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Complete Future Lifetime Distribution Under Constant Force

Learn to build and use the survival function and density of complete future lifetime when mortality changes at integer ages.

Intermediate 12 min read

What You'll Learn

  • Express survival from a fractional age using the time spent in each integer age year.
  • Construct the distribution function and density of complete future lifetime.
  • Calculate survival and death probabilities across one or more age-year boundaries.
  • Convert annual death probabilities into the corresponding constant forces.
  • Check density calculations against the force applying at the attained age.

Introduction

When mortality is modeled with a separate constant force in each integer age year, complete future lifetime has a continuous distribution whose rate of change can differ from year to year. A calculation from a fractional age must account for how much time is spent under each force. Extending the force from the starting age across the entire calculation is a common, consequential error.

This tutorial constructs the survival function and density of complete future lifetime under this model. It focuses on the distribution, not moments of the lifetime. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For a life currently aged \(x\), the complete future lifetime \(T_x\) is the nonnegative continuous time from age \(x\) until death. Its survival function is \({}_tp_x=P(T_x>t)\), and its distribution function is \(F_{T_x}(t)=P(T_x\le t)=1-{}_tp_x\), for \(t\ge 0\).

Piecewise Constant Force and Survival

Assume that, for every integer age \(j\) considered, the force of mortality is the finite constant \(\mu_j\) throughout the age interval \([j,j+1)\). Thus a life aged \(j+s\), where \(0\le s<1\), experiences force \(\mu_j\) until reaching age \(j+1\). At that boundary the applicable force changes to \(\mu_{j+1}\). This is the constant-force-within-each-year model; the force need not be the same in different age years.

The force \(\mu_j\) is a rate per year. If the life spends \(a_j(t)\) years in age interval \([j,j+1)\) during the next \(t\) years, the total cumulative force over that period is the sum of the force-times-duration terms. Here \(a_j(t)\) is the length of the intersection of \([x,x+t]\) with \([j,j+1)\).

Theorem: Under finite, piecewise constant forces, the survival function and distribution function are
$$ {}_tp_x=\exp\left(-\sum_j \mu_j a_j(t)\right), \qquad F_{T_x}(t)=1-\exp\left(-\sum_j \mu_j a_j(t)\right),\quad t\ge 0. $$
The sum includes the age years that the life occupies during the \(t\) years after age \(x\). For \(t=0\), all \(a_j(0)=0\), so \({}_0p_x=1\).

For example, if \(x=k+r\), where \(k\) is an integer and \(0\le r<1\), then \(1-r\) years remain in age year \(k\). For \(0\le t\le 1-r\), the life has not yet reached age \(k+1\), so

$$ {}_tp_x=e^{-\mu_k t}. $$

If \(t>1-r\), the time spent in age year \(k\) is \(1-r\), not \(t\); the remaining time is allocated to later age years. Writing the interval lengths explicitly is a reliable way to keep track of a calculation that crosses several boundaries.

Worked Example: Survival and Density Across an Age Boundary

A hypothetical life is aged \(40.25\). Assume \(\mu_{40}=\ln 2\) and \(\mu_{41}=\ln 4\), each constant over its stated age year. Find the probability of surviving \(1.25\) years and the density of \(T_{40.25}\) at \(t=1\).

Method. Split elapsed time at age 41. From age 40.25 to 41 is \(41-40.25=0.75\) years, and the remaining time to \(t=1.25\) is \(1.25-0.75=0.50\) years in age year 41. Therefore,

$$ {}_{1.25}p_{40.25} =\exp\left(-0.75\ln 2-0.50\ln 4\right) =\exp(-1.75\ln 2) =2^{-1.75} \approx 0.2973. $$

To find the density at \(t=1\), first find survival to that time. The life has spent \(0.75\) years in age year 40 and \(1-0.75=0.25\) years in age year 41:

$$ {}_1p_{40.25} =\exp\left(-0.75\ln 2-0.25\ln 4\right) =2^{-1.25} \approx 0.4204. $$

At \(t=1\), the attained age is \(41.25\), so the applicable force is \(\mu_{41}=\ln 4\). The density is therefore \((\ln 4){}_1p_{40.25}\approx 1.386294(0.420448)=0.5829\) per year. The survival probability is between 0 and 1, and the density is a rate, not a probability. The common mistake avoided here is applying \(\mu_{40}\) after age 41.

The Distribution Function and Density

The distribution function follows directly from the complement rule: death by time \(t\) is the event that the life does not survive beyond \(t\). Within any one age-year segment, the force is constant, so the density is the force in force at the attained age multiplied by the probability of surviving to that time.

