Why Combining the Two Periods Requires Care
A select-and-ultimate mortality table uses different mortality rates at first, depending on how recently a life was selected. After a specified number of years, the rates depend only on attained age. A probability calculation that crosses this boundary must therefore use select rates for the years before the transition and ultimate rates afterward.
The main challenge is indexing: the rate for a year is determined by the life’s duration since selection at the start of that year. An interval that begins at duration 2, for example, does not use the rates for durations 0 and 1, even if those rates belong to the same life. A second challenge is conditioning. “Death between durations 3 and 5, given survival to duration 1” includes both surviving from duration 1 to duration 3 and dying before duration 5.
The notation \(q_{[x]+j}\) refers to a select rate at duration \(j\) for a life selected at age \(x\). It is different from the ultimate rate \(q_{x+j}\), even though both apply to a life who is age \(x+j\) at the start of the year. At duration \(s\), the select period has ended, so the applicable rate is \(q_{x+s}\).
Survival Across the Select-to-Ultimate Boundary
Let \({}_t p_{[x]}\) denote the probability that a life selected at age \(x\) survives \(t\) complete years from selection. For integer \(t\le s\), all the rates used are select rates. For \(t>s\), the calculation uses all \(s\) select rates followed by the ultimate rates for the attained ages reached after the select period.
An empty product, as in the expression when \(t=0\), equals 1. The formula applies to complete-year intervals when the table supplies one-year death probabilities. For an interval beginning or ending at a fractional age, additional assumptions about mortality between integer ages would be needed.
Worked Example: Survival Through a Two-Year Select Period
A hypothetical life is selected at age 30. The select period is two years, with \(q_{[30]}=0.02\) and \(q_{[30]+1}=0.03\). The ultimate rates at ages 32 and 33 are \(q_{32}=0.04\) and \(q_{33}=0.05\). Find the probability of surviving four years from selection.
Method. There are two select-period years, durations 0 and 1, followed by two ultimate-period years, at attained ages 32 and 33. The one-year survival probabilities are the complements of the corresponding death probabilities:
Multiply the four year-specific survival probabilities:
Thus the probability of surviving four years is 0.8669 to four decimal places. As a check, the probability of dying within four years is \(1-0.8669472=0.1330528\), and the two probabilities sum to 1. The common mistake is to use an ultimate rate at age 30 or to omit an ultimate-period year; the select period ends at duration 2, not at the end of the requested four-year interval.
Probabilities Over an Interval That Starts Later
For an interval from duration \(m\) to duration \(n\), where \(0\le m<n\), the conditional survival probability given survival to duration \(m\) uses only the rates applying from duration \(m\) through duration \(n-1\). Rates from earlier durations are not included. In terms of survival probabilities from selection, the same result is the ratio \({}_n p_{[x]}/{}_m p_{[x]}\), because the probability of surviving to duration \(n\) requires first surviving to duration \(m\).
Here \(T_{[x]}\) is the future lifetime, measured in years from selection. The conditional probability of dying after duration \(m\) and by duration \(n\), given survival to \(m\), is one minus this conditional survival probability. If instead the condition is survival to an earlier duration \(r<m\), the event also requires survival from \(r\) to \(m\):
Worked Example: Conditional Death Probability Across the Transition
A hypothetical life is selected at age 55 for a two-year select period. The select rates are \(q_{[55]}=0.03\) and \(q_{[55]+1}=0.04\), and the ultimate rates are \(q_{57}=0.05\) and \(q_{58}=0.06\). Find the probability of death between durations 1 and 4, given survival to duration 1.
Method. Since the life is already known to have survived to duration 1, start with the rate for duration 1. Survival from duration 1 to 4 requires survival through three years: the final select year, then two ultimate years. Their survival probabilities are \(1-0.04=0.96\), \(1-0.05=0.95\), and \(1-0.06=0.94\).
Therefore the requested conditional death probability is
The answer is 0.1427 to four decimal places. As a check, the survival and death probabilities from duration 1 to 4 add to 1. The common mistake is to include \(q_{[55]}=0.03\), which applies before duration 1 and is irrelevant once survival to duration 1 is given.
Why the Product and Conditional Formulas Work
Each one-year survival factor is conditional on being alive at the beginning of that year. To survive multiple years, a life must survive each year in sequence. The multiplication rule for conditional probabilities therefore gives the product of those one-year factors. This reasoning does not require the successive survival events to be independent.
For a probability conditional on survival to duration \(m\), begin with the rate for duration \(m\), not the rate at selection.
Use select rates for durations below \(s\), and ultimate rates from duration \(s\) onward.
For each year in the requested interval, use \(1-q\) for that year’s applicable death probability.
