Finding the Failure in a Treatment History
In a two-period study, the treatment prescribed at the second decision can be unavailable for a particular history, even when the first treatment and that history occur in the study. The relevant question is not simply whether both treatment values appear somewhere in the data. It is whether the prescribed second-period treatment occurs among people with the history that the regime itself would produce.
We use \(A_0\) for the first treatment, \(L_1\) for a measured variable between treatment decisions, \(A_1\) for the second treatment, and \(Y\) for the outcome. A fixed regime \((a_0,a_1)\) sets \(A_0=a_0\) and then \(A_1=a_1\), whatever value of \(L_1\) occurs. Write \(L_1^{a_0}\) for the intermediate variable under the first intervention and \(Y^{a_0,a_1}\) for the outcome under both interventions.
To connect this condition to a regime mean, suppose consistency and sequential exchangeability hold. Consistency means observed outcomes and intermediate variables agree with their counterfactual values under the treatments actually received. Sequential exchangeability means that, conditional on the observed history at each decision, treatment assignment is independent of the relevant future counterfactuals. Under these assumptions and positivity, the two-period identification formula is
For a finite-valued \(L_1\), each term combines the probability of a history under the first intervention with the outcome mean observed among people who received the second treatment at that history. If the history has positive probability under \(A_0=a_0\), but the prescribed \(A_1=a_1\) never occurs there, the corresponding conditional outcome mean has no observed-data definition. The formula then cannot supply the regime mean from the observed distribution alone.
Worked Example: Locating the Unsupported History
Worked Example: Locating the Unsupported History
Suppose \(A_0\) is binary with \(P(A_0=1)=P(A_0=0)=0.5\), and \(L_1=A_0\). Thus, under either first-treatment value, the intermediate variable equals that value. The second-treatment probabilities are as follows.
| \(A_0\) | \(L_1\) | \(P(L_1\mid A_0)\) | \(P(A_1=1\mid A_0,L_1)\) |
|---|---|---|---|
| 0 | 0 | 1 | 0.5 |
| 1 | 1 | 1 | 0 |
Consider regime \((1,1)\). Its first intervention sets \(A_0=1\), so \(L_1^{1}=1\) with probability one. The only history that needs checking is therefore \((A_0,L_1)=(1,1)\). At that history,
The regime-prescribed second treatment is absent exactly at the history produced by the first intervention. In the identification formula, the only summand is the \(l=1\) term, with weight \(P(L_1=1\mid A_0=1)=1\), but its outcome mean \(E[Y\mid A_0=1,L_1=1,A_1=1]\) is not defined by the observed distribution because that conditioning event has probability zero.
This locates the failure precisely: it is not a failure to observe \(A_1=1\) anywhere, since \(A_1=1\) occurs among people with \(A_0=0,L_1=0\). It is the failure to observe \(A_1=1\) at the relevant history \((1,1)\).
What Can Still Be Identified?
A positivity failure for one regime does not automatically remove support for every regime. In the example, under regime \((0,1)\), the intermediate value is \(L_1^{0}=0\) with probability one, and \(A_1=1\) occurs with probability \(0.5\) at that history. If the conditional outcome means are \(0.2\) for \(A_1=0\) and \(0.6\) for \(A_1=1\) there, the regime mean follows directly:
Worked Example: Calculating a Supported Regime Mean
Assume consistency and sequential exchangeability, as well as the treatment probabilities just given. Suppose the observed outcome means at \((A_0,L_1)=(0,0)\) are
For regime \((0,1)\), the first treatment produces \(L_1=0\) with probability one. The second treatment has positive probability at that history. Substitution into the identification formula gives
The observed probability of the event used for the outcome mean is \(P(A_0=0,L_1=0,A_1=1)=0.5\times1\times0.5=0.25\), so that mean is supported. In contrast, the corresponding event for regime \((1,1)\) has probability \(0.5\times1\times0=0\).
The general calculation makes this distinction explicit. For any regime \((a_0,a_1)\), let \(q_l=P(L_1=l\mid A_0=a_0)\) and \(r_l=P(A_1=a_1\mid A_0=a_0,L_1=l)\). The observed probability of the history and prescribed treatment is
When \(P(A_0=a_0)>0\) and \(q_l>0\), this probability is positive exactly when \(r_l>0\). Thus, the zero in the conditional treatment probability corresponds to a missing outcome mean for a history that contributes positive weight to the regime. A history with \(q_l=0\), by contrast, contributes zero weight and does not require an observed conditional outcome mean for this regime.
Worked Example: Two Models, One Observed Distribution
Worked Example: Two Models, One Observed Distribution
Use the treatment and intermediate-variable distributions in the first example. Let \(A_0\) be assigned by a fair coin, let \(L_1=A_0\), and assign \(A_1\) by an independent fair coin when \(A_0=0\), but set \(A_1=0\) whenever \(A_0=1\). Suppose the observed outcome is binary with these conditional distributions: when \((A_0,L_1,A_1)=(0,0,0)\), \(Y\) is Bernoulli with mean \(0.2\); when \((0,0,1)\), it has mean \(0.6\); and when \((1,1,0)\), it has mean \(0.4\). These are all the treatment-history combinations observed with positive probability.
