Recovering Survival from Fitted Discrete Hazards

From Interval Hazards to Survival

A fitted discrete-time survival model gives a probability of the event in each interval, conditional on having remained event-free until that interval. A survival probability through several intervals is therefore not one of those fitted probabilities: it is the probability of avoiding the event in every interval so far. The conversion is a product, and its interpretation depends on the conditional meaning of each hazard.

Let $T^g$ be the event time that would occur under treatment regime $g$, with event times recorded in intervals $1,\ldots,K$. Assume one event type and no event before interval 1. Define survival through interval $k$ as $S^g(k)=P(T^g>k)$, with $S^g(0)=1$. The interval-specific discrete hazard under regime $g$ is the conditional probability of an event in interval $t$, given survival through the preceding interval.

Definition: For $t=1,\ldots,K$, the discrete-time hazard under regime $g$ is $$ h_t^g=P(T^g=t\mid T^g\ge t). $$ The corresponding conditional probability of surviving interval $t$ is $1-h_t^g$. Survival through interval $k$ is $S^g(k)=P(T^g>k)$.

The conditioning event $T^g\ge t$ means that the person is at risk at the start of interval $t$. The hazard is not the unconditional probability of having the event in that interval. For example, $h_2^g$ is conditional on having survived interval 1; it is not the probability of an event in interval 2 among everyone at baseline.

Deriving the Product Formula

The product follows directly from the chain rule for conditional probabilities. For a person to survive through interval $k$, they must survive interval 1, then interval 2 conditional on surviving interval 1, and so on.

Theorem: For discrete event times $1,\ldots,K$, if $h_t^g=P(T^g=t\mid T^g\ge t)$ is defined for every interval with $P(T^g\ge t)>0$, then for $k=1,\ldots,K$, $$ S^g(k)=\prod_{t=1}^{k}(1-h_t^g). $$

Proof. For $t\ge 1$, the events $T^g\ge t$ and $T^g>t-1$ are the same for integer-valued event times. Thus $$ P(T^g>t\mid T^g>t-1) =1-P(T^g=t\mid T^g\ge t) =1-h_t^g. $$ For $k\ge 1$, repeated application of the chain rule gives $$ P(T^g>k)=P(T^g>0)\prod_{t=1}^{k}P(T^g>t\mid T^g>t-1). $$ Since event-free status before interval 1 is assumed, $P(T^g>0)=1$. Substituting the conditional probabilities above proves the formula. $\square$

This argument requires that the hazard be defined among those at risk. If no one has positive probability of reaching an interval under a regime, its conditional hazard is undefined, and the product cannot be extended through that interval without additional assumptions.

Using Fitted Pooled Logistic Hazards

In a pooled logistic model, each person-interval record represents an interval in which the person is at risk. For a fitted model with linear predictor $\hat\eta_t(g)$ under regime $g$, the fitted hazard is $$ \hat h_t(g)=\operatorname{expit}\{\hat\eta_t(g)\} =\frac{\exp\{\hat\eta_t(g)\}}{1+\exp\{\hat\eta_t(g)\}}. $$ The fitted survival estimate through interval $k$ is then $$ \hat S^g(k)=\prod_{t=1}^{k}\{1-\hat h_t(g)\}. $$

For an MSM, $\hat\eta_t(g)$ is evaluated at the treatment history specified by regime $g$, using the fitted time terms and treatment terms in the model. This produces a predicted hazard for each interval under that regime. The product combines those sequential conditional predictions; it does not average hazards across intervals or add them.

Conditions for a causal interpretation: The product calculation itself is algebraic, but interpreting $\hat S^g(k)$ as survival under an intervention requires an appropriate causal and modeling basis. In particular, the intervention must be well-defined; consistency, sequential exchangeability, and positivity must support identification; censoring must be handled under suitable assumptions; and the fitted MSM must adequately represent the regime-specific hazards. The model must also predict hazards for the intervals and regimes being reported.

Worked Example: Multiplying Three Interval Hazards

Under one regime, suppose the hazards for intervals 1, 2, and 3 are $0.10$, $0.20$, and $0.25$. Calculate survival after each interval and through interval 3.

