Positivity for Stochastic Interventions

Why Positivity for a Stochastic Policy Is a Support Condition

A stochastic intervention assigns treatment according to a probability distribution rather than prescribing one treatment value for everyone with the same history. For example, it might assign treatment with probability one half at every decision, or increase the chance of treatment relative to usual care. Its probabilities need not match the observed treatment probabilities. The key requirement is narrower: the intervention must not assign positive probability to treatment values that have zero probability under the observed treatment mechanism at the relevant histories.

This requirement is called positivity for a stochastic intervention. It matters because the intervention outcome distribution must be learned from observed data. If the intervention calls for a treatment that is never observed for a history, the data provide no outcome information for that treatment-history combination. A stochastic policy can sometimes avoid this problem by assigning zero probability to unsupported treatments.

Definition: At decision time $t$, let $H_t$ denote the observed history just before treatment $A_t$, and let $\pi_t(a\mid h)=P(A_t=a\mid H_t=h)$ be the observed treatment mechanism. Let $q_t(a\mid h)$ be the treatment distribution under a stochastic intervention. Positivity for the intervention requires $$ q_t(a\mid h)>0 \ \Longrightarrow\ \pi_t(a\mid h)>0 $$ at histories $h$ that can occur under the intervention. Equivalently, the intervention treatment distribution is supported on the observed treatment support at those histories.

For discrete treatments, the support at history $h$ is the set of treatment values having positive probability. Thus the condition is a one-way implication: every treatment value used by the intervention must be available in the observed data mechanism. It does not require $q_t(a\mid h)=\pi_t(a\mid h)$, nor does it require the intervention to use every treatment value observed.

The condition is evaluated at the histories relevant to the intervention, not indiscriminately at every imaginable history. In a longitudinal study, intervention-generated histories can differ from histories common under usual care. The treatment probabilities must therefore be checked at the histories the policy can reach. This tutorial focuses on the support requirement; it does not replace the separate assumptions of consistency and sequential exchangeability needed for a causal interpretation.

From Support Inclusion to Treatment Weights

The support condition has a direct algebraic consequence. At one decision time, if the intervention assigns probability $q(a\mid h)$ and the observed mechanism assigns probability $\pi(a\mid h)$, the treatment likelihood ratio is $q(a\mid h)/\pi(a\mid h)$. This ratio is defined for intervention-assigned treatment values precisely when the support condition holds.

Result: Suppose $H$ and $A$ have finite support, and $q(a\mid h)>0$ implies $\pi(a\mid h)>0$ for every history $h$ with positive probability. Define $W=q(A\mid H)/\pi(A\mid H)$ when $\pi(A\mid H)>0$, and set $W=0$ on observed events where $q(A\mid H)=0$. For any function $m(H,A)$ with finite expectation, $$ E\left[Wm(H,A)\right] =\sum_h P(H=h)\sum_a q(a\mid h)m(h,a). $$

Proof. Sum over the joint observed distribution of $H$ and $A$. Terms with $\pi(a\mid h)=0$ have zero observed probability; by the support condition they also have $q(a\mid h)=0$, so they contribute nothing to the right-hand side. For all remaining terms, substitute the ratio:

\begin{aligned} E[Wm(H,A)] &=\sum_h P(H=h)\sum_{a:\,\pi(a\mid h)>0} \pi(a\mid h)\frac{q(a\mid h)}{\pi(a\mid h)}m(h,a)\\ &=\sum_h P(H=h)\sum_a q(a\mid h)m(h,a). \end{aligned}

The final equality includes the omitted zero-probability terms, which are zero by support inclusion. This proves the result. $\square$

In a longitudinal setting, the analogous weight is a product across decision times. To state the result precisely, suppose the variables have finite support and write an observed trajectory as $z=(l_0,a_0,l_1,\ldots,a_K,y)$, where $H_t$ is the history before $A_t$. Factor the observed trajectory probability into treatment probabilities and the conditional distributions of the non-treatment variables. The intervention distribution defined by the g-formula replaces each $\pi_t(a_t\mid h_t)$ factor with $q_t(a_t\mid h_t)$, leaving the other conditional factors unchanged. Under consistency and sequential exchangeability, this is the identified distribution under the stochastic intervention.

