Recovering Missing Life Table Values

A life table may present only some of its entries, asking you to reconstruct the rest. The task is usually a direct application of the relationships among the number alive at an exact age, the number who die before the next age, and the one-year probabilities of death and survival. The main challenge is using the correct age and denominator for each entry.

The calculations below use a life table with one-year age intervals. No interest rate or parametric lifetime distribution is assumed; the table entries themselves provide the cohort counts or mortality probabilities used. The radix is the specified number of lives at a starting age. A radix is a scale for the table, not a probability, and the same identities apply whether counts are large, small, or fractional.

Definition: \(l_x\) is the number of lives in the table alive at exact age \(x\). \(d_x\) is the number of those lives who die between exact ages \(x\) and \(x+1\). The one-year probability of death is \(q_x\), and the one-year probability of survival is \(p_x\). Thus \(q_x\) and \(p_x\) refer to the interval beginning at age \(x\), while \(l_{x+1}\) counts survivors at the end of that interval.

The Identities That Connect the Columns

For a one-year interval, the lives at age \(x\) are divided into those who die during the interval and those who survive to age \(x+1\). Therefore, the number alive at age \(x\) equals deaths during the interval plus survivors at age \(x+1\). The probabilities \(q_x\) and \(p_x\) are calculated using the lives at age \(x\) as their denominator.

l_x=d_x+l_{x+1},\qquad d_x=l_x-l_{x+1}

q_x=\frac{d_x}{l_x},\qquad p_x=\frac{l_{x+1}}{l_x},\qquad p_x=1-q_x

These identities are equivalent ways to express the same one-year transition. In particular, if \(l_x>0\), then any one of the following forms can be used to find a missing entry:

d_x=l_xq_x,\qquad l_{x+1}=l_xp_x,\qquad l_{x+1}=l_x(1-q_x),\qquad l_x=\frac{l_{x+1}}{p_x}

The last formula requires \(p_x>0\). Likewise, calculating \(q_x=d_x/l_x\) or \(p_x=l_{x+1}/l_x\) requires \(l_x>0\). A life table cannot have negative counts, and its one-year probabilities must lie between zero and one. These are useful checks, but a plausible-looking value is not enough: the identities must agree across every completed interval.

Why the Formulas Hold

The probability interpretation follows directly from the life table cohort. Among the \(l_x\) lives alive at exact age \(x\), exactly \(d_x\) die before age \(x+1\), so the fraction dying is \(d_x/l_x\). The remaining \(l_{x+1}\) lives form the fraction surviving, \(l_{x+1}/l_x\). Since every life in the interval either dies or survives, those fractions add to one.

\frac{d_x}{l_x}+\frac{l_{x+1}}{l_x} =\frac{d_x+l_{x+1}}{l_x} =\frac{l_x}{l_x} =1

Consequently, \(q_x+p_x=1\), and because \(d_x=l_x-l_{x+1}\), we also have \(q_x=(l_x-l_{x+1})/l_x\). This derivation explains why \(l_x\), not \(l_{x+1}\), is the denominator of \(q_x\): the probability is conditional on being alive at the start of the interval.

A Reliable Reconstruction Method

Before calculating, mark the age attached to each entry. Then work one interval at a time. A missing count can often be found by subtracting deaths from the starting count or adding deaths to the ending count. Once the two adjacent survivor counts are known, both \(q_x\) and \(p_x\) follow by division.

Identify the interval.
For an interval from age \(x\) to age \(x+1\), use \(l_x\), \(d_x\), \(q_x\), and \(p_x\). Do not shift the probability to the ending age.

Use the radix and known counts.
Start from the stated \(l_x\) and use \(l_x=d_x+l_{x+1}\) to recover a missing count when possible.

Calculate probabilities from the starting lives.
Divide \(d_x\) or \(l_{x+1}\) by \(l_x\) to find \(q_x\) or \(p_x\).

Check the completed interval.
Verify that the counts add and that \(q_x+p_x=1\).

