Exact Positivity Failure and Near-Violation

Two Different Problems Hidden by “Positivity Violation”

A treatment decision can be unsupported in two importantly different ways. In one case, a treatment has exactly zero probability for a history that matters to the target regime. The observed data then contain no treated outcomes from that history, and the counterfactual mean may not be identified without additional assumptions. In the other case, the treatment probability is positive but very small. The target may remain formally identified, yet its estimate can rely heavily on a few observations and be highly unstable.

The distinction is about the population distribution, not just a fitted model or a sample table. A sample with no treated people in a subgroup might reflect a true zero probability, or it might be an unlikely realization when the true probability is small. Conversely, a fitted model that returns probabilities strictly above zero does not establish that treatment is genuinely supported in every relevant history.

Definition: For a treatment level $a$ and covariate history $H=h$, the conditional treatment probability is $g(a\mid h)=P(A=a\mid H=h)$. An exact positivity failure for $a$ occurs on a set of histories with positive probability under the target population where $g(a\mid h)=0$. A near-violation occurs when $g(a\mid h)>0$ on the relevant histories but is very small in some of them. “Very small” has no universal numerical cutoff.

For a longitudinal regime $\bar a=(a_0,\ldots,a_K)$, support must be assessed at each decision time, conditional on the history available then. The relevant histories are those that can arise under the target regime, not merely those common in the observed treatment groups. A zero probability at a relevant history is an exact failure; small positive probabilities at such histories signal weak support. This is the sequential version of the distinction.

What Exact Zero Support Can Do

Consider first one treatment decision, baseline history $H$, and outcome $Y$. Write $Y^a$ for the outcome that would be observed under treatment level $a$. Suppose consistency holds, so $Y=Y^A$, and conditional exchangeability holds, so $Y^a$ is independent of $A$ given $H$. If $P(A=a\mid H=h)>0$ for almost every $h$ in the target population, then the mean under treatment $a$ is identified by standardization.

Theorem: If consistency holds, $Y^a$ is independent of $A$ given $H$, and $P(A=a\mid H)>0$ almost surely under the target distribution of $H$, then $$ E[Y^a]=E\{E[Y\mid A=a,H]\}. $$ The expectations on the right are taken over the target distribution of $H$.

To see the argument, conditional exchangeability gives $E[Y^a\mid H]=E[Y^a\mid A=a,H]$ wherever the conditional comparison is supported. Consistency gives $Y^a=Y$ among units with $A=a$. Therefore $E[Y^a\mid H]=E[Y\mid A=a,H]$ for almost every target history. Taking expectations over $H$ yields the result. The positivity condition matters because without treated observations at a history, the observed conditional mean $E[Y\mid A=a,H=h]$ is not determined there by the observed data. $\square$

Here is the precise limitation: a zero probability does not automatically prove that every conceivable causal estimand is unidentified. Special outcome restrictions or a different target population might suffice. But absent such additional information, observed data generally cannot determine the missing counterfactual mean. The following construction shows why.

Worked Example: Two Causal Models With the Same Observed Data

Let $H$ be binary, with $P(H=0)=P(H=1)=0.5$. When $H=0$, let treatment be randomized with $P(A=1\mid H=0)=0.5$. When $H=1$, let $A=0$ with probability one. Suppose the observed outcome is zero for everyone.

Construct two possible causal models. In both, set $Y^0=0$ for every person, and set $Y^1=0$ when $H=0$. In Model I, also set $Y^1=0$ when $H=1$. In Model II, instead set $Y^1=1$ when $H=1$.

Both models produce exactly the same observed data. For $H=0$, either treatment level has outcome zero. For $H=1$, everyone receives $A=0$, so consistency reveals only $Y^0=0$; the value of $Y^1$ is never observed. Treatment is independent of the potential outcomes conditional on $H$: randomization ensures this for $H=0$, and a constant treatment is independent of them for $H=1$.

Yet the target means differ. In Model I, $E[Y^1]=0$. In Model II, $E[Y^1]=0.5(0)+0.5(1)=0.5$. Thus the observed distribution, even with consistency and conditional exchangeability, does not determine $E[Y^1]$ when the treated probability is zero in a target stratum of positive probability.

Why Small Positive Probabilities Are Different

If treatment probability is positive at every relevant history, the conditional means in the standardization formula are, in principle, defined by the observed distribution. A small probability does not by itself create the same nonidentification demonstrated above. It does, however, mean that relatively few people with that history receive the treatment of interest. The corresponding conditional outcome mean is then weakly supported by data.

The inverse-probability identity makes the practical consequence visible. Under the assumptions in the theorem, define $g(a\mid H)=P(A=a\mid H)$. For a single decision,

E\left[\frac{\mathbb{1}(A=a)Y}{g(a\mid H)}\right] =E[Y^a].

Here $\mathbb{1}(A=a)$ equals one when $A=a$ and zero otherwise. To derive the identity, condition first on $H$:

E\left[\frac{\mathbb{1}(A=a)Y}{g(a\mid H)}\middle|H\right] =\frac{P(A=a\mid H)}{g(a\mid H)}E[Y\mid A=a,H] =E[Y^a\mid H].

