When Baseline Covariates Can Suffice for Exchangeability

Why the Assignment Mechanism Matters

At each treatment decision, the question is whether people receiving different treatments are comparable with respect to outcomes they would have had under a specified regime. Sometimes measured baseline covariates suffice to make them comparable. This can happen in a randomized study, or when treatment decisions use only baseline information and previous assignments. In other settings, clinicians use updated health measurements that also predict future outcomes; then baseline adjustment alone generally leaves confounding.

The distinction concerns the assignment process, not simply which variables happen to be recorded. In particular, a good-fitting treatment model or observed balance on baseline variables does not establish that all common causes of assignment and potential outcomes have been controlled.

Baseline-Only Sequential Exchangeability

Consider decisions at times $t=1,\ldots,K$. Let $V$ denote measured baseline covariates, $A_t$ the treatment assigned at time $t$, and $\bar A_t=(A_1,\ldots,A_t)$ the treatment history. A regime is $\bar a=(a_1,\ldots,a_K)$, and $Y^{\bar a}$ is the outcome that would occur under that regime. Write $\bar a_t=(a_1,\ldots,a_t)$, with the history before the first decision empty.

Definition: Baseline-only sequential exchangeability holds if, for every regime $\bar a$ and each decision time $t$, $$ Y^{\bar a}\perp A_t\mid V,\bar A_{t-1}=\bar a_{t-1} $$ at treatment histories with positive probability. Thus, within strata of baseline covariates and the past treatment history, the current assignment carries no further information about the potential outcome under the regime.

This is an assumption about potential outcomes, not a claim that assignments are marginally independent of outcomes or of baseline covariates. It also does not say that assignment is independent of every updated covariate. When an updated covariate $L_t$ is measured before $A_t$, the conventional history-based condition instead conditions on $V,\bar L_t,\bar A_{t-1}$. Baseline-only exchangeability can hold without that fuller conditioning set being necessary; conversely, it can fail when treatment decisions respond to outcome-predictive updated information.

For causal interpretation, exchangeability is used alongside consistency: a person's observed outcome equals $Y^{\bar a}$ when that person's observed treatment history is $\bar a$. Positivity is also needed for a target regime: its treatment choices must have positive assignment probability in the baseline and past-treatment strata to which the analysis applies. Exchangeability does not repair a lack of treatment support.

A Sufficient Assignment Mechanism

One clear way baseline-only exchangeability can arise is through randomization at each decision, possibly with probabilities depending on $V$ and prior assignments. Let $R_t$ denote the randomization input at time $t$. Suppose assignments follow $A_t=f_t(V,\bar A_{t-1},R_t)$, so updated covariates do not enter the assignment rule. If, conditional on $V$, each $R_t$ is independent of the entire collection of potential outcomes and of earlier randomization inputs, then current assignment is exchangeable given baseline and past treatment.

Theorem: Suppose $A_t=f_t(V,\bar A_{t-1},R_t)$ for each $t$, and, conditional on $V$, $R_t$ is independent of $(Y^{\bar a}:\bar a\in\mathcal A)$ and $(R_1,\ldots,R_{t-1})$. Then baseline-only sequential exchangeability holds for every regime.

Proof. Fix a time $t$, a regime $\bar a$, and a past treatment history $\bar A_{t-1}=\bar a_{t-1}$. Given $V$ and the earlier randomization inputs, this past history is determined by the assignment rules. The current assignment is then a function only of those fixed quantities and $R_t$. By assumption, conditional on $V$, $R_t$ is independent of both the potential outcome $Y^{\bar a}$ and earlier randomization inputs. Conditioning on the past treatment history, which is a function of earlier inputs and $V$, therefore does not make $R_t$ informative about $Y^{\bar a}$. Thus the conditional distribution of $A_t$ is the same with or without further conditioning on $Y^{\bar a}$, giving the stated conditional independence. $\square$

The independence condition matters. It is not enough to say that a randomization input is independent of one latent variable that affects the outcome: the input must also be independent of the other causes of the potential outcome, under the conditioning used. A direct assumption that $A_t\perp Y^{\bar a}\mid V,\bar A_{t-1}$ is also sufficient, but requires substantive justification.

Worked Example: Sequential Randomization With Baseline-Dependent Probabilities

Suppose $V$ is binary and each randomization uses fresh independent random numbers, independent of all potential outcomes. The probability of treatment at the first decision is $0.4$ when $V=0$ and $0.7$ when $V=1$. At the second decision, the probability is $0.3$ when $V=0$ and $0.6$ when $V=1$, regardless of the first treatment. Both decisions use only $V$ and their randomization inputs.