Worked Example: Filling Two Consecutive Rows from a Radix

A hypothetical annual life table has radix \(l_{40}=10{,}000\). It gives \(q_{40}=0.024\) and \(q_{41}=0.025\), but omits \(d_{40}\), \(l_{41}\), \(d_{41}\), and \(l_{42}\). Reconstruct the missing entries and find \(p_{40}\) and \(p_{41}\). Treat the stated probabilities as one-year mortality probabilities for the intervals beginning at the indicated ages; no interest rate or parametric distribution is involved.

Method. Work forward from the radix. First use \(d_x=l_xq_x\), then subtract deaths to obtain the next survivor count.

For age 40,

d_{40}=l_{40}q_{40}=10{,}000(0.024)=240

Therefore,

l_{41}=l_{40}-d_{40}=10{,}000-240=9{,}760

The survival probability is \(p_{40}=1-q_{40}=1-0.024=0.976\). Check by using the adjacent counts: \(9{,}760/10{,}000=0.976\).

For age 41, use the newly recovered \(l_{41}\):

d_{41}=l_{41}q_{41}=9{,}760(0.025)=244

Then \(l_{42}=l_{41}-d_{41}=9{,}760-244=9{,}516\), and \(p_{41}=1-0.025=0.975\). Checking the count ratio gives \(9{,}516/9{,}760=0.975\).

The reconstructed rows are consistent: \(10{,}000=240+9{,}760\) and \(9{,}760=244+9{,}516\). The common mistake avoided here is using the original radix for the age-41 calculation; the correct starting count for that interval is \(l_{41}=9{,}760\).

Worked Example: Recovering an Intermediate Mortality Rate

A hypothetical table has \(l_{50}=8{,}000\), \(q_{50}=0.020\), and \(l_{52}=7{,}560\). The entries \(d_{50}\), \(l_{51}\), \(d_{51}\), \(q_{51}\), and \(p_{51}\) are missing. Recover them. The table uses one-year intervals; no interest rate or parametric distribution is assumed.

Method. Find the age-50 deaths and survivors first. The two-year drop from \(l_{50}\) to \(l_{52}\) cannot by itself determine the two annual death counts, but the given \(q_{50}\) supplies the missing information for the first interval.

Calculate the age-50 deaths:

d_{50}=l_{50}q_{50}=8{,}000(0.020)=160

Then,

l_{51}=l_{50}-d_{50}=8{,}000-160=7{,}840

The age-51 deaths are the difference between the adjacent survivor counts:

d_{51}=l_{51}-l_{52}=7{,}840-7{,}560=280

The age-51 death probability uses \(l_{51}\) as its denominator:

q_{51}=\frac{d_{51}}{l_{51}} =\frac{280}{7{,}840} =\frac{1}{28} \approx 0.035714

Thus \(p_{51}=1-q_{51}=27/28\approx0.964286\). The check is \(7{,}840(27/28)=7{,}560\), as required. The age-50 row also checks: \(8{,}000=160+7{,}840\). The common mistake avoided is dividing the 280 deaths by \(l_{52}\); the probability \(q_{51}\) is conditional on being alive at age 51, so its denominator is \(l_{51}\).

Missing Probabilities and a Consistency Trap

Sometimes the table supplies a survival probability instead of a death probability. Use \(p_x=1-q_x\), or obtain \(p_x\) from the adjacent survivor counts. If both are available, calculating both ways is an efficient consistency check. Do not round an intermediate probability if the counts are exact; retain the fraction or sufficient precision until the final reported value.

Worked Example: Completing a Row from a Survival Probability

A hypothetical table has radix \(l_{60}=12{,}000\), reports \(d_{60}=720\), and gives \(p_{61}=0.950\). Find \(l_{61}\), \(q_{60}\), \(p_{60}\), \(d_{61}\), \(q_{61}\), and \(l_{62}\). The entries concern consecutive one-year age intervals; no interest rate or parametric distribution is assumed.

