Why a Continuous Lifetime Has No Point Mass

Why an Exact Time Has Zero Probability

A lifetime model assigns probabilities to possible future lifetimes. For an absolutely continuous lifetime, probability is spread across time according to a density. A particular instant has no duration over which probability can accumulate, so the probability of death at exactly that instant is zero. This does not mean death cannot occur: an interval of time can have positive probability.

Let $T_x$ denote the future lifetime of a life aged $x$, and let $F_x(t)$, $S_x(t)$, and $f_x(t)$ denote its distribution function, survival function, and density. The earlier material on lifetime distributions establishes that probabilities across intervals can be calculated from the density or from differences in the distribution function.

Definition: $T_x$ is absolutely continuous if there is a density $f_x(t)$ such that probabilities over time intervals are obtained by integrating that density. In particular, for $a<b$, $$ P(a<T_x\le b)=\int_a^b f_x(t)\,dt. $$ For an absolutely continuous lifetime, the probability at any single time $t$ is $P(T_x=t)=0$.

The distinction is between a density and a probability. A density describes how probability accumulates locally; its value at a point is not the probability of that point. Probabilities come from integrating the density over a set of times. A density may even have units such as “per year,” whereas a probability has no units.

Why the Probability at One Time Is Zero

The result follows directly from the interval formula. A single time $t$ can be enclosed in intervals whose lengths shrink toward zero. Because a density integrates to zero over a single point, the probability assigned to that point is zero. Equivalently, for an absolutely continuous distribution, the probability of an interval ending at $t$ approaches zero as both endpoints approach $t$.

P(T_x=t)=\int_t^t f_x(s)\,ds=0.

The integral has identical lower and upper limits, so its value is zero. More generally, the interval probabilities around $t$ shrink to zero as the intervals shrink to that point. This is why an absolutely continuous lifetime has no point mass: no positive probability is concentrated at any single instant.

The distribution-function statement gives the same conclusion. A point probability is the jump in the distribution function at that time. An absolutely continuous distribution function has no jumps, so

$$ P(T_x=t)=F_x(t)-F_x(t-)=0, $$

where $F_x(t-)$ is the limit of $F_x(s)$ as $s$ approaches $t$ from below. The formula $F_x(b)-F_x(a)$ gives the probability between two times; changing whether an endpoint is included does not change the result for an absolutely continuous lifetime, because each endpoint has probability zero.

Worked Example: A Uniform Future Lifetime

A hypothetical model assigns a future lifetime $T_x$ uniformly over the interval from 0 to 12 years. Thus $f_x(t)=1/12$ for $0\le t\le12$, and the density is zero otherwise. Find the probability of death exactly 4 years from now and the probability of death after 3 years but no later than 5 years.

Method. Use a point probability for the exact time and integrate the density for the time interval. The uniform density is constant, and its total probability is $(12-0)(1/12)=1$, as required.

Exact time. The probability at exactly 4 is

P(T_x=4)=\int_4^4 \frac{1}{12}\,dt=0.

Interval. The interval has length $5-3=2$ years, so

P(3<T_x\le5)=\int_3^5 \frac{1}{12}\,dt =\frac{1}{12}(5-3) =\frac{2}{12} =\frac{1}{6} =0.1667.

The point probability is zero, while the two-year interval probability is $0.1667$. The common mistake is to multiply the density $1/12$ by nothing and report it as the probability of death at year 4. That confuses a density value with a probability. Check: the interval probability equals its length times the density, and it lies between 0 and 1.

Intervals Accumulate Probability

For an interval, probability is the area under the density over that interval. If the density is positive across an interval, its integral over that interval is positive. However, an interval having positive length does not, by itself, guarantee positive probability: a density can be zero throughout that interval. The applicable test is the integral, not just the interval's length.

The equality of endpoint conventions is also useful. Since $P(T_x=a)=P(T_x=b)=0$, these probabilities are equal for an absolutely continuous lifetime:

P(a<T_x\le b)=P(a\le T_x\le b)=P(a<T_x<b).

This endpoint result is specific to a model with no point masses. In a model that assigns positive probability to an exact time, including or excluding that time can change the probability. The focus here is the absolutely continuous case.

Worked Example: A Density That Increases with Time

For a hypothetical life, suppose $T_x$ has density $f_x(t)=t/18$ for $0\le t\le6$, and density zero otherwise. Find $P(T_x=3)$ and $P(2<T_x\le4)$. Confirm that the stated function is a density.

Method. First check that the density integrates to 1, then use a singleton integral for the exact time and an interval integral for the two-year range.

Density check.

