Why the Select-to-Ultimate Boundary Matters
A select mortality table distinguishes lives by both age at selection and time since selection. During a specified initial period, mortality depends on the age at which the life was selected and the policy duration. After that period, the calculation uses ultimate mortality, which depends on attained age. A contract lasting across this boundary therefore cannot be valued by applying one mortality rate throughout: the correct rate changes with policy duration.
The essential bookkeeping is to follow one life from issue. If a policy is issued at age \(x\), its first policy year uses the select rate for duration 0. Its next year uses the select rate for duration 1, if that duration is still within the select period. Once the select period ends, use the ultimate rate for the attained age at the start of each remaining year. Surviving the select period does not start a new select period.
For each applicable annual rate, define the corresponding one-year survival probability as \(p=1-q\). For a policy issued at age \(x\), let \({}_tp_{[x]}\) denote the probability of surviving \(t\) complete years from issue. At integer durations it is the product of the appropriate one-year survival probabilities. The duration index matters: the ultimate rate used after the select period is based on attained age \(x+s\), not on the original select age alone.
| Policy year beginning at duration \(s\) | Mortality rate used | Reason |
|---|---|---|
| \(0\le s<r\) | \(q_{[x]+s}\) | The life is still in the select period. |
| \(s\ge r\) | \(q_{x+s}\) | The life has reached the ultimate period; attained age is \(x+s\). |
Survival Across the Boundary
For an integer \(t\ge 0\), the survival probability from issue is
When \(t=0\), the empty product is 1. If \(t\le r\), every factor is select-period survival. If \(t>r\), the first \(r\) factors are select-period survival and the remaining factors use ultimate mortality. For someone who has already survived to duration \(d\), the probability of surviving a further \(t\) years is conditional on being alive at \(d\). It is the product of rates applicable at durations \(d,d+1,\ldots,d+t-1\), rather than a fresh calculation from duration 0.
This result follows from conditional probability. Death in year \(k\) requires two events: survival to the beginning of that year, followed by death during that year. Conditional on being alive then, the applicable annual death probability is \(q_k\). Multiplying the probability of reaching the year by this conditional probability gives the unconditional probability of death in that year.
Worked Examples
Worked Example: Four-Year Survival Through the Boundary
A hypothetical policy is issued at age 40. The select period is 2 years, with \(q_{[40]}=0.01\) and \(q_{[40]+1}=0.02\). Ultimate rates are \(q_{42}=0.03\) and \(q_{43}=0.04\). Find the probability of surviving four years from issue.
Method. Identify the rate at the start of each policy year. Durations 0 and 1 are select durations. Durations 2 and 3 use ultimate rates at attained ages 42 and 43, respectively. The annual survival probabilities are therefore \(0.99\), \(0.98\), \(0.97\), and \(0.96\).
First, \(0.99(0.98)=0.9702\). Next, \(0.9702(0.97)=0.941094\). Finally, \(0.941094(0.96)=0.90345024\). The probability of surviving four years is \(0.9035\) to four decimal places.
Check. The probability is below each of the individual annual survival probabilities, as it must be. The calculation uses two select rates followed by two ultimate rates.
Common mistake avoided. Repeating the first select rate for all four years would give \(0.99^4=0.96059601\), which incorrectly treats later policy years as if the life were still at duration 0. The correct probability is \(0.90345024\).
Worked Example: Present Value of Death Benefits Across the Boundary
A hypothetical four-year term contract pays \$100,000 at the end of the policy year of death, if death occurs during the four years. The insured is age 50 at issue, the select period is 2 years, and the annual rates are \(q_{[50]}=0.02\), \(q_{[50]+1}=0.04\), \(q_{52}=0.06\), and \(q_{53}=0.08\). Assume an effective annual interest rate of \(i=0.05\), with payment at each year-end. Find the expected present value of the death benefit.
Method. Find the probability of death in each policy year, multiply each probability by the benefit discounted to that year-end, and add the four expected present values. The discount factor is \(v=1/(1+i)=1/1.05=0.9523809524\).
The year 1 death probability is \(0.02\). The year 2 probability is \(0.98(0.04)=0.0392\). The survival probability to year 3 is \(0.98(0.96)=0.9408\), so the year 3 death probability is \(0.9408(0.06)=0.056448\). Survival to year 4 is \(0.9408(0.94)=0.884352\), so the year 4 death probability is \(0.884352(0.08)=0.07074816\).
The discount factors are \(v=0.9523809524\), \(v^2=0.9070294785\), \(v^3=0.8638375985\), and \(v^4=0.8227024748\). The present-value contributions, before rounding, are approximately \$1,904.7619, \$3,555.5556, \$4,876.1905, and \$5,820.4688. Their sum is \$16,156.9768, so the expected present value is \$16,156.98, rounded to the nearest cent.
Check. The four death probabilities sum to \(0.18639616\), which is less than 1 because the contract ends after four years and some lives survive to maturity without a death benefit. The expected present value is also below the undiscounted expected death payment, \(100{,}000(0.18639616)=\$18{,}639.616\), as expected at a positive interest rate.
Common mistake avoided. Using ultimate rates from the first year would misclassify the first two policy years. The correct sequence is select rates \(0.02,0.04\), then ultimate rates \(0.06,0.08\), applied only after survival to the start of each year.
