Why an Integrated Guarantee Case Needs a Clear Timeline
A universal life no-lapse guarantee can keep coverage in force even when the policy account can no longer support deductions. That protection does not make the account balance irrelevant: the balance determines how much support is needed, while premium history determines whether the guarantee applies. Mortality and lapse assumptions then determine how long the insurer may remain exposed.
The calculations below use a deliberately simplified hypothetical policy. The assumptions are stated explicitly so that account mechanics, decrement probabilities, reserves and solvency measures can be checked separately. The capital formula is an illustrative measure for this case, not a prescribed regulatory formula. The integrated approach is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
The annual decrement convention is death first, followed by lapse among those who survive the year. If the annual death probability is \(q\) and the conditional lapse probability among survivors is \(w\), then the probability a policy active at the start of a year remains active at the next start is
This ordering matters. The probability of a death claim during the year is \(q\), not \(q(1-w)\), because lapse is applied only after survival under the stated convention.
Worked Example: Flexible Premiums and Account Depletion
Worked Example: Flexible Premiums and Account Depletion
Consider a hypothetical policy with \(A_1=0\), an annual credited rate of \(-10\%\), and an annual account deduction of $100. The policyholder pays $200 at issue and nothing at the start of year 2. The guarantee requires cumulative gross premiums of at least $100 per completed policy year. Assume annual mortality \(q=0.02\) and a 20% lapse probability among survivors. Find the account values, guarantee status through year 2, and the probability the policy is still active at the start of year 2.
Method. Project the account year by year, apply the premium test at the relevant anniversary, and then calculate the probability of surviving the death-first, lapse-second decrement sequence.
At the end of year 1, the premium less the deduction is \(200-100=100\). Applying the 10% account loss gives
At the end of year 2, the account available after the deduction would be \(90-100=-10\). The account is therefore depleted:
By the second anniversary, cumulative premiums are $200 and the required amount is \(2(100)=$200\). The policy meets the stated guarantee test through year 2, despite the zero account. This conclusion does not establish eligibility at the third anniversary: with no further premium, cumulative premiums remain $200 while the requirement becomes $300.
For the decrement calculation, the probability of surviving year 1 is \(1-0.02=0.98\). Conditional on survival, the probability of not lapsing is \(1-0.20=0.80\). Thus the probability of being active at the start of year 2 is
The mutually exclusive year-1 outcomes are death with probability 0.0200, lapse after survival with probability \(0.98(0.20)=0.1960\), and continuation with probability 0.7840. They sum to 1.0000, providing a check. Common mistake avoided: treating a zero account as automatic termination without first testing the contractual guarantee condition.
From Decrements to a Guarantee Reserve
A prospective reserve is the value at a valuation date of future benefits and expenses, less future premiums and other specified income, under the valuation assumptions. When account assets or charges are included in a real product valuation, they must be treated consistently on both sides of that calculation. The next example isolates the guarantee exposure after account depletion: there are no future premiums or account assets available, so the insurer funds the stated death benefit and administration expense.
The reason for the formula is that, conditional on being active at the start of a year, the expense \(e\) is paid immediately. The death benefit is paid at year-end with probability \(q\), so its expected present value for that year is \(v qB\). To reach the next year, the policy must both survive and avoid lapse, with probability \(s\); discounting one year as well gives the factor \(vs\). Repeating that argument for each future year produces the sum.
For expected profit emergence, define profit in a period as income less outgo, plus the opening reserve, less the closing reserve. The opening reserve is held for the future obligations; a release of reserve as exposure ends is therefore part of profit. With a reserve calculated on exactly the same cashflows and assumptions as the projection, expected profit is zero when there is no margin or mismatch in the basis.
Worked Example: Reserve Under a Persistency Stress
At the start of a two-year valuation horizon, a hypothetical policy has a depleted account and an active no-lapse guarantee. There are no future premiums. The death benefit is $1,000, the annual start-of-year administration expense is $10, the effective annual interest rate is 0%, and mortality is 2% per year. Under a persistency stress, conditional lapse is 5% per year. Calculate the reserve per active policy at the start, the expected aggregate reserve at the end of year 1 per initial policy, and expected year-1 profit.
Method. First calculate the probability of reaching the second year, then value the first-year and second-year expected outgo. The annual expected outgo per active policy is the $10 expense plus the expected death benefit \(0.02(1{,}000)=$20\), for a total of $30.
The probability of remaining active to the start of year 2 is
With zero interest and two years remaining, the start reserve per active policy is $30 for year 1 plus \(0.931($30)\) for year 2:
At the end of year 1, a policy that has survived and not lapsed has one year remaining. Its conditional reserve is $30. Per initial policy, the probability that such a policy remains active is 0.931, so the expected aggregate closing reserve is
Expected year-1 outgo is the $10 expense plus expected claims of $20, or $30. There is no income. Expected profit is therefore
The zero result is a check: reserves were calculated on the same assumptions and cashflows as the expected projection. Common mistake avoided: subtracting the closing reserve without first recognizing that it is held only for policies still active.
