<p>My understanding is that the classical test for serial correlation is actually conditional to the validity of the strict exogeneity assumption: <span class="math-container">$E[u_t|X]=0,$</span> or, as the requirement applies to any <span class="math-container">$t$</span>, <span class="math-container">$E[u_{t-1}|X]=0$</span>. 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).</p>
<p>The classical serial correlation approach considers that a type of conditional mean independence wrt <span class="math-container">$X$</span> holds:
<span class="math-container">$$ E[u_t|X,u_{t-1}]=E[u_t|u_{t-1}]=\rho u_{t-1}. $$</span>
The extension discussed by Wooldridge considers instead that:
<span class="math-container">$$ E[u_t|X,u_{t-1}]=\rho u_{t-1}+x_t^T\delta. $$</span>
This does not imply (nor require) strict exogeneity, by the law of iterated expectations, we have <span class="math-container">$E[u_t|X] \neq 0$</span> in general, and OLS estimates in the model <span class="math-container">$y_t=x_t^T\beta +u_t$</span> are biased for <span class="math-container">$\beta$</span> in this case.</p>

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2022-01-07T13:25:48.840Z
Thank you for the answer. 
I am a little surprised Wooldridge does this though, because $\hat{u_t}$ is uncorrelated with $x_t$ by properties of OLS residuals. Is it more the principle that the regression is motivated by the theory of the true error, $u_t$?