The theorem applies: within each $V$ stratum and past treatment history, the assignment is generated by an input independent of every regime-specific potential outcome. Consequently, baseline-only sequential exchangeability holds. For regime $(1,1)$, the probability of following the regime is $0.4(0.3)=0.12$ for $V=0$, and $0.7(0.6)=0.42$ for $V=1$. Both are positive, so this regime also has support in each baseline stratum.

The corresponding inverse-probability weights for people who follow $(1,1)$ are $1/0.12=8.333\ldots$ in the $V=0$ stratum and $1/0.42=2.381\ldots$ in the $V=1$ stratum. The weights differ because the assignment probabilities differ, not because the exchangeability assumption differs across strata.

What the Assumption Buys: A Weighting Identity

The next result makes one consequence explicit. Define $g_t(V,\bar a_{t-1})=P(A_t=a_t\mid V,\bar A_{t-1}=\bar a_{t-1})$ for a fixed target regime $\bar a$. Assume these probabilities are positive almost surely in the target population. Let $I(\bar A_K=\bar a)$ be one if the observed history follows the regime and zero otherwise.

Theorem: Under baseline-only sequential exchangeability, consistency, and positivity for $\bar a$, $$ E\left[\frac{I(\bar A_K=\bar a)Y}{\prod_{t=1}^K g_t(V,\bar a_{t-1})}\right]=E[Y^{\bar a}]. $$

Derivation. First use the counterfactual outcome in place of the observed outcome. Define $I_t=I(\bar A_t=\bar a_t)$. At time $t$, conditional on $V$, the past treatment history, and $Y^{\bar a}$, baseline-only exchangeability gives $$ E\left[\frac{I(A_t=a_t)}{g_t(V,\bar a_{t-1})}\,\middle|\,V,\bar A_{t-1}=\bar a_{t-1},Y^{\bar a}\right]=1. $$ Indeed, the conditional probability in the numerator is $g_t(V,\bar a_{t-1})$, so division by that positive probability gives one. On histories not equal to $\bar a_{t-1}$, $I_{t-1}=0$. Consequently, conditioning successively at $t=K,K-1,\ldots,1$ removes one treatment indicator and its corresponding probability at each step: $$ E\left[\left.\frac{I_KY^{\bar a}}{\prod_{t=1}^K g_t(V,\bar a_{t-1})}\right|V\right]=E[Y^{\bar a}\mid V]. $$ Integrating over $V$ gives the target mean. Finally, consistency gives $I_KY^{\bar a}=I_KY$, because everyone with observed history $\bar a$ has observed outcome $Y^{\bar a}$. Substitution proves the identity. $\square$

The derivation shows why both assumptions matter: exchangeability justifies the conditional assignment probabilities in the expectation, while positivity makes their reciprocals finite. This identity does not assert that the weights are stable in a finite sample; probabilities near zero can still produce highly variable weights.

Worked Example: Baseline Covariates Capture the Relevant Risk

Let $V$ be a binary baseline risk category, with $P(V=1)=P(V=0)=0.5$. Suppose a latent risk variable is exactly $U=V$, and potential outcomes are $Y^{\bar a}=2V+a_1+a_2+\epsilon$, where $\epsilon$ has mean zero and is independent of the assignment randomization inputs. At the first decision, treatment probabilities are $0.2$ for $V=0$ and $0.8$ for $V=1$; at the second decision they are $0.3$ and $0.6$, respectively, regardless of $A_1$. Fresh randomization inputs are independent of $\epsilon$, of each other, and of $V$.

Since $U$ is determined by $V$, all variation in this latent risk factor is controlled by conditioning on $V$. The randomization inputs are also independent of the outcome noise. Thus, conditional on $V$ and past treatment, neither assignment reveals anything further about $Y^{\bar a}$, and baseline-only exchangeability holds.

Baseline adjustment is nevertheless necessary here. For $A_1$, the marginal treatment probability is $0.5(0.2)+0.5(0.8)=0.5$. Bayes' rule gives $P(V=1\mid A_1=1)=0.8$ and $P(V=1\mid A_1=0)=0.2$. Since $E[Y^{\bar a}\mid V]=2V+a_1+a_2$, the difference in conditional means between these two assignment groups is $2(0.8-0.2)=1.2$. Marginal assignment is associated with potential outcomes even though assignment is exchangeable after conditioning on $V$.

