Complete Future Lifetime Density Under UDD

Why a Density Is Needed

A life table gives annual death probabilities, but complete future lifetime is measured in years and can take fractional values. Under the uniform distribution of deaths assumption (UDD), the annual probability of death determines a constant density within each integer-age year. The conditioning event matters: a density measured from age \(x\) is not generally the same as one measured after survival to a later age.

This tutorial derives the density for complete future lifetime, with particular attention to the current year of age. It also distinguishes three quantities that are easy to confuse: density from age \(x\), density conditional on reaching the start of a year, and density conditional on dying during that year.

Definition: For a life known to be alive at exact age \(x\), the complete future lifetime \(T_x\) is the time, in years, from age \(x\) until death. Write \(q_{x+k}\) for the probability that a life alive at integer age \(x+k\) dies before age \(x+k+1\), and \(p_{x+k}=1-q_{x+k}\). A density for \(T_x\) is measured per year of elapsed lifetime.

UDD Within the Current Year

Under UDD, conditional on being alive at age \(x\), deaths during the year from age \(x\) to age \(x+1\) are uniformly distributed over that year. Thus, for \(0\leq t\leq 1\), the probability of dying within the first \(t\) years is \(tq_x\), and the probability of surviving \(t\) years is \(1-tq_x\).

{}_tp_x=P(T_x>t)=1-tq_x,\qquad 0\leq t\leq 1.

The density is the rate at which the distribution function increases, or equivalently the negative rate at which the survival function decreases. Differentiating the survival expression gives the density during this year.

f_{T_x}(t)=-\frac{d}{dt}{}_tp_x=q_x,\qquad 0<t<1.

This is a density conditional on being alive at age \(x\). Its value is \(q_x\), not 1, because only a fraction \(q_x\) of lives alive at age \(x\) die during the year. The density has units of inverse years; since the time unit here is one year, it is customary to report the numerical value as \(q_x\) per year.

Key distinction: From age \(x\), the density of death at a duration \(t\) in the first year is \(q_x\). If instead we condition on death occurring somewhere in that year, the location of death within the year is uniform, so its conditional density is \(1\) per year on \(0<t<1\).

Deriving the Density in Later Years

For a later year, two events are required for death at a specified duration: the life must first survive from age \(x\) to the start of that year, and then die at the specified point within the year. Under UDD, the second part has density \(q_{x+k}\) per year in the year beginning at age \(x+k\).

Reach the year.
For an integer \(k\geq 0\), the probability of surviving \(k\) years from age \(x\) is \({}_kp_x\).

Use UDD within that year.
Conditional on reaching age \(x+k\), the density of death at any duration \(s\) with \(0<s<1\) is \(q_{x+k}\).

Combine the probabilities.
The density from age \(x\) is the probability of reaching the year multiplied by the within-year conditional density.

f_{T_x}(t)={}_kp_xq_{x+k}, \qquad k<t<k+1,\quad k=0,1,2,\ldots

Here \({}_kp_x\) is the probability of surviving \(k\) years from age \(x\), and the annual probability \(q_{x+k}\) applies to the year beginning at age \(x+k\). For example, in the second year, the density from age \(x\) is \(p_xq_{x+1}\). Conditional on having survived to age \(x+1\), however, the density at a duration of 0.4 years after that age is \(q_{x+1}\).

The formula also gives the probability of death during a whole year when integrated over its one-year duration:

\int_k^{k+1}f_{T_x}(t)\,dt ={}_kp_xq_{x+k}.

This agrees with the probability of surviving to age \(x+k\) and then dying before age \(x+k+1\). At an integer boundary, the left- and right-hand density formulas can differ if adjacent annual mortality probabilities differ. There is no point mass at the boundary under UDD: the probability of death at any one exact duration is zero. A density's value at an individual boundary can therefore be assigned either adjacent value without changing probabilities.