And if the theory is that $x_t$ may be correlated with $u_t$, then contemporaneous exogeneity fails - which really seems like an odd assumption to base a test on.
Michael Gmeiner
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2022-01-07T14:18:37.997Z
I agree with you on the fact that, if there is violation of strict exogeneity, then it is better (in order to avoid inconsistent OLS estimates), to obtain the $\widehat{u}_t$ using IV or GMM. What happens if $\widehat{u}_t$ is obtained after an inconsistent OLS? Is the approach still valid? Hmm. I cannot give a definite answer, but just note that strict exogeneity $E[u_t|X]=0$ does not imply that $E[u_t|X,u_{t-1}]=0.$ So it might still be justified to consider OLS residuals...
Bertrand
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2022-01-07T14:50:19.200Z
@Bertrand: You clearly know what you're talking about but I was just wondering if your answer addresses any of my confusion regarding the initial assumption then becoming invalid if the test for serial correlation rejects the null ? Thanks and, if not, don't worry about it. I've always found Wooldridge's writings to be somewhat confusing so I'll just remain confused !!!!!!!!!
mark leeds
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2022-01-06T15:03:01.727Z
2022-02-07T07:51:44.653Z
2022-02-07T07:51:44.653Z
Bertrand
<p>Note that your statement</p>
<blockquote>
<p><span class="math-container">$\hat{u}_{t}$</span> is uncorrelated with <span class="math-container">$X_{t}$</span> by properties of OLS residuals.</p>
</blockquote>
<p>is <em>sample-dependent</em>, in the sense that OLS normal equation only says <span class="math-container">$\sum_{t=1}^{T}X_{t}\hat{u}_{t}=0$</span>, meaning the full statement should be</p>
<blockquote>
<p><span class="math-container">$\{\hat{u}_{t}\}_{t=1}^{T}$</span> is uncorrelated with <span class="math-container">$\{X_{t}\}_{t=1}^{T}$</span> by properties of OLS residuals.</p>
</blockquote>
<p>But what is the value <span class="math-container">$\sum_{t=2}^{T}X_{t}\hat{u}_{t}$</span> or <span class="math-container">$\sum_{t=3}^{T}X_{t}\hat{u}_{t}$</span>? I have no idea. As a result, will <span class="math-container">$\{\hat{u}_{t}\}_{t=2}^{T}$</span> and <span class="math-container">$\{X_{t}\}_{t=2}^{T}$</span>, <span class="math-container">$\{\hat{u}_{t}\}_{t=3}^{T}$</span> and <span class="math-container">$\{X_{t}\}_{t=3}^{T}$</span> still be uncorrelated? I don't think so.</p>
<p>Now go back to your question.
The general auxiliary regression for testing up to <span class="math-container">$p$</span>-th order auto correlation is
<span class="math-container">$$\hat{u}_{t}=\gamma^{\top}X_{t}+\sum_{j=1}^{p}\alpha_{j}\hat{u}_{t-j}+\varepsilon_{t}.$$</span>
Since you include the lag terms of <span class="math-container">$\hat{u}_{t}$</span>, which <strong>forces the sample indices of above regression equation to be</strong> <span class="math-container">$t=p+1,p+2,\ldots,T$</span>. As mentioned before, generally, this will cause the regression result of <span class="math-container">$\gamma$</span> deviating from 0. (Although under null <span class="math-container">$\mathbb{E}[u_{t}|X_{t},u_{t-1},\ldots,u_{t-p}]=0$</span>, these coefficients should converge to 0, but that's exactly what we are testing here.)</p>
<p>You can also verify it by a simple numerical experiment. Run a simple OLS regression, then i) regress <span class="math-container">$\hat{u}_{i}$</span> on <span class="math-container">$X_{i}$</span> using all the samples, you will get 0. ii) do the same regression but delete the first sample <span class="math-container">$\hat{u}_{1}$</span> and <span class="math-container">$X_{1}$</span>, the result is no longer 0.</p>
<hr />
<p>Updates.</p>
<p>To cut a long story short, <strong>in the auxiliary regression, omitting <span class="math-container">$X_{t}$</span> won't cause inconsistency, but will contaminate the asymptotical distribution, hence the inference.</strong></p>
<p>If we use <span class="math-container">$u_{t}$</span> in the regressor, then your claim is true, omitting <span class="math-container">$X_{t}$</span> is totally fine. However, the situation becomes different when we replace <span class="math-container">$u_{t}$</span> by <span class="math-container">$\hat{u}_{t}$</span>.</p>
<p>Denote the original regression as
<span class="math-container">$$y_{t}=\beta^{\top}X_{t}+u_{t},$$</span>
under contemporaneous exogeneity, the infeasible auxiliary regression is
<span class="math-container">$$u_{t}=\tilde{\delta} u_{t-1}+\tilde{\varepsilon}_{t}. $$</span>
Since <span class="math-container">$u_{t}$</span> is not observable, we replace them by <span class="math-container">$\hat{u}_{t}$</span>. Note that <span class="math-container">$\hat{u}_{t}=u_{t}-(\hat{\beta}-\beta)^{\top}X_{t}$</span>, therefore despite the true parameter before <span class="math-container">$X_{t}$</span> was exactly 0, now it becomes weird
<span class="math-container">$$\hat{u}_{t}=-(\hat{\beta}-\beta)^{\top}X_{t}+\delta\hat{u}_{t-1}+\varepsilon_{t}.$$</span>
This equation is kind of beyond traditional econometric setting since the pseudo-true value <span class="math-container">$(\hat{\beta}-\beta)$</span> itself is changing and converging to 0. For now, let's first take sample size as fixed.</p>
<p>By direct computation, if <em><span class="math-container">$X_{t}$</span> is omitted in above regression</em>, we have <span class="math-container">$\hat{\delta}=\delta-A_{T}+(\sum_{t=2}^{T}\hat{u}_{t-1}^{2})^{-1}\sum_{t=2}^{T}\hat{u}_{t-1}\varepsilon_{t}$</span>, where the bias term of <span class="math-container">$\hat{\delta}$</span> is
<span class="math-container">$$A_{T}=\biggl(\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}^{2}\biggr)^{-1}\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}X_{t}^{\top}(\hat{\beta}-\beta).$$</span></p>
<p>Now let's see what happens when <span class="math-container">$T\to\infty$</span>, follow some standard procedure, we have
<span class="math-container">$$\begin{align*}
\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}^{2}&amp;=\mathbb{E}[u_{t}^{2}]+o_{p}(1)\\
\frac{1}{T}\sum_{t=2}^{T}\hat{u}_{t-1}X_{t}^{\top}&amp;=\mathbb{E}[u_{t-1}X_{t}^{\top}]+o_{p}(1)\\
\sqrt{T}(\hat{\beta}-\beta)&amp;\overset{\mathcal{L}}{\to}Z_{1}.
\end{align*}$$</span>
Combine above result, the bias term<sup>*</sup>
<span class="math-container">$$A_{T}=O_{p}(1)O_{p}(1)O_{p}(T^{-1/2})=\color{red}{O_{p}(T^{-1/2})},$$</span>
which indicates <span class="math-container">$\hat{\delta}$</span> is <strong>still consistent</strong>. The problem arises when we apply the traditional <span class="math-container">$t$</span>-test for <span class="math-container">$\delta$</span>, CLT gives
<span class="math-container">$$\sqrt{T}(\hat{\delta}-\delta+A_{T})\overset{\mathcal{L}}{\to}Z_{2}.$$</span>
Here <span class="math-container">$Z_{2}$</span> is the normal distribution that <span class="math-container">$t$</span>-test based on. However, when normalized by <span class="math-container">$\sqrt{T}$</span>, the bias term <span class="math-container">$A_{T}$</span> <strong>will show up in the limiting distribution of <span class="math-container">$\sqrt{T}(\hat{\delta}-\delta)$</span></strong> since <span class="math-container">$\sqrt{T}A_{T}=O_{p}(1)$</span>, hence the <span class="math-container">$t$</span>-test no longer works.</p>
<hr />
<p><sup>*: If you strengthen contemporaneous exogeneity <span class="math-container">$\mathbb{E}[u_{t}|X_{t}]$</span> to <em>strict exogeneity</em> <span class="math-container">$\mathbb{E}[u_{t}|X_{1},\ldots,X_{T}]=0$</span>, that <span class="math-container">$X_{t}$</span> is not allowed to be correlated with past innovation, then <span class="math-container">$A_{T}=o_{p}(T^{-1/2})$</span> since the second term becomes <span class="math-container">$o_{p}(1)$</span>, and <span class="math-container">$t$</span>-test is still valid. This is what the textbook says.</sup></p>

