Constructing Life Table Entries From qx

Introduction

A sequence of annual mortality probabilities becomes a life table only after we choose a starting number of lives and carry the survivor count forward one year at a time. The calculation is simple, but it is easy to use the wrong age’s probability or to apply a probability to the original radix instead of the current number of survivors. Both errors can make an otherwise orderly table inconsistent.

This tutorial builds the \(l_x\) and \(d_x\) columns from annual \(q_x\) values. The calculations use one-year mortality probabilities at integer ages; they do not require an assumption about the timing of deaths within a year. This construction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

What the Entries Mean

Definition: In a life table, \(l_x\) is the number of lives surviving to exact age \(x\), and \(d_x\) is the number dying between exact ages \(x\) and \(x+1\). The one-year mortality probability \(q_x\) is the probability that a life aged exactly \(x\) dies before reaching age \(x+1\). The corresponding one-year survival probability is \(p_x=1-q_x\).

A radix is the chosen starting value of \(l_x\) at a specified age. For example, a table might begin with \(l_{50}=5{,}000\). The radix is a scale for the table: it need not equal the number of people in a real population. Once the radix is selected, each \(q_x\) applies to the survivors at age \(x\), not to the initial radix unless \(x\) is the starting age.

For a constructed life table based on probabilities, the entries represent expected counts. Thus \(l_xq_x\) gives expected deaths during the year of age \(x\), and those entries can be fractional. For actual observed cohort counts, \(d_x\) is the realized number of deaths; the observed rate is \(d_x/l_x\), which need not equal a prescribed probability \(q_x\).

The One-Year Recursion

At exact age \(x\), the proportion expected to survive to age \(x+1\) is \(p_x=1-q_x\). Therefore, multiplying the number at age \(x\) by this survival probability gives the next survivor entry. The deaths during the year are the complement of those survivors.

l_{x+1}=l_xp_x=l_x(1-q_x)

d_x=l_xq_x=l_x-l_{x+1}

These equations are the basic construction rules. The \(q_x\) used to calculate \(d_x\) and \(l_{x+1}\) is the rate for the interval from age \(x\) to age \(x+1\). After obtaining \(l_{x+1}\), use that updated entry with \(q_{x+1}\) to continue.

The death identity follows directly from the survivor recursion. Since \(l_{x+1}=l_x(1-q_x)\), subtracting from \(l_x\) gives \(l_x-l_{x+1}=l_xq_x\). Over several consecutive years, the deaths telescope:

\sum_{j=x}^{x+n-1}d_j =\sum_{j=x}^{x+n-1}(l_j-l_{j+1}) =l_x-l_{x+n}.

So the total deaths over the interval equal the initial number of survivors minus the number remaining at the end. This is a useful check on a completed table.

Set the radix.
Record the starting age and its survivor entry, such as \(l_{50}=5{,}000\).

Use the rate for the current age.
For each age \(x\), take the supplied \(q_x\) and calculate \(d_x=l_xq_x\).

Advance the survivor entry.
Calculate \(l_{x+1}=l_x-d_x=l_x(1-q_x)\). This value is the starting survivor count for the next row.

Check the entries.
For every row, verify \(l_x=d_x+l_{x+1}\). Across multiple rows, verify that total deaths equal the decrease in survivors.

Worked Constructions

Worked Example: Building Three Rows From a Radix

A hypothetical table starts with \(l_{30}=1{,}000\). Assume the annual mortality probabilities are \(q_{30}=0.10\), \(q_{31}=0.20\), and \(q_{32}=0.25\). Construct the \(l_x\) and \(d_x\) entries through age 33. Treat the supplied values as probabilities for the respective one-year age intervals.

Method. Start with the radix and apply the one-year recursion in increasing age order. This avoids applying each mortality probability to the original 1,000 lives.

At age 30, the expected deaths are

d_{30}=l_{30}q_{30}=1{,}000(0.10)=100.

The number surviving to age 31 is

l_{31}=1{,}000-100=900.

At age 31, use \(l_{31}=900\), not the original radix:

d_{31}=900(0.20)=180,\qquad l_{32}=900-180=720.

At age 32, use the updated survivor entry \(l_{32}=720\):

d_{32}=720(0.25)=180,\qquad l_{33}=720-180=540.

The constructed entries are \(l_{30}=1{,}000\), \(d_{30}=100\), \(l_{31}=900\), \(d_{31}=180\), \(l_{32}=720\), \(d_{32}=180\), and \(l_{33}=540\). The total deaths are \(100+180+180=460\). The endpoint check gives \(l_{30}-l_{33}=1{,}000-540=460\), so the cumulative entries agree.

Common mistake avoided: Calculating \(d_{31}=1{,}000(0.20)\) would incorrectly apply the age-31 rate to the original radix. The relevant number exposed to age-31 mortality is \(l_{31}=900\).

Worked Example: Constructing Entries With a Non-Unit Radix

A hypothetical life table has radix \(l_{50}=5{,}000\), with \(q_{50}=0.02\), \(q_{51}=0.04\), and \(q_{52}=0.08\). Construct entries through age 53 and find the total expected deaths from age 50 to age 53. These annual probabilities are assumed to apply to the intervals \([50,51)\), \([51,52)\), and \([52,53)\), respectively.

Method. Use the current \(l_x\) for both the death calculation and the next survivor count at each age.

For age 50:

d_{50}=5{,}000(0.02)=100,\qquad l_{51}=5{,}000-100=4{,}900.

For age 51:

d_{51}=4{,}900(0.04)=196,\qquad l_{52}=4{,}900-196=4{,}704.

