Deriving the Pooled Logistic Survival Likelihood

From Conditional Hazards to a Likelihood

A pooled logistic survival model is built from a sequence of interval-specific event probabilities. Its likelihood is therefore not a product of probabilities for every calendar interval for every person. Each person contributes only the conditional probabilities for intervals in which they are known to be at risk: event-free factors up to an event, followed by one event factor, or event-free factors through the last interval observed before censoring.

The key derivation is the chain rule for conditional probabilities. It explains both why a pooled logistic likelihood can be written using person-interval records and why rows after an event must not appear. The same logic also clarifies what a censored person's observed data say: the person was event-free through the last interval with known status, but their later event status is unknown.

Data and Conditional-Hazard Notation

Let $T_i$ be the interval in which person $i$ experiences the event, and let $C_i$ be the last interval through which event status is observed if the person is censored first. Define $K_i$ as the last interval with known event status. Let $\Delta_i=1$ if an event is observed by $K_i$, and $\Delta_i=0$ if the observation ends without an event. When $\Delta_i=1$, $K_i=T_i$; when $\Delta_i=0$, the person is known to be event-free through $K_i$.

For interval $t$, define the at-risk indicator $R_{it}=I(K_i\ge t)$, and define $D_{it}=I(\Delta_i=1,K_i=t)$. Thus, $R_{it}=1$ exactly when person $i$'s event status is observed for interval $t$, and $D_{it}=1$ only in that person's event interval. Let $H_{i,t-1}$ denote the information used to condition the hazard through the start of interval $t$, including relevant observed history and model covariates.

Definition: The discrete-time conditional hazard for person $i$ in interval $t$ is $$ h_{it}=P(T_i=t\mid T_i\ge t,H_{i,t-1}). $$ For someone who is event-free at the start of interval $t$, $h_{it}$ is the conditional probability of an event in that interval. The conditional probability of remaining event-free through the interval is $1-h_{it}$.

In a pooled logistic model, one specifies $h_{it}=\operatorname{expit}(\eta_{it})$, where $\operatorname{expit}(u)=1/(1+\exp(-u))$ and $\eta_{it}$ is a modeled linear predictor. The likelihood derivation below applies to this hazard model; it does not require the linear predictor to have any particular form.

Deriving One Person's Contribution

Assume first that person $i$ has an event in interval $k$. To observe that event time, the person must be event-free through intervals $1,\ldots,k-1$, then experience the event in interval $k$. Apply the chain rule, conditioning each interval on having reached it event-free and on the relevant history. The contribution is

L_i=\left\{\prod_{t=1}^{k-1}(1-h_{it})\right\}h_{ik}.

There is one event factor, $h_{ik}$, and it appears once. The factors before it are the conditional probabilities of surviving the earlier intervals. No probability after $k$ belongs in the observed event-time contribution: after the event, the person is no longer at risk for a first event.

If instead the person is censored after interval $c$, with no event observed, the observed event information is that the person survived through every observed interval. The likelihood contribution from the event process is

L_i=\prod_{t=1}^{c}(1-h_{it}).

There is no event factor for an unobserved event time after censoring. Treating the person as event-free after $c$ would add information that the data do not contain.

Theorem: Suppose each person's event process has conditional hazards $h_{it}$ given the history at the start of interval $t$. For observed data $(K_i,\Delta_i)$, the event-process likelihood contribution is $$ L_i=\prod_{t=1}^{K_i} h_{it}^{D_{it}}(1-h_{it})^{R_{it}-D_{it}}, $$ where factors are included only for intervals with $R_{it}=1$. Equivalently, $$ L_i=\left\{\prod_{t=1}^{K_i}(1-h_{it})\right\} \left\{\frac{h_{iK_i}}{1-h_{iK_i}}\right\}^{\Delta_i}. $$

Proof. If $\Delta_i=0$, then $D_{it}=0$ in every observed interval and $R_{it}=1$ for $t=1,\ldots,K_i$. The first product in the theorem is therefore $\prod_{t=1}^{K_i}(1-h_{it})$, and the second expression gives the same value because its event-odds factor is raised to zero.

