Conditional Independence Forms of Exchangeability

Why the Conditioning History Must Match the Decision

Sequential exchangeability is a statement about treatment assignment at each decision time. Its central question is whether the treatment being assigned at time $t$ is conditionally independent of the potential outcomes of interest, given the information available just before that assignment. The conditioning history is therefore not a generic collection of covariates: it must be the history observed by the time $A_t$ is assigned.

Use times $t=0,\ldots,K$. Let $L_t$ denote covariates measured before treatment $A_t$, and let $Y$ be the outcome measured after the final treatment. Write $\bar A_{t-1}=(A_0,\ldots,A_{t-1})$ for prior treatments, and define the observed history at the time-$t$ decision by

H_t=(L_0,A_0,L_1,\ldots,A_{t-1},L_t).

At $t=0$, this means $H_0=L_0$. In particular, $H_t$ includes $L_t$, which is measured before $A_t$, but does not include $L_{t+1}$, which has not yet been measured. Let $Y^{\bar a}$ denote the potential outcome under a fixed treatment sequence $\bar a=(a_0,\ldots,a_K)$.

Definition: The time-$t$ conditional independence form of sequential exchangeability is that, at supported histories compatible with the treatment prefix $\bar a_{t-1}$, $$ Y^{\bar a}\perp\!\!\!\perp A_t\mid H_t, $$ for each treatment sequence $\bar a$ relevant to the causal question. The condition is imposed at every decision time $t=0,\ldots,K$. Here “compatible” means that the observed past treatment values in the history agree with $(a_0,\ldots,a_{t-1})$. At baseline there is no past-treatment restriction. Equivalently, one can formulate the condition using potential outcomes under future treatment choices while holding the observed past treatment prefix fixed.

This formulation isolates the assignment at one decision: $A_t$, rather than the entire vector $\bar A$. For a finite-valued treatment, the conditional independence statement means that, given $H_t$, the conditional distribution of $Y^{\bar a}$ is the same for each possible value of $A_t$ that has positive conditional probability. It does not say that potential outcomes are constant across people, nor that treatment is independent of the history itself.

Conditional Independence Without Point-Event Shortcuts

For a general outcome, conditional independence is best expressed using conditional distributions. Suppose $A_t$ takes values in a finite set and $Y^{\bar a}$ takes values in a standard Borel space, so regular conditional distributions exist. Then $Y^{\bar a}\perp\!\!\!\perp A_t\mid H_t$ means that for every measurable set $B$ of outcome values and every treatment value $a_t$,

P(Y^{\bar a}\in B,A_t=a_t\mid H_t) = P(Y^{\bar a}\in B\mid H_t)P(A_t=a_t\mid H_t) \quad\text{almost surely}.

An equivalent version is

P(A_t=a_t\mid Y^{\bar a},H_t) = P(A_t=a_t\mid H_t) \quad\text{almost surely}.

The equality is an almost-sure statement about conditional random variables; it does not require conditioning on a particular outcome value that may have probability zero. If the outcome is discrete, this condition can also be checked using point probabilities with positive conditional mass. For continuous outcomes, point probabilities are generally zero, so point-by-point probability ratios are not a valid general formulation.

Theorem: If $A_t$ has a finite state space and the conditional distributions exist, then $Y^{\bar a}\perp\!\!\!\perp A_t\mid H_t$ if and only if, for every treatment value $a_t$, $$ P(A_t=a_t\mid Y^{\bar a},H_t)=P(A_t=a_t\mid H_t) \quad\text{almost surely}. $$

Proof. Write $Z=Y^{\bar a}$. First suppose $Z\perp\!\!\!\perp A_t\mid H_t$. For any measurable set $B$, the defining conditional factorization gives

E\!\left[\mathbf 1\{Z\in B\}\mathbf 1\{A_t=a_t\}\mid H_t\right] = P(Z\in B\mid H_t)P(A_t=a_t\mid H_t).

The defining property of conditional expectation then implies that the conditional expectation of $\mathbf 1\{A_t=a_t\}$, given $Z$ and $H_t$, equals $P(A_t=a_t\mid H_t)$ almost surely. This is the stated conditional-probability equality. Conversely, suppose that equality holds. Taking conditional expectations given $H_t$ after multiplying both sides by $\mathbf 1\{Z\in B\}$ gives the factorization for every measurable $B$ and every $a_t$. That factorization is conditional independence. $\square$

When $Z$ is discrete, the same result can be written using $P(A_t=a_t\mid Z=z,H_t=h)$, but only for outcome values $z$ with positive conditional probability given $H_t=h$. The almost-sure formulation above avoids that restriction and applies to continuous outcomes as well.