For age 52:

d_{52}=4{,}704(0.08)=376.32,\qquad l_{53}=4{,}704-376.32=4{,}327.68.

The total expected deaths are

d_{50}+d_{51}+d_{52} =100+196+376.32 =672.32.

The endpoint check gives \(l_{50}-l_{53}=5{,}000-4{,}327.68=672.32\), as required. The fractional entry \(d_{52}=376.32\) is appropriate for a constructed expected-count table; it is not a claim that a realized group must contain a fractional person.

Common mistake avoided: Do not round each survivor entry to a whole number before continuing unless the problem specifically requires it. Rounding would alter later products and could make the total deaths fail to match the difference between the radix and the final survivor entry.

Worked Example: Continuing a Table With Fractional Expected Counts

A hypothetical projection begins with \(l_{60}=20{,}000\). Suppose \(q_{60}=0.0125\), \(q_{61}=0.025\), \(q_{62}=0.04\), and \(q_{63}=0.05\). Construct the entries through age 64, then calculate the expected number dying from age 61 to age 64. Keep full precision during the recursion and report counts to two decimal places.

Method. Apply each annual probability to the number at the start of that age interval, and retain unrounded intermediate entries.

At age 60:

d_{60}=20{,}000(0.0125)=250,\qquad l_{61}=20{,}000-250=19{,}750.

At age 61:

d_{61}=19{,}750(0.025)=493.75,\qquad l_{62}=19{,}750-493.75=19{,}256.25.

At age 62:

d_{62}=19{,}256.25(0.04)=770.25,\qquad l_{63}=19{,}256.25-770.25=18{,}486.

At age 63:

d_{63}=18{,}486(0.05)=924.30,\qquad l_{64}=18{,}486-924.30=17{,}561.70.

The requested interval starts at age 61, so it includes \(d_{61}\), \(d_{62}\), and \(d_{63}\), but not \(d_{60}\):

$$ 493.75+770.25+924.30=2{,}188.30. $$

As a check, the same count is \(l_{61}-l_{64}=19{,}750-17{,}561.70=2{,}188.30\). The entries reported to two decimal places are therefore consistent.

Common mistake avoided: The requested deaths begin at age 61, not age 60. Including \(d_{60}\) would give deaths over the longer interval from age 60 to age 64.

Checks, Scope, and Exam Traps

A reliable construction has both row-level and interval-level checks. For each age, verify that \(d_x=l_xq_x\), that \(l_{x+1}=l_x(1-q_x)\), and that \(l_x=d_x+l_{x+1}\). For a multi-year interval, add the relevant death entries and compare their sum with the decrease in survivor entries at the interval endpoints. These checks can detect arithmetic errors and incorrect interval boundaries.

A frequent error is to confuse an assumed probability with an observed rate. In a constructed life table, \(q_x\) is the model probability and \(l_xq_x\) is the expected number of deaths. In an observed cohort, the actual number of deaths is an integer, and the observed rate is calculated from that realized count. The two quantities should not be described as identical merely because a life table uses the same symbols.

Another trap is the age index. The rate \(q_x\) belongs to the year beginning at age \(x\); it takes \(l_x\) to \(l_{x+1}\). It does not describe the year beginning at age \(x+1\), and the rate \(q_{x+1}\) cannot be applied until the table has advanced to \(l_{x+1}\). A radix by itself does not determine later entries: the sequence of age-specific probabilities is also needed.

Key takeaway: Start with the specified radix, calculate \(d_x=l_xq_x\), and then set \(l_{x+1}=l_x-d_x\). Each new row uses the updated survivor entry. In a constructed table, fractional entries are expected counts; for actual observations, deaths are realized counts and the observed mortality rate is \(d_x/l_x\).

Check Your Understanding

For each problem, use the stated annual mortality probabilities for the corresponding one-year age intervals. Constructed entries are expected counts.

A hypothetical table has radix \(l_{30}=800\), with \(q_{30}=0.125\) and \(q_{31}=0.25\). Find \(d_{30}\), \(l_{31}\), \(d_{31}\), and \(l_{32}\). (Answer: \(d_{30}=100\), \(l_{31}=700\), \(d_{31}=175\), \(l_{32}=525\))
A table begins with \(l_{40}=2{,}500\), and \(q_{40}=0.04\), \(q_{41}=0.08\), and \(q_{42}=0.10\). Find \(l_{43}\) and total expected deaths from age 40 to age 43. (Answer: \(l_{43}=1{,}987.20\); total deaths \(=512.80\))
A projected cohort has \(l_{55}=12{,}000\), with \(q_{55}=0.02\), \(q_{56}=0.03\), and \(q_{57}=0.05\). Find \(l_{58}\) and the total expected deaths over the three intervals. (Answer: \(l_{58}=10{,}836.84\); total deaths \(=1{,}163.16\))
A hypothetical life table starts with \(l_{70}=4{,}000\), and has \(q_{70}=0.10\), \(q_{71}=0.20\), and \(q_{72}=0.25\). Find \(l_{73}\), the total expected deaths from age 70 to 73, and the proportion of the radix surviving to age 73. (Answer: \(l_{73}=2{,}160\); total deaths \(=1{,}840\); proportion surviving \(=0.54\))
A table has \(l_{80}=5{,}000\), \(q_{80}=0.06\), \(q_{81}=0.04\), and \(l_{83}=4{,}200\). Construct \(l_{81}\) and \(l_{82}\), then determine the missing \(q_{82}\), rounding the probability to four decimal places. (Answer: \(l_{81}=4{,}700\), \(l_{82}=4{,}512\), \(q_{82}=0.0691\))

What You'll Learn