Why Calculate Term Insurance Backward?
A term insurance value depends on the possible death years during the coverage period. Directly adding the contribution from every year works, but it can become cumbersome when a problem asks for several durations or when mortality and interest vary by age. A backward calculation organizes the work: first value the final year of coverage, then use that value to calculate the preceding year, continuing until the required starting age.
The method relies on a simple boundary condition. With no years of coverage remaining, there is no future death benefit, so the insurance value is zero. This zero-duration value anchors the recursion and prevents a common error: treating survival to the end of the term as if it generated a benefit.
This definition uses annual, discrete benefits: a death during a policy year produces payment at that year’s end. The mortality and interest assumptions apply to each successive year of the contract. Interest rates are deterministic, and \(q_x\) is the probability of death during the year conditional on being alive at age \(x\). If a contract instead pays continuously at death, pays on survival to maturity, or has a different benefit amount, this particular recursion does not describe its value without modification.
The Backward Recursion
For \(k\geq 1\), separate the possibilities during the first year. Death during that year has probability \(q_x\) and produces a unit payment one year from the valuation date. Survival has probability \(p_x\); conditional on survival, the remaining coverage is a \(k-1\)-year term insurance beginning at age \(x+1\). Both outcomes are valued at age \(x\).
This is a backward recursion because the value for \(k\) years at age \(x\) is calculated from the value for \(k-1\) years at age \(x+1\). To find an \(n\)-year value at age \(x\), start with the zero-duration value at age \(x+n\), then work back to age \(x\). The subscript \(k\) always means years remaining; it is not the attained age.
At the end of the coverage period, assign \(T_{x+n,0}=0\). There is no term insurance benefit after the term has ended.
For one year remaining, \(T_{x+n-1,1}=v_{x+n-1}q_{x+n-1}\), since the continuation value is zero.
At each step, use the mortality and interest assumptions for that year, together with the value one age later and one year shorter.
The recursion follows directly from conditioning on first-year survival. The first-year death contribution is \(v_xq_x\): the benefit is paid one year later, so it is discounted by \(v_x\). The survival contribution is \(v_xp_xT_{x+1,k-1}\): the insured must survive the year, and the continuation value is also one year away. Adding these contributions gives the stated formula. In particular, the discount factor multiplies both terms because both are values at age \(x+1\) brought back one year.
Worked Calculations
Each calculation below works from the final covered year toward the initial age. Intermediate values are kept unrounded so that rounding does not accumulate across recursion steps.
Worked Example: Three-Year Term Insurance With Varying Assumptions
A hypothetical policy provides a unit benefit at the end of the year of death during the three years after age 40. The mortality probabilities are \(q_{40}=0.08\), \(q_{41}=0.12\), and \(q_{42}=0.20\). The effective annual interest rates for those years are \(i_{40}=0.04\), \(i_{41}=0.05\), and \(i_{42}=0.06\). Calculate the value at age 40.
Method. Use the zero-duration boundary at age 43 and apply the age-specific recursion in order from age 42 back to age 40. The common mistake to avoid is using one interest rate for all three years.
At age 43 there is no coverage remaining, so \(T_{43,0}=0\). For the year from age 42 to 43, \(p_{42}=1-0.20=0.80\), and \(v_{42}=1/1.06\). Therefore,
For the year from age 41 to 42, \(p_{41}=1-0.12=0.88\), and \(v_{41}=1/1.05\):
For the year from age 40 to 41, \(p_{40}=1-0.08=0.92\), and \(v_{40}=1/1.04\):
The value is \(0.3179\) per unit of benefit when rounded to four decimal places. As a check, the death probabilities in the first, second, and third years are \(0.08\), \(0.92(0.12)=0.1104\), and \(0.92(0.88)(0.20)=0.16192\), respectively. Discounting each using its own sequence of annual rates gives the same value, to the displayed precision.
Worked Example: Building a Four-Year Value From the Terminal Year
A hypothetical four-year term policy starts at age 60. The annual death probabilities at ages 60 through 63 are \(0.10, 0.15, 0.20,\) and \(0.25\). The effective annual interest rate is \(0.05\) in every year. Find the value at age 60 and show the backward calculation.
Method. Begin with zero remaining coverage at age 64, then calculate the one-, two-, three-, and four-year values. The common mistake to avoid is beginning at age 60 and treating the future value at age 61 as already known.
