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 $Y_1\perp D$.

  • $\begingroup$ I Think you cannot answer either question (i.e. marginal or joint indpendence) without more information about the functions $g$ and $h$ $\endgroup$ – Sebastian Oct 10 '18 at 16:01
  • $\begingroup$ @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. $\endgroup$ – user3285148 Oct 10 '18 at 16:03
  • $\begingroup$ 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 $\endgroup$ – Sebastian Oct 10 '18 at 16:19

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 $Y_1$.


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