For death during an interval conditional on being alive at its start, subtract the interval survival probability from 1. If conditioning starts earlier, also account for reaching the beginning of the death interval.
For example, \({}_n p_{[x]}={}_m p_{[x]}P(T_{[x]}>n\mid T_{[x]}>m)\) when \(m<n\). Dividing both sides by \({}_m p_{[x]}\) gives the conditional survival ratio. The probability of dying between the two durations is the complement of that ratio. This also explains why a probability conditional on survival to an earlier duration must include the chance of reaching the later interval.
Worked Example: Death in a Later Interval Given Earlier Survival
A hypothetical life is selected at age 65 with a two-year select period. The select rates are \(q_{[65]}=0.05\) and \(q_{[65]+1}=0.06\). The ultimate rates are \(q_{67}=0.07\), \(q_{68}=0.08\), and \(q_{69}=0.09\). Find the probability of death between durations 3 and 5, given survival to duration 1.
Method. Given survival to duration 1, the life must first survive from duration 1 to duration 3. Those years have death probabilities \(q_{[65]+1}=0.06\) and \(q_{67}=0.07\). If the life reaches duration 3, death between durations 3 and 5 means dying during one of the next two years, with ultimate rates \(q_{68}=0.08\) and \(q_{69}=0.09\).
The probability of reaching duration 3, conditional on survival to duration 1, is
Conditional on reaching duration 3, the probability of dying by duration 5 is
Both parts are required. Multiply them to obtain the probability conditional on survival to duration 1:
The answer is 0.1423 to four decimal places. The answer is less than the probability \(0.1628\) of dying during durations 3 to 5 given survival to duration 3, because the stated condition is only survival to duration 1; some lives who survive to duration 1 die before duration 3. A common incorrect calculation is to report \(0.1628\) without multiplying by \(0.8742\). That calculation answers a different conditional question.
Common Indexing and Conditioning Traps
Select-and-ultimate calculations are especially vulnerable to off-by-one errors. Duration \(j\) is the year beginning \(j\) years after selection. Thus the select rate at duration \(s-1\) applies in the last select year, and the ultimate rate \(q_{x+s}\) applies in the first year after the select period. A useful check is to list the durations and rates before multiplying.
| Year in the calculation | Rate to use | Survival factor |
|---|---|---|
| Duration \(j\), where \(j<s\) | \(q_{[x]+j}\) | \(1-q_{[x]+j}\) |
| Duration \(j\), where \(j\ge s\) | \(q_{x+j}\) | \(1-q_{x+j}\) |
- Using attained age alone during selection. A life can have attained age \(x+j\) while still in the select period. Use the select rate when \(j<s\).
- Starting with the wrong duration. If survival to duration \(m\) is given, the first rate used is the one for duration \(m\), not duration 0.
- Forgetting survival to the requested interval. If the condition is survival to \(r<m\), death between \(m\) and \(n\) requires survival from \(r\) to \(m\) as well as death before \(n\).
- Using a death rate as a survival factor. A one-year survival factor is \(1-q\), not \(q\).
Check Your Understanding
For each problem, identify the applicable rates by duration before calculating. Give probabilities to four decimal places.
- A life selected at age 40 has a two-year select period, with \(q_{[40]}=0.02\), \(q_{[40]+1}=0.04\), and \(q_{42}=0.05\). Find the probability of surviving three years from selection. (Answer: 0.8938)
- A life selected at age 30 has a two-year select period, with \(q_{[30]}=0.02\), \(q_{[30]+1}=0.03\), \(q_{32}=0.04\), and \(q_{33}=0.05\). Find the probability of death between durations 1 and 4, given survival to duration 1. (Answer: 0.1427)
- A life selected at age 50 has a two-year select period, with select rates 0.04 and 0.05, and ultimate rates \(q_{52}=0.06\), \(q_{53}=0.07\), and \(q_{54}=0.08\). Find the probability of death between durations 2 and 5, given survival to selection. (Answer: 0.1785)
- A life selected at age 40 has a three-year select period, with select rates 0.01, 0.02, and 0.04, followed by ultimate rates \(q_{43}=0.03\) and \(q_{44}=0.05\). Find the probability of surviving from duration 1 to duration 5, given survival to duration 1. (Answer: 0.8669)
- A life selected at age 30 has a three-year select period, with select rates 0.01, 0.02, and 0.03, followed by ultimate rates \(q_{33}=0.04\), \(q_{34}=0.06\), \(q_{35}=0.08\), and \(q_{36}=0.10\). Find the probability of death between durations 4 and 7, given survival to duration 2. (Answer: 0.2064)