Constructing Life Table Entries From A Force

From a Force of Mortality to Life-Table Entries

A life table records the survival and death counts of a hypothetical cohort as it moves through successive ages. Sometimes the mortality information is given as annual probabilities, such as $q_x$. In other problems, it is given as a force of mortality, which describes the instantaneous rate of mortality at each point in time. To construct life-table entries from a force, first accumulate that force over the interval; the accumulated value determines the probability of surviving the interval.

This distinction matters because a force is not itself a probability. A force may vary during a year, and even a constant force does not equal the probability of dying in that year. The conversion must account for the full interval.

Definition: The force of mortality at age $x+t$, written $\mu_{x+t}$, is the instantaneous rate of mortality at that age, conditional on survival to that age. If the force is nonnegative and integrable over the interval, the probability of surviving from age $x$ to age $x+t$ is $$ {}_tp_x=\exp\left(-\int_0^t \mu_{x+s}\,ds\right). $$ For a one-year interval, $p_x={}_1p_x$ and $q_x={}_1q_x=1-p_x$.

Why the Integrated Force Gives Survival

Let $S(t)={}_tp_x$, the probability that a life aged $x$ survives for at least $t$ more years. By the definition of the force, for a small positive interval $h$, the conditional probability of death between $t$ and $t+h$, given survival to $t$, is approximately $\mu_{x+t}h$. Therefore, the conditional probability of surviving that short interval is approximately $1-\mu_{x+t}h$. Multiplying by the probability of reaching time $t$ gives

S(t+h)\approx S(t)\bigl(1-\mu_{x+t}h\bigr).

Subtract $S(t)$, divide by $h$, and take the limit as $h$ approaches zero. This gives the differential equation $S'(t)=-\mu_{x+t}S(t)$, with $S(0)=1$. Separating variables and integrating yields

\frac{S'(t)}{S(t)}=-\mu_{x+t}, \qquad \ln S(t)=-\int_0^t\mu_{x+s}\,ds, \qquad S(t)=\exp\left(-\int_0^t\mu_{x+s}\,ds\right).

This argument also explains why the force must be integrated over time. The exponent is the accumulated force, not the force at a single instant. If the force is constant at $\mu$ throughout an interval of length $t$, then ${}_tp_x=e^{-\mu t}$. If the force changes, its integral must reflect each portion of the interval.

Constructing the Life-Table Columns

Once an interval probability is known, the life-table count calculations use the same identities as when the input is $q_x$. Choose a radix $l_0$, the starting number of lives in the hypothetical cohort. For each age $x$, use the force over the year from age $x$ to age $x+1$ to find $p_x$ and $q_x$. Then apply $d_x=l_xq_x$ and $l_{x+1}=l_xp_x=l_x-d_x$.

H_x=\int_0^1\mu_{x+s}\,ds, \qquad p_x=e^{-H_x}, \qquad q_x=1-e^{-H_x}, \qquad d_x=l_xq_x, \qquad l_{x+1}=l_xp_x.

Here $H_x$ is the accumulated force during the year. These formulas assume the force specified in the problem applies throughout the interval being considered. A force specified only at one age is not enough to determine a one-year probability unless an assumption about its behavior over the year is also given. For several years, integrate over the whole period, or calculate each annual probability and multiply the annual survival probabilities.

Worked Example: Constant Force and One-Year Counts

A hypothetical cohort has $l_{50}=10{,}000$. Assume a constant force of mortality $\mu_{50+t}=0.04$ per year throughout the year from age 50 to 51. Find $p_{50}$, $q_{50}$, and the life-table counts at age 51. Report probabilities to four decimal places and counts to the nearest whole life.

Method. Integrate the constant force over one year, convert the accumulated force to a probability, and apply the life-table identities.

The accumulated force is

H_{50}=\int_0^1 0.04\,ds=0.04.

Thus

p_{50}=e^{-0.04}=0.960789\ldots\approx0.9608, \qquad q_{50}=1-p_{50}=0.039210\ldots\approx0.0392.

Using the unrounded probability for the count calculation,

d_{50}=10{,}000(1-e^{-0.04})=392.1056\ldots\approx392, \qquad l_{51}=10{,}000e^{-0.04}=9{,}607.8944\ldots\approx9{,}608.

The rounded entries check: $9{,}608+392=10{,}000$. The method avoids treating 0.04, the force, as the one-year death probability; the actual probability is $1-e^{-0.04}$.

Worked Example: A Force That Changes Halfway Through the Year

For a hypothetical cohort aged 60, suppose the force is constant at $2\ln(10/9)$ per year during the first half-year and constant at $2\ln(5/4)$ per year during the second half-year. The radix is $l_{60}=50{,}000$. Find the one-year $p_{60}$, $q_{60}$, $d_{60}$, and $l_{61}$. Report the probability to four decimal places and counts to the nearest whole life.

Method. Integrate each constant force over the length of its subinterval, then add the two accumulated forces.

H_{60} =\frac12\left[2\ln(10/9)\right] +\frac12\left[2\ln(5/4)\right] =\ln(10/9)+\ln(5/4) =\ln(25/18) =0.328504\ldots.

The one-year survival probability is the exponential of the negative accumulated force:

p_{60}=e^{-H_{60}}=e^{-\ln(25/18)}=\frac{18}{25}=0.7200, \qquad q_{60}=1-0.7200=0.2800.

The counts are

d_{60}=50{,}000(0.2800)=14{,}000, \qquad l_{61}=50{,}000(0.7200)=36{,}000.

