The Small-Interval Definition Of A Cause-Specific Hazard

From a Short-Interval Probability to a Rate

A cause-specific hazard describes how quickly a particular first-event cause occurs locally in time among individuals who have not yet had an event. Its definition is a limit: take the conditional probability of cause $k$ in a short interval beginning at time $t$, divide by the interval width, and let that width shrink toward zero. The limit, when it exists, is a rate rather than a probability.

Let $T$ be the time to the first event, and let $J$ record its cause. The causes are mutually exclusive: if a first event occurs, it has one cause $k$ among the possible causes. Write $S(t)=P(T\ge t)$ for the probability of remaining event-free up to time $t$, and let $F_k(t)=P(T\le t,J=k)$ be the probability that the first event has occurred by $t$ from cause $k$. In a continuous-time setting, the distinction between $T\ge t$ and $T>t$ does not affect probabilities when there is no event-time mass at $t$.

Definition: If $S(t)>0$, the cause-specific hazard for cause $k$ at time $t$ is $$ \lambda_k(t)=\lim_{h\downarrow 0}\frac{P(t\le T<t+h,\ J=k\mid T\ge t)}{h}, $$ provided this limit exists and is finite. Here $h$ is a positive interval width measured in the same time units as $T$.

The numerator is a conditional probability for an interval of width $h$. Dividing by $h$ expresses that probability per unit of time; taking the limit removes the dependence on a particular interval width. Consequently, $\lambda_k(t)$ has units of inverse time, such as per year. The definition requires positive probability of being event-free at $t$; otherwise the conditional probability is not defined.

Deriving the Density-Over-Survival Form

The limit has a useful expression when the cause-specific event-time distribution is smooth. Suppose $F_k$ is absolutely continuous around $t$, with density $f_k(u)=F_k'(u)$, and that $f_k$ is continuous at $t$. Also suppose $S(t)>0$ and $S$ is continuous at $t$, as it is in the usual continuous-time setting without an event-time atom there.

For $h>0$, the event in the numerator has unconditional probability $F_k(t+h)-F_k(t)$. Because it is restricted to times at or after $t$, it is already contained in the event $T\ge t$. Therefore, by the definition of conditional probability,

P(t\le T<t+h,\ J=k\mid T\ge t) =\frac{F_k(t+h)-F_k(t)}{S(t)}.

Substituting this identity into the limit and using differentiability of $F_k$ gives the result directly:

\begin{aligned} \lambda_k(t) &=\lim_{h\downarrow 0} \frac{F_k(t+h)-F_k(t)}{hS(t)}\\ &=\frac{1}{S(t)} \lim_{h\downarrow 0}\frac{F_k(t+h)-F_k(t)}{h}\\ &=\frac{f_k(t)}{S(t)}. \end{aligned}

The denominator is the probability of still being event-free at the start of the interval; the numerator in the density form is the instantaneous cause-$k$ event density in the original population. The ratio is local in time: both $f_k(t)$ and $S(t)$ are evaluated at $t$. This derivation depends on the stated smoothness conditions. Without a suitable derivative, the small-interval limit need not equal a finite density-over-survival ratio.

Worked Example: Calculating a Local Hazard From a Density

Suppose a hypothetical population is followed for time measured in years. At $t=2$ years, assume $S(2)=0.60$, and suppose the cause-$k$ event density is continuous there with $f_k(2)=0.075$ per year. Find the cause-specific hazard at 2 years.

The survival probability is positive, and the stipulated continuity of the density supports the limit calculation. By the density-over-survival result, $$ \lambda_k(2)=\frac{f_k(2)}{S(2)} =\frac{0.075\ \text{year}^{-1}}{0.60} =0.125\ \text{year}^{-1}. $$

Thus the local cause-$k$ rate at 2 years is $0.125$ per year. As a small-interval check, if the density is approximately constant over the next $0.04$ years, the conditional probability is approximately $$ \frac{0.075(0.04)}{0.60}=0.005, $$ or $0.5\%$. Dividing by $0.04$ years gives $0.005/0.04=0.125$ per year, matching the local rate under that constant-density approximation. The probability is conditional on being event-free at 2 years; it is not itself the hazard.

Checking the Limit With Shrinking Intervals

The definition can also be understood by examining conditional probabilities for successively smaller intervals. If the cause-$k$ density changes smoothly near $t$, the probability over an interval is approximately its width times the local hazard. Dividing by the width should therefore yield values that approach a stable limit as the width decreases. This check is useful for understanding the definition, though it does not by itself establish that a model or estimate is appropriate.

Worked Example: Seeing Convergence as the Interval Shrinks

Consider hypothetical event-time information near $t=4$ years. Let $S(4)=0.50$, and suppose the cause-$k$ density in the neighborhood is $$ f_k(u)=0.04+0.02(u-4)\quad \text{per year},\qquad u\ge4. $$ This density is continuous at 4, and survival at the interval's start is positive. For an interval of width $h$, the conditional probability is the integral of the density over that interval divided by $S(4)$: $$ \begin{aligned} P(4\le T<4+h,\ J=k\mid T\ge4) &=\frac{1}{0.50}\int_4^{4+h}\{0.04+0.02(u-4)\}\,du\\ &=\frac{0.04h+0.01h^2}{0.50}. \end{aligned} $$

After dividing by $h$, the interval rate is $0.08+0.02h$ per year. For $h=0.10$ years, the conditional probability is $0.0082$, and the quotient is $0.0082/0.10=0.082$ per year. For $h=0.05$, the probability is $0.00405$, giving $0.081$ per year. For $h=0.025$, the probability is $0.0020125$, giving $0.0805$ per year. These quotients approach $$ \lim_{h\downarrow0}(0.08+0.02h)=0.08\ \text{per year}. $$

The density-over-survival expression gives the same answer: $f_k(4)/S(4)=0.04/0.50=0.08$ per year. The numerical sequence illustrates why a finite-interval quotient is generally an approximation to the limiting hazard rather than the definition itself.

