Constant-Force Life-Table Calculations

Using Annual Life-Table Values at Fractional Ages

A life table usually gives survivor counts \(l_x\) at integer ages and annual death probabilities \(q_x\). Many calculations, however, begin or end at a fractional age, such as age 42.4. Under the constant-force assumption within each year of age, the annual survival probability determines survival at every point in that year. The main practical issue is to apply the force for the correct part of each year, especially when an interval crosses an integer age.

This tutorial uses life-table values to calculate fractional-age survivor counts and survival or death probabilities. These are model-based interpolations: they use the stated annual table values and the assumption that the force is constant within each year, not an assertion that actual deaths occur uniformly or at a constant rate in every population.

Definition: For an integer age \(x\), \(l_x\) is the life-table number surviving to exact age \(x\), and \(q_x\) is the probability that a life aged \(x\) dies before reaching age \(x+1\). The one-year survival probability is \(p_x=1-q_x=l_{x+1}/l_x\), when \(l_x>0\). Under a constant force of mortality during that year, the force is \(\mu_{x+t}=-\ln p_x\) for \(0\le t<1\), provided \(p_x>0\).

Fractional-Age Survivor Counts

Suppose the table gives \(l_x\) and \(q_x\), and the model assumes a constant force from age \(x\) to \(x+1\). A life aged \(x\) survives a fraction \(r\) of that year with probability \(p_x^r=(1-q_x)^r\), where \(0\le r\le1\). Thus the life-table survivor count at age \(x+r\) is \(l_x\) multiplied by that fractional survival probability.

{}_rp_x=(1-q_x)^r,\qquad l_{x+r}=l_x(1-q_x)^r

The expression \(l_{x+r}\) is a life-table count on the same scale as \(l_x\); it need not be a whole number. If only adjacent \(l\)-values are supplied, calculate \(p_x=l_{x+1}/l_x\) first. That ratio and \(1-q_x\) should agree when the table is internally consistent.

Important condition: A separate constant force is assumed in each integer-age year. The force in the year from age \(x\) to \(x+1\) is determined by \(q_x\); it can differ from the force in the next year. If \(q_x=1\), no life survives to age \(x+1\), and the finite-force expression \(-\ln(1-q_x)\) is not defined. For ordinary fractional-age calculations using a finite force, take \(q_x<1\).

Worked Example: Fractional Counts and Survival Across an Age Boundary

A hypothetical life table has \(l_{50}=12{,}000\), \(q_{50}=0.19\), and \(q_{51}=0.36\). Assume a constant force separately in each year of age. Find \(l_{50.5}\) and \(l_{51.5}\), then find the conditional probability of death between ages 50.5 and 51.5.

Method. Convert each annual death probability to a survival probability, apply the relevant fraction of each year's force, and compare the survivor counts at the two fractional ages.

For age 50 to 51, \(p_{50}=1-0.19=0.81\). The half-year survival factor is \(0.81^{0.5}=0.9\), so \(l_{50.5}=12{,}000(0.9)=10{,}800\).

The integer-age count at 51 is \(l_{51}=12{,}000(0.81)=9{,}720\). For age 51 to 52, \(p_{51}=1-0.36=0.64\), whose half-year survival factor is \(0.64^{0.5}=0.8\). Therefore, \(l_{51.5}=9{,}720(0.8)=7{,}776\).

The probability of surviving from age 50.5 to 51.5, conditional on being alive at 50.5, is \(7{,}776/10{,}800=0.72\). The conditional death probability is \(1-0.72=0.28\).

Check. The interval contains half a year under the age-50 force and half a year under the age-51 force. Its survival probability is \(0.81^{0.5}0.64^{0.5}=0.9(0.8)=0.72\), matching the survivor-count ratio. The common mistake avoided here is using just one year's \(q_x\) for an interval that crosses into a new age year.

Intervals Within or Across Integer Ages

For a conditional calculation from age \(a\) to a later age \(b\), the survival probability is the product of the survival contributions from every year segment the interval traverses. In the year of integer age \(j\), only the portion of the interval lying between \(j\) and \(j+1\) uses \(q_j\). If the interval remains within one year, this reduces to a single fractional power.

One concise way to express the calculation is to let \(h_j\) be the length of the overlap between the age interval \([a,b]\) and the year \([j,j+1]\). Only years with positive overlap contribute. Then:

P(\text{survive from }a\text{ to }b\mid\text{alive at }a) =\prod_j(1-q_j)^{h_j}, \qquad P(\text{die between }a\text{ and }b\mid\text{alive at }a) =1-\prod_j(1-q_j)^{h_j}

The exponents \(h_j\) are exposure lengths measured in years, and their sum is \(b-a\). This product is conditional on survival to age \(a\). It is not, in general, the proportion of the original cohort at some earlier age that dies in the interval. To obtain that cohort proportion, multiply the conditional death probability by the probability of surviving from the cohort's starting age to \(a\).

Worked Example: Conditional Death Probability From a Life Table

A hypothetical table has \(l_{60}=5{,}000\), \(q_{60}=0.20\), and \(q_{61}=0.10\). Under a constant force within each year, calculate the conditional probability of death from age 60.6 to age 61.3, to four decimal places.

Method. Divide the interval at age 61. From age 60.6 to 61, the exposure under \(q_{60}\) is \(0.4\) years. From 61 to 61.3, the exposure under \(q_{61}\) is \(0.3\) years.

The survival probability over the first segment is \(0.80^{0.4}\), and over the second it is \(0.90^{0.3}\). The full conditional survival probability is therefore \(0.80^{0.4}0.90^{0.3}=0.88615307\ldots\).

Subtracting from one gives the conditional death probability: \(1-0.88615307\ldots=0.11384693\ldots\). To four decimal places, the answer is 0.1138.

