Why a One-Year Recursion Works
A whole-life insurance value includes payments that may occur many years in the future. Yet its value at age \(x\) can be related to a single year of mortality and the value at age \(x+1\). The key is to separate two possibilities during the first year: death, which triggers a benefit, or survival, which leaves the insurance in force.
The recursion is useful because it expresses a long-term expected present value in terms of one-year probabilities and a value at the next age. Its meaning depends on careful attention to when the benefit is paid and when the continuation value is measured.
The benefit is one dollar, but \(A_x\) is a present value, not a probability. It accounts for both the chance that a payment is made and the discounting of that payment. The same definition applies at age \(x+1\): \(A_{x+1}\) measures the value at that age of the continuing whole-life insurance on the same benefit basis.
For the standard recursion, interest is earned at a fixed effective annual rate, mortality is described by the one-year probabilities at each attained age, and the coverage continues after the first year if the insured survives. The death benefit is paid at the end of the year of death. The formula below does not apply unchanged to a benefit payable immediately on death, a contract that ends after one year, or a first-year benefit amount different from the continuing benefit.
Deriving the Recursion by Conditioning
Condition on whether the insured dies during the first year or survives to age \(x+1\). In the death branch, the unit benefit is paid at time 1, so its present value at age \(x\) is \(v\). In the survival branch, the insurance continues and has expected present value \(A_{x+1}\) at time 1. Discount that continuation value one year to obtain its value at age \(x\).
The first expression is the probability-weighted value of each branch. The death branch has probability \(q_x\) and present value \(v\); the survival branch has probability \(p_x\) and present value \(vA_{x+1}\). Factoring out the common one-year discount factor gives the familiar form.
This is an application of conditional expectation: the expected present value is the probability of each event multiplied by the expected present value conditional on that event, summed over the events. The two branches are mutually exclusive and exhaustive because, by the end of the year, the insured has either died or survived.
Death within the year has probability \(q_x\); survival to age \(x+1\) has probability \(p_x\).
The death benefit is one unit, and the continuation value after survival is \(A_{x+1}\).
Each is one year in the future, so each is multiplied by \(v\).
Multiply by the relevant first-year probabilities and obtain \(A_x=v(q_x+p_xA_{x+1})\).
The factor \(v\) applies to the entire sum. The term \(q_x\) represents a benefit at the end of year 1, while \(p_xA_{x+1}\) represents a value located at time 1 after survival. Both are one year ahead of the valuation time at age \(x\).
Worked Example: Applying the Two First-Year Branches
A hypothetical whole-life contract pays one dollar at the end of the year of death. At age \(x\), the one-year death probability is \(q_x=0.02\), and the value of the continuing insurance at age \(x+1\) is \(A_{x+1}=0.30\). The effective annual interest rate is \(i=5\%\). Find \(A_x\), showing the value contributed by each branch.
Method and assumptions. Use first-year conditioning for a unit end-of-year-of-death benefit. The survival probability is \(p_x=1-q_x=1-0.02=0.98\), and the discount factor is \(v=1/(1.05)=0.95238095\).
Death branch. The probability-weighted value at time 1 is \(0.02(1)=0.020\). At age \(x\), its present value contribution is \(0.020(0.95238095)=0.01904762\).
Survival branch. The probability-weighted continuation value at time 1 is \(0.98(0.30)=0.294\). Discounting it to age \(x\) gives \(0.294(0.95238095)=0.28000000\).
Adding the contributions gives \(A_x=0.01904762+0.28000000=0.29904762\). Equivalently, \(A_x=(0.020+0.294)/1.05=0.314/1.05=0.29904762\). Rounded to four decimal places, \(A_x=0.2990\).
Check. The result is below the undiscounted time-1 expected value \(0.314\), as it should be at a positive interest rate. The calculation avoids the common mistake of discounting the first-year death benefit but not the continuation value.
Using Life-Table Counts
The one-year probabilities can be obtained from a life table. If \(\ell_x\) lives are alive at exact age \(x\) and \(\ell_{x+1}\) are alive at age \(x+1\), then \({}_1p_x=\ell_{x+1}/\ell_x\) and \({}_1q_x=(\ell_x-\ell_{x+1})/\ell_x\), assuming the table counts describe the same cohort over that year. Once these probabilities are found, they enter the recursion in exactly the same way as directly provided probabilities.
Worked Example: Converting Life-Table Counts to Probabilities
A hypothetical life table has \(\ell_x=5{,}000\) and \(\ell_{x+1}=4{,}900\). For a unit whole-life benefit payable at the end of the year of death, suppose \(A_{x+1}=0.35\) and \(i=4\%\). Calculate \(A_x\), rounding to four decimal places.
Method and assumptions. First convert the life-table counts to one-year probabilities. Then apply the whole-life recursion, with fixed effective annual interest and a benefit paid at the end of the year of death.
The number dying between the two exact ages is \(5{,}000-4{,}900=100\). Therefore \(q_x=100/5{,}000=0.02\), while \(p_x=4{,}900/5{,}000=0.98\). These sum to 1.
The discount factor is \(v=1/(1.04)=0.96153846\). The survival-weighted continuation value at time 1 is \(p_xA_{x+1}=0.98(0.35)=0.343\). Adding the death probability gives \(q_x+p_xA_{x+1}=0.020+0.343=0.363\).
Thus \(A_x=0.363/1.04=0.34903846\), so the answer is \(A_x=0.3490\).