The probability of surviving interval 1 is $1-0.10=0.90$, so $S^g(1)=0.90$. The conditional probability of surviving interval 2, given survival to its start, is $1-0.20=0.80$. Therefore $$ S^g(2)=0.90\times0.80=0.72. $$ For interval 3, the conditional survival probability is $1-0.25=0.75$, giving $$ S^g(3)=0.90\times0.80\times0.75=0.54. $$

Thus the probability of surviving all three intervals is $0.54$, and the probability of having the event by the end of interval 3 is $1-0.54=0.46$. The interval-3 hazard $0.25$ is not itself the probability of the event by the end of interval 3.

Worked Example: Calculating Survival Under Two Regimes

Suppose a fitted pooled logistic model has an interval-specific baseline logit $\alpha_t$ and a treatment coefficient $\beta=\log(1.5)$, so that $\hat\eta_t(a)=\alpha_t+\beta a$ for treatment value $a$. Let the fitted hazards under $a=0$ be $0.05$, $0.12$, and $0.20$ in three intervals. Calculate survival through interval 3 for regimes $a=0$ and $a=1$.

For $a=0$, the interval survival probabilities are $0.95$, $0.88$, and $0.80$. Hence $$ \hat S^{0}(3)=0.95\times0.88\times0.80=0.6688. $$

For $a=1$, add $\log(1.5)$ to each baseline logit, then apply the inverse-logit. For interval 1, $\alpha_1=\log(0.05/0.95)$, so $$ \hat h_1(1)=\operatorname{expit}\{\log(0.05/0.95)+\log(1.5)\} =\frac{3}{41}\approx0.0731707. $$ For interval 2, $\alpha_2=\log(0.12/0.88)$, giving $$ \hat h_2(1)=\operatorname{expit}\{\log(0.12/0.88)+\log(1.5)\} =\frac{9}{53}\approx0.1698113. $$ For interval 3, $\alpha_3=\log(0.20/0.80)$, giving $$ \hat h_3(1)=\operatorname{expit}\{\log(0.20/0.80)+\log(1.5)\} =\frac{3}{11}\approx0.2727273. $$ The corresponding survival factors are $38/41$, $44/53$, and $8/11$. Thus $$ \hat S^{1}(3)=\frac{38}{41}\times\frac{44}{53}\times\frac{8}{11} =\frac{13376}{23903}\approx0.5596. $$

The fitted survival estimates are approximately $0.6688$ under $a=0$ and $0.5596$ under $a=1$. Each estimate uses the hazard for its own regime in every interval. The calculations do not establish that the difference is causal; that interpretation depends on the assumptions and model adequacy stated above.

Plug-In Accuracy and a Simple Check

A useful property of the product calculation is that small errors in fitted hazards cannot create an unbounded change in survival over a fixed, finite horizon. The following statement makes this precise.

Result: Fix $k<\infty$. If, under a regime $g$, $\hat h_t(g)\xrightarrow{p}h_t^g$ for every $t=1,\ldots,k$, then $$ \prod_{t=1}^{k}\{1-\hat h_t(g)\}\xrightarrow{p}S^g(k). $$

Proof. Define $f(x_1,\ldots,x_k)=\prod_{t=1}^{k}(1-x_t)$, a continuous function on $[0,1]^k$. Because $k$ is finite, convergence in probability of each fitted hazard implies joint convergence in probability of the vector $(\hat h_1(g),\ldots,\hat h_k(g))$ to $(h_1^g,\ldots,h_k^g)$. The continuous mapping theorem therefore gives $$ f\{\hat h_1(g),\ldots,\hat h_k(g)\} \xrightarrow{p} f(h_1^g,\ldots,h_k^g) =\prod_{t=1}^{k}(1-h_t^g)=S^g(k). $$ No independence among the fitted hazards is needed. $\square$

This is a plug-in consistency statement, not a guarantee that the fitted hazards are consistent. If the MSM is misspecified, the product can converge to the survival curve implied by the wrong hazards. The product step preserves the model’s predictions; it does not repair them.

Worked Example: Recovering Survival From Fitted Linear Predictors

Suppose a fitted model has linear predictor $\hat\eta_t(a)=\alpha_t+0.7a$, with $\alpha_1=-2$ and $\alpha_2=-1.5$. Calculate fitted survival through interval 2 under $a=0$ and $a=1$.

