Computing Probabilities Within The Select Period

Introduction

A select life table distinguishes between lives recently selected at an age and lives observed at the same attained age after more time has passed since selection. Within the select period, the mortality rate can depend on both the age at selection and the duration since selection. A multi-year probability therefore requires the rate for each successive duration, not repeated use of the first rate.

This tutorial focuses on calculating survival and death probabilities entirely within the select period. The essential steps are to identify the selection age, track the duration at the start of each year, and multiply the corresponding annual survival probabilities. No interest assumption is needed for these mortality probabilities.

Definition: For a life selected at age \(x\), \(q_{[x]+j}\) is the probability of death during the year beginning at duration \(j\), conditional on survival to the start of that year. The attained age at that time is \(x+j\). The corresponding probability of surviving that year is \(p_{[x]+j}=1-q_{[x]+j}\). The brackets identify the selection age; \(j\) identifies duration since selection.

Building a Multi-Year Probability

Suppose the select period lasts \(s\) years. The select rates apply at durations \(j=0,1,\ldots,s-1\). If \(n\leq s\), then surviving \(n\) years requires surviving each of the first \(n\) annual intervals. At duration \(j\), the relevant rate is \(q_{[x]+j}\), so the probability of surviving that interval is \(p_{[x]+j}\).

{}_n p_{[x]}=\prod_{j=0}^{n-1}p_{[x]+j} =\prod_{j=0}^{n-1}(1-q_{[x]+j}), \qquad 0\leq n\leq s.

The probability of death within the first \(n\) years is the complement of survival through all \(n\) years. If \(T_{[x]}\) denotes the future lifetime, measured in years from selection, this is \(P(T_{[x]}\leq n)\). The probability of death during a particular year is different: it includes survival through all earlier years and then death in the year of interest.

P(T_{[x]}\leq n)=1-{}_n p_{[x]}, \qquad P(k<T_{[x]}\leq k+1) ={}_k p_{[x]}q_{[x]+k}.

These equations use annual mortality probabilities. In particular, the second equation assumes the life has been followed from selection and asks for the unconditional probability of death during year \(k+1\). If instead the question conditions on survival to duration \(k\), the probability of death during the next year is simply \(q_{[x]+k}\).

Worked Example: Three-Year Survival After Selection

A hypothetical select table gives a four-year select period for a life selected at age 42. Its annual rates are \(q_{[42]}=0.020\), \(q_{[42]+1}=0.030\), \(q_{[42]+2}=0.050\), and \(q_{[42]+3}=0.070\). Find the probability of surviving three years and the probability of dying within those three years.

Model and method. Use annual select life-table mortality, with the listed conditional rates applying to the four successive durations. No interest rate or separate parametric distribution is assumed; the future-lifetime probabilities are determined by these annual rates. Three years are within the four-year select period, so use durations 0, 1, and 2.

Step 1: Convert rates to survival probabilities.

p_{[42]}=1-0.020=0.980,\qquad p_{[42]+1}=1-0.030=0.970,\qquad p_{[42]+2}=1-0.050=0.950.

Step 2: Multiply the successive survival probabilities.

{}_3p_{[42]}=(0.980)(0.970)(0.950) =0.9506(0.950)=0.90307.

Thus the three-year survival probability is \(0.9031\), rounded to four decimal places. The probability of death within three years is the complement:

P(T_{[42]}\leq 3)=1-0.90307=0.09693\approx 0.0969.

Check. The probabilities add to one before rounding: \(0.90307+0.09693=1\). The common error avoided here is applying \(q_{[42]}=0.020\) in all three years; the second and third years use their own duration-specific rates.

Why the Product Formula Holds

The product is a direct application of conditional probability. To survive two years, a life must first survive the year beginning at duration 0 and then survive the year beginning at duration 1, conditional on having reached that year. The same reasoning extends to any number of years in the select period.

Write the survival event.
Survival through \(n\) years means surviving each annual interval from duration 0 through duration \(n-1\).

Apply conditional probability.
The probability of the joint event is the probability of surviving the first year multiplied by the conditional probability of surviving the second year given survival through the first, and so on.

Substitute the select rates.
Each conditional annual survival probability is \(p_{[x]+j}=1-q_{[x]+j}\). Multiplying those terms gives \({}_n p_{[x]}=\prod_{j=0}^{n-1}(1-q_{[x]+j})\).

For an interval that begins after selection, first condition on survival to its start. Given survival to duration \(r\), the chance of dying by duration \(n\), where \(r<n\leq s\), is one minus the probability of surviving every intervening year. This conditional form is useful when a problem specifies that the life has already survived part of the select period.

P(r<T_{[x]}\leq n\mid T_{[x]}>r) =1-\prod_{j=r}^{n-1}p_{[x]+j} =1-\prod_{j=r}^{n-1}(1-q_{[x]+j}).

Worked Example: Death During a Specified Year

A hypothetical five-year select period for a life selected at age 55 has rates \(q_{[55]}=0.040\), \(q_{[55]+1}=0.050\), \(q_{[55]+2}=0.060\), \(q_{[55]+3}=0.080\), and \(q_{[55]+4}=0.090\). Find the unconditional probability that the life dies during the third year after selection.