Result: If $q_t(a\mid h)>0$ implies $\pi_t(a\mid h)>0$ at every history reached under the intervention, then for any trajectory with positive intervention probability, $$ \frac{P_q(z)}{P(z)} =\prod_{t=0}^{K}\frac{q_t(a_t\mid h_t)}{\pi_t(a_t\mid h_t)}. $$ Consequently, for any function $m$ of the trajectory, $$ E_q[m(Z)] =E\left[m(Z)\prod_{t=0}^{K}\frac{q_t(A_t\mid H_t)}{\pi_t(A_t\mid H_t)}\right], $$ where the expectation on the right is under the observed distribution, with zero-weight terms interpreted in the usual way.

Proof. Under the observed factorization, the probability of $z$ is the product of the initial-history probability, the treatment factors $\pi_t(a_t\mid h_t)$, and the conditional factors for subsequent covariates and the outcome. By the definition of the intervention g-formula, $P_q(z)$ has the same non-treatment factors and replaces each treatment factor with $q_t(a_t\mid h_t)$. For a trajectory with positive intervention probability, each replacement factor is positive. Its history is reached under the intervention, so positivity makes the corresponding observed treatment factor positive. Dividing the two factorizations cancels the initial and non-treatment factors and gives the product ratio. Summing $m(z)P_q(z)$ over trajectories and substituting $P_q(z)=P(z)\prod_t q_t/\pi_t$ proves the expectation identity. $\square$

The support condition makes this change of measure possible; it does not guarantee that the resulting weights are stable in a finite sample. In particular, positivity permits a small but positive $\pi_t(a\mid h)$, which can produce a large ratio. The distinction between zero support and small probabilities is important when assessing an analysis.

Worked Examples

Worked Example: A Randomized Binary Policy Within Both Strata

Suppose $H$ is a baseline risk group, low or high. Among low-risk individuals, the observed probability of treatment is $\pi(1\mid \text{low})=0.8$; among high-risk individuals it is $\pi(1\mid \text{high})=0.2$. Consider an intervention assigning treatment with probability $q(1\mid h)=0.5$ in both groups. Check positivity and calculate the treatment ratios.

For low-risk individuals, the observed probabilities are $\pi(1\mid\text{low})=0.8$ and $\pi(0\mid\text{low})=0.2$. The intervention assigns probability 0.5 to each value. Both intervention probabilities are positive, and both observed probabilities are positive. The ratios are $$ \frac{q(1\mid\text{low})}{\pi(1\mid\text{low})} =\frac{0.5}{0.8}=0.625,\qquad \frac{q(0\mid\text{low})}{\pi(0\mid\text{low})} =\frac{0.5}{0.2}=2.5. $$

For high-risk individuals, $\pi(1\mid\text{high})=0.2$ and $\pi(0\mid\text{high})=0.8$. The corresponding ratios are $0.5/0.2=2.5$ for treatment and $0.5/0.8=0.625$ for no treatment. Thus the intervention satisfies positivity in both strata. Its treatment distribution differs from the observed one, but its support is contained within the observed support.

Worked Example: An Intervention That Uses an Unsupported Treatment

In a particular history stratum, suppose the observed mechanism is deterministic: $\pi(0\mid h)=1$ and $\pi(1\mid h)=0$. An intervention assigns treatment with probability $q(1\mid h)=0.3$ and no treatment with probability $q(0\mid h)=0.7$. Does it satisfy positivity?

No. The intervention assigns positive probability to treatment value 1, while the observed mechanism assigns it probability zero: $$ q(1\mid h)=0.3>0,\qquad \pi(1\mid h)=0. $$ The ratio $q(1\mid h)/\pi(1\mid h)=0.3/0$ is undefined. More substantively, there are no observed outcomes for treated individuals in this stratum from which to learn the outcome distribution required by the intervention. Assigning a finite weight by convention would not create that missing information. The policy would need to be changed, or the target would require additional assumptions or data.

Worked Example: Checking a Two-Time Stochastic Policy

At time 0, the observed mechanism assigns $A_0=1$ with probability 0.4. The intervention assigns it with probability 0.6. At time 1, suppose the observed probability of $A_1=1$ is 0.2 after $A_0=1$, and 0.7 after $A_0=0$. The intervention assigns $A_1=1$ with probability 0.5 after either value of $A_0$. Check the support condition and calculate the four trajectory weights.