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2022-02-07T09:12:26.557Z
This helps a lot. So the concept is that the regression in step (2) excludes the first observation?
Michael Gmeiner
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2022-02-07T11:49:28.863Z
Exactly. You see the first available regression sample would be $\hat{u}_{2}=\delta_{0}+\delta_{1}\hat{u}_{1}+\cdots$, on the LHS, we can't use $\hat{u}_{1}$ as dependent variable, since otherwise RHS will contain a non-existing term $\hat{u}_{0}$.
Q9y5
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2022-02-07T12:38:51.397Z
Okay, I understand that, but a questions still lingers.
If $x_t$ is correlated with $u_t$, then our original OLS estimates are biased anyway. I feel like this entire procedure relies on contemporaneous exogeneity, in which case there is a theoretical reason that omitting $x_t$ can't cause bias in the step 2 regression.
Am I missing something?
Michael Gmeiner
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2022-02-08T05:38:35.287Z
@MichaelGmeiner See my new updates. The point is noting that $\mathbb{E}[\hat{u}_{t}X_{t}]=\mathbb{E}[u_{t}X_{t}]-\mathbb{E}[(\hat{\beta}-\beta)^{\top}X_{t}]$. This means even under contemporaneous exogeneity, the "true parameter" of regressing $\hat{u}_{t}$ on $X_{t}$ is no longer 0 but something converging to 0 at a root-$T$ rate. Omitting this kind of variable won't cause inconsistency but causes problems in doing inference.
Q9y5
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2022-02-06T16:02:05.647Z
2022-02-08T05:15:20.090Z
2022-02-08T05:15:20.090Z
Q9y5
<p>This is from Introductory Econometrics (Wooldridge) 5th Edition page 420.</p>
<p>Consider:
<span class="math-container">$$y_t =\beta_0 +\beta_1 x_{1t}+\beta_2 x_{2t}+...+\beta_k x_{kt}+u_t$$</span></p>
<p>where <span class="math-container">$Cov(u_t, x_{jt})=0$</span> for all <span class="math-container">$j$</span> but perhaps <span class="math-container">$Cov(u_t, x_{jt-1}) \ne 0$</span> for some <span class="math-container">$j$</span>. We are concerned about autocorrelation, <span class="math-container">$$u_t =\rho u_{t-1}+v_t$$</span></p>
<p>To test for autocorrelation, the method proposed in-text is:</p>
<p>(1) Estimate the regression equation of interest and obtain residuals, <span class="math-container">$\hat{u_t}$</span>.</p>
<p>(2) Run the regression: <span class="math-container">$$\hat{u_t}=\delta_0 +\delta_1 \hat{u}_{t-1}+\delta_2 x_{1t}+...+\delta_{k+1} x_{kt} +\varepsilon_t$$</span></p>
<p>(3) Perform a standard t-test for significance of <span class="math-container">$\delta_1$</span>.</p>
<p>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.</p>
<p>The text says the inclusion is because of the potential failure of strict exogeneity (i.e., because <span class="math-container">$Cov(u_t, x_{jt-1}) \ne 0$</span>) and that including the <span class="math-container">$x$</span> is necessary to attain a correct t-distribution of our t-stat. This is stated without justification or proof.</p>
<p>What is the underlying math of this idea?</p>

