Testing for serial correlation with General Regressors

Question

This is from Introductory Econometrics (Wooldridge) 5th Edition page 420.

Consider: $$y_t =\beta_0 +\beta_1 x_{1t}+\beta_2 x_{2t}+...+\beta_k x_{kt}+u_t$$

where $Cov(u_t, x_{jt})=0$ for all $j$ but perhaps $Cov(u_t, x_{jt-1}) \ne 0$ for some $j$. We are concerned about autocorrelation, $$u_t =\rho u_{t-1}+v_t$$

To test for autocorrelation, the method proposed in-text is:

(1) Estimate the regression equation of interest and obtain residuals, $\hat{u_t}$.

(2) Run the regression: $$\hat{u_t}=\delta_0 +\delta_1 \hat{u}_{t-1}+\delta_2 x_{1t}+...+\delta_{k+1} x_{kt} +\varepsilon_t$$

(3) Perform a standard t-test for significance of $\delta_1$.

My question is, why do we need to include the x variables in the regression from step 2? It is a property of OLS that regressors are uncorrelated with estimated residuals.

The text says the inclusion is because of the potential failure of strict exogeneity (i.e., because $Cov(u_t, x_{jt-1}) \ne 0$) and that including the $x$ is necessary to attain a correct t-distribution of our t-stat. This is stated without justification or proof.

What is the underlying math of this idea?

Answer 1

Note that your statement

$\hat{u}_{t}$ is uncorrelated with $X_{t}$ by properties of OLS residuals.

is sample-dependent, in the sense that OLS normal equation only says $\sum_{t=1}^{T}X_{t}\hat{u}_{t}=0$, meaning the full statement should be

$\{\hat{u}_{t}\}_{t=1}^{T}$ is uncorrelated with $\{X_{t}\}_{t=1}^{T}$ by properties of OLS residuals.

But what is the value $\sum_{t=2}^{T}X_{t}\hat{u}_{t}$ or $\sum_{t=3}^{T}X_{t}\hat{u}_{t}$? I have no idea. As a result, will $\{\hat{u}_{t}\}_{t=2}^{T}$ and $\{X_{t}\}_{t=2}^{T}$, $\{\hat{u}_{t}\}_{t=3}^{T}$ and $\{X_{t}\}_{t=3}^{T}$ still be uncorrelated? I don't think so.

Now go back to your question. The general auxiliary regression for testing up to $p$-th order auto correlation is $$\hat{u}_{t}=\gamma^{\top}X_{t}+\sum_{j=1}^{p}\alpha_{j}\hat{u}_{t-j}+\varepsilon_{t}.$$ Since you include the lag terms of $\hat{u}_{t}$, which forces the sample indices of above regression equation to be $t=p+1,p+2,\ldots,T$. As mentioned before, generally, this will cause the regression result of $\gamma$ deviating from 0. (Although under null $\mathbb{E}[u_{t}|X_{t},u_{t-1},\ldots,u_{t-p}]=0$, these coefficients should converge to 0, but that's exactly what we are testing here.)

You can also verify it by a simple numerical experiment. Run a simple OLS regression, then i) regress $\hat{u}_{i}$ on $X_{i}$ using all the samples, you will get 0. ii) do the same regression but delete the first sample $\hat{u}_{1}$ and $X_{1}$, the result is no longer 0.

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Updates.

To cut a long story short, in the auxiliary regression, omitting $X_{t}$ won't cause inconsistency, but will contaminate the asymptotical distribution, hence the inference.

If we use $u_{t}$ in the regressor, then your claim is true, omitting $X_{t}$ is totally fine. However, the situation becomes different when we replace $u_{t}$ by $\hat{u}_{t}$.

