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Specifically, in Technical point 3.1 of the book “Causal inference What If” by Hernán and Robins, the authors note that when positivity does not hold the standardised mean is not well defined however the IP weighting is (albeit biased). They note that this is caused by f[a|L] = 0 in E(I(A=a)Y/f[a|L]) (for the standardised mean) but f[A|L] can never equal 0 in E(I(A=a)Y/f[A|L]) (for the IP weighting).

I understand why f[a|L] = 0 but I am unsure why f[A|L] can never be 0?

Thanks in advance.

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This is actually clarified in Fine Point 2.2. $f[A|L]$ refers to the probability of being in the treatment group one is actually in given the covariates one has. It can never be zero because the probability of observing a value that has been observed cannot be zero. The pair $\{A,L\}$ refers to the observed distribution of the treatment and covariates in the population.

To be more concrete, consider a population of one unit that has $A=1$ and $L=1$. There are no units with $A=0$ and $L=1$, indicating a positivity violation. We can say that $f[a|L=1] = 0$ for $a = 0$. But $f[A|L]$ is not zero because there is someone with an $\{A,L\}$ pair, namely, our one unit.

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  • $\begingroup$ Makes absolute sense - thank you $\endgroup$
    – JoshML
    Aug 1, 2021 at 8:46

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