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

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 $$E\left ( y|z=0 \right )=E\left ( y_{0}|z=0 \right )$$?

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, $$E(y|z=0)=E(y_0|z=0)$$.