When Updated History Is Needed

Worked Example: An Updated Confounder Predicts the Next Assignment

Let baseline $V$ be constant. Before the second treatment decision, measure a binary health status $L_1$, with $P(L_1=1)=P(L_1=0)=0.5$. Let $A_1$ be randomized independently of $L_1$ and of the randomization used for $A_2$, with $P(A_1=1)=0.5$. Let the second treatment assignment obey $P(A_2=1\mid L_1=1)=0.8$ and $P(A_2=1\mid L_1=0)=0.2$. Suppose the outcome under every regime is $Y^{\bar a}=L_1$; treatment has no effect in this example, but health status predicts the outcome.

The marginal assignment probability is $P(A_2=1)=0.5(0.8)+0.5(0.2)=0.5$. The joint probability $P(L_1=1,A_2=1)=0.5(0.8)=0.4$, so $P(Y^{\bar a}=1\mid A_2=1)=0.4/0.5=0.8$. Similarly, $P(L_1=1,A_2=0)=0.5(0.2)=0.1$, giving $P(Y^{\bar a}=1\mid A_2=0)=0.1/0.5=0.2$. These differ from the marginal potential-outcome mean $0.5$; baseline-only exchangeability fails. Since $A_1$ is independent of $L_1$ and the randomization for $A_2$, these same conditional probabilities hold within each $A_1$ stratum, so the association persists after conditioning on $A_1$.

Conditioning on $L_1$, however, makes $A_2$ independent of the potential outcome in this construction: within either value of $L_1$, $Y^{\bar a}$ is fixed. This is the kind of assignment process for which updated history is needed. More generally, if clinicians use updated measurements that predict future outcomes, those measurements must be included in a defensible exchangeability condition unless another argument rules out the resulting confounding.

Practical Checks, Comparisons, and Limitations

A baseline-only assumption is plausible when a protocol uses baseline strata and past treatment assignments to randomize each decision, or when a well-supported substantive model says that treatment decisions depend on no additional outcome-predictive information. It is less plausible when clinicians respond to evolving symptoms, lab results, adherence, or prior response. In the latter situation, ask whether each such measurement predicts both the next assignment and the potential outcome. If so, omitting it from the assignment history leaves a confounding path that baseline adjustment does not close.

Compare the main approaches by the information they require. Baseline-only exchangeability conditions on $V$ and prior treatments; history-based sequential exchangeability conditions additionally on measured covariates updated before each decision. Sequential randomization is a design-based route to exchangeability, whereas in observational studies the claim relies on adequate measurement and substantive knowledge of the assignment process. Neither approach guarantees positivity, and neither eliminates bias from unmeasured common causes.

Empirical diagnostics can expose problems but cannot prove exchangeability. Useful checks include documenting what information was available to decision-makers, comparing assignment frequencies across baseline and updated-history strata, and checking whether important pre-decision predictors were omitted. Poor overlap in treatment probabilities signals a separate positivity concern. Conversely, observed balance on measured baseline variables is not evidence that unmeasured prognostic factors are balanced.

Mistake: “Treatment is independent of $V$, so baseline exchangeability holds.” Independence from $V$ says nothing by itself about dependence between treatment and potential outcomes through other causes. Correction: justify conditional independence of assignment and potential outcomes, not merely balance on one covariate.
Mistake: “A randomized first treatment makes all later decisions exchangeable.” Later assignments may depend on updated prognostic information. Correction: assess the assignment mechanism separately at every decision time.
Mistake: “If baseline-only exchangeability holds, conditioning on any updated covariate must also preserve it.” Conditional independence need not survive additional conditioning. Correction: state the conditioning set required by the assignment mechanism; do not infer a stronger conditional independence without justification.

Key takeaway: Baseline covariates suffice when, at every treatment decision, assignment is independent of the relevant potential outcomes conditional on baseline and past treatment. Treatment decisions that respond to updated prognostic history generally require that history in the exchangeability assumption.

Check Your Understanding

Answer each question, then check the solutions below.

A trial uses $V$ and past assignments to set treatment probabilities. Each decision uses a new randomization input independent of all potential outcomes and previous inputs, conditional on $V$. Which exchangeability condition is supported?
In the first worked example, what is the probability of following regime $(1,1)$ when $V=1$, and what is its inverse-probability weight?
In the updated-confounder example, calculate $P(Y^{\bar a}=1\mid A_2=1)$. Does baseline-only exchangeability hold?
Does balance on measured baseline variables establish exchangeability in an observational study? Give the reason.

Solutions. (1) Baseline-only sequential exchangeability, by the independent-input assignment condition. (2) The probability is $0.7(0.6)=0.42$, so the weight is $1/0.42=2.381\ldots$. (3) It is $0.4/0.5=0.8$, not the marginal value $0.5$, so baseline-only exchangeability fails. (4) No. Balance on measured variables cannot rule out association through unmeasured causes or prove the required conditional independence.