At age 63, one year remains. Since \(v=1/1.05\),
At age 62, two years remain and \(p_{62}=0.80\):
At age 61, three years remain and \(p_{61}=0.85\):
Finally, at age 60, four years remain and \(p_{60}=0.90\):
The four-year value is \(0.4757\) per unit of benefit, rounded to four decimal places. For a check, the probabilities of death in years 1 through 4 are \(0.10\), \(0.90(0.15)=0.135\), \(0.90(0.85)(0.20)=0.153\), and \(0.90(0.85)(0.80)(0.25)=0.153\). Discounting those four contributions by \(v,v^2,v^3,v^4\), respectively, gives \(0.4757\) to four decimal places.
Worked Example: Avoiding a Single-Rate Discounting Error
A hypothetical two-year term policy begins at age 70. The first-year death probability is \(q_{70}=0.10\), the second-year probability is \(q_{71}=0.20\), and the effective annual rates are \(i_{70}=0.10\) and \(i_{71}=0.20\). Calculate the value at age 70.
Method. Apply the backward recursion with the rate for each policy year. The trap is to discount both possible death benefits using the first-year rate, even though the second-year rate differs.
For the final covered year, \(T_{71,1}=0.20/1.20=0.1666666667\). In the first year, \(p_{70}=0.90\) and \(v_{70}=1/1.10\), so
The incorrect single-rate calculation would use \(1.10\) for both years:
That result is wrong under the stated assumptions because the second-year discount factor is \(1/1.20\), not \(1/1.10\). The correct value is \(0.2273\) per unit of benefit, rounded to four decimal places. The backward result also equals \(0.10/1.10+0.90(0.20)/(1.10\cdot1.20)\), which checks the use of both annual rates.
Common Errors and Practical Checks
Backward calculation is systematic, but only if the age, duration, mortality, and interest indices remain aligned. At a step for age \(y\) with \(k\) years remaining, the continuation value must be \(T_{y+1,k-1}\), not a value at the same age or with the same duration. The mortality probability and discount factor belong to the year from age \(y\) to age \(y+1\).
- Giving a maturity benefit by mistake. For pure term insurance, survival through the whole term generates no benefit. The terminal condition is zero, not one.
- Discounting only the death branch. The continuation value is at age \(x+1\), so it too must be discounted one year when expressed at age \(x\).
- Using the wrong year’s assumptions. If \(i_x\) varies by age, use \(v_x\) at that recursion step. Do not replace the schedule with a single rate unless the problem states that rates are constant.
- Mixing up age and duration. A value with two years remaining at age \(x+1\) is \(T_{x+1,2}\); it is not the value at age \(x+1\) with one year remaining unless the original duration was two years.
- Rounding too early. Retain several guard digits in each continuation value and round only the final result, as in the worked examples.
A useful final check is to expand the recursion into its death-year contributions. Each term should use the probability of surviving to that year and then dying during it, with all preceding annual discount factors applied. This check can reveal a misplaced survival probability or a missing discount factor without changing the backward method.
Check Your Understanding
For each hypothetical policy, use a unit death benefit payable at the end of the year of death, and give the final value to four decimal places.
- A two-year policy starts at age 30. Let \(q_{30}=0.10\), \(q_{31}=0.20\), \(i_{30}=0.05\), and \(i_{31}=0.10\). Find the value at age 30. (Answer: \(0.2511\))
- A three-year policy starts at age 40. The death probabilities at ages 40, 41, and 42 are \(0.05, 0.10,\) and \(0.15\). The annual effective rate is \(0.05\) throughout. Find the value at age 40. (Answer: \(0.2446\))
- A two-year policy starts at age 50. Let \(q_{50}=0.08\), \(q_{51}=0.12\), \(i_{50}=0.04\), and \(i_{51}=0.06\). Find the value at age 50. (Answer: \(0.1771\))
- A three-year policy starts at age 60. The death probabilities are \(0.20, 0.25,\) and \(0.30\) at ages 60, 61, and 62. The annual effective rate is \(0.10\) throughout. Find the value at age 60. (Answer: \(0.4823\))
- A three-year policy starts at age 70. The death probabilities are \(0.10, 0.15,\) and \(0.20\) at ages 70, 71, and 72. The effective annual rates for those years are \(0.05, 0.04,\) and \(0.03\). Find the value at age 70. (Answer: \(0.3549\))
This backward-recursion approach is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.