\int_0^6 \frac{t}{18}\,dt =\left[\frac{t^2}{36}\right]_0^6 =\frac{36}{36}-0 =1.

Exact time.

P(T_x=3)=\int_3^3 \frac{t}{18}\,dt=0.

Interval. An antiderivative of $t/18$ is $t^2/36$, so

P(2<T_x\le4) =\int_2^4 \frac{t}{18}\,dt =\left[\frac{t^2}{36}\right]_2^4 =\frac{16}{36}-\frac{4}{36} =\frac{12}{36} =\frac{1}{3} =0.3333.

The density at 3 is $3/18=1/6$, but $1/6$ is not the probability of death exactly at year 3. Check: the interval probability is the increase in the distribution function from 2 to 4, and it is less than 1. The mistake avoided is treating the density height at an endpoint or interior point as a point probability.

A Density Can Change at a Time Without Creating a Point Mass

A density need not have the same value on both sides of every time. It may jump at a boundary and still describe an absolutely continuous lifetime. A change in density height affects how much probability accumulates in nearby intervals; it does not, on its own, place probability at the boundary.

Worked Example: A Piecewise-Constant Density

A hypothetical future lifetime has density $f_x(t)=0.10$ for $0\le t<4$, $f_x(t)=0.15$ for $4\le t\le8$, and zero otherwise. Find the probability of death exactly at 4 years and the probability of death after 3.5 but no later than 4.5 years.

Method. Check the total area, split an interval at the density change, and integrate each constant part separately.

Density check. The first section has width 4 and height 0.10; the second has width 4 and height 0.15. Therefore,

\int_0^8 f_x(t)\,dt =(4)(0.10)+(4)(0.15) =0.40+0.60 =1.

Exact time. Although the density changes at 4, the probability of the exact time is

P(T_x=4)=\int_4^4 f_x(t)\,dt=0.

Interval crossing the change. From 3.5 to 4, the width is $4-3.5=0.5$ years at density 0.10. From 4 to 4.5, the width is $4.5-4=0.5$ years at density 0.15. Thus

P(3.5<T_x\le4.5) =(0.5)(0.10)+(0.5)(0.15) =0.05+0.075 =0.1250.

The point probability remains zero even though the density has different values on either side of year 4. Check: the two interval pieces sum to $0.1250$, and the total density area over all possible lifetimes is 1. The mistake avoided is interpreting a jump in density as a jump in the distribution function; only a point mass creates such a jump.

Common Mistakes and Exam Traps

Reading the density as a point probability. A value such as $f_x(3)=1/6$ is a density, not $P(T_x=3)$. The point probability is zero in an absolutely continuous model.
Using an interval's length without the density. Probability is an integral. Length alone gives the probability only for a specified constant density, and even then it must be multiplied by that density.
Assuming every positive-length interval has positive probability. If the density is zero throughout the interval, its probability is zero. Check the integral over the stated range.
Letting endpoint notation change the answer. For an absolutely continuous lifetime, adding or removing a single endpoint does not change the probability. This is not generally true when a model has point masses.
Confusing a density discontinuity with a point mass. A density can change abruptly while the probability at the change point remains zero.

Key takeaway: For an absolutely continuous future lifetime, every exact time has probability zero. Probability accumulates over intervals by integrating the density; a positive interval probability depends on the density having positive integral there.

Check Your Understanding

For each model, distinguish the probability at an exact time from the probability over an interval. Round probabilities to four decimal places where needed.

A future lifetime is uniform from 0 to 20 years. Find $P(T_x=10)$ and $P(8<T_x\le12)$. (Answer: $0$ and $0.2000$)
A lifetime has density $f_x(t)=t/8$ for $0\le t\le4$, and zero otherwise. Find $P(T_x=2)$ and $P(1<T_x\le3)$. (Answer: $0$ and $0.5000$)
A density equals 0.20 for $0\le t<2$, equals 0.10 for $2\le t\le8$, and is zero otherwise. Find $P(T_x=2)$ and $P(1.5<T_x\le2.5)$. (Answer: $0$ and $0.1500$)
A lifetime density is zero for $0\le t<2$, equals 0.25 for $2\le t\le6$, and is zero otherwise. Find $P(1<T_x<2)$, $P(T_x=2)$, and $P(2<T_x\le4)$. (Answer: $0$, $0$, and $0.5000$)
A lifetime has density 0.50 for $0\le t\le1$, density 0.25 for $2\le t\le4$, and zero otherwise. Find $P(T_x=1)$ and $P(0.5<T_x\le2.5)$. (Answer: $0$ and $0.3750$)

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