The Density f_x(t)

Introduction

The probability that a person’s future lifetime falls within a time interval is a probability. The density at a particular time is different: it describes how concentrated probability is near that time, per unit of time. Confusing these quantities can lead to answers that are not probabilities at all, especially when a density value is greater than 1.

Fix an attained age $x$, and let $T_x$ denote the person’s future lifetime measured from age $x$. This tutorial defines the density $f_x(t)$, explains its units and local interpretation, and shows how to calculate probabilities over intervals from areas under a density curve. It does not treat a density value as the probability of dying at exactly a specified time.

Definition: If the distribution of $T_x$ has a continuous density, $f_x(t)$ is its probability density at future time $t$. At an interior point where the relevant limit exists, it can be defined by $$ f_x(t)=\lim_{h\downarrow 0}\frac{P(t<T_x\le t+h)}{h}. $$ Here $h$ is a positive time interval. Thus the density is probability per unit of time, not a probability by itself.

Reading a Density Correctly

The definition compares the probability in a short interval immediately after $t$ with the interval’s length. For small $h$, the probability is approximately $f_x(t)h$. The approximation is most useful when the density does not change much across that short interval. For a longer interval, the density may vary enough that a single value times the interval length is inaccurate.

For a continuous-density model, the probability that future lifetime lies between $a$ and $b$, where $a<b$, is the area under the density between those times:

P(a<T_x\le b)=\int_a^b f_x(t)\,dt.

The density has units reciprocal to the time unit used for $t$. If time is measured in years, $f_x(t)$ is measured per year; if time is measured in months, it is measured per month. Multiplying a density by a time interval gives a dimensionless quantity, as a probability must be. A density can exceed 1, because it is a rate of probability per unit time rather than a probability.

A density must be nonnegative and must have total area 1 over the possible future lifetimes. These conditions allow it to describe a probability distribution, but neither condition says that $f_x(t)$ itself is a probability. Also, an ordinary density describes a distribution without point masses; if a model assigns positive probability to a particular exact lifetime, a density alone does not describe the whole distribution.

Why Interval Probability Is an Area

The area rule follows from the local definition of density. Divide an interval $[a,b]$ into many short pieces. On a sufficiently short piece of length $\Delta t$ near $t$, its probability is approximately $f_x(t)\Delta t$. Adding these piece probabilities approximates the probability over the whole interval. As the pieces become shorter, the sum approaches the integral:

P(a<T_x\le b) \approx \sum_j f_x(t_j)\Delta t \quad\longrightarrow\quad \int_a^b f_x(t)\,dt.

This also explains the difference between a density value and an interval probability. A density is the height of the curve at a time; a probability is an area across a range of times. A narrow interval around a high density can have a small probability, while a wider interval at a lower density can have a larger probability.

Worked Example: Exponential Lifetime and a Short Interval

For a hypothetical group at age $x$, suppose future lifetime is exponential with rate $\lambda=0.2$ per year, so its density is $f_x(t)=0.2e^{-0.2t}$ for $t\ge 0$. Find the density at $t=3$ years and the probability that future lifetime is between 3 and 3.5 years. Compare the interval probability with the short-interval approximation using the density at 3.

Method and units. Evaluate the density at the specified time for the local concentration. For the interval probability, integrate the density over the interval. The rate is per year, so both $t$ and the interval length must be in years.

Density at 3 years.

f_x(3)=0.2e^{-0.2(3)} =0.2e^{-0.6} =0.109762\ldots\ \text{per year}.

The density is $0.1098$ per year to four decimal places. It is not a probability of $0.1098$ at the single time $t=3$.

Probability from 3 to 3.5 years. The interval has length $0.5$ year. Integrating the density gives

\begin{aligned} P(3<T_x\le 3.5) &=\int_3^{3.5}0.2e^{-0.2t}\,dt\\ &=\left[-e^{-0.2t}\right]_3^{3.5}\\ &=e^{-0.6}-e^{-0.7}\\ &=0.548812-0.496585\\ &=0.052226\ldots \end{aligned}

The interval probability is $0.0522$ to four decimal places. The short-interval approximation based on the density at 3 is

f_x(3)(0.5)=0.109762\ldots(0.5)=0.054881\ldots.

This approximation is close, but it is not the exact interval probability: the density decreases over the interval. The exact answer is the area, $0.0522$. The check is that the probability is less than 1 and is smaller than the approximate value based on the interval’s relatively higher starting density. A common mistake is to report $f_x(3)$ as the interval probability; that omits the interval length and the change in density across it.

Density Height and Interval Width

A useful way to check an interpretation is to track units. If $f_x(t)$ is in units of probability per year and the interval is measured in years, their product is dimensionless. If an interval is stated in months while the density is per year, first convert the interval to years or convert the density to per month. Mixing these units gives an incorrect area.

The following hypothetical example illustrates both the unit check and why a density can be greater than 1.

Worked Example: A Density Greater Than One

Suppose a hypothetical future lifetime is uniformly distributed over the first $0.4$ years, with density $f_x(t)=2.5$ per year for $0\le t\le0.4$, and zero otherwise. Find the density at $t=0.15$ years and the probability that lifetime is between $0.1$ and $0.2$ years.

Check the model. The density is nonnegative. Its total area is

\int_0^{0.4}2.5\,dt=2.5(0.4)=1.

