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$\textbf{Question:}$ Suppose we have ${(Y_i, X_i,Z_i,W_i)^{n}_{i=1}}$ which is a random sample from the joint distribution of $(Y,X,Z,W)$ that satisfies the following relation: $$Y_i=\beta_0+\beta_1(X_i)+\beta_2(W_i)+\mu_i$$

where $\mu_i$ is the unobserved error term, $X_i$ is endogenous and $W_i$ is exogenous. I assume that there is no linear dependency among $X_i,W_i,Z_i$ and that their variances are positive.

$Z_i$ is the instrumental variable (IV) and $X_i$ is related to it like this:

$$X_i=\pi_0+\pi_1Z_i+\nu_i$$

Now the question states that I need to identify $\beta_1$ (basically write an equation for $\beta_1$).

$\textbf{Now here is my thought process and work:}$

What I did is plug in the second equation into the first get the following:

$$Y_i=\beta_0+\beta_1(\pi_0+\pi_1Z_i+\nu_i)+\beta_2W_i+\mu_i$$ and then I identified a new variable like this:

$$X_i(\pi):\pi_0+\pi_1Z_i+\nu_i$$ and substituted $X_i(\pi)$ into the equation above to get this:

$$Y_i=\beta_0+\beta_1X_i(\pi)+\beta_2W_i+\nu_i$$

Here what I do is take the expectation of this equation, multiply by $(Z_i-E(Z_i))$, then subtract the third equation. Then when we take the expectation of both sides again we have covariance terms:

$$Cov(Y_i,Z_i)=\beta_1Cov(X_i(\pi),Z_i)+\beta_2Cov(W_i,Z_i)$$ because the $E(\nu_i)$ and $E(\mu_i)$ are both zero.

Now depending on the assumptions we have made about $Cov(W_i,Z_i)$, then we have $\beta_1$ identified.

Is this thought process correct? I feel as though my identification is incorrect and help would be greatly appreciated!

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The thought process is right but the proof can be achieved in an easier way (see here) and as it stands it's not quite complete. In order to identify a parameter you're looking for a closed form solution for $\beta_1$, so it should be on one side of the equation and all the rest on the other.

Also as long as $Cov(W_i,Z_i)=0$ you are also not identifying a partial effect, i.e. the effect of $Z_i$ through $X_i$ on $Y_i$ holding $W_i$ fixed. In order to achieve that you can first regress $$Z_i = \alpha + \gamma W_i + \tilde{Z}_i$$ and work with the residual $\tilde{Z}_i$ in your proof above. By construction this makes sure that $Cov(W_i,\tilde{Z}_i)=0$ and you will be left with a $\widehat{\beta}_1$ which is going to be equal to the reduced form $Cov(Y_i,\tilde{Z}_i)$ divided by the first stage $Cov(X_i,\tilde{Z}_i)$ holding other exogenous variables fixed.

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  • $\begingroup$ Thank you for your answer. If we assume $Cov(W_i,Z_i)=0$ then $B_2$ is gone and only $B_1$ remains in the equation so we can rearrange it and identify it. So wouldn't that make my method above correct? I understand your answer though, quite nicely explained. $\endgroup$ – nicefella Feb 8 '15 at 21:19
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    $\begingroup$ That's right, you get the same results if $Cov(W_i,Z_i)=0$. I just thought you maybe wanted to have a more general solution :-) $\endgroup$ – Andy Feb 8 '15 at 21:24

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