What Zero Support Leaves Unknown
Sequential exchangeability can remove confounding only when the treatment decisions provide information about the counterfactual outcomes of interest. If a treatment history never occurs, there are no observed outcomes under that history to provide that information. The consequence is not simply an unstable estimate: without additional restrictions, the target mean is not determined by the observed-data distribution.
The key argument is constructive. Specify two causal models that agree on the treatment mechanism and on every outcome that can be observed. Let them differ only in the potential outcome for a history that has probability zero. Since that potential outcome is never revealed by consistency, the observed distributions agree, while the mean under the unsupported regime differs.
For two treatment times, write \(A_1,A_2\) for treatment and \(Y^{a_1,a_2}\) for the outcome that would occur under treatment history \((a_1,a_2)\). The target mean under the fixed regime \((a_1,a_2)\) is \(E[Y^{a_1,a_2}]\). Consistency says that if the observed treatments equal \((a_1,a_2)\), then the observed outcome equals \(Y^{a_1,a_2}\). Consistency does not reveal that potential outcome for people whose observed history is different.
A Constructive Nonidentification Result
The following result makes the role of zero support precise. It concerns the nonparametric model: apart from consistency and the stated treatment mechanism, no assumption forces potential outcomes at different treatment histories to be linked.
Proof. Start with any compatible full-data model producing the observed distribution. Construct a second model that preserves the distribution of all observed variables, the treatment mechanism, and every potential outcome for histories other than \((a_1,a_2)\). In the second model, set \(Y^{a_1,a_2}=1-Y^{a_1,a_2}\), leaving all other variables unchanged.
Because \(P(A_1=a_1,A_2=a_2)=0\), consistency never requires the observed outcome to equal \(Y^{a_1,a_2}\). Thus changing this potential outcome cannot change the observed-data distribution. The target mean changes from \(E[Y^{a_1,a_2}]\) to \(1-E[Y^{a_1,a_2}]\). These means differ unless the original mean is \(1/2\). If it is \(1/2\), instead set \(Y^{a_1,a_2}=0\) for everyone in one compatible model and \(Y^{a_1,a_2}=1\) for everyone in another. Neither change affects observed data, and the two target means are then 0 and 1. Hence the observed-data distribution does not determine the target mean. \(\square\)
The construction proves nonidentification, not merely a failure of one estimation method. A formula based on observed conditional means cannot supply the missing outcome distribution. In a discrete two-time setting, identifying a regime mean requires observed outcomes within the histories used to evaluate that regime, as well as the relevant exchangeability and consistency assumptions. If a treatment decision has probability zero at a required history, the corresponding conditional outcome distribution is not learned from the data.
Worked Examples
Worked Example: Two Models With the Same Observed Data
Let \(U\sim\operatorname{Bernoulli}(0.3)\). Set \(P(A_1=1)=0.5\). If \(A_1=0\), let \(A_2\sim\operatorname{Bernoulli}(0.5)\), independently of \(U\); if \(A_1=1\), set \(A_2=0\). Treatment assignment is independent of \(U\). Consider two models. In both, \(Y^{0,0}=Y^{0,1}=Y^{1,0}=U\). In Model A, set \(Y^{1,1}=U\); in Model B, set \(Y^{1,1}=1-U\). The observed outcome is the potential outcome corresponding to the observed treatment history.
First calculate the history probabilities. The probability of \((0,0)\) is \(P(A_1=0)P(A_2=0\mid A_1=0)=(0.5)(0.5)=0.25\). Similarly, \(P(A_1=0,A_2=1)=(0.5)(0.5)=0.25\). For \((1,0)\), the probability is \(P(A_1=1)P(A_2=0\mid A_1=1)=(0.5)(1)=0.50\). The history \((1,1)\) has probability \((0.5)(0)=0\). These probabilities sum to 1.
On each supported history the observed outcome is \(U\), so \(P(Y=1\mid A_1,A_2)=0.3\) and \(P(Y=0\mid A_1,A_2)=0.7\). The joint probabilities are therefore:
The six probabilities sum to \(0.15+0.50+0.50=1.15\) if grouped incorrectly; grouping correctly by histories, the total is \(0.25+0.25+0.50=1\). Equivalently, summing the displayed joint probabilities gives \(0.075+0.175+0.075+0.175+0.15+0.35=1\). The unsupported history contributes no observed-data probability in either model. Thus the full observed distributions are identical.
But under regime \((1,1)\), Model A gives \(E[Y^{1,1}]=E[U]=0.3\), while Model B gives \(E[Y^{1,1}]=E[1-U]=1-0.3=0.7\). The history \((1,0)\) is observed with probability \(0.50\), not \(0.25\); the unsupported history is \((1,1)\). This difference in target means, despite identical observed data, establishes nonidentification.
In this example, treatment is independent of \(U\), and hence independent of the potential outcomes, at the randomized decision. At the second decision after \(A_1=1\), treatment is constant, so exchangeability at that history does not create a comparison between \(A_2=0\) and \(A_2=1\). The missing comparison is precisely what zero support prevents.
