Why the Term Boundary Matters
A term insurance value depends on how many years of coverage remain. Each year of coverage creates a possible death benefit; when no years remain, there is no covered period in which a benefit can be paid. That endpoint is essential when using the one-year recursion: it determines where the repeated calculation stops and prevents an extra year of coverage from entering the result.
This tutorial considers a discrete, unit-benefit term insurance. The benefit is paid at the end of the year of death if death occurs within the stated term. It does not pay a maturity benefit to someone who survives the term. The recursion and boundary below are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
For each year, \(q_x\) is the probability that a life aged \(x\) dies within one year, and \(p_x=1-q_x\) is the probability of surviving that year. The annual effective interest rate is \(i>-1\), and \(v=1/(1+i)\) is the one-year discount factor. The mortality probabilities used at each age are conditional on being alive at that age.
The definition assumes level unit coverage during the term, death benefits paid at the end of the year of death, and no payment for survival to the end of the term. The recursion below is for this benefit timing. A continuous benefit, a different payment time, or a survival benefit would require a different present-value expression.
Deriving the One-Year Recursion
For a term of \(n\geq 1\) years, consider the first year separately. Death during that year has probability \(q_x\) and produces a unit benefit at the end of the year, with present value \(v\). Survival has probability \(p_x\). After survival, the life is aged \(x+1\), has \(n-1\) years of coverage remaining, and the value at that time is \(A^1_{x+1:n-1}\). Its value at age \(x\) is discounted by one year.
This is a conditioning argument: the first term is the expected discounted benefit if death occurs in year one, while the second term is the probability of reaching the remaining term multiplied by its discounted value. When \(n=1\), the surviving policy has zero years left. Substituting the boundary makes the one-year value \(v q_x\), as it must.
The boundary \(A^1_{x:0}=0\) does not say that survival for zero years is impossible or that mortality at age \(x\) is zero. It says that a contract with no remaining coverage makes no payment. In particular, the boundary is not \(v q_x\): that is the value of a one-year term, not a zero-year term.
Record the integer number \(n\) of years of coverage at the current age.
If \(n=0\), use \(A^1_{x:0}=0\). Do not apply the positive-term recursion at zero years.
For \(n\geq1\), use \(v(q_x+p_xA^1_{x+1:n-1})\), reducing the remaining term by one after survival.
The first-year death benefit is discounted once. A benefit in later years is discounted for each year from issue to payment.
Worked Calculations
Worked Example: Building a Three-Year Value from the Boundary
A hypothetical policy pays 1 at the end of the year of death if death occurs within three years. The annual effective interest rate is \(i=25\%\), and the one-year death probabilities at ages \(x\), \(x+1\), and \(x+2\) are \(0.10\), \(0.20\), and \(0.30\). Calculate \(A^1_{x:3}\). Use the recursion, then check by enumerating the possible payment years.
Method: work from the zero-term boundary. The discount factor is \(v=1/1.25=0.8\), and the survival probabilities are \(p_x=0.90\), \(p_{x+1}=0.80\), and \(p_{x+2}=0.70\).
First calculate the one-year value at age \(x+2\). Its remaining term after survival is zero:
Next calculate the two-year value at age \(x+1\):
Then calculate the three-year value at age \(x\):
To check, the probabilities of death in years one, two, and three are \(0.10\), \(0.90(0.20)=0.18\), and \(0.90(0.80)(0.30)=0.216\). The corresponding present values are \(0.8(0.10)=0.08\), \(0.8^2(0.18)=0.1152\), and \(0.8^3(0.216)=0.110592\). Their sum is \(0.305792\), matching the recursion. Rounded to four decimal places, \(A^1_{x:3}=0.3058\). The common mistake avoided here is stopping at the last listed \(q\) without discounting that final year's possible payment.
Worked Example: A Two-Year Term and Its Zero-Year Remainder
A hypothetical unit-benefit term insurance has \(i=10\%\), \(q_x=0.10\), and \(q_{x+1}=0.20\). Calculate its two-year value. Show explicitly how the zero-year boundary enters, and verify the result by separating the two possible payment years.
