$\textbf{Question:}$ 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:


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:


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!


1 Answer 1


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.

  • $\begingroup$ 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, 2015 at 21:19
  • 1
    $\begingroup$ 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, 2015 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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