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Exam ALTAM · Universal Life Profit and Reserves · Tutorial 632 of 999

Deriving the Universal Life Reserve Recursion

Learn to derive and apply the universal life reserve recursion while keeping policy account movements, insurer cash flows and valuation interest consistent.

Advanced 10 min read

What You'll Learn

  • Define a prospective reserve as the value of future insurer outgo less future income, conditional on the policy being in force.
  • Derive the one-year recursion using the timing of premiums, expenses, death claims and survival payments.
  • Rearrange the recursion to solve for the next-period reserve.
  • Track how account-value changes affect benefits and future valuation inputs.
  • Identify timing and mortality-weighting errors that produce an incorrect recursion.

Why a Reserve Recursion Is Useful

A prospective reserve is calculated from expected future cash flows. Recalculating all those cash flows at every policy duration can be cumbersome. A one-period recursion offers an equivalent approach: separate the cash flows over the next year from the reserve needed at the end of that year. The result is useful for projecting reserves and checking valuation calculations for hypothetical universal life policies.

The key is to specify the timing before writing the equation. In this tutorial, the reserve is measured at the start of an annual period, a premium is received and an expense is paid at that time, and any death or survival benefit is paid at the end of the period. Interest is earned at the effective annual valuation rate \(i\). Deaths during the period are represented by an annual probability, but their benefits are assumed paid at year-end.

Definition: For a policy in force at time \(t\), the prospective reserve \(V_t\) is the expected present value, at time \(t\), of future insurer outgo less future insurer income, conditional on the policy being in force at time \(t\). A positive reserve represents expected net future outgo. The valuation specifies which payments count as outgo or income, their timing, and the assumptions used to value them.

Set the One-Year Cash-Flow Model

Consider a policy in force at time \(t\). Let \(P_t\) be the premium received at time \(t\), and let \(E_t\) be the expense paid at time \(t\). Let \(q_t\) be the probability of death during the year and \(p_t=1-q_t\) the probability of survival to time \(t+1\). Let \(B_t\) be the death benefit paid at time \(t+1\) if death occurs during the year. If the policy survives, let \(S_t\) be any survival benefit paid at time \(t+1\), and let \(V_{t+1}\) be the prospective reserve immediately after that payment.

All quantities are conditional on the policy being in force at time \(t\). The one-year probabilities satisfy \(p_t+q_t=1\). For the annual effective valuation rate \(i\), the discount factor for a payment one year later is \(v=1/(1+i)\). The cash-flow definition treats premiums as income and expenses and benefits as outgo.

$$ V_t=-P_t+E_t+v\left[q_t B_t+p_t(S_t+V_{t+1})\right]. $$

The negative sign on \(P_t\) reflects its status as insurer income, while \(E_t\) is an outgo. At the end of the year, the death benefit is paid only on death, so its expected amount is \(q_tB_t\). The survival benefit and next reserve occur only on survival, so their expected amount is \(p_t(S_t+V_{t+1})\).

Deriving the Recursion

The recursion follows by dividing the future cash flows in the prospective reserve into two parts: those at time \(t\), and those at time \(t+1\) or later. The time-\(t\) premium and expense need no discounting. Conditional on the policy’s status at \(t+1\), the expected end-of-year amount is the probability-weighted sum of death and survival outcomes.

1
Separate immediate cash flows.
The premium is received and the expense is paid at time \(t\), giving net outgo \(E_t-P_t\).
2
Form the end-of-year expectation.
Death, with probability \(q_t\), leads to payment \(B_t\). Survival, with probability \(p_t\), leads to payment \(S_t\) and the continuation reserve \(V_{t+1}\).
3
Discount one year.
Multiply the expected end-of-year amount by \(v\), then add the time-\(t\) net outgo to obtain \(V_t\).

To solve for the next reserve, first subtract the time-\(t\) cash flows and multiply by \(1+i\). Then remove the death-state amount and divide by the survival probability. This rearrangement requires \(p_t>0\).

$$ V_{t+1} = \frac{(1+i)(V_t+P_t-E_t)-q_tB_t}{p_t}-S_t. $$

This is the one-period reserve recursion under the stated timing. The survival benefit is subtracted after division because it is paid in the survival state alongside \(V_{t+1}\). If there is no survival benefit, set \(S_t=0\). If premiums or expenses are paid at a different time, or claims are paid at the moment of death rather than at year-end, their present values must reflect that timing; the displayed equation does not apply unchanged.