Worked Examples

Worked Example: Density and Probability in the First Year

A hypothetical life table has \(q_x=0.12\). For a life alive at exact age \(x\), find (a) the density of \(T_x\) at duration 0.4 years, (b) the probability that death occurs between durations 0.2 and 0.7 years, and (c) the density conditional on death occurring during the first year. Assume UDD.

Method. Apply the first-year UDD density for the unconditional-from-age-\(x\) quantity, and normalize by the probability of death in the year for the death-year-conditional quantity.

(a) Since \(0<0.4<1\), the density from age \(x\) is constant throughout the first year:

f_{T_x}(0.4)=q_x=0.12\text{ per year}.

(b) The interval has length \(0.7-0.2=0.5\) years. Its probability is its length multiplied by the constant density:

$$ P(0.2<T_x<0.7)=(0.7-0.2)(0.12)=0.5(0.12)=0.06. $$

(c) The probability of death during the first year is \(q_x=0.12\). Conditional on death in that year, the density is the original density divided by this probability:

f_{T_x\mid T_x<1}(t)=\frac{0.12}{0.12}=1, \qquad 0<t<1.

Thus the three answers are \(0.12\) per year, \(0.06\), and \(1\) per year. Check: the conditional probability of death between 0.2 and 0.7, given death in the first year, is \(0.06/0.12=0.5\), which equals the fraction of the year covered by the interval. The common mistake is to report \(q_x\) as the density after conditioning on death during the year; that conditioning changes the density to 1.

Worked Example: Separating Three Second-Year Densities

A hypothetical life table has \(q_x=0.08\) and \(q_{x+1}=0.05\). Find (a) the density of \(T_x\) at duration 1.40 years, (b) the density at duration 0.40 years measured from age \(x+1\), conditional on survival to age \(x+1\), and (c) the density of the time within the second year conditional on dying during that second year.

Method. First calculate survival to the start of the second year; then keep the conditioning event explicit for each part.

(a) The life must survive the first year to reach age \(x+1\):

{}_1p_x=1-q_x=1-0.08=0.92.

Duration 1.40 lies in the second year, so the density from age \(x\) is:

f_{T_x}(1.40)={}_1p_xq_{x+1}=(0.92)(0.05)=0.046\text{ per year}.

(b) Conditional on already surviving to age \(x+1\), the within-year density is simply:

f(0.40\text{ years after }x+1\mid\text{alive at }x+1)=q_{x+1}=0.05\text{ per year}.

(c) Conditional on dying during the second year, the time of death within that year is uniform. Its density is 1 per year on the one-year interval. Equivalently, the density from age \(x\) at a second-year duration is \(0.046\), while the probability of dying in the second year from age \(x\) is \(0.92(0.05)=0.046\); their ratio is \(0.046/0.046=1\).

Therefore the answers are \(0.046\), \(0.05\), and \(1\), respectively. The values differ because the conditioning differs. In particular, \(0.05\) is conditional on reaching age \(x+1\), but not conditional on dying during the second year.

Worked Example: An Interval Crossing an Integer Age

For a hypothetical life, let \(q_x=0.10\) and \(q_{x+1}=0.20\). Under UDD, find (a) \(P(0.5<T_x<1.5)\), (b) the density at duration 1.25 years from age \(x\), and (c) the probability in part (a) conditional on surviving to age \(x+0.5\).

Method. Split the interval at duration 1, because the density changes at the integer-age boundary.

(a) From duration 0.5 to 1, the interval length is \(0.5\) and the density is \(q_x=0.10\). From duration 1 to 1.5, survival to age \(x+1\) is \(p_x=1-0.10=0.90\), so the density from age \(x\) is \(0.90(0.20)=0.18\). Therefore:

$$ P(0.5<T_x<1.5) =(1-0.5)(0.10)+(1.5-1)(0.90)(0.20) =0.05+0.09=0.14. $$

(b) Since 1.25 lies in the second year:

f_{T_x}(1.25)=p_xq_{x+1}=(0.90)(0.20)=0.18\text{ per year}.