Check. The two exposure lengths add to \(0.4+0.3=0.7\) years, the full interval from 60.6 to 61.3. The two annual survival factors are both included, each with its own segment length. The common mistake avoided is applying \(q_{60}\) or \(q_{61}\) to the whole interval instead of using the correct year-specific exposure.

Why the Fractional-Power Formula Holds

The constant-force interpolation follows from the survival model within a single integer-age year. If the force is \(\mu\), survival over \(t\) years is \(e^{-\mu t}\). At one year, the survival probability must match the table value \(p_x=1-q_x\). Equating the two expressions determines the force and hence survival at any fractional duration.

e^{-\mu}=p_x \quad\Longrightarrow\quad \mu=-\ln p_x \quad\Longrightarrow\quad e^{-\mu r}=e^{r\ln p_x}=p_x^r=(1-q_x)^r

This derivation also explains why the model treats separate age years separately. The annual values may differ, so the force obtained from \(q_x\) need not equal the one obtained from \(q_{x+1}\). Across a boundary, survival through the first segment is multiplied by survival through the next segment; this multiplication follows from conditional survival over successive periods.

Worked Example: Recovering Annual Probabilities From Survivor Counts

A hypothetical life table gives \(l_{40}=10{,}000\), \(l_{41}=8{,}100\), and \(l_{42}=5{,}184\). Assume a constant force within each year. Find the fractional-age counts \(l_{40.5}\) and \(l_{41.5}\), and calculate the number of the original age-40 cohort expected to die between ages 40.5 and 41.5.

Method. First obtain each annual survival probability from the adjacent survivor counts. Then interpolate within the relevant year and take the difference in fractional-age counts for the cohort deaths.

For the year from 40 to 41, \(p_{40}=l_{41}/l_{40}=8{,}100/10{,}000=0.81\), so \(q_{40}=1-0.81=0.19\). Its half-year survival factor is \(0.81^{0.5}=0.9\), giving \(l_{40.5}=10{,}000(0.9)=9{,}000\).

For the year from 41 to 42, \(p_{41}=l_{42}/l_{41}=5{,}184/8{,}100=0.64\), so \(q_{41}=1-0.64=0.36\). Its half-year survival factor is \(0.64^{0.5}=0.8\), giving \(l_{41.5}=8{,}100(0.8)=6{,}480\).

The number of the original age-40 cohort alive at 40.5 but not alive at 41.5 is \(9{,}000-6{,}480=2{,}520\). As a proportion of the original cohort, that is \(2{,}520/10{,}000=0.252\). As a conditional probability given survival to 40.5, it is \(2{,}520/9{,}000=0.28\).

Check. The interval survival probability is \(0.9(0.8)=0.72\), so the conditional death probability is \(1-0.72=0.28\), in agreement with \(2{,}520/9{,}000\). The corresponding proportion of the original cohort is \(0.9(0.28)=0.252\). The common mistake avoided is reporting the conditional death probability as though it were the proportion of the original cohort that dies.

Checks and Common Exam Traps

Fractional-age calculations are short, but they are sensitive to endpoint placement and to the probability being requested. A reliable solution identifies the starting age, splits the interval at every integer age, and records the correct annual \(q_x\) and exposure length for each segment before multiplying.

Using a full-year probability for a partial year. For exposure \(r\), use \((1-q_x)^r\) for survival, not \(1-q_x\) unless \(r=1\).
Using the wrong annual value after a boundary. The segment after age \(x+1\) uses \(q_{x+1}\), not \(q_x\).
Confusing deaths with survival. Multiply survival factors first; the conditional death probability is one minus their product.
Confusing conditional and cohort probabilities. A conditional probability starts with a life already alive at the stated starting age. A cohort proportion uses the original cohort as its denominator.
Ignoring a life-table consistency check. Adjacent survivor counts should give \(l_{x+1}/l_x=1-q_x\). A disagreement usually signals a transcription or interpretation error.

These checks are useful because the same interpolation can be expressed through \(q_x\), \(p_x\), or \(l_x\). The method should give consistent results regardless of which form is supplied. When a calculation asks for a number of lives, use the appropriate \(l\)-count difference; when it asks for a probability conditional on being alive at a fractional age, divide by the fractional-age count at that starting age.

Key takeaway: Under a constant force within each age year, survival over a fraction \(r\) of year \(x\) is \((1-q_x)^r\). For an interval crossing integer ages, multiply the factors for the separate year segments, then take one minus the product for the conditional death probability.

Check Your Understanding

For each problem, assume a constant force within each integer-age year and retain unrounded values until the requested final answer.

A hypothetical table has \(l_{45}=8{,}000\) and \(q_{45}=0.19\). Find \(l_{45.5}\). (Answer: \(7{,}200\))
A hypothetical table has \(q_{70}=0.20\) and \(q_{71}=0.10\). Find the conditional probability of death from age 70.5 to 71.5, to four decimal places. (Answer: \(0.1515\))
A hypothetical table has \(l_{30}=10{,}000\), \(l_{31}=8{,}100\), and \(l_{32}=5{,}184\). Find \(l_{30.5}\) and \(l_{31.5}\). (Answer: \(l_{30.5}=9{,}000,\ l_{31.5}=6{,}480\))
A hypothetical table has \(q_{60}=0.20\) and \(q_{61}=0.10\). Find the conditional probability of death from age 60.6 to 61.3, to four decimal places. (Answer: \(0.1138\))
A hypothetical table has \(l_{50}=5{,}000\), \(l_{51}=4{,}000\), and \(l_{52}=3{,}200\). Find the conditional probability of death from age 50.25 to 51.75, to four decimal places. (Answer: \(0.2845\))

What You'll Learn