I am asking about the balancing property of the Propensity Score,

$X ~\bot ~T~\vert~ p(X)$,

where $X$ is a vector of covariates, $T$ is the treatment indicator and $p(X):=P(T=1\vert X)$ is the Propensity Score.

In my intuition, this balancing property holds only for $0<p(X)<1$. If you condition on $p(X)=1$, that is on those $X$ for which units are never assigned to the control group, then clearly

$P(X\vert T=0,p(X)=1)=0\neq P(X\vert T=1, p(X)=1)=1$.

Im aware that conditioning on $\{T=0,p(X)=1\}$ is problematic, since $P(T=0,p(X)=1)=0$, and $P(X\vert T=0,p(X)=1)$ is not a probability measure.

However, the intuition that for Propensity Score values of 1 or 0 the Propensity Score does not balance covariates makes sense to me. In other words, the balancing property of the Propensity Score holds only in the region of common support.

But in the original Paper of Rosenbaum & Rubin (1983), nothing is stated about the balancing property not to hold for Propensity Score values of 1 or 0.

My question is if my intuition above is correct, and if so, how to proove it mathematically.


1 Answer 1


It's not that the theorem doesn't hold, it's that it's trivial.

$T$ can only take 1 value if $P(T=1|X) = 1$; $P(T=1|X) = 1$ implies $T=1$, so the theorem can be rewritten as $X \perp T | T=1$, which is trivially true. If $P(T=1|X) = 1$ or $0$, $T$ is a constant, so $X$ will be independent of $T$ regardless of Rosenbaum & Rubin's finding.


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