Assume, the model we are trying to estimate is:


where x and u are correlated: $Cov(X,U)\neq 0$

Then OLS is inconsistent:

$\beta_1^{OLS}=\frac{Cov(Y,X)}{Var(X)}=\frac{Cov(\beta_0+\beta_1X+U,X)}{Var(X)}=\beta_1\frac{Cov(X,X)}{Var(X)}+\frac{Cov(X,U)}{Var(X)}=\beta_1+\frac{Cov(X,U)}{VaR(X)}\neq 0$

If we introduce an instrumental variable, for which it holds:

$Cov(Z,X)\neq 0$

$Cov(Z,U) = 0$

One can show, that OLS is consitent, so that


I wanted to proof this. Therefore I tried the following: $\beta_1^{OLS}=\frac{Cov(Y,Z)}{Var(Z)}=\frac{Cov(\beta_0+\beta_1X+U,Z)}{Var(Z)}=\beta_1\frac{Cov(Z,X)}{Var(Z)}+\frac{Cov(Z,U)}{Var(Z)}=?$

This does not lead to a nice solution (first of all $\frac{Cov(Z,X)}{Var(Z)}\neq 1$ and I have $Var(Z)$ in the denominator and not $Cov(X,Z)$)? Why, where is my mistake and how can I proof this?


I think you tried to prove consistency of the IV estimator but started the proof with the OLS estimator instead. If you write: $$\newcommand{\Cov}{\operatorname{Cov}}\beta_{1}^{\mathrm{IV}} = \frac{\Cov(Z,Y)}{\Cov(Z,X)} = \frac{\Cov(Z,\beta_{0}+\beta_{1}X+U)}{\Cov(Z,X)}$$ you should get your desired result $$\beta_{1}\frac{\Cov(Z,X)}{\Cov(Z,X)} + \frac{\Cov(Z,U)}{\Cov(Z,X)}$$ and then by the assumption that $\Cov(Z,U) = 0$ you will be left with $\beta_{1}$.

Have a look here on page 3 just to double-check.

Concerning the question about the formula of the IV estimator:
For your model $$Y = \beta_{0} + \beta_{1}X_{1} + U$$ take the covariance of all terms with the instrument $Z$, which gives $$\Cov(Z,Y) = \beta_{1}\Cov(Z,X) + \Cov(Z,U)$$ Then $\Cov(Z,U) = 0$ by assumption and dividing by $\Cov(Z,X)$ gives $$\beta_{1}^{\mathrm{IV}} = \frac{\Cov(Z,Y)}{\Cov(Z,X)}$$

  • $\begingroup$ +1 and thanks a lot for your answer, but why do you use Cov(Z,Y)/(Cov(Z,X)) and not Cov(Z,Y)/(Var(X)) ? Where did you get this first equation, that beta = Cov(Z,Y)/(Cov(Z,X)) ? $\endgroup$ – Stat Tistician Jul 14 '13 at 8:23
  • $\begingroup$ I edited the answer and added the explanation you asked for in the comment. $\endgroup$ – Andy Jul 14 '13 at 8:35

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.