It seems that if the proportion of always-takers in the control group (to whom eligibility was not assigned) is much smaller than the proportion of compliers in the treatment group (to whom eligibility was assigned), then ATT would be similar to LATE.

Is this correct? If so, why?

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    $\begingroup$ I doubt introducing an additional average-treatment-efefct tag is helpful. treatment-effect incorporates it and it is not too heavily used, so there should not be a problem attracting the right kind of attention to the question. $\endgroup$ May 8, 2020 at 19:03

1 Answer 1


No, this is not correct. Let's walk through the basics to see why, and to see under what other assumptions ATT = LATE.

Let us call treatment assignment $Z$, and actual treatment taken $D$. Compliers have $D(Z = 1) = 1$ and $D(Z = 0) = 0$: If assigned treatment, they take, if assigned control, they do not take the treatment. These are counterfactual variables. Without further assumptions, we cannot tell whether a given person is a complier, because we do not observe what she would have done under a different assignment.

The LATE equals the ATT in the case of an experiment with "one-sided non-compliance". That is, everyone not eligible ($Z = 0$) cannot take the treatment $D$, but those assigned ($Z = 1$) may or may not. Think of a medical trial with a new drug, where you cannot possible take it if you are in the control group, but you may refuse it when you are told to. Formally, this means that for everyone the counterfactual variable $D(Z = 0)$ is 0.

Then, for those who took treatment ($D = 1$), by design, $Z = 1$ (there is no other way to receive treatment). So for these this means that $D(Z = 1) = 1$. Since everyone has $D(Z = 0) = 0$, this means that the treated are the compliers, and so ATT = LATE. The other remaining group are the "never-takers".

Regarding your specific question, if we are talking about a (general) design where $Z$ is randomized, then the proportions of always-takers, compliers, etc. is the same for $Z = 0$ and $Z = 1$. This is because these types are like background variables, and randomization makes $Z$ independent from such variables.

This is also means that if $Z$ is not randomized, then you could have a situation that you describe, where $P(AT|Z = 0) < P(C|Z = 1)$ (AT are always-takers, C compliers). However, this would mean that $Z$ is not a valid instrument. Perhaps you could solve this problem by conditioning on further confounders $X$.

Lastly, the situation you describe does not imply that ATT equals LATE. This is because the $D = 1$ group (the treated) is made up of always-takers, compliers, and possibly defiers. $P(AT|Z = 0) < P(C|Z = 1)$ is not sufficient to make sure that this group consists only of compliers.

It would be sufficient to assume that everyone is a complier (this is also testable, because given randomization of $Z$, it implies $P(D = 1|Z = 1) = 1$ and $P(D = 0|Z = 0) = 1$). Then ATE = ATT = LATE = ATC. This is because then the experiment is actually perfect: $Z$ is the same variable as $D$. All the confounding of $D$ and $Y$ is killed by the experimental manipulation. Accordingly, units don't choose $D$ depending on potential outcomes of $Y$, so ATE = ATT = ATC. Furthermore, $P(C) = 1$, so LATE = ATE (because the population and the compliers are the same units).


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