Suppose we have the following simple regression model (time series framework)"

$$y_1=\beta_0+\beta_1 y_2+\beta_2 z_1 +\beta_3 z_2 +u,$$

where $z_1$ and $z_2$ are exogenous and $y_2$ is either exogeneous or endougenous (this is what we want to check). In order to determine whether or not $y_2$ is endogenous, we can apply endogeneity test (Hausman test), which follows the following procedure:

Estimate the reduced form for $y_2$, i.e. estiamte the following equation:

$$y_2=\alpha_0+\alpha_1z_1+\alpha_2z_2+\alpha_3z_3+\alpha_4z_4+\nu,$$ where $z_3$ and $z_4$ are instruments. Since each $z_j$ is uncorrelated with $u$, $y_2$ is uncorrelated with $u$ if and only if $v$ is uncorrelated with $u$; this is what we want to test. The easiest way to test this is to include $v$ as an additional regressor in structural eqaution and to do a $t$ test, i.e. estimate the model

$$y_1=\beta_0+\beta_1 y_2+\beta_2 z_1 +\beta_3 z_2 +\gamma_1\widehat\nu+error.$$ I don't understad this part. We want to determine whether $u$ and $\nu$ are correlated, but how we unleash it by including $\widehat \nu$ at the structural equation? Doing so, we esimate the impact of $\widehat \nu$ on $y_1$, rather than on $u$. Please explain the intuition.

P.S. In my understanding in order to determine whether $u$ and $\nu$ are correlated, we can apply the following steps:

  • Estimate $\widehat u$ from the structural equation,
  • Estimate $\widehat \nu$ from the reduced eqaution,
  • Regress $\widehat u$ on $\widehat \nu$.
  • $\begingroup$ Hi: I think it goes like this but hopefully someone else will give a better explanation. Think of $u$ as the same as $\gamma_{1} \hat{v}$ + error. This is allowed because the rest of the first and third equations are the same. So, if $\gamma_1$ ends up being significant, then the "error" term doesn't do enough to soak up all the noise. Therefore $v$ must soak some up also so it has to be correlated with $u$. Note that this only works because, if you disregard noise terms, then, besides the $\gamma$ term, the third equation is the same as the first equation. $\endgroup$
    – mlofton
    Commented Nov 7, 2020 at 16:08

1 Answer 1


The equation

$$ y_2 = \alpha_0 + \alpha_1z_1 + \alpha_2z_2 + \alpha_3z_3 + \alpha_4z_4 + v $$

is usually called the first stage. The rationale for this test is that if $z_3$ and $z_4$ are good instruments, then $\hat{v}$ contains the potentially bad variation in $y_2$, i.e., the part that might be endogenous. If you are able to exclude $\hat{v}$ from the structural equation, you cannot reject that $y_2$ is exogenous.

Your intuition is not bad, but think about it as checking the correlation between $y_2$ and and $u$ instead. Residuals are uncorrelated with the regressors that made them (if unfamiliar with this check here) so the correlation between $\hat{u}$ and $y_2$ is zero by construction. Same problem applies to correlation between $\hat{u}$ and $\hat{v}$.

Notice that the structural equation including $\hat{v}$ returns the IV estimate. Another way to understand this test is that if we interpret IV as OLS controlling for $\hat{v}$ and if we can exclude $\hat{v}$ from this equation then it is, arguably, uneccessary to control for $\hat{v}$; i.e, to use IV.

However, note that this is rarely a relevant test.

First, it assumes that the intrument(s) is(are) good. Second, if you do have a good(ish) instrument the test could fail to reject simply because the IV estimate is not very precise. In this case, I would still pick IV over OLS any day of the week since there is probably some theoretical reason to suspect endogeneity.

Also check here for more on the point about good instruments.

  • $\begingroup$ Thank you for the asnwer! First, could you please explain why in order to determine whether $u$ and $v$ are coorelated we include $\widehat v$ in our structural equation? Doing this do we determine whether $u$ and $v$ are correlated or not, and how? This is what I don't uderstand. As I usggested in my question, I am inclide to do it by regressing $\widehat u$ on $\widehat \nu$. Second, if this test is not relevant (as you highlighted in your answer), what is the alternative to this which yield to more precise results? Thanks again! $\endgroup$
    – Duo
    Commented Nov 7, 2020 at 22:05
  • $\begingroup$ Np. Updated answer. There is no alternative and my point is that this is not something that is usually relevant to test for. To test for endogeneity you must first have a credible solution to the problem (that is IV). If the solution is credible, you don’t havet a problem anymore that you need test for. If solution is not credible, then you can’t (credibly) test for problem. $\endgroup$
    – Jonathan
    Commented Nov 8, 2020 at 6:07

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