So the stated function has total probability 1 over its support.

Density at the specified time. Since $0.15$ is within the support, $f_x(0.15)=2.5$ per year. The value exceeds 1, but this is permitted: it is probability per year, not probability.

Interval probability. The interval length is $0.2-0.1=0.1$ year, and the density is constant there:

P(0.1<T_x\le0.2) =\int_{0.1}^{0.2}2.5\,dt =2.5(0.1) =0.25.

The result is a valid probability. The check is that $0.25$ is between 0 and 1 and equals the fraction $0.1/0.4$ of the uniform support’s length. A common mistake is to call $2.5$ a probability because it is the density value. The correction is to multiply by the interval width, here in years, to obtain an interval probability.

A Changing Density: Exact Area Versus Approximation

When the density changes with time, multiplying its value at one endpoint by the interval length is only an approximation. The exact probability requires the area across the full interval. The next example includes a normalization check and compares an exact interval probability with a local approximation.

Worked Example: A Linearly Increasing Lifetime Density

For a hypothetical lifetime measured in years, suppose the density is $f_x(t)=t/8$ for $0\le t\le4$, and zero outside this range. Find the density at 2 years, the probability of a lifetime between 1 and 3 years, and the probability between 2 and 2.1 years. Use the density at 2 years to form a short-interval approximation for the last probability.

Check the density. It is nonnegative on its support, and its total area is

\int_0^4\frac{t}{8}\,dt =\left[\frac{t^2}{16}\right]_0^4 =\frac{16}{16}=1.

Thus the specified function is normalized as a density.

Density at 2 years.

f_x(2)=\frac{2}{8}=0.25\ \text{per year}.

Probability from 1 to 3 years.

\begin{aligned} P(1<T_x\le3) &=\int_1^3\frac{t}{8}\,dt\\ &=\left[\frac{t^2}{16}\right]_1^3\\ &=\frac{9-1}{16}\\ &=0.5. \end{aligned}

The area is $0.5$, which is a valid probability.

Probability from 2 to 2.1 years.

\begin{aligned} P(2<T_x\le2.1) &=\int_2^{2.1}\frac{t}{8}\,dt\\ &=\left[\frac{t^2}{16}\right]_2^{2.1}\\ &=\frac{4.41-4}{16}\\ &=0.025625. \end{aligned}

To four decimal places, this probability is $0.0256$. The approximation using the density at 2 is

$$ f_x(2)(2.1-2)=0.25(0.1)=0.0250. $$

The exact area is slightly larger because the density increases over the interval. This comparison checks both the interval width and the direction of change. A common mistake is to use an endpoint density times the interval width as though it were exact for a changing density; integration gives the exact result.

Common Exam Traps and Practical Checks

Density calculations often fail because the candidate uses the right function with the wrong interpretation or units. Before accepting a result, make these checks:

Check whether the question asks for a height or an area. “Find $f_x(t)$” asks for a density value. “Find the probability that lifetime is between $a$ and $b$” asks for an interval area.
Keep time units consistent. A density per year must be paired with an interval measured in years. For example, a two-month interval has width $2/12$ years.
Do not reject a density just because it exceeds 1. A probability cannot exceed 1, but a density measured per unit time can.
Check the support. A density formula may apply only over a specified range. Outside that range, use the stated density, often zero, rather than extending the formula automatically.
Use area for a nonconstant density. A single density value multiplied by interval length is a local approximation, not generally the exact interval probability.
Check normalization when a density is proposed. Its total area over all possible future lifetimes should be 1.

These checks are useful in applications as well as exam calculations: they distinguish a local concentration from a probability and expose unit or support errors before they affect a result.

Key takeaway: $f_x(t)$ measures probability concentration per unit of future time. To obtain a probability over an interval, calculate the area under the density on that interval, using consistent time units.

Check Your Understanding

For each problem, distinguish a density height from an interval probability and keep the stated time units throughout.

A hypothetical future lifetime has exponential density $f_x(t)=0.1e^{-0.1t}$ for $t\ge0$, with $t$ measured in months. Find $f_x(5)$ and $P(5<T_x\le7)$. Give the density to four decimal places per month and the probability to four decimal places. (Answer: $f_x(5)=0.0607$ per month; probability $=0.1099$.)
A hypothetical lifetime is uniform from 0 to 2 years. Find its density at 1 year and the probability that it lies between 0.4 and 1.3 years. (Answer: $0.5$ per year; probability $=0.45$.)
A hypothetical density is $f_x(t)=3t^2/8$ for $0\le t\le2$, with time in years, and zero otherwise. Find $f_x(1)$ and the probability of a lifetime between 0.5 and 1.5 years. (Answer: $0.375$ per year; probability $=0.40625$.)
A hypothetical density is $0.4$ per year for $0\le t<1$, $0.2$ per year for $1\le t\le4$, and zero otherwise. Find the density at $t=0.75$ years and the probability of a lifetime between 0.5 and 2 years. (Answer: $0.4$ per year; probability $=0.4$.)
A hypothetical future lifetime is uniform over the first $0.25$ years. Find its density per year and the probability that lifetime is between $0.05$ and $0.15$ years. (Answer: $4$ per year; probability $=0.4$.)

What You'll Learn