Worked Example: Solving for the Next Reserve

A hypothetical policy has \(V_t=\$4{,}000\), a premium of \(P_t=\$1{,}200\), and an expense of \(E_t=\$100\), all with the premium and expense occurring at the start of the year. The annual effective valuation rate is \(i=5\%\). The annual death probability is \(q_t=0.02\), and the end-of-year death benefit is \(B_t=\$20{,}000\). There is no survival benefit. Find \(V_{t+1}\).

Method. Use \(p_t=1-q_t=0.98\) and rearrange the prospective reserve equation for \(V_{t+1}\). The end-of-year death benefit contributes \(q_tB_t\), not the full benefit with certainty.

$$ V_{t+1} = \frac{(1.05)(4{,}000+1{,}200-100)-(0.02)(20{,}000)}{0.98}. $$

The amount inside the first parentheses is \(4{,}000+1{,}200-100=\$5{,}100\). Accumulating it one year gives \(1.05(5{,}100)=\$5{,}355\). The expected death benefit is \(0.02(20{,}000)=\$400\). Thus

$$ V_{t+1} = \frac{5{,}355-400}{0.98} = \frac{4{,}955}{0.98} = \$5{,}056.12. $$

Check. Substituting this value into the forward equation gives \(V_t=-1{,}200+100+[400+0.98(5{,}056.1224)]/1.05 =-1{,}100+5{,}355/1.05 =\$4{,}000\), as required. The result is rounded to the nearest cent. The common mistake avoided is treating the death benefit as certain instead of weighting it by the death probability.

Account Values and Benefits in the Recursion

In a universal life valuation, the account value can influence the amount of a future benefit. Its projection may depend on premiums allocated to the account, deductions from the account, and the credited rate. For example, under a simplified annual account convention, the account value might evolve according to \(A_{t+1}=(A_t+P_t-C_t)(1+j)\), where \(A_t\) is the account value at time \(t\), \(C_t\) is an account deduction, and \(j\) is the annual credited rate. This is an account projection rule, not a reserve recursion.

If the death benefit is defined using the account value, project the account under the policy terms and use the resulting benefit in the death state. Do not automatically add the account value as a separate insurer cash flow: it is not a payment merely because the insurer tracks it. It enters the reserve calculation when it determines an actual benefit, survival payment, or other cash flow specified by the contract. Also distinguish the credited rate \(j\) used in the account projection from the valuation rate \(i\) used to discount insurer cash flows; they need not be equal.

Worked Example: Account Projection and Reserve Recursion

A hypothetical policy has account value \(A_t=\$10{,}000\) and receives a premium of \(P_t=\$1{,}000\) at the start of the year. An account deduction of \(C_t=\$200\) is taken before the account earns the annual credited rate \(j=4\%\). The death benefit at year-end is the projected account value plus \(\$3{,}000\). For the reserve calculation, use \(i=5\%\), \(q_t=0.02\), \(p_t=0.98\), and \(E_t=\$100\). Suppose the prospective reserve at time \(t\) is \(V_t=\$771.085714\), and there is no survival benefit. Find the projected account value, death benefit, and \(V_{t+1}\).

Method. First project the account with the credited rate, then use the specified benefit definition. Finally, apply the reserve recursion using the valuation rate and the death probability.

The account amount before crediting is \(10{,}000+1{,}000-200=\$10{,}800\). Therefore,

$$ A_{t+1}=10{,}800(1.04)=\$11{,}232, \qquad B_t=11{,}232+3{,}000=\$14{,}232. $$

Now solve for the next reserve:

$$ V_{t+1} = \frac{1.05(771.085714+1{,}000-100)-0.02(14{,}232)}{0.98}. $$

The amount \(771.085714+1{,}000-100\) is \(\$1{,}671.085714\). After accumulation, \(1.05(1{,}671.085714)=\$1{,}754.64\). The expected death benefit is \(0.02(14{,}232)=\$284.64\). Hence

$$ V_{t+1} = \frac{1{,}754.64-284.64}{0.98} = \frac{1{,}470}{0.98} = \$1{,}500.00. $$

Check. The forward equation gives \(-1{,}000+100+(284.64+0.98(1{,}500))/1.05=-900+1{,}754.64/1.05=\$771.085714\), matching the stated \(V_t\). The account projection uses \(j=4\%\), while the reserve recursion uses \(i=5\%\); substituting one rate for the other would change the calculation.

Prospective Valuation Across More Than One Year

The recursion can be applied repeatedly because each next-period reserve represents the future net outgo conditional on survival to that period. The following example calculates a reserve forward from a terminal value of zero. That terminal value means there are no further insurer net cash flows after the second year; it is an assumption of this hypothetical contract, not a general property of universal life policies.

