# Marginal independence sufficient for showing that ATE is identified?

I have a basic question on the relation between counterfactual outcomes and treatment. In what follows ATE means average treatment effect.

Notation: Let $$D$$ denote the treatment, taking value in $$\{0,1\}$$ where $$D=0$$ means untreated and $$D=1$$ means treated.

Let $$Y_0, Y_1$$ denote the counterfactual outcomes.

Let $$Y=DY_1+(1-D)Y_0$$ denote the factual outcome.

Perfect randomisation assumption: $$(Y_0, Y_1) \perp D$$

Under this assumption we can show that ATE is identified (i.e., it is a function of known probability distributions). Indeed,

$$\text{ATE}\equiv E(Y_1)-E(Y_0)\overbrace{=}^{(Y_0, Y_1) \perp D \Rightarrow Y_1\perp D, Y_0\perp D} E(Y_1|D=1)-E(Y_0|D=0)= \underbrace{E(Y|D=1)-E(Y|D=0)}_{\text{known in an identification exercise}}$$

Question: it seems that to show that ATE is identified it is sufficient to have marginal independence (i.e., $$Y_1\perp D, Y_0\perp D$$) rather than joint independence (i.e., $$(Y_0, Y_1) \perp D$$). Is this correct?

• $Y_{1,i}$ and $Y_{0,i}$ are not observed simultaneously for subject $i$ (fundamental problem of causal inference). We observe, however, $(Y_i,D_i)$ for all $i$. Then the probabilistic assumption should be $(Y_{1,i},Y_{0,i})\perp D_i$. – MauOlivares Oct 10 '18 at 21:11