Why the Select-Period Boundary Matters
A select-and-ultimate mortality table distinguishes between lives recently selected at underwriting and lives classified by attained age after a specified select period. To read the table correctly, track two quantities at once: how long the life has been selected and the age reached. The boundary is an indexing point, not an extra year: the first year after the select period uses the ultimate rate for the age at which that year begins.
This tutorial focuses on locating that boundary and choosing the correct one-year rate there. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Read the Boundary by Duration and Age
The select period counts years since selection, not years since birth and not the number of rows one happens to see in a table. If the select period is three years, the selected life has select durations 0, 1, and 2. At duration 3, the select period has ended. Because the life is then \(x+3\) years old, the relevant ultimate rate is \(q_{x+3}\).
In standard notation, \(q_{[x]+k}\) denotes the select rate for a life selected at age \(x\), at select duration \(k\). The square brackets identify the selection age; the \(+k\) tracks duration. By contrast, \(q_{x+k}\) is an ultimate rate indexed by attained age. The two rates can differ even when they refer to the same age, because a select rate also reflects the life’s select status.
| Time since selection | Attained age | Rate with an \(r\)-year select period |
|---|---|---|
| Duration \(0\) | \(x\) | Select \(q_{[x]}\) |
| Duration \(1\) | \(x+1\) | Select \(q_{[x]+1}\), if \(1<r\) |
| Duration \(k<r\) | \(x+k\) | Select \(q_{[x]+k}\) |
| Duration \(r\) | \(x+r\) | Ultimate \(q_{x+r}\) |
| Duration \(k>r\) | \(x+k\) | Ultimate \(q_{x+k}\) |
The rate at duration \(k\) describes the one-year interval beginning at that duration. Thus, duration 2 refers to the year from time 2 to time 3 since selection, and duration 3 refers to the next year, from time 3 to time 4. The switch occurs between those intervals when \(r=3\): duration 2 is still select, and duration 3 is ultimate.
Why the First Ultimate Rate Has This Index
The indexing follows directly from the timeline. At selection, the life is age \(x\) and duration 0. Each elapsed year increases both the duration and attained age by one. After \(r\) complete years, duration is \(r\) and attained age is \(x+r\). The following one-year mortality rate is therefore the ultimate rate for age \(x+r\).
Here \(q^{(x)}_k\) is simply a convenient label for the rate applicable to this life at duration \(k\); it does not replace the table’s standard select or ultimate notation. The piecewise rule assumes a fixed select period of \(r\) years and a table that assigns select rates during those first \(r\) years. It does not say that select rates must be lower than ultimate rates. Their relative sizes depend on the table.
The year numbering is a common source of error. If a question asks about the “fourth policy year,” that year starts at duration 3, not duration 4. For a three-year select period, the fourth policy year therefore uses the first ultimate rate. If a question instead asks for the rate at duration 3, it is also the first ultimate rate, because duration 3 marks the start of that year.
Worked Example: Finding the First Ultimate Rate
A hypothetical table has a three-year select period. For a life selected at age 42, the relevant entries are \(q_{[42]}=0.0080\), \(q_{[42]+1}=0.0095\), and \(q_{[42]+2}=0.0110\). The ultimate rates are \(q_{45}=0.0140\) and \(q_{46}=0.0160\). Identify the rate for each of the first five policy years, and find the rate for the first year governed by ultimate mortality.
Method. Policy year \(k+1\) starts at duration \(k\). Compare that duration with the three-year select period, then use the attained age \(42+k\) to identify an ultimate rate if \(k\geq3\).
Policy year 1: it starts at duration \(0\), when attained age is \(42+0=42\). Since \(0<3\), use \(q_{[42]}=0.0080\).
Policy year 2: it starts at duration \(1\), when attained age is \(42+1=43\). Since \(1<3\), use \(q_{[42]+1}=0.0095\).
Policy year 3: it starts at duration \(2\), when attained age is \(42+2=44\). Since \(2<3\), use \(q_{[42]+2}=0.0110\).
Policy year 4: it starts at duration \(3\), when attained age is \(42+3=45\). The select period has ended, so use \(q_{45}=0.0140\). This is the first ultimate rate.
Policy year 5: it starts at duration \(4\), when attained age is \(42+4=46\). Use the ultimate rate \(q_{46}=0.0160\).
The sequence is \(0.0080,\ 0.0095,\ 0.0110,\ 0.0140,\ 0.0160\). The boundary check is consistent: the first three policy years use durations 0 through 2, and the fourth begins at duration 3 and attained age 45. Common mistake avoided: carrying the final select rate \(0.0110\) into policy year 4. That year is not at duration 2; it begins at duration 3.
Apply the Rule to a Requested Policy Year
Questions may specify a policy year rather than a duration. Convert the policy-year number to its starting duration before choosing a rate: policy year \(n\) starts at duration \(n-1\). Then calculate the attained age at that duration. This two-step conversion prevents confusing the year’s label with the duration at its beginning.
