Exchangeability With Time-Varying Confounders

Why the Relevant History Changes Over Time

In a longitudinal study, treatment is assigned more than once. Between decisions, a participant’s health may change. That changing health can be affected by earlier treatment, can predict the next treatment, and can also predict the outcome. It is therefore not enough to ask whether the first treatment was exchangeable given baseline covariates. We must ask whether each treatment decision is exchangeable given the history available at that decision.

Consider two treatment decisions. Let $L_0$ denote baseline covariates, $A_1$ the first treatment, $L_1$ a covariate measured after $A_1$ but before the next decision, $A_2$ the second treatment, and $Y$ the final outcome. The observed history before the second treatment is $(L_0,A_1,L_1)$. Here, treatments are binary for ease of illustration; the exchangeability statements also apply to other treatment spaces.

Definition: For a treatment history $(a_1,a_2)$, let $Y^{a_1,a_2}$ be the potential outcome under that history, and let $L_1^{a_1}$ be the intermediate covariate that would be observed under first treatment $a_1$. Sequential exchangeability requires, at the first decision, $$ A_1 \perp \{L_1^{a_1},Y^{a_1,a_2}:a_1,a_2\}\mid L_0, $$ and, at the second decision, $$ A_2 \perp \{Y^{a_1,a_2}:a_1,a_2\}\mid L_0,A_1,L_1. $$ Each statement is understood for the treatment histories under consideration, almost surely on histories where the relevant conditional distributions are defined. The second statement concerns the potential outcomes under all regimes, not merely the potential outcome matching the treatment actually received.

These are assumptions about treatment assignment conditional on the relevant observed history and potential outcomes. Consistency connects observed outcomes and covariates to the potential outcomes under the treatment actually received; it does not itself establish either exchangeability statement. Nor does the fact that $L_1$ predicts $A_2$ and $Y$ prove the second statement: it indicates why $L_1$ may be an important part of the history, but other common causes may remain.

A Covariate That Carries Forward Confounding

Suppose a clinical study follows patients with a chronic condition. At baseline, clinicians record disease severity $L_0$. Patients receive an initial therapy $A_1$. At a later visit, clinicians measure current symptom control $L_1$, decide whether to continue or intensify treatment $A_2$, and eventually record a health outcome $Y$. Poor symptom control can prompt intensification and predict a poor outcome. It can also be influenced by the initial therapy.

Worked Example: Baseline Adjustment Misses Later Confounding

Consider a simple data-generating model. Let $U$ be uniformly distributed on $(0,1)$, and let the first treatment $A_1$ be a fair random assignment independent of $U$. Define the potential symptom-control indicator by $L_1^0=1\{U\leq 0.3\}$ and $L_1^1=1\{U\leq 0.7\}$. The observed covariate is $L_1=L_1^{A_1}$. A value of 1 represents poor symptom control. The second treatment has probability $0.8$ when $L_1=1$ and probability $0.2$ when $L_1=0$. Finally, set $Y^{a_1,a_2}=L_1^{a_1}$: the final outcome is determined by symptom control under the initial treatment, and the second treatment has no causal effect in this constructed example.

At the first decision, $A_1$ is independent of $U$, and all the potential covariates and outcomes above are functions of $U$. Therefore $A_1$ is independent of those potential quantities given the constant baseline history. At the second decision, conditional on $L_1$, the assignment of $A_2$ uses a fresh random draw independent of $U$. Since the potential outcomes are functions of $U$, $A_2$ is independent of the potential outcomes given $A_1,L_1$. Thus the relevant sequential exchangeability statements hold in this model.

Now restrict attention to patients with $A_1=1$, and adjust only for baseline. Then $P(L_1=1\mid A_1=1)=0.7$. The probability of receiving the second treatment is $$ P(A_2=1\mid A_1=1)=0.7(0.8)+0.3(0.2)=0.62. $$ Because $Y^{1,a_2}=L_1^1$, the joint probability of $Y^{1,a_2}=1$ and $A_2=1$, conditional on $A_1=1$, is $0.7(0.8)=0.56$. Hence $$ P(Y^{1,a_2}=1\mid A_2=1,A_1=1)=\frac{0.56}{0.62}\approx 0.903. $$ For $A_2=0$, the corresponding joint probability is $0.7(0.2)=0.14$, and the probability of $A_2=0$ is $0.38$. Thus $$ P(Y^{1,a_2}=1\mid A_2=0,A_1=1)=\frac{0.14}{0.38}\approx 0.368. $$ The large difference between these conditional probabilities is not an effect of $A_2$; by construction, $A_2$ has no effect on $Y$. It arises because symptom control $L_1$ predicts both the later treatment and outcome. Conditioning on the decision-time history removes this particular association; conditioning on baseline alone does not.