Check. The probability inputs add to 1, and the discounted value \(0.3490\) is less than the time-1 weighted amount \(0.363\). A common mistake is to use \(100/4{,}900\) as the death probability; the denominator must be the number alive at the start of the year, \(\ell_x\).
Interpreting the Discounting Carefully
The recursion can also be read as a two-stage valuation. First calculate the expected amount, measured at the end of the first year, using the two possible first-year outcomes. Then discount that amount one year. In symbols, the time-1 expected value is \(q_x+p_xA_{x+1}\), and its value at age \(x\) is \(v(q_x+p_xA_{x+1})\).
This interpretation helps distinguish two quantities that are easy to confuse. The continuation value \(A_{x+1}\) is already a present value at age \(x+1\); it is not yet a present value at age \(x\). Therefore it needs one additional year of discounting when used in the age-\(x\) calculation. Likewise, the first-year death benefit is paid at time 1, not at time 0, so it too is discounted once.
Worked Example: Checking the Timing of the Continuation Value
A hypothetical whole-life policy has \(q_x=0.025\), \(p_x=0.975\), \(A_{x+1}=0.60\), and an effective annual interest rate of \(i=4\%\). The benefit is one dollar at the end of the year of death. Calculate \(A_x\) by separating the discounted branch contributions, then identify the error in a calculation that leaves the continuation value undiscounted.
Method and assumptions. Both branch values are measured at time 1 before they are discounted to age \(x\). The discount factor is \(v=1/1.04=0.96153846\).
The death-branch contribution is \(q_xv=0.025(0.96153846)=0.02403846\). The survival-branch contribution is \(p_xA_{x+1}v=0.975(0.60)(0.96153846)\). First, \(0.975(0.60)=0.585\), and then \(0.585(0.96153846)=0.56250000\).
Adding the two present value contributions gives \(A_x=0.02403846+0.56250000=0.58653846\). Equivalently, the time-1 weighted amount is \(0.025+0.585=0.610\), and \(0.610/1.04=0.58653846\). Rounded to four decimal places, \(A_x=0.5865\).
Common mistake and correction. Leaving the continuation value undiscounted gives \(0.025v+0.975(0.60)=0.02403846+0.585=0.60903846\), which incorrectly combines a time-0 amount with a time-1 amount. The continuation value is located at age \(x+1\), so it must be multiplied by \(v\) just like the first-year death benefit. The corrected value is \(0.5865\).
Check. The probability-weighted amount at time 1 is \(0.610\); discounting it at a positive rate produces \(0.5865\), a smaller value.
Common Errors and When the Formula Applies
The most reliable way to avoid timing errors is to label the valuation time for every amount before adding it. In this recursion, \(A_x\) is measured at age \(x\), whereas the death payment and the continuation value after survival are both valued at time 1. The first-year probabilities determine how often each time-1 outcome occurs; they do not remove the need to discount.
- Omitting the one-year discount. The expression \(q_x+p_xA_{x+1}\) is the expected amount at time 1, not the value at age \(x\). Multiply the entire expression by \(v\).
- Discounting only one branch. Both outcomes are valued at time 1. The correct branch values are \(vq_x\) and \(vp_xA_{x+1}\).
- Using the wrong survival probability. The survival branch has probability \(p_x=1-q_x\), not \(q_x\). If using life-table counts, divide survivors and deaths by \(\ell_x\), the starting number alive.
- Changing the benefit timing without changing the model. A benefit paid immediately on death is not represented by the same first-year benefit value \(v\); this recursion is for payment at the end of the year of death.
- Treating the contract as if it ends after one year. The term \(A_{x+1}\) appears because the whole-life coverage continues after survival. If coverage ends at year 1, the survival branch does not have that whole-life continuation value.
The recursion is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because it tests the connection between mortality probabilities, benefit timing, discounting, and whole-life present values. The essential discipline is to keep the probability, payment time, and valuation time distinct at every step.
Check Your Understanding
For each problem, assume a unit whole-life benefit payable at the end of the year of death, unless stated otherwise. Give present values to four decimal places.
- A hypothetical contract has \(i=3\%\), \(q_x=0.01\), and \(A_{x+1}=0.20\). Find \(p_x\) and \(A_x\). (Answer: \(p_x=0.99,\ A_x=0.2019\))
- A life table gives \(\ell_x=5{,}000\) and \(\ell_{x+1}=4{,}900\). If \(i=5\%\) and \(A_{x+1}=0.35\), calculate \(q_x\), \(p_x\), and \(A_x\). (Answer: \(q_x=0.0200,\ p_x=0.9800,\ A_x=0.3457\))
- For a hypothetical insured, \(q_x=0.06\), \(A_{x+1}=0.40\), and \(i=4\%\). Find the first-year death and survival branch contributions to \(A_x\), then find \(A_x\). (Answer: death contribution \(0.0577\), survival contribution \(0.3615\), \(A_x=0.4192\))
- A calculation uses \(q_x=0.03\), \(p_x=0.97\), \(A_{x+1}=0.50\), and \(i=6\%\), but discounts only the death branch. What value does that incorrect method give, and what is the correct \(A_x\)? (Answer: incorrect value \(0.5133\); correct \(A_x=0.4858\))
- A hypothetical whole-life contract has \(q_x=0.025\), \(A_{x+1}=0.60\), and \(i=2\%\). Calculate \(p_x\), the weighted amount at time 1, and \(A_x\). (Answer: \(p_x=0.975,\) time-1 amount \(0.6100,\ A_x=0.5980\))