Method: calculate the final covered year first. Here \(v=1/1.10=10/11\), while \(p_x=0.90\) and \(p_{x+1}=0.80\). The one-year term at age \(x+1\) has a zero-year remainder:
Use that one-year value in the two-year recursion at age \(x\):
For the direct check, the year-one contribution is \((10/11)(0.10)=1/11\). The probability of death in year two is \(0.90(0.20)=0.18\), so its contribution is \((10/11)^2(0.18)=18/121\). Thus \(1/11+18/121=11/121+18/121=29/121\), as above. Rounded to four decimal places, the value is \(0.2397\). The mistake avoided is treating the two-year contract as if it paid only for death in the first year, or treating the zero-year remainder as another covered year.
Worked Example: Repeated Mortality with a Discount-Factor Trap
A hypothetical three-year term has a unit death benefit, \(i=25\%\), and a constant one-year death probability of \(0.20\) at each covered age. Calculate its value by recursion and check the result by payment year.
Method: apply the recursion at each remaining duration. The discount factor is \(v=0.8\), the survival probability is \(p=0.8\), and the zero-year value is \(A^1_{x+3:0}=0\). The one-year value at age \(x+2\) is
The two-year value at age \(x+1\) is
The three-year value at age \(x\) is
The probabilities of death in years one, two, and three are \(0.20\), \(0.80(0.20)=0.16\), and \(0.80^2(0.20)=0.128\). Discounting each payment to issue gives \(0.8(0.20)=0.16\), \(0.8^2(0.16)=0.1024\), and \(0.8^3(0.128)=0.065536\). The sum is \(0.327936\). Rounded to four decimal places, the value is \(0.3279\). The check catches a frequent error: discounting the later payment by only one year instead of by its full time from issue.
Interpreting and Checking the Result
For a unit-benefit term insurance, the recursion is a weighted expected value of two first-year outcomes: a death benefit and a surviving life with a shorter term. Each iteration moves one year forward in age and removes one year of coverage. This is why the age and duration indices must change together. After survival through the first year, the continuation is \(A^1_{x+1:n-1}\), not \(A^1_{x+1:n}\).
If \(i\geq0\), then \(0<v\leq1\). Since a unit benefit is paid at most once and is discounted by at least one year, its present value is between zero and one, so \(0\leq A^1_{x:n}\leq1\). This upper bound depends on the nonnegative-interest assumption. If \(-1<i<0\), then \(v>1\), and a unit benefit paid in the future can have present value greater than one. Nonnegativity of the value still follows from nonnegative benefits and probabilities, but the upper bound of one does not follow without \(i\geq0\).
A reliable calculation check is to write the death probability in each covered year and multiply it by the discount factor for that payment year. Under the stated model, death in year \(k\) requires survival through the previous \(k-1\) years and death in the next year. Its probability is \({}_{k-1}p_x q_{x+k-1}\), and its present-value contribution is \(v^k{}_{k-1}p_xq_{x+k-1}\). Adding these contributions for \(k=1,\ldots,n\) checks the recursive calculation without adding a survival payment at maturity.
Check Your Understanding
For each problem, use the specified payment timing and zero-year boundary. Probabilities are given as annual conditional death probabilities.
- A unit-benefit two-year term has \(i=0\), \(q_x=0.12\), and \(q_{x+1}=0.25\). Calculate its value using the one-year recursion. (Answer: \(0.3400\))
- A unit-benefit two-year term has \(i=25\%\), \(q_x=0.20\), and \(q_{x+1}=0.30\). Find the value, using the zero-year boundary for the final remaining term. (Answer: \(0.3520\))
- A unit-benefit three-year term has \(i=10\%\) and death probabilities \(0.10\), \(0.20\), and \(0.30\) at ages \(x\), \(x+1\), and \(x+2\). Calculate its value, rounding to four decimal places. (Answer: \(0.4020\))
- A unit-benefit two-year term has \(i=4\%\), \(q_x=0.07\), and \(q_{x+1}=0.11\). Calculate its value by recursion, starting with the one-year value at age \(x+1\), and round to four decimal places. (Answer: \(0.1619\))
- A one-year unit-benefit term has \(i=25\%\) and \(q_x=0.16\). Calculate its value, then state the value of the same contract with zero years of coverage. (Answer: \(A^1_{x:1}=0.1280\) and \(A^1_{x:0}=0\))