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2022-01-05T17:23:17.167Z
what sort of proof you are looking for? The auxiliary regression (step 2) is regression like any other regression so the Gauss-Markov assumptions must hold, if you do not control for $\mathbf{x}$ then you will have omitted variable bias since some of the error is explained by $\mathbf{x}$. I guess one can provide proof for why the GM assumptions are needed but I feel that is not what you are looking for
1muflon1
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2022-01-05T17:43:12.157Z
that procedure seems odd to me because of the following: Suppose that that $u_t = \rho u_{t-1} + v_t$ does hold. Then this implies ( atleast according to the problem statement ) that $Cov(u_t, x_{j (t-1)}) \neq 0$. But it also implies that   $Cov(u_{t-1}, x_{j (t-1)}) \neq 0$ which is the same as $Cov(u_t, x_{j t}) \neq 0$. But  $Cov(u_t, x_{j t})$  was assumed to be zero in the first place !!!! Wooldridge of course knows what he's talking about so I must be missing something.
mark leeds
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2022-01-06T08:22:28.613Z
@1muflon1. I am not convinced there is OVB. $x$ is uncorrelated with $\hat{u}_t$ as a property of $\hat{u}_t$ being residuals from a regression of $y$ on $x$. If a variable is uncorrelated with the outcome, it can't cause OVB.
Michael Gmeiner
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2022-01-07T18:44:13.167Z
@mark leeds: A serially correlated error term does not imply anything regarding the dependence between $u_t$ and $x_s$ and is compatible with either $cov(u_t,x_s)=0$ or $cov(u_t,x_s) \neq 0$ for any subscript $s.$
Bertrand
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2022-01-08T16:16:23.287Z
@Bertrand: If the $u_t$ is serially correlated, then this means that $u_t$ is correlated with $u_{t-1}$. So, if $u_t$ is correlated with $x_{j(t-1)}$ and also correlated with $u_{t-1}$, then this should imply that $u_{t-1}$ and $x_{j(t-1)}$ are correlated, no ? If this is not true, can you give me a reference.  Intuitively,( to me atleast ), if A is correlated with B and A is correlated with C, then shouldn't B be correlated with C ? Thanks.
mark leeds
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2022-01-08T16:31:30.950Z
@mark leeds: correlation is not "transitive" in the sense that if A is correlated with B and B is correlated with C, then A is not necessarily correlated with C. Consider for instance the case of $X$ endogenous, so that $cov(X,u) \neq 0$, wlg take $cov(X,u) > 0$. A good instrument $Z$ is typically such that $cov(X,Z) >0$ (or the other way round, but correlated with $X$) and such that $cov(u,Z)=0$.  So $u$ is correlated with $X$ and $X$ is correlated with $Z$, but this does not imply that $u$ is correlated with $Z$. The textbook of Wooldridge is an excellent reference.
Bertrand
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2022-01-08T18:59:32.087Z
@Bertrand: It's appreciated. I will check out Wooldridge's text. You are referring to his black text titled "cross section and panel data" or something close to that, right. Thanks a lot for wisdom and enlightenment.
mark leeds
<econometrics><time-series><autocorrelation>
Testing for serial correlation with General Regressors
Michael Gmeiner
# Testing for serial correlation with General Regressors

(Asked by: username_0 on 2022-01-05)

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?