The check is $14{,}000+36{,}000=50{,}000$. A common incorrect calculation reports the accumulated force as 1.388; that does not follow from the two half-year contributions. Each contribution is its force multiplied by $1/2$, so the correct accumulated force is $0.3285\ldots$, giving a death probability of 0.2800 rather than approximately 0.750. The interval lengths must be included before converting the accumulated force to a probability.

Worked Example: Building Several Consecutive Ages

A hypothetical life table starts with $l_{70}=20{,}000$. Assume the force is constant within each age year, with $\mu_{70+t}=\ln(20/19)$ for $0\le t<1$, $\mu_{71+t}=\ln(10/9)$ for $0\le t<1$, and $\mu_{72+t}=\ln(20/19)$ for $0\le t<1$. Construct $p_x,q_x,d_x,l_{x+1}$ for ages 70, 71, and 72, and find the three-year survival probability. Report probabilities to four decimal places and counts to the nearest whole life.

Method. Convert each age-specific constant force separately, then carry the survivors forward as the radix for the next age.

At age 70, $H_{70}=\ln(20/19)$, so $p_{70}=19/20=0.9500$ and $q_{70}=0.0500$. Therefore $d_{70}=20{,}000(0.05)=1{,}000$, and $l_{71}=20{,}000-1{,}000=19{,}000$.

At age 71, $H_{71}=\ln(10/9)$, so $p_{71}=9/10=0.9000$ and $q_{71}=0.1000$. Using the age-71 survivors,

d_{71}=19{,}000(0.1000)=1{,}900, \qquad l_{72}=19{,}000-1{,}900=17{,}100.

At age 72, the force is again $\ln(20/19)$, so $p_{72}=0.9500$ and $q_{72}=0.0500$. Hence

d_{72}=17{,}100(0.0500)=855, \qquad l_{73}=17{,}100-855=16{,}245.

The three-year survival probability is the product of the annual survival probabilities:

{}_3p_{70}=0.95(0.90)(0.95)=0.81225\approx0.8123.

As a count check, $l_{73}/l_{70}=16{,}245/20{,}000=0.81225$, which agrees with the product. This calculation avoids applying the same annual rate at every age: the age-71 force differs from the forces at ages 70 and 72.

Practical Checks and Common Traps

A force-based life-table calculation has two distinct stages: integrate the force over the correct time interval, then convert the accumulated force into survival and death probabilities. Keeping those stages separate makes it easier to identify errors.

Do not equate force and probability. For a constant force $\mu$ over one year, $q_x=1-e^{-\mu}$, not $\mu$. The approximation $q_x\approx\mu$ may be close for a small force, but it is not the exact conversion.
Use the correct interval length. If a force applies for half a year, its contribution to the accumulated force is one-half of its annual value. For a $t$-year interval with constant force $\mu$, the accumulated force is $\mu t$.
Combine segments before exponentiating. For a piecewise force, add the integrated contributions. Equivalently, calculate each segment's survival factor and multiply those factors. Do not average the segment forces without accounting for their durations.
Carry survivors forward. The death count at age $x+1$ is calculated from $l_{x+1}$, not from the original radix, unless the question explicitly asks for deaths as a fraction of that radix.
Check the table identities. Within each year, $l_x=d_x+l_{x+1}$. Across multiple years, the ratio of the final survivor count to the initial count must equal the product of the annual survival probabilities.

When rounding is required, retain unrounded probabilities and counts during intermediate calculations, then round the requested final quantities. Rounding each annual probability too early can create small inconsistencies in the final table.

Key takeaway: Integrate the force over the interval to obtain $H$; then use $p=e^{-H}$ and $q=1-p$. Given a radix or current survivor count, calculate $d=lq$ and the next survivor count $l_{\text{next}}=lp=l-d$.

Check Your Understanding

For each problem, assume time is measured in years and use the stated force throughout each stated interval.

A hypothetical cohort has $l_{40}=8{,}000$, and the force is constant at $0.03$ per year from age 40 to 41. Find $p_{40}$, $q_{40}$, $d_{40}$, and $l_{41}$. Give probabilities to four decimals and counts to the nearest whole life. (Answer: $p_{40}=0.9704$, $q_{40}=0.0296$, $d_{40}=236$, $l_{41}=7{,}764$.)
A hypothetical force is constant at $\ln(8/7)$ per year for one year, and $l_{55}=5{,}600$. Find the one-year death probability and the death and survivor counts. (Answer: $q_{55}=0.1250$, $d_{55}=700$, $l_{56}=4{,}900$.)
For a hypothetical life aged 60, the force is $2\ln(6/5)$ for the first half-year and $2\ln(10/9)$ for the second half-year. If $l_{60}=12{,}000$, find the one-year survival and death probabilities and the counts at age 61. (Answer: $p_{60}=0.7500$, $q_{60}=0.2500$, $d_{60}=3{,}000$, $l_{61}=9{,}000$.)
A hypothetical table has $l_{40}=50{,}000$. The force is constant within each age year: $\mu_{40+t}=\ln(25/24)$ and $\mu_{41+t}=\ln(50/47)$, for $0\le t<1$. Find $l_{42}$ and ${}_2q_{40}$. (Answer: $l_{41}=48{,}000$, $l_{42}=45{,}120$, and ${}_2q_{40}=0.0976$.)
For a hypothetical life aged 30, let $\mu_{30+t}=6t\ln(5/4)$ for $0\le t\le1$, and let $l_{30}=25{,}000$. Find the accumulated force, one-year survival and death probabilities, and the two counts at age 31. (Answer: $H_{30}=3\ln(5/4)=0.6694\ldots$, $p_{30}=0.5120$, $q_{30}=0.4880$, $d_{30}=12{,}200$, $l_{31}=12{,}800$.)

What You'll Learn