Alternative Binary Links · Tutorial 372 of 981

Deriving The Discrete-Time Hazard Connection

Connect proportional hazards to interval-specific complementary log-log regression, and identify when the connection does—and does not—hold.

Intermediate 10 min read

What You'll Learn

Derive an interval event probability from a proportional-hazards survival model
State the regularity conditions needed for the integrated-hazard formula
Express the interval probability using the complementary log-log link
Interpret interval-specific baseline hazards as model intercepts
Recognize why covariates that vary within an interval can break the simple form

From a Continuous-Time Hazard to an Interval Probability

A complementary log-log regression for binary outcomes has a direct connection to survival analysis. Suppose the binary response records whether an event occurred during a particular time interval, among individuals who were still event-free at the interval's start. If the underlying event time follows a proportional-hazards model and the covariates remain fixed during that interval, the event probability has an exact complementary log-log form. The interval-specific baseline cumulative hazard appears as an intercept; it is not generally a single common intercept across intervals.

The distinction between a hazard rate and an interval probability is important. A hazard rate describes the instantaneous event rate among those who have survived to a given time. An interval probability accumulates that rate over time, accounting for the fact that an event earlier in the interval prevents the individual from remaining at risk later in it.

Definition: Let $T$ be an event time and $x$ a fixed covariate vector. A proportional-hazards model specifies the hazard rate as $$ \lambda(t\mid x)=\lambda_0(t)\exp(x^\mathsf{T}\beta), $$ where $\lambda_0(t)$ is a nonnegative baseline hazard and $\beta$ is a coefficient vector. For an interval $(a,b]$, the interval event probability among those event-free at $a$ is $$ h(a,b\mid x)=P(a<T\leq b\mid T>a,x). $$

Conditions for the Exact Connection

For the derivation, assume that, conditional on $x$, the event-time distribution is absolutely continuous over the time range considered, its survival function is positive at $a$, and its hazard is measurable and locally integrable. Assume the proportional-hazards specification holds over $(a,b]$, with $x$ fixed there. These conditions allow the survival function $S(t\mid x)=P(T>t\mid x)$ to satisfy the hazard equation $S'(t\mid x)=-\lambda(t\mid x)S(t\mid x)$ almost everywhere.

Define the baseline cumulative hazard over the interval by $\Lambda_0(a,b)=\int_a^b\lambda_0(u)\,du$. It is finite under the stated integrability condition. The corresponding cumulative hazard conditional on $x$ is $\exp(x^\mathsf{T}\beta)\Lambda_0(a,b)$, because $x$ and $\beta$ do not vary with time within the interval.

Theorem: Under the conditions above, the conditional probability of an event during $(a,b]$, given survival to $a$, is $$ h(a,b\mid x) =1-\exp\left\{-\Lambda_0(a,b)\exp(x^\mathsf{T}\beta)\right\}. $$ If $\Lambda_0(a,b)>0$, its complementary log-log transform is $$ \log\{-\log[1-h(a,b\mid x)]\} =\log\{\Lambda_0(a,b)\}+x^\mathsf{T}\beta. $$ If $\Lambda_0(a,b)=0$, the interval event probability is zero, so the displayed transformed expression is not finite.

Proof. For fixed $x$, write $S(t)=S(t\mid x)$ and $c=\exp(x^\mathsf{T}\beta)$, which is a positive constant. The hazard equation gives $S'(t)=-c\lambda_0(t)S(t)$ almost everywhere. Since $S(t)>0$ on the interval of interest, the absolutely continuous function $\log S(t)$ has derivative $S'(t)/S(t)=-c\lambda_0(t)$ almost everywhere. Integrating from $a$ to $b$, $$ \log S(b)-\log S(a) =-c\int_a^b\lambda_0(u)\,du =-c\Lambda_0(a,b). $$ Exponentiating yields $S(b)/S(a)=\exp\{-c\Lambda_0(a,b)\}$. By conditional probability, $$ P(a<T\leq b\mid T>a,x) =1-\frac{S(b)}{S(a)} =1-\exp\{-c\Lambda_0(a,b)\}. $$ Substitute $c=\exp(x^\mathsf{T}\beta)$. If the resulting probability is $h$ and $\Lambda_0(a,b)>0$, then $1-h=\exp\{-\Lambda_0(a,b)\exp(x^\mathsf{T}\beta)\}$. Taking the negative logarithm and then the logarithm gives the stated complementary log-log equation. $\square$

