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Fitting MSMs for Survival Outcomes · Tutorial 197 of 501

Why Fitted Odds Are Not Discrete-Time Risks

Learn to convert pooled-logistic fitted odds into discrete-time risks and quantify the error from interpreting odds directly as probabilities.

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What You'll Learn

  • Distinguish an interval event risk from its odds among those at risk.
  • Convert a fitted log-odds linear predictor into a probability.
  • Quantify how much odds used as probabilities overstate risk.
  • Explain why an odds ratio alone does not determine an absolute risk.
  • Propagate converted interval risks into survival and cumulative risk.
  • Identify common interpretation errors in pooled logistic MSM predictions.

What a Pooled Logistic Fit Predicts

In a discrete-time survival analysis, each interval contributes an event indicator for individuals who are event-free at its start. A pooled logistic model uses these records to model the conditional probability of an event in the interval. The word logistic refers to the link: the model is linear in the log odds, not in the risk itself. Confusing these quantities changes predicted probabilities, sometimes substantially.

Consider a single event type and intervals indexed by \(t\). Let \(R_t=1\) indicate that a person is at risk at the start of interval \(t\), and let \(Y_t=1\) indicate that the event occurs during that interval. For a specified treatment regime \(a\), define the interval risk among those still at risk as

$$ p_t(a)=P(Y_t=1\mid R_t=1,\text{ regime }a). $$

This is a conditional risk: it is the probability of an event during the interval, given survival to its start. It is not the unconditional probability of having had the event by the end of the interval. A pooled logistic model represents the log odds of this conditional event as a linear predictor \(\eta_t(a)\):

$$ \log\left\{\frac{p_t(a)}{1-p_t(a)}\right\}=\eta_t(a). $$

In a weighted marginal structural model, the weights and model specification determine the estimand and the fitted linear predictor. They do not change the meaning of the logistic link. Under the assumptions required for the causal interpretation of the weighted model, the fitted predictor describes the log odds in the target pseudo-population. In all cases, converting the fitted log odds to a probability is a mathematical step, not an additional causal assumption.

Definition: The odds corresponding to an interval risk \(p\) are \(o=p/(1-p)\). Conversely, if the odds are \(o\), the risk is \(p=o/(1+o)\). In a pooled logistic model, the linear predictor \(\eta\) is the log odds, so the fitted odds are \(\exp(\eta)\), while the fitted risk is \(\exp(\eta)/(1+\exp(\eta))\).

Converting Fitted Odds to Interval Risks

The conversion follows directly from the definition. If \(o=p/(1-p)\), then \(o(1-p)=p\). Rearranging gives \(o=p(1+o)\), hence \(p=o/(1+o)\). Since \(\eta=\log(o)\), substituting \(o=\exp(\eta)\) gives the familiar inverse-logit formula. This conversion applies at each time interval and for each treatment regime or covariate pattern for which a prediction is required.

Result: For any finite fitted linear predictor \(\eta\), the corresponding fitted interval risk is
$$ \widehat p=\frac{\exp(\eta)}{1+\exp(\eta)} =\frac{1}{1+\exp(-\eta)}. $$
The fitted odds are \(\exp(\eta)\), not \(\widehat p\).

This result is an identity, not an approximation. It holds for any value of \(\eta\), regardless of whether the model is correctly specified. Correct specification matters for whether the fitted quantity estimates the intended population quantity; it does not alter the conversion from the model's linear predictor to its implied probability.

1
Compute the linear predictor.
Insert the interval, treatment regime, and any other model inputs into the fitted pooled-logistic model to obtain \(\hat\eta_t(a)\).
2
Exponentiate to get odds, if needed.
The fitted odds are \(\exp\{\hat\eta_t(a)\}\). An odds value is not bounded above by 1.
3
Apply the inverse-logit.
Calculate \(\hat p_t(a)=\exp\{\hat\eta_t(a)\}/[1+\exp\{\hat\eta_t(a)\}]\). This is the fitted interval risk.

Worked Example: Converting an Odds Value of 0.25

Suppose a pooled logistic fit gives an interval odds of \(0.25\) for a specified regime and interval. The corresponding risk is

$$ \widehat p=\frac{0.25}{1+0.25} =\frac{0.25}{1.25} =0.20. $$

Thus, among individuals at risk at the start of the interval, the fitted probability of an event during the interval is \(0.20\), or 20%. Treating the odds \(0.25\) as a risk would report 25%, an overstatement of 5 percentage points. The odds exceed the risk because the odds divide the event probability by the probability of no event: \(0.20/(1-0.20)=0.20/0.80=0.25\).