(c) Survival to duration 0.5 has probability \(1-(0.5)(0.10)=0.95\). The event in part (a) is contained within survival to duration 0.5, so divide its probability by 0.95:

P(0.5<T_x<1.5\mid T_x>0.5) =\frac{0.14}{0.95}=0.147368\ldots\approx 0.1474.

The answers are \(0.14\), \(0.18\) per year, and \(0.1474\). The check is that the conditional probability exceeds the unconditional probability, as expected after restricting to lives known to have survived the first half-year. The common mistake is to use the second-year density over the entire interval; the portion before duration 1 uses the first-year density.

Common Conditioning and Boundary Errors

Density calculations are especially vulnerable to omitted conditioning. Before using a value, identify the population to which it applies: all lives alive at age \(x\), lives that reached age \(x+k\), or lives that died during a specified year. Those populations have different density scales.

Common exam trap: With \(q_x=0.08\) and \(q_{x+1}=0.05\), the density at duration 1.40 from age \(x\) is not \(0.05\). That value conditions on survival to age \(x+1\). From age \(x\), the correct density is \((1-0.08)(0.05)=0.046\). If the question instead conditions on dying during the second year, the density of the location within that year is \(1\) per year.

A second error is treating the annual death probability as a probability at a point. A density is not itself the probability of death at an exact duration; probabilities come from integrating density over an interval. For example, in the first year, \(P(0.2<T_x<0.7)=0.5q_x\), not \(q_x\) and not \(0.5\).

A third error is ignoring the integer-age boundary. Under UDD, the density from age \(x\) is constant within each year, but it can change at duration 1, 2, and so on, because both survival to that year and the annual mortality probability can change. An interval crossing one of these boundaries must be split into year segments.

Why the Density Representation Matters

The density gives a continuous-time description consistent with annual life-table probabilities. It makes fractional-year probabilities straightforward and provides the distributional building block for later calculations involving complete future lifetime. The key result can be summarized as follows.

Theorem: Under UDD in every integer-age year, for a life alive at age \(x\), the complete future lifetime density is constant within each year. In the year from age \(x+k\) to \(x+k+1\), its value from age \(x\) is \({}_kp_xq_{x+k}\). Conditional on survival to age \(x+k\), the density within that year is \(q_{x+k}\); conditional on death during that year, the within-year location has density 1 per year.

These are not competing formulas. They answer different conditional questions. Writing the conditioning event before calculating is the most reliable way to choose the correct one. This distinction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each question, assume UDD within every stated year. Keep the conditioning event explicit.

A life table has \(q_x=0.14\). Find the density of \(T_x\) at duration 0.35, the probability of death between durations 0.2 and 0.6, and the density conditional on death in the first year. (Answer: \(0.14\) per year; \(0.056\); \(1\) per year.)
Let \(q_x=0.10\) and \(q_{x+1}=0.25\). Find the density at duration 1.30 from age \(x\), the density 0.30 years after age \(x+1\) conditional on reaching age \(x+1\), and the density conditional on dying during the second year. (Answer: \(0.90(0.25)=0.225\) per year; \(0.25\) per year; \(1\) per year.)
Suppose \(q_x=0.08\). Find the probability of death between durations 0.2 and 0.8 and the conditional probability of that interval given death in the first year. (Answer: \(0.6(0.08)=0.048\); \(0.048/0.08=0.6\).)
Let \(q_x=0.12\) and \(q_{x+1}=0.10\). Find \(P(0.75<T_x<1.25)\), splitting the interval at duration 1. (Answer: \(0.25(0.12)+0.25(0.88)(0.10)=0.052\).)
Let \(q_x=0.20\) and \(q_{x+1}=0.10\). Find \(P(0.5<T_x<1.5\mid T_x>0.5)\). (Answer: \(\{0.5(0.20)+0.5(0.80)(0.10)\}/\{1-0.5(0.20)\}=0.14/0.90\approx 0.1556\).)

What You'll Learn