Method. First complete the age-60 interval from its count and deaths. Then apply the age-61 survival probability to the resulting \(l_{61}\).

At age 60,

l_{61}=l_{60}-d_{60}=12{,}000-720=11{,}280

Therefore,

q_{60}=\frac{720}{12{,}000}=0.060,\qquad p_{60}=\frac{11{,}280}{12{,}000}=0.940

The two probabilities add to \(1.000\), as required. At age 61, use the supplied survival probability:

l_{62}=l_{61}p_{61}=11{,}280(0.950)=10{,}716

The remaining age-61 values are

d_{61}=l_{61}-l_{62}=11{,}280-10{,}716=564

and

q_{61}=1-p_{61}=1-0.950=0.050

Check: \(11{,}280(0.050)=564\), and \(11{,}280=564+10{,}716\). The common mistake avoided is applying \(p_{61}\) to \(l_{60}\); it applies to lives alive at age 61, namely \(l_{61}\).

Common Errors and Final Checks

A completed table should satisfy the identities interval by interval. These checks are especially useful when a problem gives scattered entries at several ages, because an error in one reconstructed \(l_x\) can affect every later row.

Using the wrong denominator. Both \(q_x=d_x/l_x\) and \(p_x=l_{x+1}/l_x\) use the number alive at the start of the interval.
Shifting the age index. \(d_x\), \(q_x\), and \(p_x\) describe the interval from age \(x\) to age \(x+1\), not the interval beginning at age \(x+1\).
Forgetting the radix. A starting \(l_x\) sets the scale for subsequent counts. Do not replace it with a standard radix or rescale partway through the calculation.
Rounding too soon. Keep exact counts and fractions when possible, and round only the final reported probability.
Accepting inconsistent entries. Verify both \(l_x=d_x+l_{x+1}\) and \(q_x+p_x=1\). If the given values do not satisfy these relationships, recheck the arithmetic and the age labels.

Key takeaway: For each one-year interval, begin with the lives at age \(x\): \(l_x=d_x+l_{x+1}\), \(q_x=d_x/l_x\), and \(p_x=l_{x+1}/l_x=1-q_x\). Reconstruct one interval at a time and check both the count and probability identities.

Check Your Understanding

For each problem, use one-year intervals and the stated entries as exact. No interest rate or parametric lifetime distribution is assumed.

A hypothetical table has radix \(l_{30}=20{,}000\) and \(q_{30}=0.015\). Find \(d_{30}\), \(l_{31}\), and \(p_{30}\). (Answer: \(d_{30}=300\), \(l_{31}=19{,}700\), \(p_{30}=0.985\).)
A table gives \(l_{45}=6{,}400\) and \(l_{46}=6{,}208\). Find \(d_{45}\), \(q_{45}\), and \(p_{45}\). (Answer: \(d_{45}=192\), \(q_{45}=0.030\), \(p_{45}=0.970\).)
A table has \(l_{70}=9{,}000\), \(d_{70}=450\), and \(q_{71}=0.040\). Find \(l_{71}\), \(d_{71}\), and \(l_{72}\). (Answer: \(l_{71}=8{,}550\), \(d_{71}=342\), \(l_{72}=8{,}208\).)
A table gives \(l_{80}=5{,}000\), \(q_{80}=0.040\), and \(l_{82}=4{,}560\). Recover \(d_{80}\), \(l_{81}\), \(d_{81}\), \(q_{81}\), and \(p_{81}\). (Answer: \(d_{80}=200\), \(l_{81}=4{,}800\), \(d_{81}=240\), \(q_{81}=0.050\), \(p_{81}=0.950\).)
A table has \(l_{90}=1{,}250\), \(p_{90}=0.920\), and \(q_{91}=0.040\). Find \(l_{91}\), \(d_{90}\), \(l_{92}\), and \(d_{91}\). (Answer: \(l_{91}=1{,}150\), \(d_{90}=100\), \(l_{92}=1{,}104\), \(d_{91}=46\).)

These reconstruction techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Recovering Missing Life Table Values

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