Worked Example: Converting a Linear Predictor

Suppose the fitted linear predictor for an interval is \(\hat\eta=\log(0.75)\). Exponentiating gives fitted odds of \(\exp\{\hat\eta\}=0.75\). The fitted risk is therefore

$$ \widehat p=\frac{0.75}{1+0.75} =\frac{0.75}{1.75} =\frac{3}{7} \approx 0.4286. $$

The correct prediction is about 42.9%, not 75%. If instead the linear predictor were \(0\), the fitted odds would be \(\exp(0)=1\), and the fitted risk would be \(1/(1+1)=0.5\). A linear predictor of zero therefore does not mean zero risk.

Why Odds Used as Risks Overstate the Event Probability

For any positive odds \(o\), the risk is \(p=o/(1+o)\). Subtracting the risk from the odds gives an exact expression for the error made by reporting \(o\) as if it were a probability.

Result: If positive odds \(o\) are mistakenly interpreted as a risk, the absolute overstatement is \(o^2/(1+o)\), and the overstatement relative to the true risk is \(o\). The error is positive for every \(o>0\), and grows as the odds increase.

To verify the result, subtract \(p=o/(1+o)\) from \(o\):

$$ o-p =o-\frac{o}{1+o} =\frac{o(1+o)-o}{1+o} =\frac{o^2}{1+o}>0. $$

For the relative error, divide this difference by the true risk \(p\):

$$ \frac{o-p}{p} = \frac{o^2/(1+o)}{o/(1+o)} =o. $$

For example, odds of \(0.10\) overstate the risk by 10% relative to the risk; odds of \(1\) overstate it by 100%. When the risk is small, odds and risk are numerically close, which can make the mistake appear harmless. But closeness is an approximation that depends on the event being rare, not an identity. At \(p=0.5\), the odds are \(1\), twice the risk.

Worked Example: The Error at a Higher Interval Risk

Suppose the fitted odds are \(1.5\). The correct interval risk is

$$ \widehat p=\frac{1.5}{1+1.5} =\frac{1.5}{2.5} =0.60. $$

Using \(1.5\) as a probability would claim a risk of 150%, which is impossible. Even if a program or report truncates such a value, truncation does not repair the interpretation. The exact absolute error is \(1.5-0.60=0.90\), matching \(o^2/(1+o)=2.25/2.5=0.90\). The relative error is \(0.90/0.60=1.5\), equal to the odds.

From Interval Risks to Survival and Cumulative Risk

A fitted interval risk is conditional on still being event-free at the interval's start. For consecutive intervals, the probability of surviving through all of them is the product of the corresponding conditional probabilities of surviving each interval. The conversion from odds to risks must therefore happen before combining intervals.

Result: Let \(p_t(a)\) be the event risk in interval \(t\), conditional on being at risk at its start under regime \(a\). For \(K\) intervals, survival through interval \(K\) is \(\prod_{t=1}^{K}\{1-p_t(a)\}\), and cumulative event risk is \(1-\prod_{t=1}^{K}\{1-p_t(a)\}\).

For the derivation, write \(S_K(a)=P(R_{K+1}=1\mid a)\), the probability of remaining event-free through interval \(K\). By the multiplication rule,

$$ P(R_{K+1}=1\mid a) = \prod_{t=1}^{K}P(R_{t+1}=1\mid R_t=1,a). $$

For each interval, no event occurs with probability \(1-p_t(a)\), so \(P(R_{t+1}=1\mid R_t=1,a)=1-p_t(a)\). Substitution yields the survival product. Since survival through \(K\) intervals and an event by the end of interval \(K\) are complementary events, cumulative event risk is one minus that product. This derivation assumes a single event type and intervals defined so that remaining at risk means no event has yet occurred.

Worked Example: Two Intervals and Cumulative Risk

Suppose the fitted odds are \(1/9\) in the first interval and \(1/4\) in the second. Convert each odds value separately:

$$ \widehat p_1=\frac{1/9}{1+1/9}=\frac{1}{10}=0.10, \qquad \widehat p_2=\frac{1/4}{1+1/4}=\frac{1}{5}=0.20. $$

The fitted probability of surviving both intervals is

$$ (1-\widehat p_1)(1-\widehat p_2) =(0.90)(0.80)=0.72, $$

so cumulative event risk by the end of the second interval is \(1-0.72=0.28\), or 28%. If the odds were incorrectly used as interval risks, the calculation would instead give \(1-(1-1/9)(1-1/4)=1-(8/9)(3/4)=1-2/3=1/3\), or about 33.3%. That is not the risk implied by the fitted logistic model.