Under $a=0$, the hazards are $$ \hat h_1(0)=\operatorname{expit}(-2)\approx0.1192,\qquad \hat h_2(0)=\operatorname{expit}(-1.5)\approx0.1824. $$ Thus $$ \hat S^0(2)=(1-0.1192)(1-0.1824) \approx0.8808\times0.8176 \approx0.7201. $$

Under $a=1$, the linear predictors are $-2+0.7=-1.3$ and $-1.5+0.7=-0.8$. The hazards are $$ \hat h_1(1)=\operatorname{expit}(-1.3)\approx0.2142,\qquad \hat h_2(1)=\operatorname{expit}(-0.8)\approx0.3100. $$ Therefore $$ \hat S^1(2)=(1-0.2142)(1-0.3100) \approx0.7858\times0.6900 \approx0.5422. $$

The fitted survival under $a=1$ is lower through interval 2 in this example. The calculation first converts each regime-specific linear predictor to a hazard, then multiplies the interval survival probabilities.

Diagnostics, Alternatives, and Limitations

Several checks help identify when the calculation or the model is going wrong:

Check the scale. Each fitted hazard should lie in $[0,1]$. A logistic inverse link guarantees this, but a hand-entered quantity such as a linear predictor or an odds value is not a hazard.
Check the survival sequence. With hazards in $[0,1]$, fitted survival starts at 1 and cannot increase: $\hat S^g(t)=\hat S^g(t-1)\{1-\hat h_t(g)\}\le\hat S^g(t-1)$. An increase signals a coding or calculation error.
Check regime and interval indexing. A treatment value or history entered for the wrong interval changes the predicted hazards and therefore every later survival value.
Check support and censoring. Sparse risk sets, poor treatment support, or informative censoring can make a model-based curve unreliable even when every prediction is between 0 and 1.
Check model structure. A product can be calculated precisely from an inadequate model. Compare predicted hazards with observed interval patterns in relevant parts of the data, and assess whether the time and treatment terms can represent those patterns.

A weighted empirical survival estimator, such as a weighted product-limit estimate, is a main alternative. It builds interval survival from weighted event and risk-set counts rather than from a pooled logistic regression model. The regression approach can provide a compact, structured set of regime-specific hazards, but relies on its functional form; the empirical approach is less dependent on that form but can be unstable when weighted risk sets are sparse. Neither removes the need for appropriate causal assumptions, treatment support, and censoring assumptions.

The product is most straightforward for a single event type observed over a fixed set of discrete intervals. It does not by itself handle competing events, define what to do with delayed entry, or justify extrapolation beyond observed follow-up. It also does not provide uncertainty intervals: uncertainty in the fitted hazards must be carried through the survival calculation using an appropriate inferential procedure.

Common pitfall: Multiplying $1-\hat h_t(g)$ is correct because each hazard is conditional on reaching its interval. Multiplying $1-\hat h_t(g)$ when the inputs are unconditional interval event probabilities is not generally correct; those probabilities have a different conditioning set. First verify that the fitted quantity is the at-risk conditional hazard for the specified regime and interval.

Check Your Understanding

Answer each question, then check the solutions below.

Hazards under a regime are $0.08$, $0.10$, and $0.20$. What is survival through interval 3?
A fitted model predicts hazards $0.15$ and $0.25$ in two intervals under a regime. What is the probability of the event by the end of interval 2?
Why can survival not increase from one interval to the next when hazards are valid probabilities?
What additional assumption is needed to interpret a fitted product as survival under an intervention, rather than simply as survival implied by a statistical model?

Solutions. (1) $S^g(3)=0.92\times0.90\times0.80=0.6624$. (2) Survival is $0.85\times0.75=0.6375$, so the event probability by interval 2 is $1-0.6375=0.3625$. (3) Since $0\le\hat h_t(g)\le1$, $\hat S^g(t)=\hat S^g(t-1)(1-\hat h_t(g))\le\hat S^g(t-1)$. (4) The causal interpretation requires identification assumptions, including consistency, sequential exchangeability, positivity, and suitable assumptions for censoring, as well as an adequate fitted hazard model.