Reading the Result as a Binary Regression Model

For interval $j$, write $h_j(x)$ for the event probability among those event-free at the start of that interval, and let $\Lambda_{0j}$ denote the baseline cumulative hazard over it. If $\Lambda_{0j}>0$, the theorem gives $$ \log\{-\log[1-h_j(x)]\}=\alpha_j+x^\mathsf{T}\beta, \qquad \alpha_j=\log(\Lambda_{0j}). $$

Thus the complementary log-log link arises exactly, not approximately, under the stated assumptions. If the intervals have different baseline cumulative hazards, each interval has its own intercept $\alpha_j$. The covariate coefficient vector $\beta$ is shared across intervals when the proportional-hazards assumption holds. An analysis that uses one common intercept across intervals adds a restriction: it requires the interval baseline cumulative hazards to be equal, or some other model structure that justifies that restriction.

Derived result: For two covariate profiles $x_1$ and $x_0$ observed over the same interval with positive baseline cumulative hazard, $$ \log\{-\log[1-h_j(x_1)]\} -\log\{-\log[1-h_j(x_0)]\} =(x_1-x_0)^\mathsf{T}\beta. $$ The interval-specific baseline term cancels. This is the complementary log-log scale counterpart of the shared proportional-hazards effect.

Derivation. Apply the interval equation to each profile: $\log\{-\log[1-h_j(x_k)]\}=\log(\Lambda_{0j})+x_k^\mathsf{T}\beta$, for $k=0,1$. Subtract the equation for $x_0$ from the one for $x_1$. The common term $\log(\Lambda_{0j})$ cancels, leaving $(x_1-x_0)^\mathsf{T}\beta$. $\square$

This result concerns the conditional event probability among individuals at risk at the interval's start. It does not say that successive interval indicators for an individual are independent: after an event, the individual is no longer at risk for later intervals. Nor does the probability identity, by itself, settle how censoring or incomplete observation should be handled in an analysis.

Worked Example: Computing an Interval Risk from a Hazard

Suppose the baseline cumulative hazard over a one-unit interval is $0.16$, and the hazard model is $\lambda(t\mid x)=\lambda_0(t)\exp(0.7x)$, where $x$ is fixed within the interval. Calculate the interval event probability at $x=0$ and $x=1$.

At $x=0$, the covariate multiplier is $\exp(0)=1$. The integrated hazard is $0.16$, so $$ h(0)=1-\exp(-0.16)\approx1-0.852144=0.147856. $$

At $x=1$, the multiplier is $\exp(0.7)\approx2.0137527$. The integrated hazard is $0.16\exp(0.7)\approx0.3222004$, and therefore $$ h(1)=1-\exp\{-0.16\exp(0.7)\} \approx1-0.72455 =0.27545. $$

The event probability increases with $x$, consistent with the positive hazard coefficient. The probability difference is about $0.27545-0.14786=0.12759$. The calculation uses the interval event formula; the coefficient $0.7$ is a log hazard ratio, not a direct additive change in probability.

Worked Example: Different Intervals Produce Different Intercepts

Suppose there are two successive intervals. Their baseline cumulative hazards are $0.10$ and $0.30$, and a binary covariate $x$ has coefficient $\beta=\log 2$. Find the event probabilities for $x=0$ and $x=1$ in each interval, conditional on being event-free at that interval's start.

For $x=0$, $\exp(x\beta)=1$. In the first interval the integrated hazard is $0.10$, giving $h_1(0)=1-\exp(-0.10)\approx0.09516$. In the second interval it is $0.30$, giving $h_2(0)=1-\exp(-0.30)\approx0.25918$.

