Event-Time Distributions And Survival Functions

Representing the Distribution of an Event Time

Duration analysis begins by specifying what counts as the event and when measurement starts. Let \(T\) denote the time from a defined origin to that event. For example, \(T\) might be the number of days from entering a program until leaving employment, or the number of months from a loan’s issuance until repayment. The origin, event definition, and time units are part of the model: changing any of them can change the distribution being described.

An event time is nonnegative. Its distribution describes how probability is allocated across possible times, not merely an average duration. Two groups can have the same mean event time but different probabilities of an event occurring early, or different upper quantiles. In econometrics, the distribution may describe a population or a population conditional on observed covariates. In either case, the definitions below apply to the relevant population.

Definition: The cumulative distribution function (CDF) of an event time \(T\) is \(F(t)=P(T\leq t)\). The survival function is \(S(t)=P(T>t)\), the probability that the event time exceeds \(t\). For any \(t\), \(S(t)=1-F(t)\). Both functions are defined for real \(t\); for a nonnegative event time, \(F(t)=0\) when \(t<0\) and \(S(t)=1\) when \(t<0\).

The CDF accumulates probability up to and including time \(t\). The survival function gives the remaining probability beyond that time. They contain the same distributional information because one is one minus the other. For \(a<b\), the probability that the event occurs after \(a\) but no later than \(b\) is

P(a<T\leq b)=F(b)-F(a)=S(a)-S(b).

The endpoint convention matters if \(T\) can take an exact value with positive probability. The formula above excludes an event exactly at \(a\) and includes one exactly at \(b\). If the distribution is continuous, the probability of either exact endpoint is zero, so alternative endpoint conventions give the same interval probability.

Discrete Event Times and Probability Masses

A discrete event time takes values in a finite or countable set, such as integer days or weeks. Its probability mass function (PMF) assigns a probability to each possible value: \(p(t)=P(T=t)\). Each mass is nonnegative, and the masses across all possible event times sum to one. The CDF is obtained by adding the masses at times no greater than \(t\). Consequently, it is a step function, with a jump of size \(p(t)\) at each possible time \(t\).

Definition: For a discrete event time, \(p(t)=P(T=t)\). The CDF and survival function satisfy \(F(t)=\sum_{u\leq t}p(u)\) and \(S(t)=\sum_{u>t}p(u)\), where the sums range over possible event times \(u\). In particular, \(F(t)-F(t^-)=p(t)\), where \(F(t^-)\) is the limit of \(F(s)\) as \(s\) approaches \(t\) from below.

Worked Example: A Discrete Duration Distribution

Suppose, hypothetically, that a program’s completion time \(T\), measured in weeks, takes values \(1,2,4,\) or \(6\), with probabilities \(0.20,0.35,0.25,\) and \(0.20\), respectively. These probabilities are nonnegative and sum to \(0.20+0.35+0.25+0.20=1.00\), so they define a valid discrete distribution.

The CDF at three weeks adds the probabilities for completion by week 3:

F(3)=P(T\leq 3)=P(T=1)+P(T=2)=0.20+0.35=0.55.

Thus \(S(3)=P(T>3)=1-F(3)=1-0.55=0.45\). In this example, that probability is the sum of the masses at weeks 4 and 6: \(0.25+0.20=0.45\). The probability of completion after week 2 and by week 6 is

P(2<T\leq 6)=F(6)-F(2)=1.00-0.55=0.45.

The median is week 2 under the lower-quantile convention defined below: \(F(1)=0.20<0.50\), while \(F(2)=0.55\geq0.50\). A cumulative probability does not need to equal exactly 0.50 at an attainable event time. Here, 2 weeks is the first time by which at least half of the population has experienced the event.

Continuous Event Times and Densities

A continuous event time can take any value in an interval. Its probability density function (PDF), written \(f(t)\), describes how probability accumulates locally over time. A density is nonnegative and has total area one. Probabilities are areas under the density, not the density’s height at a particular point. In particular, for a continuous distribution, \(P(T=t)=0\) at any single exact time \(t\), even when \(f(t)\) is positive.