Result: If \(t\) lies within a segment for which the attained age \(x+t\) is in \([j,j+1)\), then, away from an integer-age boundary,
$$ f_{T_x}(t)=\frac{d}{dt}F_{T_x}(t) =\mu_j\,{}_tp_x. $$
For \(x+t\in[j,j+1)\), the same statement can be written \(f_{T_x}(t)=\mu_j\,{}_tp_x\). The density has units of inverse years when \(t\) is measured in years.

At an integer-age boundary the force may change abruptly, so the left- and right-hand density values may differ. The value assigned to a density at one isolated point does not affect probabilities. Since each force is finite, survival is continuous at each boundary, and the model assigns zero probability to death at any exact time \(t\). In particular, it has no point mass at an integer age.

The density formula can be justified directly. On a short interval of length \(h>0\) contained in age year \(j\), conditional survival over the interval is \(e^{-\mu_jh}\). Thus the conditional probability of death in that interval is \(1-e^{-\mu_jh}\). Given survival to time \(t\), the unconditional probability of death between \(t\) and \(t+h\) is

$$ {}_tp_x\left(1-e^{-\mu_jh}\right). $$

Divide by \(h\) and let \(h\) decrease to zero. Since \(\lim_{h\to 0}(1-e^{-\mu_jh})/h=\mu_j\), the limiting rate of probability per unit time is \(\mu_j{}_tp_x\), which is the density on that segment. This also explains why the force used in the density is the one at the attained age, not necessarily the one at the starting age.

1
Mark the age boundaries.
Starting from age \(x\), find the elapsed times at which the life reaches each next integer age.
2
Allocate elapsed time.
For each occupied age year, record the portion of the requested interval spent there. The portions must add to the total elapsed time.
3
Build survival, then the requested quantity.
Use the force-duration sum in the exponential for survival. Use its complement for death by time \(t\), a difference for death in an interval, and the attained-age force times survival for density.

Death Probabilities Over an Interval

For \(0\le s<t\), the probability of death after \(s\) but by \(t\), measured from age \(x\), is

$$ P(s<T_x\le t)={}_sp_x-{}_tp_x. $$

The first term is the probability of being alive at time \(s\); subtract from it the probability of still being alive at time \(t\). This subtraction works even when the interval crosses one or more age boundaries, provided each survival probability is calculated with the correct force-duration segments.

Worked Example: Death Probability Across Two Boundaries

A hypothetical life is aged \(58.60\). Assume \(\mu_{58}=\ln(4/3)\), \(\mu_{59}=\ln 2\), and \(\mu_{60}=\ln 4\), with each force constant over its corresponding age year. Find the probability of death between \(0.70\) and \(1.70\) years from now.

Method. Age 59 is reached after \(59-58.60=0.40\) years. At \(s=0.70\), the life has spent \(0.40\) years in age year 58 and \(0.30\) years in age year 59. Hence

$$ {}_{{0.70}}p_{58.60} =\exp\left(-0.40\ln(4/3)-0.30\ln 2\right) \approx 0.7240. $$

At \(t=1.70\), the life has spent \(0.40\) years in age year 58, \(1.00\) year in age year 59, and \(0.30\) years in age year 60. Therefore,

$$ {}_{{1.70}}p_{58.60} =\exp\left(-0.40\ln(4/3)-1.00\ln 2-0.30\ln 4\right) \approx 0.2940. $$

Subtract the two survival probabilities:

$$ P(0.70<T_{58.60}\le 1.70) ={}_{0.70}p_{58.60}-{}_{1.70}p_{58.60} \approx 0.7240-0.2940 =0.4300. $$

The answer is a probability, so it is dimensionless. The interval includes portions of age years 59 and 60, but the survival calculation at its earlier endpoint does not include any time in age year 60. The mistake to avoid is using the force at age 58.60 for the whole interval or omitting the \(0.30\) years in age year 60 at the later endpoint.

Using Annual Death Probabilities

Sometimes the given information is an annual death probability \(q_j\), rather than a force. Under the constant-force assumption for age year \(j\), one-year survival is \(e^{-\mu_j}\). Equating this to \(1-q_j\) gives the conversion below. The conversion is valid only when the force is constant throughout that year; it does not assert that the force is constant across different age years.

