Converting a Density into a Distribution Function

From a Lifetime Density to a Cumulative Probability

A lifetime density describes how probability is distributed across possible durations, but a density value at one instant is not the probability of death at that instant or by that time. To find the probability that a life aged $x$ dies within a specified duration, accumulate the density over all durations from zero up to the specified time. This is the role of the distribution function $F_x(t)$.

The time variable $t$ measures future lifetime after age $x$, not age attained. For example, $t=5$ means five years after the life is age $x$. The calculation depends on using the same time unit in the density and in the integration limits.

Definition: Let $T_x$ be the future lifetime of a life aged $x$. Its distribution function is $F_x(t)=P(T_x\le t)$, the probability of death within $t$ units of time. If $T_x$ has a continuous density $f_x(t)$ over nonnegative time, then, for $t\ge 0$, $$ F_x(t)=\int_0^t f_x(u)\,du. $$ Here $u$ is a dummy variable of integration. The density $f_x(u)$ is measured in inverse time units, so integrating it over a time interval produces a dimensionless probability.

This formula assumes the future lifetime follows a continuous distribution described by the specified density over its full range. In particular, the density must be nonnegative and its integral over that range must equal one. The formula does not, by itself, account for a model that assigns positive probability to a particular lifetime as a separate point mass.

Why Integration Gives the Distribution Function

The distribution function asks for the probability of all outcomes with future lifetime between zero and $t$. Divide that time range into many small intervals. For an interval of width $\Delta u$ near time $u$, the probability represented by the density is approximately $f_x(u)\Delta u$. Adding these contributions over the range from zero to $t$, and taking the limit as the intervals become arbitrarily small, gives the integral.

P(T_x\le t) =\lim_{\max \Delta u\to 0}\sum f_x(u_j)\Delta u =\int_0^t f_x(u)\,du.

The integration starts at zero because future lifetime cannot be negative. It ends at $t$ because $F_x(t)$ includes every lifetime up to and including that duration. For a continuous model, including or excluding the single endpoint does not change the integral. The resulting value must be between zero and one. As $t$ increases, the range being accumulated grows, so the cumulative probability cannot decrease.

A useful check is to integrate the density over its entire support. The result should be one: the model assigns total probability one to all possible future lifetimes. If it does not, either the density has been copied incorrectly, the support has been misread, or the proposed function is not a valid density as stated.

Worked Examples

Worked Example: Integrating a Decreasing Density

A hypothetical lifetime model measures future lifetime in years and specifies $f_x(t)=(8-t)/32$ for $0\le t\le 8$, with density zero outside that interval. Calculate the probability of death within 3 years and within 6 years.

Method. Integrate the density from zero to each requested duration. First check that the stated function is normalized:

\int_0^8 \frac{8-u}{32}\,du =\left[\frac{u}{4}-\frac{u^2}{64}\right]_0^8 =2-1=1.

The density is nonnegative on the specified support, since $8-t\ge 0$ for $0\le t\le8$. Its cumulative distribution function on that support is

F_x(t)=\int_0^t\frac{8-u}{32}\,du =\frac{t}{4}-\frac{t^2}{64}.

At 3 years,

F_x(3)=\frac{3}{4}-\frac{9}{64} =\frac{48-9}{64} =\frac{39}{64} =0.609375.

At 6 years,

F_x(6)=\frac{6}{4}-\frac{36}{64} =1.5-0.5625 =0.9375.

Therefore, the probabilities of death within 3 years and within 6 years are $0.6094$ and $0.9375$, respectively, to four decimal places. The second probability exceeds the first, as expected for a cumulative probability. Common mistake avoided: treating $f_x(3)$ as the probability of death within 3 years. The requested probability is the area from zero to 3, not the density’s value at 3.

Worked Example: Building the CDF for a Piecewise Density

For a hypothetical group of lives aged $x$, future lifetime is measured in years and has density

f_x(t)= \begin{cases} 0.1, & 0\le t<2,\\ 0.2, & 2\le t<4,\\ 0.1, & 4\le t\le8,\\ 0, & \text{otherwise}. \end{cases}

Find the distribution function on the support and calculate the probabilities of death within 3 years and within 6 years.

