Sequential Positivity Versus Joint Positivity

Two Ways to Describe Treatment Support

In a longitudinal study, treatment is assigned at several decision times. At each time, the available treatment choices may depend on what has happened so far. Positivity can therefore be described at two levels: the probability of a treatment choice conditional on the current observed history, or the probability of observing a complete treatment sequence. These statements concern related probabilities, but they do not make the same claim.

Write \(A_t\) for treatment at decision time \(t\), \(X\) for baseline covariates, and \(H_t\) for the observed information available immediately before \(A_t\) is assigned. Thus \(H_t\) includes \(X\), prior treatments, and relevant covariate history. For a finite number \(T\) of decision times, write \(\bar A=(A_1,\ldots,A_T)\) for the treatment sequence. The discussion below uses discrete variables so that histories and probabilities can be written as sums.

Definition: Sequential positivity for a specified set of treatment choices requires that, at each time \(t\), each required treatment value \(a_t\) has positive probability conditional on every history \(h_t\) in the relevant support: \(P(A_t=a_t\mid H_t=h_t)>0\). The support must be specified for the treatment sequences and population being studied. Joint treatment-sequence positivity, conditional on baseline covariates, requires \(P(\bar A=\bar a\mid X=x)>0\) for each treatment sequence \(\bar a\) of interest and each relevant baseline value \(x\).

The qualification “in the relevant support” matters. A conditional probability at a history with zero probability is not estimated from the observed data, and it is not a meaningful empirical requirement there. When assessing a proposed treatment sequence, sequential positivity needs to hold at histories that can arise after its preceding treatments, not merely at histories that happen to be common overall.

How the Probabilities Relate

For two decision times, let \(L\) denote the covariate history observed between \(A_1\) and \(A_2\), and suppose it is included in \(H_2\). The chain rule gives

P(A_1=a_1,A_2=a_2\mid X=x) = P(A_1=a_1\mid X=x) \sum_l P(L=l\mid A_1=a_1,X=x) P(A_2=a_2\mid A_1=a_1,L=l,X=x).

The sum over \(l\) is essential: a complete treatment sequence can occur through any intermediate covariate history that has positive probability after the first treatment. The sequence probability is not generally the product of two treatment probabilities evaluated at one fixed history. For more decision times, repeated use of the chain rule gives a sum over covariate-history paths, with the treatment probabilities conditioned on the history at each decision.

Result: In a finite discrete setting, suppose \(P(A_1=a_1\mid X=x)>0\), and, for every \(l\) with \(P(L=l\mid A_1=a_1,X=x)>0\), suppose \(P(A_2=a_2\mid A_1=a_1,L=l,X=x)>0\). Then \(P(A_1=a_1,A_2=a_2\mid X=x)>0\).

Proof. In the displayed chain-rule expression, the first factor is positive. The sum has at least one term with \(P(L=l\mid A_1=a_1,X=x)>0\), since these conditional probabilities sum to one. For every such term, the second probability in the product is positive by assumption. Thus at least one summand is positive, all summands are nonnegative, and the whole sum is positive. Multiplying by the positive first factor proves the result. \(\square\)

Worked Example: From Decision-Level Probabilities to Sequence Probability

Suppose \(X\) is fixed, \(P(A_1=1\mid X)=0.4\), and after \(A_1=1\), the intermediate covariate \(L\) is 0 with probability 0.75 and 1 with probability 0.25. At the second decision, suppose \(P(A_2=1\mid A_1=1,L=0)=0.2\) and \(P(A_2=1\mid A_1=1,L=1)=0.8\). Find the probability of the sequence \((A_1,A_2)=(1,1)\).

First average the second-decision treatment probability over the histories that can follow \(A_1=1\):

P(A_2=1\mid A_1=1,X) =0.75(0.2)+0.25(0.8) =0.15+0.20=0.35.

Then apply the chain rule:

P(A_1=1,A_2=1\mid X) =0.4(0.35)=0.14.

Equivalently, the two history-specific contributions are \(0.4(0.75)(0.2)=0.06\) and \(0.4(0.25)(0.8)=0.08\); their sum is \(0.14\). Both second-decision probabilities are positive, and the complete sequence consequently has positive probability.

Sequential Positivity Is Stronger Than Sequence Positivity

For a fixed baseline value and specified treatment sequences, sequential positivity implies joint treatment-sequence positivity under a support condition: at each decision, the required treatment must have positive probability at every history that can arise after the preceding treatments. In finite discrete settings, this follows by induction over decision times.

Result: Let \(X=x\) be fixed and \(\bar a=(a_1,\ldots,a_T)\) a specified sequence. Suppose \(P(A_1=a_1\mid X=x)>0\). At every later time \(t\), assume \(P(A_t=a_t\mid H_t=h_t)>0\) for each history \(h_t\) having positive probability conditional on \(X=x,\bar A_{t-1}=\bar a_{t-1}\). Then \(P(\bar A=\bar a\mid X=x)>0\).

Proof. The claim holds for the one-treatment prefix because \(P(A_1=a_1\mid X=x)>0\). Suppose the prefix through time \(t-1\) has positive probability conditional on \(X=x\). Conditional on that prefix and \(X=x\), the possible histories \(H_t\) have probabilities summing to one. There is at least one such history with positive conditional probability. By the assumption, the conditional probability of \(A_t=a_t\) is positive at every history with positive conditional probability. Therefore

P(A_t=a_t\mid X=x,\bar A_{t-1}=\bar a_{t-1}) = \sum_{h_t} P(H_t=h_t\mid X=x,\bar A_{t-1}=\bar a_{t-1}) P(A_t=a_t\mid H_t=h_t)>0.