**Tags:** econometrics, time-series, autocorrelation

> **Comments:**
> **username_1**: what sort of proof you are looking for? The auxiliary regression (step 2) is regression like any other regression so the Gauss-Markov assumptions must hold, if you do not control for $\mathbf{x}$ then you will have omitted variable bias since some of the error is explained by $\mathbf{x}$. I guess one can provide proof for why the GM assumptions are needed but I feel that is not what you are looking for  
> **username_2**: that procedure seems odd to me because of the following: Suppose that that $u_t = \rho u_{t-1} + v_t$ does hold. Then this implies ( atleast according to the problem statement ) that $Cov(u_t, x_{j (t-1)}) \neq 0$. But it also implies that   $Cov(u_{t-1}, x_{j (t-1)}) \neq 0$ which is the same as $Cov(u_t, x_{j t}) \neq 0$. But  $Cov(u_t, x_{j t})$  was assumed to be zero in the first place !!!! Wooldridge of course knows what he's talking about so I must be missing something.  
> **username_0**: @username_1. I am not convinced there is OVB. $x$ is uncorrelated with $\hat{u}_t$ as a property of $\hat{u}_t$ being residuals from a regression of $y$ on $x$. If a variable is uncorrelated with the outcome, it can't cause OVB.  
> **username_3**: @username_2: A serially correlated error term does not imply anything regarding the dependence between $u_t$ and $x_s$ and is compatible with either $cov(u_t,x_s)=0$ or $cov(u_t,x_s) \neq 0$ for any subscript $s.$  
> **username_2**: @username_3: If the $u_t$ is serially correlated, then this means that $u_t$ is correlated with $u_{t-1}$. So, if $u_t$ is correlated with $x_{j(t-1)}$ and also correlated with $u_{t-1}$, then this should imply that $u_{t-1}$ and $x_{j(t-1)}$ are correlated, no ? If this is not true, can you give me a reference.  Intuitively,( to me atleast ), if A is correlated with B and A is correlated with C, then shouldn't B be correlated with C ? Thanks.  
> **username_3**: @username_2: correlation is not "transitive" in the sense that if A is correlated with B and B is correlated with C, then A is not necessarily correlated with C. Consider for instance the case of $X$ endogenous, so that $cov(X,u) \neq 0$, wlg take $cov(X,u) > 0$. A good instrument $Z$ is typically such that $cov(X,Z) >0$ (or the other way round, but correlated with $X$) and such that $cov(u,Z)=0$.  So $u$ is correlated with $X$ and $X$ is correlated with $Z$, but this does not imply that $u$ is correlated with $Z$. The textbook of Wooldridge is an excellent reference.  
> **username_2**: @username_3: It's appreciated. I will check out Wooldridge's text. You are referring to his black text titled "cross section and panel data" or something close to that, right. Thanks a lot for wisdom and enlightenment.  

---

## 2 answers

### Answer by username_4 on 2022-02-06. Score: 3 | Accepted answer

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.

---

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.

---

\*: 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.

> **Comments:**
> **username_0**: This helps a lot. So the concept is that the regression in step (2) excludes the first observation?  
> **username_4**: Exactly. You see the first available regression sample would be $\hat{u}_{2}=\delta_{0}+\delta_{1}\hat{u}_{1}+\cdots$, on the LHS, we can't use $\hat{u}_{1}$ as dependent variable, since otherwise RHS will contain a non-existing term $\hat{u}_{0}$.  
> **username_0**: Okay, I understand that, but a questions still lingers. If $x_t$ is correlated with $u_t$, then our original OLS estimates are biased anyway. I feel like this entire procedure relies on contemporaneous exogeneity, in which case there is a theoretical reason that omitting $x_t$ can't cause bias in the step 2 regression. Am I missing something?  
> **username_4**: @username_0 See my new updates. The point is noting that $\mathbb{E}[\hat{u}_{t}X_{t}]=\mathbb{E}[u_{t}X_{t}]-\mathbb{E}[(\hat{\beta}-\beta)^{\top}X_{t}]$. This means even under contemporaneous exogeneity, the "true parameter" of regressing $\hat{u}_{t}$ on $X_{t}$ is no longer 0 but something converging to 0 at a root-$T$ rate. Omitting this kind of variable won't cause inconsistency but causes problems in doing inference.  

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### Answer by username_3 on 2022-01-06. Score: 5

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.

> **Comments:**
> **username_0**: Thank you for the answer.  I am a little surprised Wooldridge does this though, because $\hat{u_t}$ is uncorrelated with $x_t$ by properties of OLS residuals. Is it more the principle that the regression is motivated by the theory of the true error, $u_t$?  And if the theory is that $x_t$ may be correlated with $u_t$, then contemporaneous exogeneity fails - which really seems like an odd assumption to base a test on.  
> **username_3**: I agree with you on the fact that, if there is violation of strict exogeneity, then it is better (in order to avoid inconsistent OLS estimates), to obtain the $\widehat{u}_t$ using IV or GMM. What happens if $\widehat{u}_t$ is obtained after an inconsistent OLS? Is the approach still valid? Hmm. I cannot give a definite answer, but just note that strict exogeneity $E[u_t|X]=0$ does not imply that $E[u_t|X,u_{t-1}]=0.$ So it might still be justified to consider OLS residuals...  
> **username_2**: @username_3: You clearly know what you're talking about but I was just wondering if your answer addresses any of my confusion regarding the initial assumption then becoming invalid if the test for serial correlation rejects the null ? Thanks and, if not, don't worry about it. I've always found Wooldridge's writings to be somewhat confusing so I'll just remain confused !!!!!!!!!  

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