Tutorials › Marginal Structural Models › Unmeasured Confounding at a Later Decision

Sequential Exchangeability, Consistency and Positivity · Tutorial 42 of 501

Unmeasured Confounding at a Later Decision

See why exchangeability at the first decision cannot compensate for unmeasured confounding at the later decision, using a fully specified counterexample.

Intermediate 12 min read

What You'll Learn

  • State first- and later-decision exchangeability conditions for a two-period treatment history.
  • Construct a data-generating process where an unmeasured variable confounds the later treatment.
  • Calculate the observed treatment–outcome association when the later treatment has no causal effect.
  • Prove nonidentification by comparing two models with identical observed distributions and different causal effects.
  • Distinguish checks for measured-data support from evidence about unmeasured confounding.

Why a Good First Decision Is Not Enough

In a longitudinal study, treatment is assigned more than once. Randomization, or a credible exchangeability assumption, at the first decision protects comparisons made at that time. But it does not automatically protect the comparison at a later decision. A variable that predicts both the later treatment and the outcome can make people receiving different later treatments incomparable, even if the first treatment was assigned independently of their potential outcomes.

The important distinction is between an assumption at each decision and an assumption about the study as a whole. Let \(A_1\) and \(A_2\) be binary treatments at two decision times, and let \(Y(a_1,a_2)\) denote the outcome that would be observed under treatment history \((a_1,a_2)\). First-decision exchangeability concerns \(A_1\). Later-decision exchangeability concerns \(A_2\), conditional on the observed history available before that decision. The first condition does not imply the second.

Definition: First-decision exchangeability holds if \(A_1\) is independent of the relevant potential outcomes (and any relevant potential covariates) under consideration. Later-decision exchangeability holds if, for every treatment history \((a_1,a_2)\), the later treatment \(A_2\) is independent of \(Y(a_1,a_2)\) conditional on the measured history available before \(A_2\). If \(A_1\) is the only measured history, the latter condition is \(A_2 \perp Y(a_1,a_2)\mid A_1\). An unmeasured common predictor of \(A_2\) and the potential outcome can violate this condition.

This tutorial focuses on that specific failure. The counterexample will satisfy consistency, have treatment support at both decisions, and make \(A_1\) exchangeable. Nevertheless, its observed data will not determine the causal effect of \(A_2\). Consistency means that the observed outcome equals the potential outcome under the treatment history actually received: \(Y=Y(A_1,A_2)\). Treatment support means that each treatment value occurs with positive probability in the histories considered.

A Two-Period Counterexample

Consider a population with no measured covariates. Let \(A_1\) be a fair random treatment, \(P(A_1=1)=1/2\). The later outcome is binary. We compare two causal models, each of which generates the same observed distribution of \((A_1,A_2,Y)\). In the first model, an unmeasured variable \(U\), present just before the second decision, predicts both \(A_2\) and \(Y\). In the second, the later treatment is randomized and has a causal effect.

Worked Example: Generating the Observed Data With Unmeasured Confounding

In Model I, let \(U\sim\operatorname{Bernoulli}(1/2)\), independently of \(A_1\). Given \(U\), assign the later treatment according to

$$ P(A_2=1\mid U=1)=0.8,\qquad P(A_2=1\mid U=0)=0.2. $$

Let the outcome equal \(U\), regardless of either treatment. Thus \(Y(a_1,a_2)=U\) for all \(a_1,a_2\). The probability of each \((A_2,Y)\) pair follows directly:

$$ \begin{aligned} P(A_2=1,Y=1)&=P(U=1)P(A_2=1\mid U=1)=0.5(0.8)=0.4,\\ P(A_2=0,Y=1)&=P(U=1)P(A_2=0\mid U=1)=0.5(0.2)=0.1,\\ P(A_2=1,Y=0)&=P(U=0)P(A_2=1\mid U=0)=0.5(0.2)=0.1,\\ P(A_2=0,Y=0)&=P(U=0)P(A_2=0\mid U=0)=0.5(0.8)=0.4. \end{aligned} $$

The resulting observed table is:

\(Y=1\)\(Y=0\)Total
\(A_2=1\)0.40.10.5
\(A_2=0\)0.10.40.5
Total0.50.51

The same table holds separately for \(A_1=0\) and \(A_1=1\), because \(A_1\) is independent of \(U\), and the later-treatment mechanism does not depend on \(A_1\). In this model, intervening to set \(A_2\) does not change \(Y\): for either value of \(a_2\), \(P(Y(a_1,a_2)=1)=P(U=1)=0.5\).

