Why Age Indexing Matters
A mortality rate is meaningful only when its age and time interval are read correctly. In a select-and-ultimate table, the select age identifies when a life was selected, while select duration counts the time elapsed since selection. The attained age changes as time passes. Confusing any two of these quantities can lead to choosing a rate for the wrong year, even when the arithmetic is otherwise correct.
The key habit is to identify the interval first. A one-year rate for a life aged \(x\) at the start of an interval applies to the interval from age \(x\) to age \(x+1\). For select rates, the age at selection and the duration must both be tracked to identify that interval’s rate.
The brackets in \(q_{[x]+j}\) distinguish the age at selection \(x\) from the duration \(j\). They do not mean that the rate applies to an interval beginning at age \(x\) for every value of \(j\). For example, \(q_{[40]+2}\) applies to a selected life who has reached age 42, at duration 2; it covers the year from age 42 to age 43. If the select period has ended by then, the relevant rate is instead the ultimate rate \(q_{42}\).
A Reliable Indexing Procedure
For each policy year in a problem, write down the attained age at the start of the year, the select duration at that time, and the table entry that covers the year. This prevents the common error of using the selection age as though it were the current age. It also makes the transition from select to ultimate mortality explicit.
For a life selected at age \(x\), keep \(x\) fixed throughout the calculation.
At the start of policy year \(j+1\), the select duration is \(j\), and the attained age is \(x+j\).
Use \(q_{[x]+j}\) while duration \(j\) is within the select period. After that period, use the ultimate rate for the attained age at the start of the interval.
For successive one-year intervals, multiply the relevant \(1-q\) factors. To include death in a specified year, multiply by that year’s \(q\) after the preceding survival factors.
Why the Product Formula Works
The indexing procedure supplies the appropriate one-year conditional rates. The probability calculation follows from conditioning on survival to the start of each interval. If the successive one-year death rates are \(r_0,r_1,\ldots,r_{n-1}\), then surviving all \(n\) years requires survival in every interval. The conditional probability of surviving interval \(j\), given survival to its start, is \(1-r_j\).
To see why, the probability of surviving the first two intervals is \(P(\text{survive interval 1})P(\text{survive interval 2}\mid\text{survive interval 1})=(1-r_0)(1-r_1)\). Repeating this conditional-probability decomposition for each interval gives the product. If death is required in the final interval, multiply the survival factors for all earlier intervals by the final interval’s death rate. This calculation assumes that each listed rate is the conditional one-year death probability for the interval to which it is assigned. It does not apply if a rate refers to a different duration, age, or time period.
Worked Examples
Worked Example: Death in the Third Select Year
A hypothetical table has a three-year select period. For a life selected at age 40, the rates are \(q_{[40]}=0.010\), \(q_{[40]+1}=0.020\), and \(q_{[40]+2}=0.030\). Find the probability that the life dies during the third policy year, given survival to selection.
Method. The third policy year begins after two years have elapsed. The select duration is therefore 2, and the attained age is \(40+2=42\). The applicable rate is \(q_{[40]+2}=0.030\), but the life must first survive the first two select years.
The first-year survival probability is \(1-0.010=0.990\). The second-year survival probability, conditional on surviving the first year, is \(1-0.020=0.980\). Thus the probability of reaching the start of the third year is
The probability of then dying during the third year is
The probability is \(0.0291\) to four decimal places. The calculation uses the third-year rate only after accounting for survival to that year. A common mistake is to treat \(q_{[40]+2}\) as the rate for the first year simply because the selection age is 40.
Worked Example: Select Period Ends Before the Next Interval
A hypothetical table has a three-year select period. For a life selected at age 40, \(q_{[40]}=0.010\), \(q_{[40]+1}=0.020\), and \(q_{[40]+2}=0.030\). The ultimate rate at age 43 is \(q_{43}=0.030\). Find the probability that the life survives the three select years and dies during the following year.
Method. The three select-year intervals are ages 40 to 41, 41 to 42, and 42 to 43. After surviving them, the life is age 43 and the select period has ended. The next interval uses ultimate \(q_{43}\), not another select rate.
The probability of surviving the three select years is
Multiplying by the ultimate death rate for age 43 gives
The required probability is \(0.02823\) to five decimal places. As a check, the probability is less than \(q_{43}=0.030\), since it also requires survival through the preceding three years. The indexing trap is to use an ultimate rate for age 42 in the third select year, or to use a select rate after the select period has ended.