This example illustrates the temporal structure that matters: $L_1$ occurs after $A_1$, so it is not simply another baseline covariate. Its role at the second decision is to describe the history on which $A_2$ is assigned. The fact that it was affected by earlier treatment does not make it irrelevant to assessing the later assignment mechanism.

What Conditional Randomization Establishes

A study protocol can make exchangeability at a decision plausible by randomizing treatment within the recorded history. The randomization mechanism must be conditionally independent of the potential outcomes—not merely independent of them before conditioning on the history.

Result: Suppose a binary treatment at a decision time is assigned by $A=1\{R\leq p(H)\}$, where $H$ is the observed history, $p$ is a measurable assignment-probability function, and $R$ is uniform on $(0,1)$ and independent of the joint vector $(H,\mathbf{Y})$ of history and all potential outcomes of interest. Then $A\perp\mathbf{Y}\mid H$.

Proof. For any measurable set $B$ of potential-outcome vectors and almost every history value $h$, independence of $R$ from the joint vector $(H,\mathbf{Y})$ gives $$ P(A=1,\mathbf{Y}\in B\mid H=h) =P(R\leq p(h),\mathbf{Y}\in B\mid H=h) =p(h)P(\mathbf{Y}\in B\mid H=h). $$ Also, $P(A=1\mid H=h)=p(h)$. The displayed factorization is precisely conditional independence for the event $A=1$; the event $A=0$ follows by subtraction from $P(\mathbf{Y}\in B\mid H=h)$. Therefore $A\perp\mathbf{Y}\mid H$. $\square$

Worked Example: Randomization Within Each Recorded History

Suppose that at the second decision the recorded history is $H=(L_0,A_1,L_1)$. In a particular history stratum $h$, the protocol assigns $A_2=1$ with probability $0.65$. Let $B$ be the event that a specified potential outcome equals 1, and suppose $P(B\mid H=h)=0.40$. If the randomization draw is independent of the joint history and the potential-outcome vector, then $$ P(A_2=1,B\mid H=h)=0.65(0.40)=0.26. $$ The conditional probability of the potential-outcome event among those assigned treatment is $0.26/0.65=0.40$, the same as $P(B\mid H=h)$. This verifies the required factorization for this event. The argument applies to every measurable event of the potential-outcome vector, not just this numerical illustration.

The joint independence condition matters. Marginal independence between a random draw and a potential outcome is not enough to guarantee independence after conditioning on history.

Why Marginal Independence Can Mislead

A common mistake is to reason that a treatment draw is independent of potential outcomes overall, and therefore independent of them within every observed history. Conditioning can change dependence. In particular, a history can be associated with both the draw and the potential outcome, creating a conditional association even when there was no marginal association.

Worked Example: Independence Before Conditioning Is Not Enough

Let $Y$ and $R$ be independent Bernoulli random variables with success probability $1/2$. Define the history $H=Y\mathbin{\mathrm{XOR}}R$, where XOR is 1 exactly when its two inputs differ, and let the treatment be $A=R$. Marginally, $A$ is independent of $Y$. Also, each value of $H$ has probability $1/2$, and $P(A=1\mid H)=1/2$.

However, when $H=0$, the XOR definition implies $Y=R=A$. Thus $P(Y=1\mid A=1,H=0)=1$, while $P(Y=1\mid A=0,H=0)=0$. When $H=1$, $Y=1-R$, so these two conditional probabilities are 0 and 1, respectively. Treatment and potential outcome are therefore associated within either history, despite their marginal independence. This is why a valid conditional-randomization argument requires independence from the joint history and potential outcomes, or a direct assumption of conditional independence given history.

Common pitfall: “The treatment draw is independent of the potential outcome, so it is independent within each history” is not a valid inference. The correction is to establish independence conditional on the observed history directly, or to specify a randomization mechanism whose draw is independent of the joint history and potential outcomes.

