How to prove that one-sided non-compliance implies $E\left ( y|z=0 \right )=E\left ( y_{0}|z=0 \right )$?

Question

Let $d \in \left \{ 0,1 \right \}$ denote the treatment status with unity indicating that the individual has been treated and let $z \in \left \{ 0,1 \right \}$ denote a binary indicator for whether or not the individual has been randomized into the treatment group with unity indicating that they have been randomized into the treatment group.

If the non-compliance is one-sided or restricted to people with $z=1$ then the instrumental variable is an randomized control trial that will recover the average treatment effect on the treated.

That is $z=0 \Rightarrow d = 0 \Leftrightarrow d=1 \Rightarrow z= 1$.

From this, how can I prove that $E\left ( y|z=0 \right )=E\left ( y_{0}|z=0 \right )$?

Answer 1

Since $y=y_0+(y_1-y_0)d$ you have $E(y|z=0)=E(y_0|z=0)+E((y_1-y_0)d|z=0)$. The result follows if one can show that $E((y_1-y_0)d|z=0)=0$:

For simplicity, let $w\equiv y_1-y_0$. Then $E((y_1-y_0)d|z=0)=E(wd|z=0)$. Now, $$E(wd|z=0)=E(wd|z=0,d=0)P(d=0|z=0)+E(wd|z=0,d=1)P(d=1|z=0).$$

Since $z=0$ implies $d=0$, $P(d=0|z=0)=1$ and so $P(d=1|z=0)=0$. So $E(wd|z=0)=E(wd|z=0,d=0)$. But $E(wd|z=0,d=0)=E(w0|z=0,d=0)=0$. Consequently, $E(y|z=0)=E(y_0|z=0)$.