Construct two causal models with those same assignments, intermediate variables, and observed-outcome distributions. In both models, let the potential outcome \(Y^{1,0}\) be Bernoulli with mean \(0.4\), and let \(Y^{0,0}\) and \(Y^{0,1}\) be Bernoulli with means \(0.2\) and \(0.6\), respectively. Make potential outcomes independent of the treatment-assignment coins. In Model A, let \(Y^{1,1}\) be Bernoulli with mean \(0.1\); in Model B, let it be Bernoulli with mean \(0.9\). In each model, set \(Y=Y^{A_0,A_1}\).
The observed data never have \(A_0=1,A_1=1\). Therefore, the changed distribution of \(Y^{1,1}\) affects no observed outcomes. Both models give exactly the same probabilities for the observed treatment histories and the same conditional outcome distribution in every observed history. Both also satisfy consistency by construction. Because assignments are independent of the potential outcomes, sequential exchangeability holds wherever treatment assignment varies; at the history where it does not vary, independence holds as well.
Yet the target regime mean differs:
Thus no function of the observed-data distribution can equal \(E[Y^{1,1}]\) in both models: the observed distribution is identical, but the target values are different. The regime mean is not nonparametrically identified under these assumptions.
Worked Example: Comparing Regimes by Their Histories
Worked Example: Comparing Regimes by Their Histories
In the same setup, consider the four fixed regimes. Under \(A_0=0\), \(L_1=0\) with probability one, and both second-treatment values have positive probability. Hence both \((0,0)\) and \((0,1)\) have support at their required history. Under \(A_0=1\), \(L_1=1\) with probability one, but only \(A_1=0\) occurs. Therefore \((1,0)\) has support and \((1,1)\) does not.
Using the stated outcome means, the identified means are \(E[Y^{0,0}]=0.2\), \(E[Y^{0,1}]=0.6\), and \(E[Y^{1,0}]=0.4\). The mean \(E[Y^{1,1}]\) is not identified. A contrast such as \(E[Y^{1,1}]-E[Y^{0,1}]\) is consequently not identified either, because its first component is not identified while its second is.
How to Diagnose and Respond
A practical audit should follow the regime in temporal order. First determine which values of \(L_1\) occur under the regime’s first intervention. Then, separately for each such history, check the probability of the regime-prescribed \(A_1\). This prevents two common errors: treating overall exposure frequency as sufficient evidence of support, and treating an observed history as relevant merely because it occurs in the sample even when the regime cannot produce it.
Write down \(a_0\) and the prescribed second treatment \(a_1\).
Identify each \(l\) with \(P(L_1^{a_0}=l)>0\); under the assumptions used for identification, this distribution is represented by \(P(L_1=l\mid A_0=a_0)\).
Evaluate \(P(A_1=a_1\mid A_0=a_0,L_1=l)\) for every history from Step 2.
A zero at a positive-probability history leaves a required conditional outcome mean unsupported for this regime.
The main alternative is to change the target, for example by considering a regime whose prescribed treatment is supported at the relevant histories. Another possibility is to impose a model that extrapolates the outcome mean into the unsupported treatment-history combination. Such extrapolation is an additional modeling assumption, not information recovered from the observed data. A third response is to report that the regime mean is not identified under the stated assumptions rather than produce a numerical estimate.
There are limits to what the example proves. A zero treatment probability means the observed data do not determine the missing conditional outcome mean without further restrictions. It does not imply that every causal model gives different regime means: a known structural restriction, such as a justified model fixing the missing mean, could make the target identifiable with additional assumptions. Nor does the algebra alone diagnose how a real data set was generated. The treatment probabilities must be assessed for the histories relevant to the target, and empirical zeros in a finite sample should not be confused automatically with population probabilities equal to zero.
Common Mistake
Incorrect reasoning: “The study contains people who received \(A_1=1\), so regime \((1,1)\) has treatment support.” This ignores the history at which treatment was assigned. In the example, all observed \(A_1=1\) individuals have \(A_0=0,L_1=0\), whereas regime \((1,1)\) requires \(A_1=1\) at \(A_0=1,L_1=1\).
Correction: Check the conditional probability of the prescribed treatment at each history reached under the regime. Overall treatment prevalence cannot replace this history-specific check. Also, a zero-weight history under the regime does not itself cause this particular failure: only a zero treatment probability at a history with positive regime probability removes a required term.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Under regime \((1,1)\), suppose \(P(L_1^{1}=0)=0.3\), \(P(L_1^{1}=1)=0.7\), and the probabilities of \(A_1=1\) at those histories are \(0.4\) and \(0\), respectively. Which history causes the positivity failure?
- If \(P(L_1^{a_0}=l)=0\), does a zero value of \(P(A_1=a_1\mid A_0=a_0,L_1=l)\) by itself violate positivity for that regime? Explain.
- In the first numerical example, what is \(P(A_0=1,L_1=1,A_1=1)\), and what does it imply about the observed conditional outcome mean at that history?
- Why does observing \(A_1=1\) among people with \(A_0=0,L_1=0\) not identify the outcome mean for \(A_0=1,L_1=1,A_1=1\)?
Solutions. (1) The history \(L_1=1\) has positive regime probability \(0.7\), but the prescribed \(A_1=1\) has conditional probability zero there. The \(L_1=0\) history is supported because its treatment probability is \(0.4\). (2) No. That history contributes zero weight to the regime mean, so this condition does not require its conditional outcome mean. (3) The probability is \(0.5\times1\times0=0\). The observed conditional outcome mean is undefined from the observed distribution at that treatment-history combination. (4) The conditional outcome distributions refer to different histories. The observed data do not link them without an additional restriction or model.