Worked Examples

Worked Example: Checking Independence Within a History Stratum

Suppose $H_t$ is binary. Within the stratum $H_t=0$, let $A_t$ be Bernoulli with probability $1/2$, and let the binary potential outcome $Y^{\bar a}$ have probability $1/4$ of being 1 for each treatment value. Suppose their joint probabilities in this stratum are

	$Y^{\bar a}=1$	$Y^{\bar a}=0$
$A_t=1$	0.125	0.375
$A_t=0$	0.125	0.375

The row totals are $0.5$, and the probability of $Y^{\bar a}=1$ is $0.125+0.125=0.25$. For the joint cell $A_t=1,Y^{\bar a}=1$, the product of the marginal probabilities is $0.5\times0.25=0.125$, equal to the observed joint probability. The other three cells also factor: $0.5\times0.75=0.375$ for each $Y^{\bar a}=0$ cell. Thus $A_t$ and $Y^{\bar a}$ are independent conditional on $H_t=0$.

For example, the conditional probability of treatment among those with $Y^{\bar a}=1$ is $0.125/(0.125+0.125)=0.5$, the same as $P(A_t=1\mid H_t=0)$. This point-probability calculation is legitimate here because the potential outcome is discrete and the event has positive probability.

Worked Example: A Continuous Outcome Requires Conditional Distributions

Let $H_t$ be constant, let a pre-treatment latent variable $C\sim\mathrm{Bernoulli}(1/2)$, and set $A_t=C$. Let $U\sim\mathrm{Uniform}(0,1)$ be independent of $C$, and define $Y^{\bar a}=C+U$ for every fixed treatment sequence $\bar a$. For every fixed real $y$, $P(Y^{\bar a}=y)=0$, so a test phrased only in terms of conditional probabilities at individual point values would say nothing.

Instead, use the measurable event $B=(1,\infty)$. If $A_t=1$, then $Y^{\bar a}=1+U>1$ with probability 1, so $P(Y^{\bar a}>1\mid A_t=1)=1$. If $A_t=0$, then $Y^{\bar a}=U\leq1$ with probability 1, so $P(Y^{\bar a}>1\mid A_t=0)=0$. Marginally, $P(Y^{\bar a}>1)=1/2$, and therefore

P(A_t=1,Y^{\bar a}>1)=\tfrac12 \ne \tfrac12\times\tfrac12 = P(A_t=1)P(Y^{\bar a}>1).

The factorization fails, so conditional independence fails. This example shows why general statements must use conditional distributions, or equivalent almost-sure conditional-probability statements, rather than probabilities of continuous outcomes taking one exact value.

Worked Example: A Time-Specific History Can Matter

Suppose $H_1$ is a binary history measured before $A_1$, with $P(H_1=1)=P(H_1=0)=1/2$. Let $P(A_1=1\mid H_1=1)=0.9$ and $P(A_1=1\mid H_1=0)=0.1$. Suppose a binary potential outcome has conditional probability $0.8$ of being 1 in the $H_1=1$ stratum and $0.2$ in the $H_1=0$ stratum, and is independent of $A_1$ within each stratum.

Within $H_1=1$, the joint probability of treatment and outcome is $0.9\times0.8=0.72$, which equals the product of the conditional probabilities. Within $H_1=0$, it is $0.1\times0.2=0.02$, again the product. Thus the time-specific conditional independence holds.

Without conditioning on $H_1$, $P(A_1=1)=0.5$, and $P(Y^{\bar a}=1)=0.5(0.8)+0.5(0.2)=0.5$. But

P(A_1=1,Y^{\bar a}=1) = 0.5(0.9)(0.8)+0.5(0.1)(0.2) = 0.37,

whereas $P(A_1=1)P(Y^{\bar a}=1)=0.5\times0.5=0.25$. The marginal independence statement fails even though the conditional one holds. The history is part of the claim, not optional notation.