If $\Delta_i=1$, the event occurs in interval $K_i$. Thus $D_{iK_i}=1$, $D_{it}=0$ for $t<K_i$, and $R_{it}=1$ through $K_i$. The first expression equals $\{\prod_{t=1}^{K_i-1}(1-h_{it})\}h_{iK_i}$. In the second expression, the factor $1-h_{iK_i}$ in the survival product is multiplied by $h_{iK_i}/(1-h_{iK_i})$, leaving $h_{iK_i}$; the earlier survival factors are unchanged. This gives the event contribution derived above. Both cases establish the result. $\square$

The Pooled Logistic Likelihood

Suppose $n$ people contribute observed intervals, and let $x_{it}$ be the model covariate vector for person $i$ in interval $t$. For a logistic hazard model, write

h_{it}=\frac{\exp(x_{it}^{\mathsf T}\beta)} {1+\exp(x_{it}^{\mathsf T}\beta)}.

Under independent sampling of people, the event-process likelihood is the product of their individual contributions. Substituting the logistic hazards gives the pooled logistic likelihood

L(\beta)=\prod_{i=1}^{n}\prod_{t:R_{it}=1} h_{it}^{D_{it}}(1-h_{it})^{1-D_{it}}.

Taking logarithms and using $\log h_{it}=x_{it}^{\mathsf T}\beta-\log\{1+\exp(x_{it}^{\mathsf T}\beta)\}$ and $\log(1-h_{it})=-\log\{1+\exp(x_{it}^{\mathsf T}\beta)\}$, the log-likelihood is

\ell(\beta)=\sum_{i=1}^{n}\sum_{t:R_{it}=1} \left[D_{it}x_{it}^{\mathsf T}\beta-\log\{1+\exp(x_{it}^{\mathsf T}\beta)\}\right].

This algebra has the form of a Bernoulli log-likelihood for the person-interval records that are at risk. That form does not mean that a person contributes independent observations after an event, or that every person has an outcome recorded in every interval. The risk indicator determines which terms are observed, and the sequential conditional-hazard interpretation is what justifies the product.

Result: If the hazard model is correctly specified and the covariates $x_{it}$ are treated as given, differentiating the log-likelihood gives the score $$ \frac{\partial\ell(\beta)}{\partial\beta} =\sum_{i=1}^{n}\sum_{t:R_{it}=1}x_{it}(D_{it}-h_{it}). $$

Derivation. For one included interval, the derivative of $D_{it}x_{it}^{\mathsf T}\beta$ is $D_{it}x_{it}$. The derivative of $\log\{1+\exp(x_{it}^{\mathsf T}\beta)\}$ is $h_{it}x_{it}$, by the chain rule and the definition of the expit function. Subtracting gives $x_{it}(D_{it}-h_{it})$; summing over all included person-intervals gives the displayed score. $\square$

Worked Likelihood Calculations

Worked Example: Event in the Third Interval

A person is observed event-free at the start of intervals 1, 2, and 3, and experiences the event in interval 3. Their fitted hazards are $h_{i1}=0.10$, $h_{i2}=0.20$, and $h_{i3}=0.40$. The individual contribution is $$ L_i=(1-0.10)(1-0.20)(0.40)=0.90\times0.80\times0.40=0.288. $$

The first two factors represent survival through intervals 1 and 2; the last factor represents the event in interval 3. There is no interval-4 factor. For comparison, the probability of reaching interval 3 event-free is $0.90\times0.80=0.72$, and $0.72\times0.40=0.288$, as required.

Worked Example: Censoring After Two Event-Free Intervals

A second person is observed event-free through interval 2 and then censored before interval 3. Suppose their hazards for the observed intervals are $0.10$ and $0.20$. The event-process contribution is $$ L_i=(1-0.10)(1-0.20)=0.90\times0.80=0.72. $$

There is no factor $h_{i3}$ or $1-h_{i3}$, because the person's event status in interval 3 is unknown. The value $0.72$ is the probability of the observed event-free history through interval 2, not the probability of remaining event-free indefinitely.

