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I have some problems understanding the difference between ATE and ATET and the Selection Bias. To explain what my understanding is I have done the following representation so you can correct me:

lsd

We want to see the effect of taking LSD, to do so we have a Group to which we assign the "treatment" so $D=1$ and a group which has no treatment assigned so $D=0$. On the top left corner we can see $Y_{1i}$ and on the right bottom corner $Y_{0i}$. The problem comes now, as far as I can understand because there is no randomization (?) in the sample we have that some guys in the control group that take LSD because they are drug-abusers (red dots) ($E[Y_{0i}|D=1]$). In addition some other guys in the treatment group do not do LSD because they are afraid ($E[Y_{1i}|D=0]$). So:

$ATE = E[Y_{1i}|D=1] - E[Y_{0i}|D=1] + E[Y_{0i}|D=1] - E[Y_{0i}|D=0]$

Where the first two terms are ATE and the last two are Selection Bias.

Why $E[Y_{0i}|D=1]$ is not observable? Is this explanation correct? Why if we apply randomization in the treatment we manage to get $E[Y_{0i}|D=1] - E[Y_{0i}|D=0] = 0$ ?

adrian1121
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    If you randomize, there is no selection bias since in expected value people belonging to the treatment group and to the control group are the same. The purpose of random sampling is to build two groups that are equivalent. If you look for a nice explanation this is a good book I've used: https://www.amazon.it/Experimental-Quasi-Experimental-Designs-Generalized-Inference/dp/0395615569 – PhDing Dec 11 '16 at 15:05
  • I think it would we useful to clarify exactly what the treatment is. Is it being handed LSD (or encouraged to seek it out), or is it actually ingesting it? Which one do you care about? From the graph it seems like it is the first one, with some non-compliance in both groups. But your written description is at odds with that. Also, what is the outcome here? – dimitriy Dec 16 '17 at 00:56
  • In addition to @Alessando's helpful suggestion, you should also check out Mostly Harmless Econometrics by Joshua Angrist and Jorn-Steffen Pischke. It's easy to read and goes over this exact problem quite in-depth, i.e. treatment effects with non-compliers and randomization. – JuliusBilly Jul 20 '18 at 17:43

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$E[Y_{0i}|D=1]$ is not observable because we cannot know the counterfactual, i.e. what outcome the individual would have if they did not receive the treatment as they always get the treatment (D=1). What randomisation does is guarantee that the treatment assignment is statistically independent of potential outcomes, therefore: $E[Y^*_0]=E[Y^*_0|D=1]=E[Y^*_0|D=0]$ and hence $E[Y^*_0]=E[Y^*_0|D=1]-E[Y^*_0|D=0]=0$, i.e. the expected outcome for a non-treated individual is the same if they are in the treated group or if they are in the control group.

Gooze
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