Worked Example: Converting a Policy-Year Label
A hypothetical select table has a four-year select period. A life is selected at age 39. The table gives \(q_{[39]+3}=0.0152\), and its ultimate table gives \(q_{43}=0.0171\) and \(q_{44}=0.0184\). Determine the mortality rate for policy years 4, 5, and 6.
Method. A policy year \(n\) begins at duration \(n-1\). The select period covers durations 0, 1, 2, and 3; ultimate rates begin at duration 4.
Policy year 4: starting duration \(=4-1=3\). The attained age is \(39+3=42\). Since \(3<4\), this remains a select duration, and the rate is the given \(q_{[39]+3}=0.0152\).
Policy year 5: starting duration \(=5-1=4\). The attained age is \(39+4=43\). Duration 4 is the first ultimate duration, so use \(q_{43}=0.0171\).
Policy year 6: starting duration \(=6-1=5\). The attained age is \(39+5=44\). This is after the select period, so use \(q_{44}=0.0184\).
Thus the rates are \(0.0152,\ 0.0171,\ 0.0184\), respectively. The age index for policy year 5 is 43 because that year begins at age 43. Common mistake avoided: treating policy year 4 as duration 4 and switching one year too early. The fourth policy year starts at duration 3.
Same Attained Age, Different Select Status
A life’s attained age alone may not determine the correct table entry if the life is still within the select period. Two lives can be the same age but have different select durations because they were selected at different ages or times. One may still use a select rate while the other has reached the ultimate table. This is why the rate must be chosen using both duration and attained age.
Worked Example: Two Lives at the Same Attained Age
A hypothetical table has a three-year select period. At attained age 64, the ultimate rate is \(q_{64}=0.0240\). The select table gives \(q_{[62]+2}=0.0210\). Life A was selected at age 61 and has now completed three years; Life B was selected at age 62 and has completed two years. Identify the rate applicable to each at the start of the current year.
Method. For each life, calculate duration and verify that attained age equals selection age plus duration. Then compare duration with the three-year select period.
Life A: duration \(=3\), selection age plus duration \(=61+3=64\). The attained age agrees with the stated age. Since \(3\geq3\), Life A has reached the ultimate period, so the rate is \(q_{64}=0.0240\).
Life B: duration \(=2\), selection age plus duration \(=62+2=64\). Again the age check agrees. Since \(2<3\), Life B remains in the select period, so the applicable rate is \(q_{[62]+2}=0.0210\).
Therefore, the two rates are \(0.0240\) for Life A and \(0.0210\) for Life B, despite both lives being age 64. The answer follows from their different durations, not from age alone. Common mistake avoided: assigning \(q_{64}\) to both lives just because they have the same attained age.
Boundary Checks and Common Traps
Before selecting a rate, write down the three quantities that determine the table entry: the select period \(r\), the duration \(k\) at the start of the year, and the attained age \(x+k\). Then apply the boundary test \(k<r\) versus \(k\geq r\). This small check is especially useful when a problem describes elapsed years in words rather than giving a duration directly.
- Off-by-one duration: policy year \(n\) starts at duration \(n-1\), not \(n\).
- Wrong age index: at the first ultimate duration \(r\), use age \(x+r\), not \(x+r-1\).
- Using select rates for \(r\) full years after the boundary: an \(r\)-year select period has durations 0 through \(r-1\), exactly \(r\) durations. Duration \(r\) is already ultimate.
- Using age alone: an attained age can correspond to a select rate for one life and an ultimate rate for another, depending on their durations.
- Assuming select rates are always smaller: the table determines the rate values. The select/ultimate distinction is about table classification, not a guaranteed ordering of rates.
Check Your Understanding
For each problem, identify the duration at the start of the stated year before choosing the rate.
- A two-year select table applies to a life selected at age 48. Which rate applies in policy year 2: \(q_{[48]+1}\) or \(q_{49}\)? (Answer: \(q_{[48]+1}\))
- A four-year select table applies to a life selected at age 55. Find the first ultimate rate and its attained-age index. (Answer: \(q_{59}\), at attained age 59)
- A three-year select table applies to a life selected at age 46. Which rate applies at duration 2, and which at duration 3? (Answer: Select \(q_{[46]+2}\) at duration 2; ultimate \(q_{49}\) at duration 3)
- A five-year select table applies to a life selected at age 60. Identify the rate category and age index for policy year 6. (Answer: Ultimate \(q_{65}\); policy year 6 starts at duration 5 and attained age 65)
- A three-year select table has \(q_{[70]+2}=0.0310\) and \(q_{73}=0.0340\). A life selected at age 70 is at duration 3. Which rate applies, and what is the incorrect choice if the final select rate is carried forward? (Answer: \(q_{73}=0.0340\) applies; the incorrect choice is \(q_{[70]+2}=0.0310\))