Worked Example: A Later Policy Year in a Select-and-Ultimate Table
A hypothetical life is selected at age 38. The select period lasts two years, with \(q_{[38]}=0.006\) and \(q_{[38]+1}=0.009\). The ultimate rates are \(q_{40}=0.012\) and \(q_{41}=0.015\). Find the probability of death during the fourth policy year, given survival to selection.
Method. The fourth policy year begins after three years. The life is then age \(38+3=41\). Since the two-year select period has ended, the rate for the fourth policy year is ultimate \(q_{41}=0.015\). The life must first survive the two select years and the next ultimate year at age 40.
The three survival factors are \(1-0.006=0.994\), \(1-0.009=0.991\), and \(1-0.012=0.988\). Their product is
The probability of death during the fourth policy year is therefore
The answer is \(0.01460\) to five decimal places. The rates appear in order as \(q_{[38]}\), \(q_{[38]+1}\), \(q_{40}\), and \(q_{41}\). In particular, the fourth-year rate is \(q_{41}\), not \(q_{[38]+3}\) if the select notation is defined only for the two-year select period.
Common Age-Indexing Errors
The word “age” can refer to different points in a mortality-table calculation. Before using a rate, ask which of these the problem supplies and which it requests: age at selection, attained age, or select duration. Then identify the one-year interval. A rate named \(q_y\) is not a probability of dying at the exact age \(y\); it is the conditional probability of dying during the year beginning at age \(y\).
| Quantity or notation | What it identifies | Interval or interpretation |
|---|---|---|
| Selection age \(x\) | Age when the life enters the select table | Fixed for the selected life |
| Select duration \(j\) | Years elapsed since selection | At the start of the interval |
| Attained age \(x+j\) | Age at the start of that interval | Changes by one each year |
| \(q_{[x]+j}\) | Select rate for selection age \(x\), duration \(j\) | Next year, beginning at attained age \(x+j\) |
| \(q_y\) | Ultimate rate at attained age \(y\) | One year, from age \(y\) to age \(y+1\) |
A useful check is to make a short timeline before multiplying rates. If a policy is in its first year, the select duration at the start of the year is 0, not 1. If a life has survived \(k\) full years since selection, the attained age is \(x+k\), and the duration is \(k\). These conventions prevent an off-by-one error when moving from a select rate to an ultimate rate.
- Using attained age as selection age. A life selected at 40 and now aged 42 is at duration 2; it is not a newly selected life at age 42.
- Counting policy years from 1 as durations. The first policy year has duration 0 at its start. The second has duration 1 at its start.
- Using a rate for the wrong interval. \(q_{41}\) covers ages 41 to 42. It does not cover the interval from 40 to 41.
- Extending select notation past its stated period. Once the select period ends, use the ultimate rate for the attained age at the start of the next year.
Check Your Understanding
For each problem, identify the correct interval before calculating. Probabilities are rounded to four decimal places unless another precision is specified.
- A life is selected at age 42. The first two select rates are \(q_{[42]}=0.008\) and \(q_{[42]+1}=0.011\). Find the probability of dying during the second policy year, given survival to selection. (Answer: \(0.0109\))
- A two-year select period applies to a life selected at age 60. The rates are \(q_{[60]}=0.020\), \(q_{[60]+1}=0.030\), and \(q_{62}=0.040\). Find the probability of surviving the select period and dying during the next year. (Answer: \(0.0380\))
- A three-year select period applies to a life selected at age 35. The select rates are \(0.010\), \(0.020\), and \(0.030\), in duration order. The ultimate rate at age 38 is \(0.025\). Find the probability of surviving the select period and dying in the following year. (Answer: \(0.0235\))
- A life selected at age 55 is at duration 1, and the select period lasts three years. If \(q_{[55]+1}=0.018\) and the ultimate rate \(q_{56}=0.021\), which rate applies to the next year, and what is the one-year death probability? (Answer: \(q_{[55]+1}=0.018\); \(0.0180\))
- A life is selected at age 47, with a two-year select period. The select rates are \(q_{[47]}=0.012\) and \(q_{[47]+1}=0.016\); the ultimate rates are \(q_{49}=0.020\) and \(q_{50}=0.024\). Find the probability of death during the fourth policy year, given survival to selection. (Answer: \(0.0229\))