Comparing Regime Survival and Risk from Pooled Logistic Fits

From Fitted Hazards to Regime Comparisons

A fitted pooled logistic model gives a hazard for each follow-up interval and treatment regime. To make an interpretable comparison, choose a common follow-up horizon, calculate survival under each regime through that horizon, and then express the complement as cumulative risk. This tutorial focuses on those calculations and their interpretation—not on treating a hazard coefficient as if it were itself a risk difference.

Let $k=1,\ldots,H$ index discrete follow-up intervals, and let $g$ denote a specified treatment regime, such as treatment at every interval or no treatment at any interval. Write $T^g$ for the event time that would be observed under regime $g$. Define the regime-specific discrete hazard as the probability of an event in interval $k$, conditional on having remained event-free through the preceding intervals under that regime.

Definition: The interval hazard under regime $g$ is $h_k^g=P(T^g\text{ falls in interval }k\mid T^g\text{ is beyond interval }k-1)$. The survival probability through interval $k$ is $S_k^g=P(T^g\text{ is beyond interval }k)$, and the cumulative risk through horizon $H$ is $R_H^g=P(T^g\text{ falls in one of intervals }1,\ldots,H)$.

The pooled logistic model supplies fitted hazards $\hat h_k(g;\hat\beta)$, obtained by substituting the treatment history prescribed by $g$ into the fitted model. For example, if the model is $\operatorname{logit}\{h_k\}=\alpha_k+\beta A_k$, then a regime specifying $A_k=a_k^g$ yields $\hat h_k(g;\hat\beta)=\operatorname{expit}(\hat\alpha_k+\hat\beta a_k^g)$. More general models can include time-by-treatment terms or other regime-history terms; the same calculation applies after making the appropriate substitutions.

These fitted quantities estimate causal regime hazards only under the causal identification and model conditions needed for the weighted analysis. In particular, the treatment regimes must be well-defined, consistency must hold, there must be no unmeasured confounding conditional on measured treatment and outcome history, and treatment positivity must hold for the histories needed by the regimes. Censoring must also be handled appropriately, for example by a valid censoring model or weights. The fitted marginal structural model must represent the relevant regime hazards adequately, and the treatment and censoring weights must support the intended target population. Without these conditions, the calculations remain predictions from a fitted model but need not have a causal interpretation.

Calculating Survival, Risk, and Risk Differences

For a fixed regime, surviving through interval $k$ means surviving through interval $k-1$ and then not experiencing the event in interval $k$. Thus, the fitted survival estimate is obtained by multiplying the interval survival probabilities. Cumulative risk is its complement.

\hat S_H^g=\prod_{k=1}^{H}\{1-\hat h_k(g;\hat\beta)\}, \qquad \hat R_H^g=1-\hat S_H^g.

For two regimes $g_1$ and $g_0$, define the risk difference as the risk under $g_1$ minus the risk under $g_0$. State this order whenever reporting a contrast: a negative value then means lower estimated cumulative risk under $g_1$.

\widehat{\mathrm{RD}}_H =\hat R_H^{g_1}-\hat R_H^{g_0} =\hat S_H^{g_0}-\hat S_H^{g_1}.

Result: If the interval hazards under a regime are $h_1^g,\ldots,h_H^g$, then the risk through $H$ is the sum of the interval-specific event probabilities: $$ R_H^g=\sum_{k=1}^{H} h_k^g\prod_{j=1}^{k-1}(1-h_j^g), $$ where the product for $k=1$ is 1.

Proof. The probability of an event in interval $k$ equals the probability of surviving the first $k-1$ intervals multiplied by the probability of an event in interval $k$ conditional on that survival. By repeated conditioning, the first probability is $\prod_{j=1}^{k-1}(1-h_j^g)$. The events “the event occurs in interval $k$” are disjoint across $k=1,\ldots,H$; summing their probabilities gives the displayed risk. Equivalently, the identity follows by telescoping: if $S_0^g=1$, then $S_{k-1}^g-S_k^g=S_{k-1}^g h_k^g$, so $$ 1-S_H^g=\sum_{k=1}^{H}(S_{k-1}^g-S_k^g) =\sum_{k=1}^{H}h_k^g\prod_{j=1}^{k-1}(1-h_j^g). $$ This also shows why adding the hazards alone is generally wrong: each later hazard applies only to those who have survived earlier intervals. $\square$

Worked Example: Comparing Two Three-Interval Regimes

Suppose fitted hazards under regime $g_1$ are $0.10, 0.20, 0.25$, and under regime $g_0$ are $0.15, 0.15, 0.20$. Calculate survival and cumulative risk through interval 3, then report the risk difference in the order $g_1-g_0$.