Odds Ratios Do Not Supply Absolute Risks

A fitted treatment coefficient in a logistic model is often exponentiated and described as an odds ratio. An odds ratio compares odds between model conditions; it is not itself a risk, a risk ratio, or a risk difference. To obtain absolute risks, one needs the relevant fitted linear predictors, or equivalently the baseline odds together with the odds ratio. This distinction matters especially when reporting regime-specific predictions or absolute treatment contrasts.

Worked Example: An Odds Ratio Needs a Baseline

Suppose the untreated interval odds are \(0.20\), and the treatment odds ratio is \(2\). The treated odds are \(2(0.20)=0.40\). Convert each to a risk:

$$ p_0=\frac{0.20}{1.20}=\frac{1}{6}\approx 0.1667, \qquad p_1=\frac{0.40}{1.40}=\frac{2}{7}\approx 0.2857. $$

The risk difference is \(2/7-1/6=12/42-7/42=5/42\approx0.1190\), or about 11.9 percentage points. The risk ratio is \((2/7)/(1/6)=12/7\approx1.714\), not \(2\). The odds ratio alone cannot produce either absolute risk: the baseline odds \(0.20\) were needed to recover them.

Diagnostics, Alternatives, and Common Pitfalls

Several checks help prevent an interpretation error. First, label model output according to what it is: a linear predictor, a log odds, an odds value, or a risk. Second, verify that every reported interval probability lies between 0 and 1; an odds value need not. Third, when presenting cumulative risks, confirm that each interval input is a converted conditional risk, rather than an odds value or an unconditional probability. For a weighted causal interpretation, these checks do not replace the assumptions and diagnostics concerning the weights and model specification.

The main alternative is to model interval risks directly using a model with a probability-scale mean structure. That approach still requires a link or other method to keep predictions in the interval from 0 to 1, and it does not make the logistic model's output a risk without conversion. The logistic link is useful because it maps any finite linear predictor to a valid probability and gives coefficients a log-odds interpretation. The cost is that those coefficients are not direct absolute-risk contrasts.

  • Wrong: “The fitted odds are \(0.25\), so the event probability is 25%.” Correction: Convert \(0.25\) to \(0.25/1.25=20\%\).
  • Wrong: “The treatment odds ratio is 2, so treatment doubles the interval risk.” Correction: An odds ratio of 2 doubles the odds; calculate both risks using their fitted odds before comparing risks.
  • Wrong: “A fitted odds above 1 shows the model is invalid.” Correction: Odds can exceed 1. The corresponding risk remains below 1 after applying \(o/(1+o)\).
  • Wrong: “Odds and risk are interchangeable when interpreting fitted values.” Correction: They are close only when risks are small; the exact conversion is required for probability predictions.

The transformation itself is exact, but a converted prediction can still be misleading if the fitted model, target population, or causal assumptions are inappropriate. It also does not address competing risks: with competing events, the outcome and risk being predicted must be defined for the relevant event process. The practical lesson is narrow but essential: interpret the logistic model on the odds scale, then convert to the probability scale before communicating interval or cumulative risk.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. A fitted interval odds is \(0.60\). What is the corresponding interval risk?
  2. A fitted linear predictor is \(\log(3)\). Give the fitted odds and the fitted risk.
  3. Two consecutive interval risks are \(0.05\) and \(0.10\). What is the cumulative event risk by the end of the second interval?
  4. Why does an odds ratio of 1.5 not, by itself, determine the treated interval risk?

Solutions. (1) The risk is \(0.60/(1+0.60)=0.60/1.60=0.375\), or 37.5%. (2) The odds are \(\exp\{\log(3)\}=3\); the risk is \(3/(1+3)=0.75\). (3) Survival through both intervals is \((1-0.05)(1-0.10)=(0.95)(0.90)=0.855\), so cumulative risk is \(1-0.855=0.145\), or 14.5%. (4) An odds ratio gives a multiplicative comparison of odds, not an absolute odds value. To obtain the treated risk, one also needs the baseline odds or the treated model's fitted linear predictor; then the odds-to-risk conversion can be applied.

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