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A logistic regression can specify each binary response conditionally on recorded past outcomes and baseline covariates. A chain rule for probabilities gives the path likelihood as the product of the conditional Bernoulli distribution masses. This permits dependence within a person's history without treating their outcomes as unconditionally independent. The model must specify the initial history and which past features enter its conditional mean.

Let $Y_{it}\in \{0,1\}$ indicate a smoking-free week, $S_i$ the recorded sex indicator, $A_i$ age in years and $T_i$ assigned treatment. The screening history supplies $Y_{i0}=0$; earlier lag initializations are also zero. A history-dependent logistic regression for the first model isHere $t=1,\dots ,10$, ${\mathcal{H}}_{i,t-1}$ is the observed past, and the five unknown regression coefficients have time-invariant values. Conditional on baseline covariates, this model has the first-order Markov property: only the immediately preceding outcome enters the current conditional probability. Distinct subjects have independent histories. There is no subject random effect or additional time trend in this fit.

Using the chain rule for probabilities, the individual conditional likelihood isThe lagged values in ${\eta }_{it}$ are the individual's actual preceding outcomes. This product is a sequential conditional likelihood, not an assertion of unconditional independence of the ten readings. The full conditional likelihood is ${\prod }_i{L}_i$; no extra Bernoulli distribution factor is attached to the fixed screening history.

Take the event to mean three smoking-free weeks in succession, $Y_{i1}=Y_{i2}=Y_{i3}=1$. The initial cumulative count is zero. Along this path it equals $0,1,2$ in weeks one, two, three, respectively. In the cumulative-response logistic model, defineThe chain rule for probabilities then givesThe three conditional probabilities are approximately $0.19891,0.27015,0.35559$. If “stop during the first three weeks” instead means at least one smoking-free week by week three, its different event has probability $1-(1-{\pi }_0{)}^3\simeq 0.48590$: along the all-failure path the cumulative count stays zero. Stating the event resolves this wording ambiguity.