The observed association is substantial, although the causal effect in Model I is zero. The reason is that the observed treatment groups differ in \(U\). Indeed, \(U=1\) is more common among those who receive \(A_2=1\), and \(U=1\) also guarantees \(Y=1\).

Worked Example: Checking the Later-Decision Confounding

Bayes' rule gives the distribution of \(U\) within each later-treatment group:

$$ P(U=1\mid A_2=1)=\frac{0.5(0.8)}{0.5}=0.8,\qquad P(U=1\mid A_2=0)=\frac{0.5(0.2)}{0.5}=0.2. $$

Since \(Y=U\), these are also the observed outcome risks:

$$ P(Y=1\mid A_2=1)=0.8,\qquad P(Y=1\mid A_2=0)=0.2. $$

The observed risk difference is \(0.8-0.2=0.6\), but the causal risk difference under intervention is \(0.5-0.5=0\). Further, conditioning on \(A_1\) does not remove the problem: \(U\) remains associated with \(A_2\) within each \(A_1\) group, because the assignment mechanism for \(A_2\) is the same in both. Thus later-decision exchangeability conditional on \(A_1\) fails.

First Exchangeability Can Still Hold

In Model I, all potential outcomes are functions of \(U\), and \(A_1\) was generated independently of \(U\). Therefore \(A_1\) is independent of the entire collection \(\{Y(a_1,a_2):a_1,a_2\in\{0,1\}\}\). This verifies first-decision exchangeability directly. At the same time, \(A_2\) depends on \(U\), and \(Y(a_1,a_2)=U\), so \(A_2\) is associated with the potential outcome even after conditioning on \(A_1\).

Worked Example: Verifying First Exchangeability and Positivity

For any \(a_1,a_2\), Model I has \(Y(a_1,a_2)=U\). Since \(A_1\) and \(U\) are independent,

$$ P\{Y(a_1,a_2)=1\mid A_1=1\} =P(U=1)=0.5 =P\{Y(a_1,a_2)=1\mid A_1=0\}. $$

Thus the first treatment is exchangeable with respect to these potential outcomes. But for the later decision,

$$ P\{Y(a_1,a_2)=1\mid A_2=1,A_1=a_1\}=0.8 \ne 0.2=P\{Y(a_1,a_2)=1\mid A_2=0,A_1=a_1\}. $$

The inequality holds because \(Y(a_1,a_2)=U\), and the conditional probabilities of \(U=1\) were calculated above. Both later treatment values have positive probability in each \(A_1\) group: \(P(A_2=1\mid A_1=a_1)=P(A_2=0\mid A_1=a_1)=0.5\). Thus this example is not explained by a lack of observed treatment support. Support does not, by itself, make treatment groups exchangeable.

Why the Causal Effect Is Not Identified

A causal quantity is identified from the observed distribution if every causal model satisfying the stated assumptions and producing that observed distribution gives the quantity the same value. Showing two compatible models with different values is enough to disprove identification. Here the first model attributes the observed association to \(U\); the second can attribute it to a causal effect of \(A_2\).

Theorem: Under consistency and first-decision exchangeability alone, the causal effect of the later treatment need not be identified from the observed distribution, even when both treatment values have positive probability at each observed first-treatment level.

To prove the claim, retain the same fair \(A_1\) in Model II. Generate \(A_2\sim\operatorname{Bernoulli}(1/2)\), independently of \(A_1\) and a random variable \(V\sim\operatorname{Uniform}(0,1)\). Define

$$ Y(a_1,a_2)=\mathbb{I}\{V\leq 0.2+0.6a_2\}. $$

Here \(A_1\) is independent of all potential outcomes, as in Model I. Under consistency, \(Y=Y(A_1,A_2)\). For \(A_2=1\), the outcome risk is \(P(V\leq0.8)=0.8\); for \(A_2=0\), it is \(P(V\leq0.2)=0.2\). Since \(A_2\) is a fair coin independent of \(V\), the four joint probabilities of \((A_2,Y)\) are \(0.4, 0.1, 0.1, 0.4\), in the same cells as Model I. Independence and the same distribution of \(A_1\) then make the full observed distribution of \((A_1,A_2,Y)\) identical in the two models.

