# Understanding the perfect randomization assumption in treatment models

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: $$(Y_0, Y_1) \perp D$$

Scenario: The treatment is offered to everyone; agents who are offered the treatment can decide whether to receive it.

Question: I understand that under the scenario above $$(Y_0, Y_1) \not\perp D$$. I would like to know whether under the scenario above we still have $$Y_0 \perp D$$ and/or $$Y_1\perp D$$ and why.

My thoughts:

• An agent is totally described by some random features $$(X,U)$$ (not necessarily entirely observed by the researcher). Hence, $$Y_0=g(\underbrace{0}_{\text{untreated}},X,U)$$, $$Y_1=g(\underbrace{1}_{\text{treated}},X,U)$$, for some unknown function $$g$$.

• The perfect randomization assumption can be rewritten as $$D\perp (X,U)$$. This, combined with $$Y_0=g(0,X,U)$$ $$Y_1=g(1,X,U)$$, implies $$D\perp (Y_0, Y_1)$$.

• If an agent can choose whether of not to received the treatment, then $$D=h(X,U)$$ for some function $$h$$ capturing the choice-behavior of the agent. This, combined with $$Y_0=g(0,X,U)$$ $$Y_1=g(1,X,U)$$, implies $$D\not \perp (Y_0, Y_1)$$.

• It remains unclear to me if with self-selection we still have $$Y_0 \perp D$$ and/or $$Y_1\perp D$$.

• I Think you cannot answer either question (i.e. marginal or joint indpendence) without more information about the functions $g$ and $h$ – Sebastian Oct 10 '18 at 16:01
• @Sebastian I believe that the failure of the joint independence is a well known fact (if agents self-select into treatment then the perfect randomization assumption fails). I've doubts for the marginal independence. – user3285148 Oct 10 '18 at 16:03
• Yeah sure empirically Yes. Why exactly so you have doubts for the marginal one? The answer also of course depends on the exact setup of the experiment – Sebastian Oct 10 '18 at 16:19

## 1 Answer

No, you do not have marginal independence, not even under restrictive parametric assumptions.

Let's ignore $$X$$ and let Y = g(U) be linear, so that

$$Y = \beta D + U.$$

Furthermore, let $$D \sim Unif[0, 1]$$, and $$D = I(U > 0.5)$$.

Then $$E[Y_1] = \beta + 0.5$$, but

$$E[Y_1|D = 1] = \beta + E[U|U > 0.5] = \beta + 0.75,$$

so $$D$$ is not independent from $$Y_1$$.