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Read something about the Hausman test that didn't sound right in some grad course handout online.

It stated that the Hausman null of OLS and IV not being statistically different, if not rejected, means there is no evidence of endogeneity.

It also states that under the null it is assumed that OLS is consistent+efficient and that IV is consistent+inefficient.

That doesn't sound right to me. If OLS is biased and IV is correctly identified, wouldn't rejecting the null be evidence of no endogeneity? And why would we assume OLS is consistent for the null?

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  • $\begingroup$ you have received an excellent answer by Dan. Please consider upvoting and/or accepting it if you are happy with his explanation. The model of this site depends on the contribution of everyone. Writing a good answer takes a good amount of time which is time that is devoted to help others. Accepting/upvoting only costs you a click with the mouse :) $\endgroup$
    – Andy
    Jul 14 '14 at 18:05
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The Hausman test is given as: $H=\left(\hat{\beta}_{OLS}-\hat{\beta}_{IV}\right)'\left[var\left(\hat{\beta}_{IV}\right)-var\left(\hat{\beta}_{OLS}\right)\right]^{-1}\left(\hat{\beta}_{OLS}-\hat{\beta}_{IV}\right)\sim\chi^{2}\left(M\right) $ where $\hat{\beta}_{IV} $ is the $M\times1 $ vector of IV estimates. The test is built with OLS being consistent under the null because: $\left[var\left(\hat{\beta}_{IV}\right)-var\left(\hat{\beta}_{OLS}\right)\right] $ will always be a positive definite matrix since $var\left(\hat{\beta}_{IV}\right)>var\left(\hat{\beta}_{OLS}\right) $. (The variance of the OLS estimator is given as: $Avar\left(\hat{\beta}_{OLS}\right)=\hat{\sigma^{2}\left(X'X\right)^{-1}}$ while the variance for the IV estimator is given as: $Avar\left(\hat{\beta}_{IV}\right)=\hat{\sigma^{2}\left(Z'X\right)^{-1}Z'Z\left(Z'X\right)^{-1}}$ ). What we want to do is to test the difference between the two estimators. If we do not have endogeneity then both the OLS and the IV estimates are consistent but the IV estimate is inefficient.The null hypothesis is that we do not have any endogeneity!! This means that under the null of no endogeneity, the OLS estimate is consistent and efficient while the IV estimator is consistent although inefficient. Under the alternative hypothesis only the IV estimate is consistent (since $E\left(X'u\right)\neq0 )$, correlation between $x_{i} $ and $u_{i} $ makes the OLS estimator inconsistent. This can be seen as:

$\hat{\beta}=\left(X'X\right)^{-1}X'y$

$\hat{\beta}=\left(X'X\right)^{-1}X'\left(X\beta+u\right)$

$\hat{\beta}=\beta+\left(X'X\right)^{-1}X'u$

We can divide through by N and take probability limits so we get

$plim\left(\hat{\beta}\right)=\beta+plim\left(N^{-1}X'X\right)plim\left(N^{-1}X'u\right) $

since we assume a random sample we can rewrite this equation by the law of large number as:

$plim\left(\hat{\beta}\right)=\beta+E\left(X'X\right)^{-1}E\left(X'u\right)=\beta $

since we assume $E\left(X'u\right)=0 $. If we have endogenity, i.e. $E\left(X'u\right)\neq0 $ then OLS will be inconsistent ($plim\left(\hat{\beta}\right)=\beta+E\left(X'X\right)^{-1}E\left(X'u\right) $). However, IV is by assumption $cov\left(z,\epsilon\right)=0 $ and $cov\left(z,x\right)\neq0 $ ($z $ is the instrument while $\epsilon $ is the error term),hence it'll be consistent. The IV estimator can be shown to be consistent from: The IV estimator can be shown to be consistent from: $plim\left(\hat{\beta_{IV}}\right)=\beta_{IV}+E\left(Z'X\right)^{-1}E\left(Z'u\right)=\beta_{IV} $. So to answer your question: Rejecting the null would indicate that we have endogeneity!

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  • $\begingroup$ +1, maybe as a caveat: not rejecting the null does not necessarily mean that we do not reject exogeneity but simply the test may not detect the difference between OLS and IV if IV is too imprecise. This can happen for instance when the first stage does not explain much in the variation of the endogenous variables, i.e. the instrument isn't very strong. Then the standard errors of IV will be larger. $\endgroup$
    – Andy
    Jul 13 '14 at 10:22

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