Worked Example: Zero Support Within a Baseline Subgroup
Let \(X\) be a baseline binary covariate with \(P(X=1)=0.4\). For \(X=0\), assign \(A=1\) with probability \(0.5\); for \(X=1\), always assign \(A=0\). Suppose \(Y^0=0\) for everyone. For \(X=0\), let \(Y^1=1\); for \(X=1\), compare Model A, where \(Y^1=0\), with Model B, where \(Y^1=1\). Assignment is randomized within \(X=0\) and deterministic within \(X=1\).
The observed-data probabilities are the same in both models. The event \((X=0,A=0,Y=0)\) has probability \((0.6)(0.5)=0.3\); \((X=0,A=1,Y=1)\) also has probability \(0.3\). The event \((X=1,A=0,Y=0)\) has probability \(0.4\). These probabilities sum to 1, and there are no other observed outcomes. In particular, for \(X=1\), treatment \(A=1\) never occurs, so the difference between the two values assigned to \(Y^1\) in that subgroup is not visible.
Now consider the regime that sets \(A=1\) for everyone. In Model A its mean is \(E[Y^1]=(0.6)(1)+(0.4)(0)=0.6\). In Model B it is \(E[Y^1]=(0.6)(1)+(0.4)(1)=1.0\). The target means differ by \(0.4\), although the observed-data probabilities agree exactly. Data with treated individuals only in \(X=0\) cannot determine the treated outcome distribution for \(X=1\).
Worked Example: A Positive Marginal Rate Does Not Fill a Missing History
Return to the treatment mechanism in the first example. Although \(P(A_2=1)=0.25\), because \(A_2=1\) occurs when \(A_1=0\), the regime history \((1,1)\) still has probability zero. Specifically,
whereas
The observed mean in history \((1,0)\) is \(E[Y\mid A_1=1,A_2=0]=0.3\), since that history has probability \(0.50\) and its observed outcome is \(U\). This is not an observed mean for \((1,1)\): the event on which that conditional mean would be based has probability zero. Substituting \(0.3\) for the missing mean would produce \(0.3\) as a prediction only by imposing an additional assumption that outcomes under \((1,1)\) behave like those under \((1,0)\). The observed-data distribution does not justify that substitution.
How to Diagnose the Failure and What It Does Not Mean
To assess a specified regime, follow its treatment decisions in order and inspect the conditional treatment probability at each history the regime requires. In the first example, \(P(A_2=1\mid A_1=1)=0\), so the fixed history \((1,1)\) is unsupported. Marginal treatment frequencies can conceal this failure: treatment \(A_2=1\) occurs overall, but never after \(A_1=1\).
A zero probability must also be distinguished from a small positive probability. With positive support, the population conditional distribution is defined, although a finite sample may contain few or no observations in that history. That is a serious estimation and precision problem, but not the same population-level nonidentification argument. Here the conditional probability is exactly zero, so increasing the sample size under the same treatment mechanism cannot produce observations from the missing history.
The main alternatives do not resolve this lack of information without adding assumptions. Inverse-probability weighting would require a treatment probability in the denominator; at a zero-probability history that weight is undefined, not a usable zero. An outcome model can return a prediction there, but that prediction is extrapolation from its modeling restrictions, not a quantity determined by the observed-data distribution. Restricting the target population to histories with support or changing the intervention can define a different target; neither identifies the original unsupported-regime mean.
| Situation | What the observed data provide | Implication for the target |
|---|---|---|
| Required history has probability zero | No outcomes under that history | Not identified without extra restrictions |
| Required history has small positive probability | Some population information, possibly few sample observations | Potentially identified, but estimation may be unstable |
| Required history has adequate positive probability | Outcomes can be observed at the required history | Identification still requires the other causal assumptions |
A common mistake is to argue, “Treatment is randomized, so exchangeability identifies the regime mean.” Randomization supports exchangeability, but it does not ensure that every treatment choice is possible after every relevant history. The correction is to check the treatment probabilities along the specified regime as well as exchangeability and consistency. Another mistake is to treat a missing conditional mean as zero. A conditional mean given an event of probability zero is not defined by the observed distribution; assigning it zero is an extra modeling choice, not a consequence of the data.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If \(P(A_1=1)=0.4\) and \(P(A_2=1\mid A_1=1)=0\), what is \(P(A_1=1,A_2=1)\)?
- In the first worked example, what is the probability of history \((1,0)\), and why is it not \(0.25\)?
- Suppose two models agree on every observed treatment history but assign different means to \(Y^{1,1}\), where \((1,1)\) never occurs. What does this establish?
- If \(P(A_2=1)>0\), does that guarantee support for the history \((1,1)\)? Explain briefly.
Solutions. (1) The joint probability is \(P(A_1=1)P(A_2=1\mid A_1=1)=(0.4)(0)=0\). (2) It is \(P(A_1=1)P(A_2=0\mid A_1=1)=(0.5)(1)=0.50\); the conditional probability of \(A_2=0\) after \(A_1=1\) is 1, not \(0.5\). (3) It establishes nonidentification: the same observed-data distribution is compatible with different target means. (4) No. \(A_2=1\) may occur only after \(A_1=0\), leaving \((1,1)\) unsupported.