<p>No, you're right, $Cov[Z_i,u_i]=0$ does not imply $E[u_i|Z_i]=0$ in general. If the author defined "instrument exogeneity" as $Cov(Z_i,u_i)=0$ previously, he/she is being careless here.</p>

<p>"Instrument exogeneity" is often defined as $E[u_i|Z_i]=0$ too, in which case all you need to do would be to understand "implies" as "means". But since exogeneity is defined as uncorrelatedness (zero covariance), this issue has to be addressed. the author'd better fix it in a later version. How about "Instrument exogeneity usually means ..." instead of "Instrument exogeneity implies ..."?</p>

CC BY-SA 4.0
2018-08-12T01:16:14.623Z
2018-08-12T01:16:14.623Z
chan1142
<p>In the following slide <a href="https://i.sstatic.net/8OMa8.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8OMa8.png" alt="Instrumental variables: 1 endogenous regressor &amp; 1 instrument"></a>
<a href="https://www.uio.no/studier/emner/sv/oekonomi/ECON4150/v14/undervisningsmateriale/lecture16_instrumental_variables.pdf" rel="nofollow noreferrer">ECON4150 - Introductory Econometrics Lecture 16: Instrumental variables, Monique de Haan</a> </p>

<p>it says that "instrument exogeneity implies $E[u_i \mid Z_i]=0$" where instrument exogeneity is defined on slide 13 as $cov(Z_i, u_i) =0$. Here $Z_i$ is the instrument and $u_i$ is the error term in the structural model with a single endogenous variable. </p>

<p>Can someone prove to me why this is true? I thought the implication only held the other way, i.e., $E[u_i \mid Z_i] = 0$ implies $cov(Z_i, u_i)=0$ but the slides suggests that $cov(Z_i, u_i)=0$ implies $E[u_i \mid Z_i] = 0$, which isn't true in general?</p>

<econometrics><regression>
Why does instrument exogeneity imply conditional mean zero?
elbarto
# Why does instrument exogeneity imply conditional mean zero?

(Asked by: username_0 on 2018-08-01)

In the following slide [![Instrumental variables: 1 endogenous regressor & 1 instrument](https://i.sstatic.net/8OMa8.png)](https://i.sstatic.net/8OMa8.png)
[ECON4150 - Introductory Econometrics Lecture 16: Instrumental variables, Monique de Haan](https://www.uio.no/studier/emner/sv/oekonomi/ECON4150/v14/undervisningsmateriale/lecture16_instrumental_variables.pdf)

it says that "instrument exogeneity implies $E[u\_i \mid Z\_i]=0$" where instrument exogeneity is defined on slide 13 as $cov(Z\_i, u\_i) =0$. Here $Z\_i$ is the instrument and $u\_i$ is the error term in the structural model with a single endogenous variable.

Can someone prove to me why this is true? I thought the implication only held the other way, i.e., $E[u\_i \mid Z\_i] = 0$ implies $cov(Z\_i, u\_i)=0$ but the slides suggests that $cov(Z\_i, u\_i)=0$ implies $E[u\_i \mid Z\_i] = 0$, which isn't true in general?

**Tags:** econometrics, regression

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## 1 answers

### Answer by username_1 on 2018-08-12. Score: 2 | Accepted answer

No, you're right, $Cov[Z\_i,u\_i]=0$ does not imply $E[u\_i|Z\_i]=0$ in general. If the author defined "instrument exogeneity" as $Cov(Z\_i,u\_i)=0$ previously, he/she is being careless here.

"Instrument exogeneity" is often defined as $E[u\_i|Z\_i]=0$ too, in which case all you need to do would be to understand "implies" as "means". But since exogeneity is defined as uncorrelatedness (zero covariance), this issue has to be addressed. the author'd better fix it in a later version. How about "Instrument exogeneity usually means ..." instead of "Instrument exogeneity implies ..."?


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