Why a Lapse Stress Can Increase the Guarantee Liability
A stress that increases lapses does not always worsen a product’s financial position. If retained policies are loss-making, more lapses may reduce future losses; if policies generate profitable charges, more lapses may remove valuable margins. For a depleted-account guarantee with no future premiums in the simplified model above, lower lapse means more policies persist into future years in which the insurer bears expenses and mortality exposure. The direction of the result must be established from the projected cashflows, not inferred from the word “stress.”
Worked Example: Reserve Movement and a Simplified Solvency Measure
Use the two-year guarantee case above, but compare a baseline conditional lapse rate of 20% with the 5% persistency stress. Keep mortality at 2%, interest at 0%, the death benefit at $1,000, and annual expense at $10. At the valuation date, hypothetical assets are $65 per policy. For this example only, define the capital requirement as 10% of the reserve plus a fixed $2 expense-risk buffer. Calculate both reserves, own funds, capital requirements, and solvency ratios.
Method. For each lapse basis, calculate the annual continuation probability \(s\), then use \(V_0=30(1+s)\). Own funds are assets less the reserve; the solvency ratio is own funds divided by the defined capital requirement.
Under the 20% baseline lapse assumption,
Baseline own funds are \(65-53.52=$11.48\). The defined requirement is \(0.10(53.52)+2=$7.352\), and the solvency ratio is
Under the 5% persistency stress, the reserve is $57.93, as calculated above. Own funds fall to \(65-57.93=$7.07\). The capital requirement becomes \(0.10(57.93)+2=$7.793\), giving
The reserve rises by \(57.93-53.52=$4.41\), and own funds fall by the same amount because assets are held fixed. Under the stated capital formula, the stressed position has a shortfall of \(7.793-7.07=$0.723\) against the requirement. These figures are specific to the illustrative formula and assumptions; they are not a regulatory solvency result.
Common mistake avoided: treating the reserve increase and capital requirement as the same quantity. The reserve is a liability estimate; the defined capital requirement is a separate amount, and own funds must be compared with that amount.
Timing and Decrement Traps
The calculations depend on the contract timeline as much as on the arithmetic. A premium paid at the start of a year can affect the account and the cumulative-premium test before that year’s deductions. A death benefit payable at year-end is not discounted as an immediate payment. A lapse probability specified among survivors is conditional, not an unconditional decrement probability.
Apply premiums and deductions in the stated order, credit interest for the stated period, and floor the account only where the contract specifies.
Compare cumulative qualifying premiums with the required cumulative amount at the contract’s test date. Do not substitute account value for premium history.
Apply mortality and lapse in their stated conditional order. For death first and lapse among survivors, the continuation probability is \((1-q)(1-w)\).
Include each cashflow once, at its correct timing. Reconcile expected profit to opening and closing reserves before interpreting the result as a margin or loss.
Calculate assets less liabilities for own funds, then compare own funds with the separately specified capital requirement.
A common decrement error is to replace the conditional sequence with \(1-q-w\). For the persistency-stress assumptions above, that shortcut gives \(1-0.02-0.05=0.93\), rather than the correct \(0.931\). It would produce a two-year reserve of \(30(1+0.93)=$57.90\), instead of $57.93. The shortcut is not the exact probability for death first and lapse conditional on survival; the product \((1-q)(1-w)\) is required.
Check Your Understanding
Use the stated order of events and show intermediate steps when solving.
- A policy starts with account value $0. Its annual interest credit is 0%, the annual deduction is $50, and it receives $150 at issue but no year-2 premium. The guarantee requires $75 of cumulative premiums per completed year. Find the account after each of the first two years and determine whether the guarantee test is met at the second anniversary. (Answer: account values $100 and $50; the year-2 test is met because $150=$150.)
- A depleted-account guarantee has two years remaining, zero interest, death benefit $500, annual start-of-year expense $8, mortality 1%, and conditional lapse 10% among survivors. There are no future premiums. Find the prospective reserve per active policy. (Answer: \(s=0.99(0.90)=0.891\), reserve \(=13(1+0.891)=$24.583\), or $24.58.)
- For a two-year guarantee, use zero interest, mortality 3%, conditional lapse 4%, annual expense $12 and death benefit $800, with no future premiums. Find the reserve per active policy. (Answer: \(s=0.97(0.96)=0.9312\); annual expected outgo \(=12+0.03(800)=36\); reserve \(=36(1+0.9312)=$69.5232\), or $69.52.)
- Using the reserve from question 3, suppose assets are $100 and the defined capital requirement is 8% of reserve plus $3. Calculate own funds and the solvency ratio. (Answer: own funds \(=100-69.52=$30.48\); requirement \(=0.08(69.5232)+3=$8.561856\); ratio \(=30.48/8.561856=356.0\%\) to one decimal place.)
- For one projection period, income is $15, outgo is $22, opening reserve is $40 and closing reserve is $28. Calculate profit using the reserve-movement convention in this tutorial. (Answer: \(15-22+40-28=$5\).)