Under $g_1$, the interval survival probabilities are $0.90,0.80,0.75$. Therefore, $$ \hat S_3^{g_1}=0.90(0.80)(0.75)=0.54, \qquad \hat R_3^{g_1}=1-0.54=0.46. $$

Under $g_0$, they are $0.85,0.85,0.80$, so $$ \hat S_3^{g_0}=0.85(0.85)(0.80)=0.578, \qquad \hat R_3^{g_0}=1-0.578=0.422. $$

The risk difference is $0.460-0.422=0.038$, or 3.8 percentage points higher cumulative risk under $g_1$ through interval 3. The survival difference in the opposite order is $0.54-0.578=-0.038$, which is consistent: lower survival under $g_1$ corresponds to higher cumulative risk.

Using a Fitted Logistic Model

A logistic coefficient does not directly give a horizon-specific risk contrast. First calculate each regime's interval hazards from the fitted model, then apply the product and complement. The next example illustrates this distinction when a treatment coefficient represents a common odds ratio across intervals.

Worked Example: Substituting Regimes Into a Logistic Fit

Suppose a fitted model has interval-specific intercepts that give untreated hazards $0.05,0.10,0.15$, and a treatment coefficient $\hat\beta=\log(2)$. The model has no time-by-treatment interaction, so treatment doubles the odds of the interval event at each time. Compare always treated ($g_1$) with never treated ($g_0$) through interval 3.

The untreated hazards are already given. For a treated interval, multiply the untreated event odds by 2 and convert back to a probability. The three untreated odds are $0.05/0.95=1/19$, $0.10/0.90=1/9$, and $0.15/0.85=3/17$. Thus, the treated hazards are $$ \hat h_1^{g_1}=\frac{2/19}{1+2/19}=\frac{2}{21}, \qquad \hat h_2^{g_1}=\frac{2/9}{1+2/9}=\frac{2}{11}, \qquad \hat h_3^{g_1}=\frac{6/17}{1+6/17}=\frac{6}{23}. $$

For $g_0$, survival and risk are $$ \hat S_3^{g_0}=(0.95)(0.90)(0.85)=0.72675, \qquad \hat R_3^{g_0}=0.27325. $$

For $g_1$, multiply the fitted interval survival probabilities: $$ \hat S_3^{g_1} =\left(\frac{19}{21}\right)\left(\frac{9}{11}\right)\left(\frac{17}{23}\right) =\frac{2907}{5313}\approx 0.54715, \qquad \hat R_3^{g_1}\approx 1-0.54715=0.45285. $$

The risk difference $g_1-g_0$ is approximately $0.45285-0.27325=0.17960$, or 18.0 percentage points. The coefficient $\log(2)$ describes an interval odds ratio in this model; it is not the 18.0-percentage-point risk difference, nor is it itself a risk ratio. The horizon-specific comparison depends on all the fitted hazards and their accumulation over follow-up.

Regimes Can Vary Across Intervals

A regime need not assign the same treatment at every interval. For a dynamic or time-varying regime, use the treatment value prescribed at each interval when obtaining that interval's fitted hazard. Keep the horizon and the target population common when comparing regimes; otherwise the resulting difference mixes distinct follow-up targets.

Worked Example: A Time-Varying Regime Through Four Intervals

Suppose the fitted hazards for treatment $A_k=0$ across four intervals are $0.04,0.08,0.12,0.16$, and for $A_k=1$ are $0.06,0.06,0.09,0.10$. Compare the alternating regime $g_1=(1,0,1,0)$ with the never-treated regime $g_0=(0,0,0,0)$.