For $x=1$, $\exp(\beta)=2$, so the interval cumulative hazards are doubled. Thus $h_1(1)=1-\exp(-0.20)\approx0.18127$, while $h_2(1)=1-\exp(-0.60)\approx0.45119$.

On the complementary log-log scale, the first interval's intercept is $\log(0.10)\approx-2.30259$, and the second's is $\log(0.30)\approx-1.20397$. The coefficient of $x$ is $\log 2$ in both intervals. The differing intercepts encode differing baseline cumulative hazards; they do not contradict the common proportional-hazards coefficient.

When Covariates Vary Within an Interval

The fixed-covariate condition is substantive. If the covariate changes over the interval, the hazard is instead $\lambda_0(t)\exp\{X(t)^\mathsf{T}\beta\}$, and the cumulative hazard becomes $\int_a^b\lambda_0(u)\exp\{X(u)^\mathsf{T}\beta\}\,du$. In general, this integral cannot be factored into a baseline interval cumulative hazard times one covariate multiplier. The interval probability still has the form one minus the exponential of the negative integrated hazard, but it need not have the simple linear predictor $\alpha_j+x_j^\mathsf{T}\beta$ for a single interval covariate value.

Worked Example: Why an Endpoint Covariate Can Give the Wrong Probability

Let $\lambda_0(t)=0.2$ on the interval $(0,2]$, let $\beta=\log 2$, and suppose $X(t)=0$ on $(0,1]$ and $X(t)=1$ on $(1,2]$. Calculate the actual interval risk, then compare it with the calculation that incorrectly treats $X=1$ as fixed throughout.

On the first unit of time, the covariate multiplier is $\exp(0\log 2)=1$, so the cumulative hazard contribution is $0.2(1)(1)=0.2$. On the second unit it is $\exp(\log 2)=2$, giving contribution $0.2(2)(1)=0.4$. The total integrated hazard is $0.2+0.4=0.6$. Therefore the actual probability is $$ 1-\exp(-0.6)\approx1-0.548812=0.451188. $$

If one incorrectly uses the endpoint value $X=1$ for the full two-unit interval, the calculated cumulative hazard is $0.2(2)(2)=0.8$, giving $1-\exp(-0.8)\approx0.550671$. This differs from the actual probability because it applies the later covariate value to the earlier part of the interval as well.

Common Mistake and Why the Connection Matters

A common mistake is to identify a discrete-time interval probability with the hazard rate itself. For example, if a hazard rate is $0.2$ per unit time, it does not follow that the probability of an event over a two-unit interval is $0.4$. Even with a constant rate and no covariate effect, that probability is $1-\exp(-0.2\times2)\approx0.32968$, because survival declines during the interval. The integrated hazard, not the instantaneous rate multiplied without adjustment, determines the interval probability.

The derivation explains why complementary log-log regression is a natural model for interval event indicators under proportional hazards. It also clarifies what its intercepts represent and which assumptions support the interpretation of its slopes. In practice, the time intervals, risk sets, and covariate timing must match the model: a correct link cannot compensate for treating people who were not at risk as though they were, or for assigning an inappropriate single covariate value to an interval with substantial within-interval change.

Key takeaway: With a fixed covariate vector over an interval and a proportional-hazards event-time model, the conditional interval probability satisfies $\log\{-\log(1-h_j(x))\}=\log(\Lambda_{0j})+x^\mathsf{T}\beta$. The baseline cumulative hazard supplies the interval intercept, and a time-varying covariate generally requires integrating its time-specific hazard contribution.

Check Your Understanding

Use the assumptions and derivation above to answer each question.

State the conditional probability represented by an interval hazard indicator and identify the population at risk for that probability.
If $\Lambda_{0j}=0.25$ and $x^\mathsf{T}\beta=0$, calculate the interval event probability.
Why can different intervals have different complementary log-log intercepts even when the covariate coefficient is shared?
For a covariate that changes within an interval, write the integrated hazard expression that replaces $\Lambda_{0j}\exp(x^\mathsf{T}\beta)$.
Explain why an instantaneous hazard rate of $0.2$ does not imply an event probability of $0.2$ over a two-unit interval.

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