Result: If the one-year death probability at integer age \(j\) is \(q_j\) under a constant force during \([j,j+1)\), then
$$ 1-q_j=e^{-\mu_j}, \qquad \mu_j=-\ln(1-q_j). $$
For \(0\le s\le 1\) within that age year, survival from age \(j\) through \(s\) years is \((1-q_j)^s\).

Worked Example: Constructing the Distribution from Annual Probabilities

A hypothetical life is aged \(52.60\). Assume annual death probabilities \(q_{52}=0.20\) and \(q_{53}=0.50\), with a separate constant force in each age year. Find the probability of death by \(1.40\) years and the density at \(t=1.10\).

Method. First convert each annual probability to its age-year force:

$$ \mu_{52}=-\ln(1-0.20)=-\ln(0.80)=\ln(1.25), \qquad \mu_{53}=-\ln(1-0.50)=\ln 2. $$

The life reaches age 53 after \(0.40\) years. By \(t=1.40\), it has spent \(0.40\) years in age year 52 and \(1.00\) year in age year 53. Thus

$$ {}_{1.40}p_{52.60} =\exp\left(-0.40\ln(1.25)-1.00\ln 2\right) =(0.80)^{0.40}(0.50) \approx 0.4573. $$

The probability of death by that time is the complement:

$$ F_{T_{52.60}}(1.40)=1-{}_{1.40}p_{52.60} \approx 1-0.4573=0.5427. $$

At \(t=1.10\), the life has spent \(0.40\) years in age year 52 and \(1.10-0.40=0.70\) years in age year 53. Therefore,

$$ {}_{1.10}p_{52.60} =\exp\left(-0.40\ln(1.25)-0.70\ln 2\right) =(0.80)^{0.40}(0.50)^{0.70} \approx 0.5630. $$

The attained age is \(53.70\), so the density uses \(\mu_{53}=\ln 2\):

$$ f_{T_{52.60}}(1.10) =(\ln 2){}_{1.10}p_{52.60} \approx 0.693147(0.5630) =0.3902 $$

per year. The density is not the probability of dying exactly at \(1.10\); it is a probability rate at that time. This calculation also avoids the trap of treating \(q_{52}=0.20\) as a force: the force is \(-\ln(0.80)\), not \(0.20\).

Common Boundary and Density Traps

A boundary is determined by attained age, not by the number of years written in a question. If a life is aged \(x=k+r\), it reaches age \(k+1\) after \(1-r\) years. Any requested survival duration longer than \(1-r\) requires information about mortality in at least one later age year.

For instance, knowing only \(\mu_{30}=\ln 2\) for a life aged \(30.25\) is enough to calculate survival for the next \(0.25\) years: it is \(e^{-0.25\ln 2}\approx 0.8409\). It is not enough to determine survival for the next \(0.50\) years, because the final \(0.25\) years are spent in age year 31 and \(\mu_{31}\) has not been specified. Extending \(\mu_{30}\) across that boundary would silently impose an additional assumption.

A second trap is confusing survival with density. Survival is a probability and is dimensionless; density is a rate and requires multiplication by the force applying at the attained age. For an interval probability, subtract survival values. Do not substitute a density value for an interval probability unless integrating the density over the interval.

Key takeaway: For each requested time, partition the elapsed duration at every integer-age boundary. Survival is the exponential of the negative sum of force times time in each age year; the density is survival to the time multiplied by the force then in effect.

Check Your Understanding

For each problem, assume the stated force is constant throughout its named integer age year. Give probabilities and densities to four decimal places.

  1. A life is aged \(33.40\) and \(\mu_{33}=\ln 2\). Find the probability of surviving the next \(0.30\) years. (Answer: \(0.8123\))
  2. A life is aged \(46.70\), with \(\mu_{46}=\ln(5/4)\) and \(\mu_{47}=\ln 2\). Find the probability of surviving \(0.80\) years. (Answer: \(0.6613\))
  3. A life is aged \(60.20\), with \(\mu_{60}=\ln 2\) and \(\mu_{61}=\ln 4\). Find the probability of death between \(0.80\) and \(1.30\) years from now. (Answer: \(0.2872\))
  4. For a life aged \(70.50\), annual death probabilities are \(q_{70}=0.25\) and \(q_{71}=0.20\), with constant force separately in each age year. Find the probability of surviving \(1.50\) years. (Answer: \(0.6928\))
  5. A life is aged \(25.60\), with \(\mu_{25}=0.10\) and \(\mu_{26}=0.30\). Find the density of complete future lifetime at \(t=0.70\), in inverse years. (Answer: \(0.2634\) per year)

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