/1.05+0.90(68)/1.05^2=$65.0340\). Their difference is \(-$18.6054\), agreeing with the discounted signature. Values are rounded to the nearest cent only at the final step. The trap is to compare the undiscounted total signature, \(-$20.425\), directly with a present value; the comparison requires discounting profits at their year-end times.

Competing Decrements and a Full Reconciliation

When more than one decrement is possible, each decrement benefit is weighted by its own conditional probability, while the reserve is weighted by the probability of persistence. Applying the survival probability to a lapse benefit, for example, would incorrectly assign that benefit to policies that remain in force.

Worked Example: Death, Lapse, and Persistence

Method. Separate the decrement benefit terms, use the modified reserve only for persisting policies, and weight the later profit by the probability of reaching that year. Consider a hypothetical two-year contract with annual effective interest \(i=4\%\). Its modified reserves are \(V_0=0\), \(V_1=$30\), and \(V_2=0\). Net premium less expense is $40 in year 1 and $25 in year 2. In year 1, death probability is 0.10 with benefit $120, lapse probability is 0.10 with surrender benefit $8, and persistence probability is 0.80. In year 2, among policies still in force, death probability is 0.25 with benefit $120 and maturity probability is 0.75 with benefit $50.

Year 1 conditional profit. Accumulated opening funds are \((0+40)(1.04)=$41.60\). Expected year-end benefits are \(0.10(120)+0.10(8)=12+0.80=$12.80\). Expected reserve for persisting policies is \(0.80(30)=$24\). Therefore, \(\Pi_0=41.60-12.80-24=$4.80\).

Year 2 conditional profit. Accumulated opening funds are \((30+25)(1.04)=$57.20\). Expected benefits are \(0.25(120)+0.75(50)=30+37.50=$67.50\), and the terminal reserve is zero. Therefore, \(\Pi_1=57.20-67.50=-$10.30\).

The total undiscounted signature is \(4.80+0.80(-10.30)=4.80-8.24=-$3.44\). To check the present-value reconciliation, the discounted signature is \(4.80/1.04+0.80(-10.30)/1.04^2=$4.6154-$7.6183=-$3.0030\). The net premium and expense cash flows have present value \(40+0.80(25)/1.04=$59.2308\). The benefit present value is \([0.10(120)+0.10(8)]/1.04+0.80[0.25(120)+0.75(50)]/1.04^2=$62.2337\). The difference is \(-$3.0030\), as required. The year-1 lapse benefit is \(0.10(8)\), not \(0.80(8)\): the lapse benefit is paid on lapse, while the 0.80 probability applies to persistence and the reserve.

Common Exam Traps

  • Mixing conditional profit and signature contribution. Calculate \(\Pi_t\) conditional on being in force first; only then multiply by the in-force probability.
  • Using the wrong probability for the reserve. Deduct \(p_tV_{t+1}\), not \(V_{t+1}\) for every policy and not a decrement probability times the reserve.
  • Applying a decrement benefit to the wrong group. A death benefit is weighted by the death probability, a surrender benefit by the lapse probability, and the next reserve by persistence.
  • Confusing timing in a signature check. An undiscounted signature is not a present value when \(i\ne0\). Discount each end-of-year profit by its year-end time.
  • Changing the reserve path mid-calculation. Use the modified reserve at the exact duration specified. An inconsistent opening and closing reserve prevents the reserve cancellation in the reconciliation.

A negative profit in a particular duration does not by itself indicate an error. It can arise because expected benefits or reserve requirements exceed the accumulated funds in that year. The important checks are that every cash flow is assigned to the correct event and time, that conditional profits are weighted by the correct in-force probabilities, and that the discounted signature reconciles with the policy cash flows under consistent assumptions. This approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, calculate conditional profits first and apply in-force weights only when forming the signature.

  1. A one-year hypothetical contract has \(i=0\%\), opening reserve zero, net premium less expense of $18, death probability 0.20 with benefit $50, survival probability 0.80, and modified reserve $10 for survivors. Find the year’s profit. (Answer: $0.00)
  2. In a one-year hypothetical contract, \(i=0\%\), opening reserve is zero, and net premium less expense is $35. Death probability is 0.10 with benefit $100; lapse probability is 0.10 with benefit $5; persistence probability is 0.80 with modified reserve $20. Find conditional profit. (Answer: $8.50)
  3. A two-year contract has conditional profits of $6 in year 1 and \(-$12\) in year 2. The probability of remaining in force through year 1 is 0.70. Find the total undiscounted profit signature. (Answer: \(-$2.40\))
  4. For one year, \(i=4\%\), \(V_0=0\), net premium less expense is $22, death probability is 0.25 with benefit $40, and persistence probability is 0.75 with modified reserve \(V_1=$12\). Find conditional profit. (Answer: $3.88)
  5. In a hypothetical two-year contract with \(i=0\%\), year-1 net premium less expense is $30. Year 1 has death probability 0.20 with benefit $50, lapse probability 0.10 with benefit $10, persistence probability 0.70, and modified reserve \(V_1=$15\). In year 2, net premium less expense is $20; death probability is 0.30 with benefit $50, and maturity probability is 0.70 with benefit $25. The terminal reserve is zero. Find both conditional profits and the total undiscounted signature. (Answer: profit vector \(($8.50,$2.50)\); total signature $10.25.)
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