Denote the original regression as $$y_{t}=\beta^{\top}X_{t}+u_{t},$$ under contemporaneous exogeneity, the infeasible auxiliary regression is $$u_{t}=\tilde{\delta} u_{t-1}+\tilde{\varepsilon}_{t}. $$ Since $u_{t}$ is not observable, we replace them by $\hat{u}_{t}$. Note that $\hat{u}_{t}=u_{t}-(\hat{\beta}-\beta)^{\top}X_{t}$, therefore despite the true parameter before $X_{t}$ was exactly 0, now it becomes weird $$\hat{u}_{t}=-(\hat{\beta}-\beta)^{\top}X_{t}+\delta\hat{u}_{t-1}+\varepsilon_{t}.$$ This equation is kind of beyond traditional econometric setting since the pseudo-true value $(\hat{\beta}-\beta)$ itself is changing and converging to 0. For now, let's first take sample size as fixed.

By direct computation, if $X_{t}$ is omitted in above regression, we have $\hat{\delta}=\delta-A_{T}+(\sum_{t=2}^{T}\hat{u}_{t-1}^{2})^{-1}\sum_{t=2}^{T}\hat{u}_{t-1}\varepsilon_{t}$, where the bias term of $\hat{\delta}$ is $$A_{T}=\biggl(\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}^{2}\biggr)^{-1}\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}X_{t}^{\top}(\hat{\beta}-\beta).$$

Now let's see what happens when $T\to\infty$, follow some standard procedure, we have $$\begin{align*} \frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}^{2}&=\mathbb{E}[u_{t}^{2}]+o_{p}(1)\\ \frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}X_{t}^{\top}&=\mathbb{E}[u_{t-1}X_{t}^{\top}]+o_{p}(1)\\ \sqrt{T}(\hat{\beta}-\beta)&\overset{\mathcal{L}}{\to}Z_{1}. \end{align*}$$ Combine above result, the bias term^* $$A_{T}=O_{p}(1)O_{p}(1)O_{p}(T^{-1/2})=\color{red}{O_{p}(T^{-1/2})},$$ which indicates $\hat{\delta}$ is still consistent. The problem arises when we apply the traditional $t$-test for $\delta$, CLT gives $$\sqrt{T}(\hat{\delta}-\delta+A_{T})\overset{\mathcal{L}}{\to}Z_{2}.$$ Here $Z_{2}$ is the normal distribution that $t$-test based on. However, when normalized by $\sqrt{T}$, the bias term $A_{T}$ will show up in the limiting distribution of $\sqrt{T}(\hat{\delta}-\delta)$ since $\sqrt{T}A_{T}=O_{p}(1)$, hence the $t$-test no longer works.

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^{*: If you strengthen contemporaneous exogeneity $\mathbb{E}[u_{t}|X_{t}]$ to strict exogeneity $\mathbb{E}[u_{t}|X_{1},\ldots,X_{T}]=0$, that $X_{t}$ is not allowed to be correlated with past innovation, then $A_{T}=o_{p}(T^{-1/2})$ since the second term becomes $o_{p}(1)$, and $t$-test is still valid. This is what the textbook says.}

Answer 2

My understanding is that the classical test for serial correlation is actually conditional to the validity of the strict exogeneity assumption: $E[u_t|X]=0,$ or, as the requirement applies to any $t$, $E[u_{t-1}|X]=0$. This is a shortcoming, because violations of the strict exogeneity assumption usually generates autocorrelation (we may find evidence for serially correlated errors because strict exogeneity is actually violated).

The classical serial correlation approach considers that a type of conditional mean independence wrt $X$ holds: $$ E[u_t|X,u_{t-1}]=E[u_t|u_{t-1}]=\rho u_{t-1}. $$ The extension discussed by Wooldridge considers instead that: $$ E[u_t|X,u_{t-1}]=\rho u_{t-1}+x_t^T\delta. $$ This does not imply (nor require) strict exogeneity, by the law of iterated expectations, we have $E[u_t|X] \neq 0$ in general, and OLS estimates in the model $y_t=x_t^T\beta +u_t$ are biased for $\beta$ in this case.