Model and method. Use the annual select rates as conditional death probabilities at the stated durations. No interest rate or additional probability distribution is assumed. The third year begins at duration 2, so the life must survive durations 0 and 1 before dying at duration 2.

Step 1: Find the first two survival probabilities.

p_{[55]}=1-0.040=0.960,\qquad p_{[55]+1}=1-0.050=0.950.

Step 2: Use the death rate for duration 2. The probability of death in the third year, conditional on reaching its start, is \(q_{[55]+2}=0.060\). Therefore the unconditional probability is

P(2<T_{[55]}\leq 3) =(0.960)(0.950)(0.060) =0.912(0.060)=0.05472.

The requested probability is \(0.0547\), rounded to four decimal places. Check. Conditional on surviving two years, the third-year death probability is \(0.060\); the unconditional probability \(0.05472\) is smaller because it also requires surviving the first two years. The calculation avoids the mistake of reporting \(0.060\) as an unconditional probability.

Conditional Probabilities Across Several Years

A question may ask for death over several years conditional on survival to a particular duration. The rates used begin at that duration, not at the selection date. For example, conditional on surviving one year, death by the end of year four is calculated using the rates at durations 1, 2, and 3. The initial year’s survival probability does not appear, because survival to its end is already given.

Worked Example: Death Over an Interval Given Prior Survival

For a hypothetical four-year select period at selection age 63, let the rates be \(q_{[63]}=0.010\), \(q_{[63]+1}=0.020\), \(q_{[63]+2}=0.030\), and \(q_{[63]+3}=0.040\). Given that the life survived the first year, calculate the probability of dying before the end of year four.

Model and method. Assume these are the annual conditional select death probabilities, with no interest rate or separate parametric distribution. The condition is survival through duration 1. The relevant subsequent years begin at durations 1, 2, and 3.

Step 1: Calculate survival at those durations.

p_{[63]+1}=1-0.020=0.980,\quad p_{[63]+2}=1-0.030=0.970,\quad p_{[63]+3}=1-0.040=0.960.

Step 2: Find the conditional probability of surviving all three years.

P(T_{[63]}>4\mid T_{[63]}>1) =(0.980)(0.970)(0.960) =0.9506(0.960)=0.912576.

Step 3: Take the complement.

P(1<T_{[63]}\leq 4\mid T_{[63]}>1) =1-0.912576=0.087424\approx 0.0874.

The answer is \(0.0874\), rounded to four decimal places. Check. The three-year conditional survival probability and the conditional probability of death over that interval sum to one. The initial rate \(q_{[63]}\) is correctly excluded because survival through the first year is stipulated.

Indexing Checks and Common Traps

A select rate has two pieces of indexing information: the selection age and the duration since selection. If a life was selected at age \(x\), the rate for the year beginning after \(j\) full years is \(q_{[x]+j}\), not a new select rate based on treating \(x+j\) as the selection age. The latter would describe a different selection cohort.

Before multiplying rates, check the event and the time interval. Survival for \(n\) years uses \(n\) survival probabilities. Death during the \(k\)th year uses \(k-1\) survival probabilities followed by one death probability. A conditional probability given survival to duration \(r\) starts with the rate at duration \(r\), because earlier survival is already known.

Common exam trap: If the rates are \(q_{[x]},q_{[x]+1},q_{[x]+2}\), using \((1-q_{[x]})^3\) assumes the same mortality rate in all three years. That is not the select-table model unless the problem explicitly gives equal rates. Use \((1-q_{[x]})(1-q_{[x]+1})(1-q_{[x]+2})\) instead.

Finally, verify that the requested interval remains within the stated select period. The formulas here apply only when every annual interval used is part of that period. If a calculation extends beyond it, the select-period rates alone do not determine the remaining probability; an appropriate rate assumption for the later duration is also required.

Key takeaway: Track duration from the selection date, convert each duration-specific death rate to a survival rate, and multiply only the terms required by the event and its conditioning. This method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated annual select rates and round probabilities to four decimal places.

A hypothetical three-year select period at age 38 has rates \(q_{[38]}=0.030\), \(q_{[38]+1}=0.040\), and \(q_{[38]+2}=0.050\). Find the probability of surviving all three years. (Answer: 0.8846)
For a life selected at age 47, let \(q_{[47]}=0.020\), \(q_{[47]+1}=0.030\), and \(q_{[47]+2}=0.040\). Find the unconditional probability of dying during the second year. (Answer: 0.0294)
A four-year select period at age 52 has rates \(0.050, 0.060, 0.070,\) and \(0.080\) at durations 0 through 3, respectively. Given survival through the first year, find the probability of dying during either the second or third year. (Answer: 0.1258)
A hypothetical four-year select period at age 60 has rates \(0.010, 0.020, 0.030,\) and \(0.040\). Find the probability of death within four years. (Answer: 0.0965)
For a five-year select period at age 70, the rates at durations 0 through 4 are \(0.020, 0.030, 0.040, 0.050,\) and \(0.060\). Given survival through the first year, find the probability that the life survives at least three years from selection but dies by year five. (Answer: 0.0996)

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