But in Model II, for either fixed \(a_1\),

$$ P\{Y(a_1,1)=1\}-P\{Y(a_1,0)=1\}=0.8-0.2=0.6, $$

whereas the same contrast is \(0.5-0.5=0\) in Model I. Both models satisfy consistency, first-decision exchangeability, and the stated treatment support. They agree on every observed-data probability but disagree on the causal contrast. Therefore that contrast is not identified under those assumptions. The differing values are possible because first-decision exchangeability does not impose later-decision exchangeability. \(\square\)

Worked Example: Comparing the Two Causal Explanations

The risk difference \(P(Y=1\mid A_2=1)-P(Y=1\mid A_2=0)\) is \(0.6\) in both models, so the observed association cannot distinguish them. Under intervention, however, Model I has risks \(P(Y(a_1,1)=1)=0.5\) and \(P(Y(a_1,0)=1)=0.5\), giving a difference of zero. Model II has corresponding risks \(0.8\) and \(0.2\), giving a difference of \(0.6\). The data alone select neither explanation: the causal interpretation requires assumptions or information beyond the observed distribution.

What Can Be Checked, and What Cannot

The example separates several issues that are easy to conflate. Consistency says that observed outcomes agree with the potential outcomes under the treatments actually received. Positivity says that the relevant treatment options occur in the observed histories. Exchangeability says that, conditional on the measured history, treatment assignment is unrelated to the potential outcomes. The first two properties do not entail the third.

Observed data can reveal some warning signs: measured histories may have very different treatment rates, outcome patterns may be sensitive to adjustment choices, or treatment may be nearly absent in some measured strata. Those findings can point to practical problems, but no diagnostic based only on the observed distribution can establish that an unmeasured common cause is absent. In the counterexample, the two models have exactly the same observed distribution, so every diagnostic computed from that distribution has the same result under both.

The main comparison is between a claim about the first decision and a claim about every decision. Randomizing \(A_1\), or adjusting for baseline variables that make \(A_1\) exchangeable, does not control a variable that predicts \(A_2\) and \(Y\) later. Adjusting for measured history can address confounding represented in that history, but it cannot directly adjust for \(U\) when \(U\) is unavailable. A design that randomizes the later treatment would remove this particular assignment pathway; otherwise, substantive knowledge, richer measurement, and sensitivity analysis are needed to assess how conclusions depend on the unmeasured factor.

Common pitfall: “The first treatment was randomized, so the treatment groups are exchangeable throughout follow-up.” This reasoning carries a valid first-decision condition forward without checking the later assignment mechanism. The correction is to assess exchangeability at the later decision conditional on the history actually measured before that decision. If an unmeasured variable predicts both later treatment and outcome, the condition can fail even when \(A_1\) was randomized and all observed treatment histories have support.

The counterexample establishes a limitation, not a method for repairing it: observed data cannot, without additional assumptions, determine which of the two causal explanations is correct. It also does not say that later exchangeability always fails. It shows why that condition must be defended separately rather than inferred from first-treatment exchangeability, consistency, or support.

Check Your Understanding

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

  1. In Model I, what are the observed risks \(P(Y=1\mid A_2=1)\) and \(P(Y=1\mid A_2=0)\), and what is the causal risk difference under intervention?
  2. Why does conditioning on \(A_1\) not remove confounding by \(U\) in Model I?
  3. What feature of the two-model construction proves that the causal effect is not identified?
  4. Does positive probability of both \(A_2\) values within each \(A_1\) group establish later exchangeability? Explain.

Solutions. (1) The observed risks are \(0.8\) and \(0.2\), from the joint table or Bayes' rule. Since \(Y(a_1,a_2)=U\) in Model I for either \(a_2\), both intervention risks are \(0.5\), so the causal difference is zero. (2) \(U\) is independent of \(A_1\), but \(A_2\) still depends on \(U\) within each \(A_1\) stratum: the conditional probabilities \(P(U=1\mid A_2=1,A_1=a_1)\) and \(P(U=1\mid A_2=0,A_1=a_1)\) are \(0.8\) and \(0.2\). (3) The models have the same observed distribution but different causal risk differences, zero and \(0.6\); hence the observed distribution and stated assumptions do not determine a unique effect. (4) No. Positive treatment probability establishes support, not independence between treatment assignment and potential outcomes. In Model I both later treatment values occur, yet \(U\) makes later treatment associated with the potential outcome.

Clinical Biostats