<p>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.</p>

<p>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 <code>probit D Z</code> to see an interesting result.</p>

<p>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.</p>

<p><strong>EDIT:</strong> 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.</p>

<p><strong>EDIT:</strong> 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.</p>

CC BY-SA 3.0
2016-12-09T16:20:50.383Z
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."?
cecefuss
CC BY-SA 3.0
2016-12-10T06:57:48.893Z
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.
chan1142
CC BY-SA 3.0
2016-12-05T23:20:29.993Z
2016-12-07T15:10:11.757Z
2016-12-07T15:10:11.757Z
chan1142
<p>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?</p>

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

(Asked by: username_0 on 2016-12-05)

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?

**Tags:** econometrics

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## 1 answers

### Answer by username_1 on 2016-12-05. Score: 4 | Accepted answer

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.

> **Comments:**
> **username_0**: 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."?  
> **username_1**: 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.  

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