The last equality follows from consistency and conditional exchangeability. Taking expectations over $H$ proves the identity. The denominator is harmless algebraically when it is positive, but when it is small the factor $1/g(a\mid H)$ is large. Thus a small number of observations can receive large weights. If $g(a\mid H)=0$, the expression is undefined at that history, reflecting the support problem rather than merely a large-weight problem. $\square$

Worked Example: A Small Positive Probability and a Large Weight

Suppose a history stratum has treatment probability $g(1\mid H=h)=0.02$. A person in that stratum who receives treatment has inverse-probability weight $1/0.02=50$. If the probability were $0.20$, the corresponding weight would be $5$. Both probabilities are positive, so they are not exact positivity failures, but the first gives each treated observation ten times the weight.

With $n=500$ people in the stratum, the expected number treated is $500(0.02)=10$, compared with $500(0.20)=100$ at probability $0.20$. A few outcomes among the ten treated people can therefore have substantial influence on the estimated mean. Increasing the weight does not create additional information about their outcomes; it amplifies the contribution of the limited information available.

Worked Example: Near-Violation and Effective Sample Size

Take two history strata, with probabilities $0.9$ and $0.1$ in the target population. In a sample of $n=1000$, suppose treatment probabilities are $0.5$ and $0.01$, respectively. For the target $A=1$, the expected treated counts are $1000(0.9)(0.5)=450$ and $1000(0.1)(0.01)=1$.

The corresponding weights are $1/0.5=2$ and $1/0.01=100$. Using these expected counts to illustrate weight concentration, the sum of weights is $450(2)+1(100)=1000$, and the sum of squared weights is $450(2^2)+1(100^2)=1800+10000=11800$. The usual effective-sample-size calculation gives

\frac{(\sum_i w_i)^2}{\sum_i w_i^2} \approx \frac{1000^2}{11800}\approx 84.7.

This is an illustration based on expected counts, not a claim that the realized sample must have exactly those counts. It shows how one high-weight observation can make a nominal sample of 1000 behave, for some weighted summaries, more like a much smaller sample. In particular, the rare history has an expected treated count of only one, so the target outcome mean in that stratum may be extremely imprecise.

Diagnosing Support and Choosing a Response

Useful diagnostics look for evidence that the treatment comparison is thinly supported. They should be interpreted as warnings, not as proofs that population positivity holds or fails.

Tabulate treatment by important histories. In discrete settings, zero or very small cell counts can expose unsupported or sparse comparisons.
Inspect estimated treatment probabilities. Very small fitted probabilities and a long upper tail of inverse weights indicate sensitivity to a small number of observations.
Check the weight distribution and effective sample size. A large gap between the nominal sample size and effective sample size warns that weighted information is concentrated.
Assess stability. If estimates change markedly when a few high-weight observations are removed or modest modeling choices change, support may be weak.

These checks cannot establish that a population probability is exactly zero. A sample cell with no treated observations could occur by chance when the true probability is positive; the chance of seeing zero treated people among $m$ independent people with probability $p$ is $(1-p)^m$. For example, at $p=0.01$ and $m=20$, this probability is $0.99^{20}\approx0.818$. A zero sample count is therefore unsurprising in a small stratum even when treatment is possible.

The appropriate response depends on the problem. If a target regime includes histories with exact zero support, one may need a different target population or regime, or additional substantive assumptions that explicitly supply information about unsupported outcomes. Restricting the target population changes the estimand and should be described as such. A parametric model can produce predictions in unsupported regions, but those predictions come from extrapolation assumptions rather than nonparametric support in the data.

For near-violations, options such as weight truncation can reduce the influence of extreme weights, but they generally trade variance for bias and do not repair exact nonidentification. Restricting to better-supported histories can improve stability but again changes the target. No numerical threshold for “too small” works in every study: the consequences depend on sample size, outcome variability, history structure, and the estimand.

Common pitfall: “Every estimated treatment probability is greater than zero, so positivity is satisfied and the estimate is reliable.” This reasoning confuses a fitted model with the population support condition and confuses formal identification with stable estimation. A model may assign small positive probabilities where data provide almost no information, or conceal true structural zeros through its functional form. Check the histories and observed support, report how concentrated the weights are, and distinguish assumptions used for extrapolation from evidence in the data.

Check Your Understanding

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

In a target stratum with $P(H=h)=0.2$, suppose $P(A=1\mid H=h)=0$. Is this an exact positivity failure for the target $A=1$? Why?
If $P(A=1\mid H=h)=0.005$ and the relevant stratum has positive target probability, does this alone imply nonidentification? What is the main practical concern?
In a sample of 15 people from a stratum with true treatment probability $0.02$, what is the probability that no one receives treatment?
Why can a regression model that predicts outcomes for an unsupported treatment-history combination fail to establish identification from the observed data?
Does truncating very large inverse-probability weights solve an exact positivity failure? Explain.

Solutions. (1) Yes. The target stratum has positive probability, and the treatment probability is exactly zero there. (2) No, not by itself: the probability is positive, so formal identification may hold under the other assumptions. The practical concern is weak support, large weights, and potentially unstable estimation. (3) The probability is $(1-0.02)^{15}=0.98^{15}\approx0.739$, so an empty treated cell is quite plausible. (4) The prediction in that region depends on the model’s extrapolation assumptions; no observed treated outcomes there determine the relevant conditional mean. (5) No. Truncation changes the influence of observed weights but cannot supply outcomes for a treatment-history combination with zero support; it may also change the estimator’s bias.

Exact Positivity Failure and Near-Violation

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