# If my IV (denoted Z) equals 1, my endogenous var D always equals 1; is this a problem?

I have a binary instrumental variable Z={0,1} and a binary endogenous variable D={0,1}. By construction, D=1 necessarily holds if Z=1. There are also cases where D=1 if Z=0, but there are no cases where D=0 if Z=1. (Side note: in the first stage of 2SLS, I get an F-value of around 3000.) Intuitively, I feel I should have more variation, i.e., that I should also have cases where D=0 if Z=1. Are my intuitive worries justified?

I am no expert in the IV literature, so please comment on this answer if I miss anything. But consider these two very nice answers here.

Usually under strong monotonicity there are three groups,

Always-taker ($D=1\,$ for $\,Z=0/1$),

Compliers ($D=1\, \& \,Z=1$ or $D=0\, \& \,Z=0$) and

Never-taker ($D=0\,$ for $\,Z=0/1$)

You do not observe the never-taker with $D=0\,\&\,Z=1$. So you have to assume that $$Pr(D=0|Z =1)=0 \text{ or } Pr(D=1|Z =1)=1$$

What does this means?

It means that those with $Z=1$, or the treatment, always receive the treatment. There are no never-taker with $Z=1$ (this a somewhat strange form of one-sided non-compliance, usually you assume that the control group never receives the treatment under non-sided non-compliance).

The formula for the Wald estimator is (copied from this answer) in this case \begin{align} \text{LATE} &= \frac{E(Y|Z=1)−E(Y|Z=0)}{Pr(D=1|Z =1)−Pr(D=1|Z =0)}\\ &= \frac{E(Y|Z=1)−E(Y|Z=0)}{1-Pr(D=1)}\\ &=\text{Average Treatment Efect On the Untreated} \end{align}

Your analysis is valid, but you make strong assumptions (in addition to independence and monotonicty) by implicitly assuming that the local average treatment effect (LATE) is equal to the average treatment effect on the untreated units.