Definition: If \(T\) has a continuous distribution with density \(f\), then \(f(t)\geq0\), \(\int_{-\infty}^{\infty} f(u)\,du=1\), \(F(t)=\int_{-\infty}^{t}f(u)\,du\), and \(S(t)=\int_t^\infty f(u)\,du=1-F(t)\). For \(a<b\), \(P(a<T\leq b)=\int_a^b f(u)\,du\). Where \(F\) is differentiable, \(f(t)=F'(t)\).

The density has units inverse to the time unit. If time is measured in months, a density value is measured per month, while probabilities and CDF values are unitless. A value such as \(f(4)=0.08\) per month is not an 8% probability of the event occurring exactly at month 4; probabilities require integration over an interval.

Worked Example: Uniform Event Times Over a Fixed Window

Suppose, hypothetically, that a service time \(T\) in months is uniformly distributed between 0 and 12. Its density is \(f(t)=1/12\) per month for \(0\leq t\leq12\), and zero outside that interval. The total area is \(\int_0^{12}(1/12)\,dt=12/12=1\), as required.

For \(0\leq t\leq12\), integrating the density gives \(F(t)=t/12\), and therefore \(S(t)=1-t/12\). The probability that service ends after month 3 but by month 9 is

P(3<T\leq9)=F(9)-F(3)=\frac{9}{12}-\frac{3}{12}=\frac{6}{12}=0.500.

The result also follows from the interval’s length: the interval from month 3 to month 9 is 6 months out of a 12-month window. The median is 6 months, since \(F(6)=6/12=0.50\). The 90th percentile is 10.8 months, since \(F(10.8)=10.8/12=0.90\). These are population quantiles under the hypothetical uniform model, not estimates from an observed sample.

Quantiles for Discrete and Continuous Distributions

A quantile gives a time threshold corresponding to a specified cumulative probability. Because a discrete CDF can jump over the target probability, a quantile convention is needed to state which time is reported. The standard lower-quantile definition handles both discrete and continuous distributions.

Definition: For \(0<p<1\), the \(p\)-quantile of \(T\) is \(q_p=\inf\{t:F(t)\geq p\}\), the earliest time at which the CDF reaches or exceeds \(p\). The 0.50-quantile is called the median. For a continuous, strictly increasing CDF, \(q_p\) solves \(F(q_p)=p\). For a discrete distribution, \(F(q_p)\) can be greater than \(p\) because the CDF jumps at \(q_p\).

The definition makes the phrase “the \(p\)-quantile” precise: at least a proportion \(p\) of the population has event time at or below \(q_p\), and the threshold is no later than necessary to reach that proportion. It does not assert that exactly \(p\) of the population has event time at or below that value when there is a mass at the quantile.

Worked Example: A Nonuniform Continuous Duration

Consider a hypothetical continuous event time \(T\) measured in days, with density \(f(t)=t/8\) for \(0\leq t\leq4\) and zero elsewhere. The density 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}-0=1.

It is therefore a valid density. Integrating from zero to \(t\) gives \(F(t)=t^2/16\) for \(0\leq t\leq4\); the corresponding survival function on that range is \(S(t)=1-t^2/16\). At day 2, \(F(2)=4/16=0.25\), so \(S(2)=0.75\). The density at day 2 is \(f(2)=2/8=0.25\) per day; this is a density value, not the probability of an event exactly on day 2.

For the interval from day 1 to day 3, the probability is

P(1<T\leq3)=F(3)-F(1)=\frac{9}{16}-\frac{1}{16}=\frac{8}{16}=0.500.

To obtain the median, solve \(F(q_{0.50})=0.50\). Since \(F(t)=t^2/16\) is increasing over the support,

\frac{q_{0.50}^2}{16}=0.50 \quad\Longrightarrow\quad q_{0.50}=\sqrt{8}\approx2.828\text{ days}.