Worked Example: Linking Two Annual Reserves

A hypothetical two-year policy has annual valuation interest \(i=5\%\). At time 1, its premium is \(P_1=\$300\), its expense is \(E_1=\$50\), its death probability is \(q_1=0.03\), and its end-of-year death benefit is \(B_1=\$12{,}000\). There is no survival benefit and \(V_2=0\). At time 0, the premium is \(P_0=\$300\), the expense is \(E_0=\$50\), the death probability is \(q_0=0.02\), and the end-of-year death benefit is \(B_0=\$10{,}000\). Find \(V_1\) and \(V_0\).

Method. First use the one-year equation at time 1, then use that reserve as the survival-state continuation value at time 0.

At time 1, \(p_1=1-0.03=0.97\), but \(V_2=0\), so the survival term is zero. The premium less the expense gives \(-300+50=-\$250\). The expected death benefit is \(0.03(12{,}000)=\$360\). Thus

$$ V_1=-250+\frac{360+0.97(0)}{1.05} =-250+342.857143 =\$92.857143. $$

At time 0, \(p_0=1-0.02=0.98\). The expected survival continuation amount is \(0.98(92.857143)=\$91\), and the expected death benefit is \(0.02(10{,}000)=\$200\). Therefore,

$$ V_0=-250+\frac{200+91}{1.05} =-250+277.142857 =\$27.142857. $$

Check. Substitution into the time-0 recursion returns \(V_0=-250+291/1.05=\$27.142857\). Rounded to the nearest cent, \(V_1=\$92.86\) and \(V_0=\$27.14\). The common mistake avoided is omitting the survival probability on \(V_1\): the policy reaches that reserve only if it survives the first year.

Timing Traps and Interpretation

The recursion is an accounting identity for the stated valuation model, not a formula that can be used without checking timing and conditioning. A frequent error is to place a full death benefit in the expected end-of-year cash flow. In the first worked example, doing so would replace \(q_tB_t=\$400\) by \(B_t=\$20{,}000\). The resulting incorrect calculation would be

$$ V_{t+1}^{\text{wrong}} = \frac{5{,}355-20{,}000}{0.98} = -\$14{,}943.88, $$

which fails because the benefit is paid only on death. The correct death-state contribution is \(0.02(20{,}000)=\$400\), giving \(V_{t+1}=\$5{,}056.12\). Other common errors include discounting a time-\(t\) premium as though it were received at year-end, using the credited rate in place of the valuation rate without justification, and treating an account balance as an insurer payment when no benefit is paid.

The recursion is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. For any application, write down the policy status condition, the payment timing, the death and survival probabilities, and the account-based benefit definition before substituting numbers.

Check Your Understanding

For each problem, use the stated annual cash-flow timing and round monetary answers to the nearest cent where needed.

  1. A policy has \(V_t=\$800\), \(P_t=\$500\), \(E_t=\$40\), \(i=4\%\), \(q_t=0.01\), and \(B_t=\$10{,}000\), with no survival benefit. Find \(V_{t+1}\). (Answer: \(\$1{,}222.63\))
  2. A policy has \(i=5\%\), \(q_t=0.02\), \(B_t=\$15{,}000\), \(V_{t+1}=\$700\), \(P_t=\$400\), \(E_t=\$60\), and no survival benefit. Find \(V_t\). (Answer: \(\$598.10\))
  3. A policy has \(A_t=\$8{,}500\), \(P_t=\$1{,}200\), an account deduction \(C_t=\$250\), and credited rate \(j=3\%\). Its year-end death benefit is the projected account value plus \(\$2{,}266.50\). If \(V_t=\$1{,}000\), \(E_t=\$80\), \(i=4\%\), and \(q_t=0.015\), find the projected account value, death benefit, and \(V_{t+1}\). (Answer: account value \(\$9{,}733.50\), death benefit \(\$12{,}000.00\), and \(V_{t+1}=\$2{,}055.63\))
  4. A two-year policy has \(i=4\%\), \(V_2=0\), and no survival benefits. At time 1, \(P_1=\$250\), \(E_1=\$30\), \(q_1=0.05\), and \(B_1=\$8{,}000\). At time 0, \(P_0=\$250\), \(E_0=\$30\), \(q_0=0.02\), and \(B_0=\$6{,}000\). Find \(V_1\) and \(V_0\). (Answer: \(V_1=\$164.62\) and \(V_0=\$50.50\))
  5. In a one-year model, a premium is received at the start of the year, an expense is paid at the start, and a benefit is paid at year-end only if the policyholder dies. If the death probability is \(0.04\) and the benefit is \(\$25{,}000\), what amount belongs in the expected year-end death-benefit term? (Answer: \(0.04(\$25{,}000)=\$1{,}000\))
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