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?


1 Answer 1


Let $\hat{\pi}_0$ and $\hat{\pi}_1$ be the first stage coefficients. That is, $\hat{D} = \hat\pi_0 + \hat\pi_1 Z$. When $Z_i D_i = Z_i$ (if $Z_i=1$ then $D_i=1$), we can show that $\hat\pi_0 + \hat\pi_1 = 1$ always and $\hat\pi_0 = (\bar{D}-\bar{Z})/(1-\bar{Z})$. (Proof is fun.) Unless $\bar{D} \simeq 1$, $\hat\pi_0$ is far from unity, which means $\hat\pi_1$ is far from zero. Not a surprise that the p-value is very small.

Another argument: Because $ZD=Z$, we have $Z(1-D)=0$. Thus, $Z$ and $1-D$ are strongly correlated unless $E(Z)=0$ or $E(D)=1$. By the way, try probit D Z to see an interesting result.

Whether you need more variability is another thing. The question is whether it is possible that $Z$ is exogenous and $D$ is endogenous. Need to think more.

EDIT: Thought about whether it is possible that $Z$ is exogenous, $D$ is endogenous, and $ZD=Z$ at the same time. It looks possible. Let $u$ be the regression error term. We can let $D=Z+(1-Z)\xi$, where $\xi$ is a Bernoulli random variable, so that $ZD=Z$. Let $E(u|Z)=0$ as we wish. Then $E(Du)=E(u|Z=1)P(Z=1) + E(\xi u|Z=0) P(Z=0) = E(\xi u|Z=0) P(Z=0)$, which can be nonzero. It is possible that $Z$ is uncorrelated with $u$, $D$ is correlated with $u$, and $ZD=Z$, I guess. So my answer is: No worries.

EDIT: But "$Z=1 \Rightarrow D=1$" means that $D$ is exogenous if $Z=1$, and is endogenous if $Z=0$. Whether this is OK depends on the context.

  • $\begingroup$ Hi chan1142! I really appreciate your help. I tried the probit. I get an error message (in Stata): Z ≠ 0 predicts success perfectly. What do you mean by your last sentence: "Whether this is OK depends on the context."? $\endgroup$
    – cecefuss
    Dec 9, 2016 at 16:20
  • 1
    $\begingroup$ Hi cecefuss. That's what happens when fitted by probit. The intercept is identified but the estimated coefficient of $Z$ is $\infty$. Anyway, what I meant by the sentence is that it depends on your model and data. In order for your estimator (using your $Z$ as instrument) to be consistent, it has to be true that $D$ is exogenous if $Z=1$ (or conditional on the event that $Z=1$). If you think that is true in your model, you can defend yourself of using your $Z$ as instrument. $\endgroup$
    – chan1142
    Dec 10, 2016 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.