Assessing the Assumption in Practice

In an observational study, sequential exchangeability is usually a substantive assumption, not a property that can be confirmed from the observed data alone. The following checks help clarify whether it is credible, but none proves that all relevant common causes were measured.

Set the time order.
Identify which covariates are measured before each treatment decision. A variable measured after $A_2$ cannot be part of the history justifying exchangeability for $A_2$.

Describe the assignment process.
Ask what clinicians, patients, or a protocol knew when assigning treatment. Consider whether those factors also predict potential outcomes.

Check the history for omissions and measurement limitations.
A recorded proxy may not capture the clinical state driving treatment. Error in an evolving confounder can leave residual dependence even when its measured value is included.

Separate plausibility from verification.
Observed treatment patterns can reveal unexpected assignment behavior, but they cannot show that treatment is independent of unobserved potential outcomes conditional on history.

For example, suppose an unrecorded acute deterioration $U$ predicts both the second treatment and the final outcome, even after conditioning on $(L_0,A_1,L_1)$. Then the second exchangeability condition fails. If, within one fixed recorded history, $P(U=1)=0.5$, the second treatment probabilities are $0.8$ for $U=1$ and $0.2$ for $U=0$, and $Y^{a_1,a_2}=U$, then $$ P(A_2=1)=0.5(0.8)+0.5(0.2)=0.5, $$ but $$ P(Y^{a_1,a_2}=1\mid A_2=1)=0.8,\qquad P(Y^{a_1,a_2}=1\mid A_2=0)=0.2. $$ The assignment is associated with the potential outcome within the recorded history. The calculation does not make $U$ observable; it shows why conditioning on measured history may be insufficient if important causes are omitted.

How This Assumption Differs From Alternatives

Baseline-only exchangeability asks whether treatment assignment is independent of potential outcomes given baseline covariates. Sequential exchangeability makes a separate conditional-independence claim at each decision, using the history available then. The two-period example showed that baseline adjustment can leave later-treatment confounding even when the first treatment was randomized.

A randomized trial can support exchangeability through its assignment mechanism, provided the relevant randomization draw is independent of the joint history and potential outcomes as specified. In observational data, exchangeability depends instead on whether the measured history adequately captures the factors governing assignment and potential outcomes. Regression adjustment and weighting are possible analytical approaches, but neither changes the underlying exchangeability requirement. In particular, fitting a model to observed data does not establish that unmeasured confounding is absent.

The assumption has limits. It may be implausible when treatment decisions depend on unrecorded clinical judgment, when covariates are measured inaccurately, or when the timing of measurement is unclear. It is also distinct from consistency and positivity: sequential exchangeability does not ensure that observed treatment matches the specified intervention, or that each treatment is represented in every relevant history. Those are separate conditions.

Key takeaway: At each decision, ask whether treatment is conditionally independent of the relevant potential outcomes given the history available immediately before that decision. An evolving covariate can be essential to this later condition even when it was affected by earlier treatment.

Check Your Understanding

Answer each question, then compare with the solutions below.

For the two-decision setup, state the history used in the second sequential-exchangeability condition and name the potential outcome on its right-hand side.
In the numerical example with $A_1=1$, compute the two risks $P(Y^{1,a_2}=1\mid A_2=1,A_1=1)$ and $P(Y^{1,a_2}=1\mid A_2=0,A_1=1)$. What explains their difference?
Why does independence of a randomization draw from potential outcomes marginally not necessarily imply conditional independence given history?
If a measured covariate predicts the next treatment but not the outcome, does that fact alone establish sequential exchangeability?

Solutions. (1) The second-decision history is $(L_0,A_1,L_1)$, and the condition is $A_2\perp\{Y^{a_1,a_2}:a_1,a_2\}\mid L_0,A_1,L_1$. (2) The risks are $0.56/0.62\approx0.903$ and $0.14/0.38\approx0.368$. Their difference is due to $L_1$, which predicts both $A_2$ and the outcome; the constructed model gives $A_2$ no causal effect. (3) Conditioning on a history associated with both variables can induce conditional dependence. In the XOR example, $A=R$ is marginally independent of $Y$, but within $H=0$, $A=Y$. (4) No. It may make the covariate relevant to describing treatment assignment, but exchangeability also requires independence from the potential outcomes given the full history; other common causes may remain.