At time 0, both treatment values have positive observed and intervention probabilities. The time-0 ratios are $0.6/0.4=1.5$ for $A_0=1$ and $0.4/0.6=2/3$ for $A_0=0$.

After $A_0=1$, the observed probabilities of $A_1=1$ and $A_1=0$ are 0.2 and 0.8. The time-1 ratios are $0.5/0.2=2.5$ and $0.5/0.8=0.625$. After $A_0=0$, the observed probabilities are 0.7 and 0.3, giving ratios $0.5/0.7=5/7$ and $0.5/0.3=5/3$. All are positive, so the intervention satisfies positivity at both times.

Multiply the time-specific ratios to obtain the trajectory weights: $$ \begin{array}{c|c|c|c} (A_0,A_1)&P(A_0,A_1)&P_q(A_0,A_1)&P_q/P\\ (1,1)&0.4(0.2)=0.08&0.6(0.5)=0.30&0.30/0.08=3.75\\ (1,0)&0.4(0.8)=0.32&0.6(0.5)=0.30&0.30/0.32=0.9375\\ (0,1)&0.6(0.7)=0.42&0.4(0.5)=0.20&0.20/0.42=10/21\\ (0,0)&0.6(0.3)=0.18&0.4(0.5)=0.20&0.20/0.18=10/9 \end{array} $$ The observed path probabilities sum to $0.08+0.32+0.42+0.18=1$, and the intervention path probabilities sum to $0.30+0.30+0.20+0.20=1$. Each displayed weight is the product of the two treatment ratios.

Diagnostics, Comparisons, and Limitations

A support check should be made at each decision time and within histories relevant to the policy. For discrete treatments, a practical audit asks which treatment values the policy assigns positive probability to, then checks whether the observed mechanism assigns positive probability to each of them. A zero in the observed mechanism is a direct failure for any policy that gives that treatment positive probability. Empirical treatment counts can flag sparse support, but a zero count alone does not prove the population probability is zero.

Approach	Positivity requirement	Key distinction
Deterministic treatment regime	The prescribed treatment must be observed with positive probability at relevant histories.	A special case in which the intervention distribution puts all mass on one treatment value.
Stochastic intervention	Every treatment value assigned positive intervention probability must have positive observed probability.	The policy can assign probability across several supported values without matching the observed probabilities.
Observed treatment mechanism	No intervention change is made.	Using the observed mechanism as the intervention satisfies support wherever that mechanism is defined.

A common mistake is to claim that positivity requires the intervention to copy the observed treatment probabilities. That is stronger than necessary: support inclusion permits different probabilities, as the first worked example shows. The correction is to compare which treatment values can occur, not whether their probabilities agree.

The opposite mistake is to treat any positive observed probability as practically adequate. Exact positivity distinguishes zero from positive probability; it does not prevent very large weights when an observed probability is small. Nor does support inclusion address unmeasured confounding, inconsistent treatment versions, or an ill-defined intervention. It is one assumption in a causal identification argument, not a guarantee that an estimate will be reliable.

Check Your Understanding

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

At a history $h$, the observed probabilities for treatment values 0 and 1 are 0.9 and 0.1. An intervention assigns probabilities 0.4 and 0.6. Does it satisfy positivity?
At a history $h$, the observed mechanism assigns probability zero to treatment 1. What must a positive intervention probability for treatment 1 imply about positivity?
In a two-time study, why must the treatment support condition be checked at both decision times?
Does satisfying stochastic-intervention positivity ensure that treatment weights are small? Explain.

Solutions. (1) Yes. Both treatment values have positive observed probabilities, so both values used by the intervention are supported. (2) If the intervention assigns positive probability to treatment 1, positivity fails because the observed probability is zero. (3) The longitudinal weight is a product of time-specific ratios; an unsupported treatment at either time makes a required ratio undefined and obstructs identification for that policy. (4) No. A positive but very small observed probability can yield a large ratio, even though the exact support condition holds.

Positivity for Stochastic Interventions

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