Assumptions, Diagnostics, and Failure Modes

The small-interval definition is mathematically appropriate under identifiable conditions. There must be a well-defined first-event time and mutually exclusive cause label; $S(t)$ must be positive so conditioning is possible; and the limit must exist. To use the density-over-survival formula, the cause-specific event distribution must additionally have a suitable density near $t$, with continuity at $t$ sufficient for the derivation above. A density that changes abruptly or is undefined at $t$ calls for additional care rather than automatic use of the ratio.

A practical diagnostic is to compare the conditional interval probability divided by $h$ for a sequence of smaller widths. Stable values are consistent with a local rate description. Strong dependence on the width can indicate that the interval is not yet small enough, that the underlying rate changes rapidly, or that the process has a discontinuity or point mass. Numerical instability can also arise when $S(t)$ is very small: the conditional quantity is defined if $S(t)>0$, but small denominators make estimates sensitive to limited information.

A continuous-time hazard and a fixed-interval conditional probability are related but not interchangeable. The probability is dimensionless and depends on the selected interval width; the limiting hazard has units of inverse time. In discrete-time settings, an interval-specific conditional probability is often the natural quantity, and dividing it by an interval width does not automatically create a meaningful continuous-time hazard. The limit requires a suitable time scale and regularity of the event-time distribution.

Worked Example: Why an Event-Time Atom Breaks the Finite-Rate Limit

Suppose, hypothetically, that $S(t)=0.80$, and that $0.10$ of the original population has a cause-$k$ event exactly at time $t$. The event-free probability is positive. But the interval in the definition includes $t$, so for every $h>0$, $$ P(t\le T<t+h,\ J=k\mid T\ge t)\ge\frac{0.10}{0.80}=0.125. $$

After dividing by $h$, the quotient is at least $0.125/h$. As $h\downarrow0$, this lower bound grows without limit. Thus there is no finite cause-specific hazard at $t$ under this definition. The issue is not a calculation error: a fixed positive event probability concentrated at one instant cannot shrink proportionally to the interval width. In a setting with scheduled or otherwise discrete event times, report and analyze interval-specific probabilities or use a framework designed for discrete time rather than assuming a finite continuous-time rate.

Common Mistakes and Scope

Stopping at a finite interval. The quotient $P(t\le T<t+h,J=k\mid T\ge t)/h$ for one selected $h$ is an interval rate, not necessarily the hazard. The hazard is the limit as $h$ shrinks.
Dropping the conditioning denominator. The cause-specific density $f_k(t)$ is not, by itself, the conditional rate among those still event-free. Under the stated conditions, the rate is $f_k(t)/S(t)$.
Ignoring units. Dividing a probability by a time width produces units of inverse time. State those units when reporting a hazard.
Using the smooth formula in the presence of an atom. A point mass can make the quotient diverge, as the example shows; a density-based derivation cannot be applied without its assumptions.
Treating the definition as an estimator. The limit defines a population quantity. Estimating it from observed data, especially when follow-up can be censored, requires additional methods and assumptions.

The main alternative representation is the density-over-survival ratio, useful when differentiable event-time distributions are available. The small-interval definition is more fundamental because it states what the local rate means without requiring the ratio as the starting point. Neither representation removes the need to check whether a continuous-time limit is appropriate.

Key takeaway: The cause-specific hazard is the limiting conditional probability of a cause-$k$ first event in a short interval, divided by that interval's width. When the cause-specific density is continuous and survival is positive, this limit equals $f_k(t)/S(t)$.

Check Your Understanding

Answer each question using the small-interval definition. Solutions are provided below the questions.

In the definition of $\lambda_k(t)$, what happens to the interval width, and what are the units of the limit?
If $S(t)=0.40$ and $f_k(t)=0.032$ per month, calculate $\lambda_k(t)$.
At time $t$, suppose $S(t)=0.50$, $f_k(t)=0.025$ per year, and the density is continuous. What is the approximate conditional probability of cause $k$ in the next $0.08$ years?
Why can a positive probability mass for cause $k$ exactly at time $t$ prevent a finite limit?

Solutions. (1) The positive interval width $h$ decreases toward zero after the conditional probability is divided by $h$. The limit has units of inverse time. (2) $\lambda_k(t)=0.032/0.40=0.080$ per month. (3) The hazard is $0.025/0.50=0.05$ per year, so the short-interval probability is approximately $0.05(0.08)=0.004$, or $0.4\%$. This approximation uses the local rate over a sufficiently short interval. (4) The conditional probability then includes a fixed positive contribution for every interval starting at $t$; dividing that contribution by a width tending to zero makes the quotient unbounded, so a finite rate limit cannot exist.