Under $g_1$, substitute the prescribed treatment at each interval. Its hazard sequence is $0.06,0.08,0.09,0.16$, so $$ \hat S_4^{g_1}=(0.94)(0.92)(0.91)(0.84)=0.66045312, \qquad \hat R_4^{g_1}=0.33954688. $$

Under $g_0$, the hazard sequence is $0.04,0.08,0.12,0.16$, giving $$ \hat S_4^{g_0}=(0.96)(0.92)(0.88)(0.84)=0.65286144, \qquad \hat R_4^{g_0}=0.34713856. $$

The risk difference is $0.33954688-0.34713856=-0.00759168$, about $-0.76$ percentage points. In this fitted example, cumulative risk through interval 4 is slightly lower under the alternating regime. This conclusion comes from the products across all four intervals, not from simply comparing or averaging the treatment-specific hazards.

Result: For any two regimes at the same horizon, the risk difference is exactly the negative of the survival difference when both differences use the same regime order: $$ R_H^{g_1}-R_H^{g_0}=-(S_H^{g_1}-S_H^{g_0}). $$ Proof. By definition, $R_H^g=1-S_H^g$ for each regime. Subtracting gives $$ R_H^{g_1}-R_H^{g_0} =(1-S_H^{g_1})-(1-S_H^{g_0}) =S_H^{g_0}-S_H^{g_1} =-(S_H^{g_1}-S_H^{g_0}). $$ Thus the signs must be opposite, and the absolute differences must be equal. $\square$

Interpretation, Diagnostics, and Alternatives

Survival and cumulative risk describe the same horizon-specific event experience on complementary scales. Reporting both can make a result easier to understand, but it does not create two independent findings. A hazard contrast instead compares conditional event probabilities among people still event-free at each interval; it is not generally equal to a risk difference or a survival difference. In particular, the risk difference accumulates interval hazards while accounting for survival up to each interval.

Before interpreting estimates, check the calculation and the fitted predictions:

Each fitted hazard must be between 0 and 1. Each resulting survival must be between 0 and 1, and survival must not increase as the horizon extends.
Cumulative risk must be between 0 and 1 and must not decrease with a longer horizon.
Confirm that the same interval definitions, horizon, and target population are used for both regimes. Apply each regime's prescribed treatment at the correct interval.
Inspect the data support for the treatment histories required by each regime. Sparse histories, extreme weights, or near-violations of positivity can make fitted regime predictions unstable even when the product calculation is correct.
Check whether the fitted time pattern and treatment terms adequately represent the hazards. A plausible-looking product cannot repair misspecification or an invalid causal identification strategy.
Ensure censoring has been addressed: a product based on fitted hazards is not automatically a valid estimate of survival if censoring makes later risk sets unrepresentative.

The main alternative is to estimate regime-specific survival by standardizing predictions from a conditional outcome model over the target population, rather than using a weighted marginal structural model. Either approach still requires justified causal assumptions and adequate models; they differ in how the outcome and covariate history enter estimation. A hazard-ratio summary can be more compact, but it answers a different question and does not by itself provide the horizon-specific absolute risk contrast calculated here. The plug-in calculations also do not quantify uncertainty: sampling variability and uncertainty from estimated weights require appropriate inferential methods.

A frequent mistake is to add interval hazards and call the result cumulative risk. For example, hazards $0.10$ and $0.20$ do not imply risk $0.30$: the second hazard applies only among those who survived the first interval, so the risk is $0.10+(0.90)(0.20)=0.28$. Another mistake is to report a risk difference without its regime order or horizon. Naming both makes the sign and the target of the comparison unambiguous.

Check Your Understanding

Answer each question, then compare your work with the solutions.

Under one regime, fitted hazards are $0.20$ and $0.10$. What are survival and cumulative risk through interval 2?
At a fixed horizon, regime A has risk $0.30$ and regime B has risk $0.22$. What is the risk difference A minus B, and what is the survival difference A minus B?
Why is the sum of interval hazards generally not the cumulative risk?
A treatment regime specifies $A_1=1$ and $A_2=0$. Which treatment predictions should be used for the two interval hazards?

Solutions. (1) Survival is $(1-0.20)(1-0.10)=0.80(0.90)=0.72$, so cumulative risk is $1-0.72=0.28$. (2) The risk difference is $0.30-0.22=0.08$. The corresponding survival probabilities are $0.70$ and $0.78$, so the survival difference A minus B is $0.70-0.78=-0.08$. (3) Each later hazard applies only to those who survived earlier intervals; cumulative risk weights each hazard by the probability of reaching its interval event-free. (4) Substitute treatment 1 into the fitted model for interval 1 and treatment 0 for interval 2, then multiply the two corresponding interval survival probabilities.