Method. Integrate each constant-density section over the portion of that section included before $t$. Check the total probability first:

$$ (0.1)(2-0)+(0.2)(4-2)+(0.1)(8-4) =0.2+0.4+0.4=1. $$

For $0\le t<2$, only the first section contributes, giving $F_x(t)=0.1t$. For $2\le t<4$, the first section contributes $0.2$, followed by density $0.2$ over a duration of $t-2$. For $4\le t\le8$, the first two sections contribute $0.2+0.4=0.6$, followed by density $0.1$ over a duration of $t-4$. Thus,

F_x(t)= \begin{cases} 0, & t<0,\\ 0.1t, & 0\le t<2,\\ 0.2+0.2(t-2), & 2\le t<4,\\ 0.6+0.1(t-4), & 4\le t\le8,\\ 1, & t>8. \end{cases}

At 3 years, the first section contributes $0.1(2)=0.2$, and the part of the second section from year 2 to year 3 contributes $0.2(1)=0.2$. Therefore, $F_x(3)=0.2+0.2=0.4000$. At 6 years, the first two sections contribute $0.2+0.4=0.6$, and the third section from year 4 to year 6 contributes $0.1(2)=0.2$. Therefore, $F_x(6)=0.6+0.2=0.8000$.

The formulas agree at the breakpoints: at $t=2$, both adjoining expressions give $0.2$; at $t=4$, both give $0.6$; and at $t=8$, the final expression gives $1$. Common mistake avoided: integrating only the density in the section containing the requested time. A cumulative probability must include the contributions from every earlier section as well.

Worked Example: Converting Months to the Density’s Time Unit

A hypothetical model gives future lifetime in years with density $f_x(t)=t/18$ for $0\le t\le6$, and zero elsewhere. Find the probability of death within 18 months and within 4 years.

Method. Confirm the model’s time unit, normalize the density, and convert any duration stated in months into years before substituting it into the CDF. The total probability is

\int_0^6\frac{u}{18}\,du =\left[\frac{u^2}{36}\right]_0^6 =\frac{36}{36}=1.

For a duration $t$ in years, integration gives

F_x(t)=\int_0^t\frac{u}{18}\,du =\frac{t^2}{36}, \qquad 0\le t\le6.

Convert 18 months to years: $18/12=1.5$ years. Then

F_x(1.5)=\frac{(1.5)^2}{36} =\frac{2.25}{36} =0.0625.

For 4 years,

F_x(4)=\frac{4^2}{36} =\frac{16}{36} =\frac{4}{9} =0.4444\ldots.

The probabilities are $0.0625$ within 18 months and $0.4444$ within 4 years, to four decimal places. Both durations fall within the model’s 6-year support. Common mistake avoided: substituting $18$ directly into a formula whose argument is measured in years. The formula requires $1.5$, not $18$.

Checks and Common Exam Traps

Before accepting a calculated value, check the density, the integration limits, and the units. These checks catch errors that can otherwise produce plausible-looking answers.

Density versus probability: $f_x(t)$ is a rate of probability per time unit. A probability over a duration comes from integrating the density over that duration.
Starting at zero: $F_x(t)$ accumulates the density from the start of future lifetime, not just from the most recent breakpoint or from the current attained age.
Using the full support: Verify that the density integrates to one over every interval where it is nonzero. A missing interval can make the resulting CDF incomplete.
Matching time units: If the density uses years, every integration limit must be in years. Convert months, days, or other units before evaluating the formula.
Respecting the support: A formula obtained by integration applies only on the interval where the density was specified. Beyond the upper endpoint of a fully supported density, the cumulative probability is one.

Key takeaway: For a continuous future lifetime with density $f_x$, the probability of death within $t$ time units is the accumulated area under that density from zero to $t$: $F_x(t)=\int_0^t f_x(u)\,du$. Check normalization, include all earlier pieces of a piecewise density, and keep the time units consistent.

Check Your Understanding

For each model, integrate over the stated duration and check that the density and time units are being used correctly.

A hypothetical future lifetime is measured in years and has density $f_x(t)=0.1$ for $0\le t\le10$, zero otherwise. What is the probability of death within 4 years? (Answer: $0.4000$)
A density is $f_x(t)=t/8$ for $0\le t\le4$, zero otherwise. Confirm it integrates to one, then find the probability of death within 2 years. (Answer: density integral $=1$; probability $=0.2500$)
A future-lifetime density is $0.2$ on $0\le t<1$, $0.3$ on $1\le t<3$, and $0.1$ on $3\le t\le5$, with time measured in years. Find the probabilities of death within 2 years and within 4 years. (Answer: $0.5000$ and $0.9000$)
A density is $f_x(t)=(6-t)/18$ for $0\le t\le6$, zero otherwise, with $t$ in years. Check normalization and calculate the probability of death within 2 years. (Answer: density integral $=1$; probability $=5/9\approx0.5556$)
A hypothetical model has density $f_x(t)=t/32$ for $0\le t\le8$, zero otherwise, where $t$ is measured in years. Find the probability of death within 30 months. (Answer: $0.0977$, to four decimal places)

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