Worked Example: A Two-Person Pooled Likelihood

Suppose person A has an event in interval 1, with hazard $0.20$. Person B is event-free through interval 2 and censored after that interval; their hazards are $0.30$ and $0.40$. Assuming independent people, person A contributes $0.20$. Person B contributes $$ (1-0.30)(1-0.40)=0.70\times0.60=0.42. $$

The joint likelihood contribution is $0.20\times0.42=0.084$. In the person-interval representation, the included outcomes are $D_{A1}=1$, $D_{B1}=0$, and $D_{B2}=0$. Thus the product is $0.20\times0.70\times0.60=0.084$. Person A has no second-interval record: their contribution ends with the event interval.

Assumptions, Diagnostics, and Common Errors

The likelihood describes the observed event history under the specified conditional hazards. Its usual use for event-model inference requires independent people (or an analysis that accounts for dependence in the sampling design), a correctly specified hazard model for the conditioning information, and censoring that is ignorable for the event-process likelihood. A typical sufficient condition is that, conditional on the modeled observed history, censoring does not depend on the future event time in a way that provides additional event information. If that condition fails, censoring may need its own model or other adjustment; simply treating censoring as event-free follow-up is not valid.

Useful checks begin with the person-interval construction. Verify that each event person has exactly one $D_{it}=1$, at $t=K_i$, and no records after that interval. A censored person should have zero event indicators and records only through the last interval with known event-free status. Also check that every included interval has a person at risk at its start, and that fitted probabilities lie strictly between zero and one for finite logistic linear predictors. These checks catch data and coding errors; they do not establish that the hazard model or censoring assumption is correct.

A common mistake is to multiply the event factor repeatedly while writing an event contribution. For an event in interval 4, the correct expression is $\left\{\prod_{t=1}^{3}(1-h_{it})\right\}h_{i4}$, not $\prod_{t=1}^{3}\{(1-h_{it})h_{i4}\}$. The latter incorrectly inserts the interval-4 event probability three times. Another mistake is to keep a person in the data after their first event; those additional rows would describe outcomes that are not at risk for a first event.

The likelihood here is a full conditional likelihood for discrete-time event histories. It differs from a Cox partial likelihood, which uses risk-set comparisons rather than multiplying discrete interval event and survival probabilities. It also differs from an approach that models only a single event indicator at a fixed study endpoint, because the pooled model uses the sequence of observed interval outcomes. The likelihood does not, by itself, resolve informative censoring, misspecified hazards, or dependence between people. In a marginal structural model, weighting changes the estimating procedure applied to these interval contributions; the unweighted derivation here establishes the likelihood factors to which that later procedure is applied.

Check Your Understanding

Work out each contribution before reading the solutions.

A person has hazards $0.25$, $0.50$, and $0.10$, and experiences the event in interval 3. Write and evaluate their likelihood contribution.
A person is censored after two intervals with hazards $0.20$ and $0.30$. What is the event-process contribution, and why is there no third-interval factor?
Write the contribution for an event in interval 4 using survival factors and an event factor, with the event factor appearing exactly once.
For a logistic hazard model, what is the score contribution from one observed interval with covariate vector $x_{it}$, event indicator $D_{it}$, and hazard $h_{it}$?

Solutions. (1) The contribution is $(1-0.25)(1-0.50)(0.10)=0.75\times0.50\times0.10=0.0375$. (2) It is $(1-0.20)(1-0.30)=0.80\times0.70=0.56$; interval 3 is unobserved, so its event status contributes no factor. (3) It is $\left\{\prod_{t=1}^{3}(1-h_{it})\right\}h_{i4}$. (4) Differentiating the interval's log-likelihood term gives $x_{it}(D_{it}-h_{it})$.