Multiplying this positive conditional probability by the positive probability of the prefix shows that the prefix through time \(t\) has positive probability. Induction through \(T\) proves the result. \(\square\)

The converse does not hold. A treatment sequence may have positive probability because it occurs in one subset of histories, even though a required treatment is impossible in another history. The sequence-level probability averages over those histories and can conceal that gap.

Worked Example: Positive Sequences Despite a Conditional Zero

Fix \(X=x\). Let \(A_1\) be 0 or 1, each with probability 0.5. Let \(L\) be 0 or 1, each with probability 0.5 after either first treatment. At the second decision, set \(A_2=L\) with probability one: \(P(A_2=1\mid A_1,L=0)=0\) and \(P(A_2=1\mid A_1,L=1)=1\).

All four complete treatment sequences have positive probability. For either \(a_1\in\{0,1\}\),

P(A_1=a_1,A_2=0\mid X=x) =0.5(0.5)=0.25, \qquad P(A_1=a_1,A_2=1\mid X=x) =0.5(0.5)=0.25.

Yet sequential positivity for \(A_2\) fails: at history \(L=0\), treatment \(A_2=1\) has probability zero, and at \(L=1\), treatment \(A_2=0\) has probability zero. Each treatment sequence is represented, but neither second-decision choice is available at every intermediate history. Thus joint positivity of the treatment vector conditional on baseline \(X\) does not establish sequential positivity.

How Small Conditional Probabilities Accumulate

Even when every required conditional probability is positive, a long treatment sequence can be rare. This is not a contradiction: sequential positivity rules out zero probabilities at the specified histories, but it does not guarantee that a complete sequence is common enough for stable estimation.

Worked Example: Positive at Each Time, Rare as a Sequence

Suppose that for a specified treatment choice, the conditional probability is 0.2 at each of four decision times at every relevant history. If, for simplicity, these conditional probabilities are constant across histories, the chain rule gives the probability of four successive specified choices as

0.2\times0.2\times0.2\times0.2=(0.2)^4=0.0016.

So each decision satisfies positivity, but only 0.16% of individuals receive the complete specified sequence. This illustrates why checking each conditional probability and checking how frequently complete sequences occur answer different diagnostic questions. The product calculation here is valid because each conditional treatment probability is fixed at 0.2; with varying probabilities and intermediate covariates, the sequence probability must instead average across history paths.

Diagnostics, Comparisons, and Limitations

A useful assessment reports both levels of support. For each decision, inspect treatment counts and estimated treatment probabilities across relevant histories. Separately, count or estimate the frequency of complete treatment sequences. Sparse histories, zero counts, and probabilities close to zero can indicate that conditional estimates are unreliable or that the treatment sequence is rare. A positive observed count alone does not establish adequate support in every history, and a fitted probability that is merely nonzero does not resolve a structural zero hidden by an overly coarse history definition.

Support statement	What it asks	What it does not establish
Sequential positivity	Is each required choice available at each relevant observed history?	That a complete treatment sequence is frequent or precisely estimable.
Joint treatment-sequence positivity	Does each specified treatment vector occur with positive probability given baseline covariates?	That each choice is available within every intermediate history.
Full history-and-treatment support	Does a specified complete covariate-and-treatment path have positive probability?	Support for other paths or adequate sample size in this path.

The phrase “joint positivity” is sometimes used for positivity of a complete observed-data history, rather than only the treatment vector conditional on baseline covariates. These are different events. For an exact full path, the chain rule factors its probability into conditional probabilities along that path; a zero conditional factor makes that path’s joint probability zero. Always state which variables are included in the joint event.

This distinction is important when applying longitudinal identification formulas: conditional support is required at the histories relevant to the treatment strategy, not merely a positive overall frequency of the treatment vector. The introductory material on positivity at each treatment decision motivates the conditional requirement; the key point here is how it relates to sequence-level support. Positivity is a support condition, not a guarantee of precise estimates, and observed data cannot by themselves verify support at histories that are absent or extremely sparse.

Common pitfall: “Every complete treatment sequence has been observed, so positivity holds.” This reasoning checks only marginal sequence support. In the second worked example, every sequence has probability 0.25, but second-decision treatment probabilities are zero in particular histories. The correction is to check the treatment probabilities conditional on relevant histories as well as the overall sequence probabilities.

Check Your Understanding

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

If every required conditional treatment probability is positive at every relevant history, must the corresponding complete treatment sequence have positive probability in a finite discrete setting? State the support qualification.
In the two-time formula, why can a positive value of \(P(A_1=a_1,A_2=a_2\mid X=x)\) fail to show that \(P(A_2=a_2\mid A_1=a_1,L=l,X=x)>0\) for every relevant \(l\)?
Suppose \(P(A_1=1\mid X)=0.5\), \(P(L=0\mid A_1=1,X)=0.6\), and the probabilities of \(A_2=1\) at \(L=0,1\) are 0.5 and 0.1, respectively. Find \(P(A_1=1,A_2=1\mid X)\), taking \(P(L=1\mid A_1=1,X)=0.4\).
At two decision times, each required treatment choice has conditional probability 0.3 at every relevant history. Under this constant-probability condition, what is the probability of the specified two-treatment sequence?

Solutions. (1) Yes, provided the required conditional probabilities are positive on histories with positive probability after each preceding treatment prefix. The induction argument above then gives a positive probability for every prefix and hence the full sequence. (2) The joint sequence probability averages over intermediate histories. Some history-specific terms can be zero while another positive term makes the sum positive. (3) The conditional probability of \(A_2=1\) after \(A_1=1\) is \(0.6(0.5)+0.4(0.1)=0.30+0.04=0.34\). Multiplying by \(P(A_1=1\mid X)=0.5\) gives \(0.17\). (4) By the chain rule, \(0.3(0.3)=0.09\).