The model therefore assigns half the population an event time no greater than approximately 2.828 days and half no less than that time. The equality at the median is appropriate here because the distribution is continuous.

Why the Distribution Is More Informative Than a Mean

A mean summarizes a distribution with one number, but it does not show how probability is distributed across the time scale. A CDF directly answers “what proportion has experienced the event by time \(t\)?” A survival function answers “what proportion remains event-free beyond time \(t\)?” Quantiles answer “by what time has a specified proportion experienced the event?” These questions are often substantively useful when comparing durations or describing predictions.

In regression settings, the distribution can depend on covariates. For example, \(F(t\mid X=x)\) denotes the conditional CDF of event time \(T\) for units whose covariate vector is \(X=x\); \(S(t\mid X=x)=1-F(t\mid X=x)\) is the corresponding conditional survival function. The conditioning set must be stated when interpreting the result. A distribution for one covariate profile need not describe the overall population, which averages across profiles.

These definitions describe distributions, whether they are specified as a model or considered as population quantities. In observed duration data, some event times may not be fully observed. The definitions of \(F\), \(S\), and quantiles still refer to the underlying event time, but calculations from incomplete observations require additional methods rather than treating every recorded follow-up time as an event time.

Deriving the CDF–Survival Relationship

The identity between the CDF and survival function follows by partitioning all possible outcomes at a fixed \(t\). The events \(T\leq t\) and \(T>t\) are mutually exclusive and together cover every possible value of \(T\). Therefore,

P(T\leq t)+P(T>t)=1, \qquad F(t)+S(t)=1, \qquad S(t)=1-F(t).

For \(a<b\), the event \(\{a<T\leq b\}\) is the part of \(\{T\leq b\}\) that is not in \(\{T\leq a\}\). Subtracting the probabilities of these nested events gives \(P(a<T\leq b)=F(b)-F(a)\). Substituting \(F(t)=1-S(t)\) yields \(S(a)-S(b)\). This reasoning applies to discrete and continuous distributions; no density assumption is needed.

Common Mistakes

Treating a density height as an event probability. The value \(f(t)\) is not \(P(T=t)\). For a continuous event time, an exact point has probability zero; an interval probability is an area, \(\int_a^b f(u)\,du\).
Reversing the survival-function inequality. Here \(S(t)=P(T>t)\), not \(P(T\leq t)\). It describes durations that extend beyond \(t\), and it equals \(1-F(t)\).
Assuming a discrete quantile must hit its target exactly. A jump in the CDF can take it from below \(p\) to above \(p\). Under the lower-quantile convention, report the first time with \(F(t)\geq p\).
Leaving time units or the event definition unstated. A statement such as “the median is 6” is incomplete without saying what event time is measured and whether the units are days, weeks, or months.

Key takeaway: The CDF and survival function describe cumulative probability on opposite sides of a time threshold. Discrete distributions allocate probability as point masses; continuous distributions allocate it through areas under a density. Quantiles turn a chosen cumulative probability into an event-time threshold.

Check Your Understanding

Use the definitions and calculations in this tutorial to answer the following questions.

A discrete event time takes values 2, 5, and 8 days with probabilities 0.30, 0.45, and 0.25. Find \(F(5)\), \(S(5)\), and the lower 0.50-quantile.
A continuous event-time density is constant at \(1/10\) per month from 0 to 10 months and zero elsewhere. Find the probability of an event after month 2 and by month 7.
In one sentence, explain why a density value of 0.12 per week at week 4 is not a 12% probability of an event exactly at week 4.
If \(F(12)=0.70\), what is the probability that the event time exceeds 12? State what that probability means in context after specifying a suitable event.
For a discrete distribution, the CDF is 0.35 just before a possible event time and 0.65 at that time. Under the lower-quantile definition, what is the 0.50-quantile, and why?