# Marginal independence sufficient for showing ATE=ATT?

I have a basic question on the relation between counterfactual outcomes and treatment.

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.

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

Under this assumption we can show that ATE=ATT (average treatment effect = average treatment effect among treated). Indeed,

$$\text{ATT}\equiv E(Y_1-Y_0|D=1)=E(Y_1|D=1)-E(Y_0|D=1)=E(Y_1)-E(Y_0)\equiv \text{ATE}$$ where the last step comes from the fact that $$(Y_0, Y_1) \perp D$$ $$\Rightarrow$$ $$Y_1\perp D, Y_0\perp D$$

Question: it seems that to show ATE=ATT 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?