Worked Example: Valuation Conditional on Reaching a Policy Duration
A hypothetical insured entered a contract at age 60 and has survived to duration 1. The select period is 3 years. The remaining select rates are \(q_{[60]+1}=0.05\) and \(q_{[60]+2}=0.07\); the next two ultimate rates are \(q_{63}=0.08\) and \(q_{64}=0.10\). The contract pays \$20,000 at the end of the next policy year if death occurs in that year, and likewise at the end of the following two years. Assume \(i=0.04\). Find the expected present value at duration 1 of these three possible payments.
Method. Because the insured is already at duration 1, use the rates for durations 1, 2, and 3. The first two are still select rates; duration 3 is the first ultimate duration, at attained age 63. Discount payments one, two, and three years from the valuation date. Here \(v=1/1.04=0.9615384615\).
The conditional probability of death in the first future year is \(0.05\). The probability of surviving that year and dying in the second is \(0.95(0.07)=0.0665\). The probability of surviving the first two years and dying in the third is \(0.95(0.93)(0.08)=0.07068\).
The discount factors are \(v=0.9615384615\), \(v^2=0.9245562130\), and \(v^3=0.8889963587\). The three discounted probability factors are approximately \(0.0480769231\), \(0.0614829882\), and \(0.0628318626\). Their sum is \(0.1723917739\). Multiplying by \$20,000 gives \$3,447.8355, so the expected present value is \$3,447.84, rounded to the nearest cent.
Check. The probability of death during these three future years is \(0.05+0.0665+0.07068=0.18718\). This is consistent with the three-year survival probability \(0.95(0.93)(0.92)=0.81282\), since \(1-0.81282=0.18718\). The present value is less than the undiscounted expected payment of \$3,743.60.
Common mistake avoided. Starting again at duration 0 would incorrectly apply \(q_{[60]}\). The valuation is conditional on having reached duration 1, so its first applicable mortality rate is \(q_{[60]+1}\).
Turning Mortality into an Expected Present Value
For a contract paying a specified amount \(B_k\) at the end of year \(k\) if death occurs in that year, the expected present value at issue is
This formula assumes annual mortality probabilities, a constant effective annual interest rate \(i\), and benefits paid at the end of the policy year of death. It follows by summing the discounted benefit over mutually exclusive death-year events: the probability of the event in year \(k\) is \({}_{k-1}p_{[x]}q_k\), and the payment at time \(k\) has present value \(v^kB_k\). It does not apply unchanged when benefits are paid immediately at death, at another time within the year, or under a different interest model; the payment timing and discount factor must then match the contract.
If a contract also makes a payment at a fixed maturity time \(n\) provided the insured is alive then, that survival-contingent payment contributes \(v^n C\,{}_np_{[x]}\), where \(C\) is its amount. This term uses the same select-to-ultimate survival calculation. The general discipline is to establish the mortality path first, then attach each payment to its correct event and payment time.
Record the duration at the start of every covered year and identify where the select period ends.
Use the select rate while duration is less than \(r\); use the ultimate rate at attained age \(x+s\) from duration \(r\) onward.
Multiply survival probabilities up to the start of each year by that year's applicable death probability.
Discount each benefit from its actual payment time, and include survival-contingent benefits only on the event that triggers them.
Common Traps and a Final Check
The most frequent error is to confuse attained age with policy duration. They move together for a policy issued at an integer age, but they answer different questions: duration determines whether the life is still select, while attained age identifies the ultimate rate once the select period has ended. A second error is to restart select status after a policy anniversary or after conditioning on survival. Select duration counts from selection; it does not restart simply because the valuation date changes.
Before accepting a calculation, check that the first \(r\) years use exactly the available select rates, and that later years use ultimate rates for the correct attained ages. Check that death probabilities for different years are unconditional probabilities from the same starting point, or clearly conditional probabilities from a later duration. For a fixed term, the sum of death probabilities over covered years should equal one minus the probability of surviving to the end of the term.
This careful alignment of mortality, duration, and payment time is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. It is especially useful in multi-year calculations, where a single off-by-one error at the select boundary changes every later survival probability and expected present value.
Check Your Understanding
For each problem, identify the mortality rate applicable in every year before calculating. Answers are given to four decimal places for probabilities and to the nearest cent for dollar amounts.
- A life issued at age 61 has a two-year select period, with \(q_{[61]}=0.03\) and \(q_{[61]+1}=0.05\). The ultimate rates are \(q_{63}=0.04\) and \(q_{64}=0.06\). Find the four-year survival probability. (Answer: \(0.8316\))
- For the mortality basis in Question 1, find the probability of death during policy year 3. (Answer: \(0.03686\))
- A life issued at age 30 has a two-year select period, with \(q_{[30]}=0.02\), \(q_{[30]+1}=0.03\), and \(q_{32}=0.05\). Find the probability of surviving three years. (Answer: \(0.9031\))
- A two-year contract has a one-year select period. For a life issued at age 44, \(q_{[44]}=0.04\) and \(q_{45}=0.06\). It pays \$10,000 at the end of the year of death. If \(i=0.10\), find the expected present value. (Answer: \$839.67)
- A life issued at age 70 has a three-year select period and has survived to duration 1. The rates for the next three years are \(q_{[70]+1}=0.05\), \(q_{[70]+2}=0.07\), and \(q_{73}=0.08\). Find the probability of surviving all three further years. (Answer: \(0.8128\))