What the Time Index Does—and Does Not—Say

At time $t$, the relevant comparison is between $A_t$ and potential outcomes for treatment sequences under consideration, conditional on $H_t$. Earlier treatment and covariate values belong in that history because they have already occurred. Later covariates do not belong in $H_t$, because they are not yet observed at the decision. This timing is essential to the meaning of the assumption.

The collection of time-specific statements is not itself a claim that the full treatment vector $(A_0,\ldots,A_K)$ is jointly independent of potential outcomes given baseline covariates. It also does not say that $A_t$ is independent of $H_t$; treatment probabilities may vary substantially across histories. The assumption concerns remaining association with potential outcomes after conditioning on the specified history.

This differs from a baseline-only exchangeability claim, which conditions only on $L_0$. As the numerical example showed, even conditional independence given the appropriate later history need not imply marginal independence. Conversely, a baseline comparison does not by itself establish that assignment at each later decision is unconfounded given the information then available. Exchangeability is also distinct from positivity: the former concerns conditional independence, while the latter concerns whether treatment options occur with adequate probability in histories of interest.

Assessment and Common Pitfalls

Sequential exchangeability is an assumption about the assignment process and counterfactual outcomes. Since only one treatment path and its observed outcome are seen for each person, the assumption is generally not directly testable from observed data. Data can reveal features that make it implausible or indicate model problems, but passing a diagnostic cannot prove it.

Using the wrong time index. Conditioning on $H_{t-1}$ omits information measured before the time-$t$ decision; conditioning on later measurements uses information unavailable at that decision.
Replacing local statements with a baseline statement. A treatment comparison that ignores the evolving history may mix people with different treatment probabilities and potential-outcome distributions.
Reading conditional independence as no treatment selection. Treatment can depend on $H_t$; the claim is that, within the specified history, assignment carries no further information about the potential outcomes in question.
Checking only point probabilities for a continuous outcome. Exact point events usually have probability zero. Check conditional distributions over measurable sets, or use the almost-sure conditional-probability formulation.
Treating diagnostics as proof. Balance on measured variables or a good treatment model can be informative, but neither establishes independence from unmeasured causes of treatment and outcome.

In practice, assessment therefore begins with the decision timeline: specify what was measured before each assignment, then ask whether those recorded variables plausibly capture the information that jointly predicts assignment and potential outcomes. Subject-matter knowledge and careful measurement matter more than a single empirical test.

Key takeaway: At each time $t$, write the exchangeability claim for $A_t$ conditional on the history $H_t$ available just before that decision. For general outcomes, interpret conditional independence through conditional distributions or almost-sure conditional probabilities, not point-event ratios.

Check Your Understanding

Answer each question, then compare with the solutions below.

With $L_0,A_0,L_1,A_1,L_2,A_2$, what is the history immediately before $A_2$?
Why is $P(A_t=1\mid Y^{\bar a}=y,H_t)=P(A_t=1\mid H_t)$ for every $y$ with positive point probability not a suitable general definition for a continuous outcome?
In the third worked example, what is $P(A_1=1,Y^{\bar a}=1)$, and why does it differ from the product of the marginal probabilities?
Does sequential exchangeability require $A_t$ to be independent of $H_t$? Explain.
Can observed-data diagnostics establish conditional independence from all potential outcomes? Why or why not?

Solutions. (1) $H_2=(L_0,A_0,L_1,A_1,L_2)$: it includes all recorded covariates and treatments before $A_2$, but not $A_2$ or later measurements. (2) For a continuous outcome, every singleton may have probability zero, so those equalities can be vacuous; conditional distributions over measurable sets or the almost-sure conditional-probability equality are valid general formulations. (3) The joint probability is $0.5(0.9)(0.8)+0.5(0.1)(0.2)=0.37$. The marginal product is $0.5\times0.5=0.25$; the association arises because both treatment and outcome distributions vary with $H_1$. (4) No. Treatment may depend on the history; exchangeability says that, conditional on that history, assignment is independent of the potential outcomes specified by the assumption. (5) No. Diagnostics can assess observed patterns and model fit, but potential outcomes are not all observed, and unmeasured common causes cannot generally be ruled out from observed data alone.

	\(Y^{\bar a}=1\)	\(Y^{\bar a}=0\)
\(A_t=1\)	0.125	0.375
\(A_t=0\)	0.125	0.375