I have a linear regression model and because of heteroskedasticity or autocorrrelation I use HAC (Newey-West) estimates. This influences also p-values of significance t-tests of estimated coefficients as they are then different than the normal p-values would be. My question is: should this influence also the p-value of intercept or should this p-value remain the same?
Solved – Influence of HAC estimates to p-value of t-test
autocorrelationhacheteroscedasticityregressiont-test
Related Solutions
HAC procedures are just about providing consistent estimates of the standard errors. They do not change the estimation of the coefficients. If you have strict exogeneity with serial correlation, your coefficients are unbiased, but the standard errors are incorrect. HAC standard errors address the latter point.
As you allude to, this does not give efficient coefficient estimates. To achieve efficiency, in economics, at least, we typically use a Cochrane-Orcutt/Prais-Whinston procedure. This requires much stronger modeling assumptions to estimate the structure of the serial correlation, however.
They are analogous to Eicker-White heteroskedasticity robust standard errors. This procedure does not alter estimation, it only changes the estimates of the standard errors to ensure that they are consistent in the presence of heteroskedasticity. The efficient fix would be weighted least squares, but this requires modeling the form of the heteroskedasticity.
Question 3)
In notation to be understood as matrix-vector, assume that the correct specification is
$$y = X\beta + \gamma y_{-1}+ e$$
(where $X$ contains the constant and the $X_1$ variable and $e$ is white noise, and $E(e\mid X) =0$), but you specify and estimate instead
$$y = X\beta + u$$ i.e. without including the LAD, and so in reality $u =\gamma y_{-1}+ e$.
Then OLS estimation will give
$$\hat \beta = (X'X)^{-1}X'y = (X'X)^{-1}X'(X\beta + \gamma y_{-1}+ e) $$ $$= \beta + (X'X)^{-1}X'y_{-1}\gamma +(X'X)^{-1}X'e$$
The expected value of the estimator is
$$E(\hat \beta) = \beta + E\Big[(X'X)^{-1}X'y_{-1}\gamma\Big] +E\Big[(X'X)^{-1}X'e\Big]$$ and using the law of iterated expectations
$$E(\hat \beta) = \beta + E\Big(E\Big[(X'X)^{-1}X'y_{-1}\gamma\Big]\mid X\Big) +E\Big(E\Big[(X'X)^{-1}X'e\Big]\mid X\Big)$$
$$= \beta + E\Big((X'X)^{-1}X'E\Big[y_{-1}\gamma\mid X\Big]\Big) +E\Big((X'X)^{-1}X'E\Big[e\mid X\Big]\Big)$$
$$=\beta + E\Big((X'X)^{-1}X'E\Big[y_{-1}\gamma\mid X\Big]\Big) + 0 $$ the last term being zero per our assumptions. But $E\Big[y_{-1}\gamma\mid X\Big] \neq 0$, because $X$ contains all the regressors (from all time periods), and so there is correlation with the LAD vector. Therefore $E(\hat \beta) \neq \beta$. In other words, ignoring the lag dependent variable will not make the estimator unbiased, as long as $\gamma \neq 0$, i.e. as long as the LAD does belong to the regression.
Question 1)
Assume now that you specify correctly, and denote $Z$ the matrix containing also the LAD.
Here (using the same steps as before)
$$\hat \beta = \beta + (Z'Z)^{-1}Z'e$$
and $$E(\hat \beta) = \beta + E\Big((Z'Z)^{-1}Z'E\Big[e\mid Z\Big]\Big)$$
But is $e$ (the vector) independent of $Z$? No, because $Z$ contains the LAD from all time periods bar the most recent, while $e$ contains the errors from all time periods bar the first. So even if $e$ is not serially correlated, it is correlated with the vector $y_{-1}$. So indeed, the last term is not zero and $$E(\hat \beta) \neq \beta$$ the OLS estimator is biased.
But the OLS estimator will be consistent if indeed the inclusion of the LAD eliminates serial correlation, because (using the properties of the plim operator)
$$\operatorname{plim}\hat \beta = \beta + \operatorname{plim}\left(\frac 1{n-1} Z'Z\right)^{-1}\cdot \operatorname{plim}\left(\frac 1{n-1}Z'e\right)$$
Part of the standard assumptions (and rather "easily" satisfied), is that the first plim of the product converges to something finite. The second plim written explicitly is (and using the stationarity assumption to invoke the LLN)
$$\operatorname{plim}\left(\frac 1{n-1}Z'\mathbf e\right) = \left[\begin{matrix} \operatorname{plim}\frac 1{n-1}\sum_{i=2}^ne_i \\ \operatorname{plim}\frac 1{n-1}\sum_{i=2}^nx_{i}e_i \\ \operatorname{plim}\frac 1{n-1}\sum_{i=2}^ny_{i-1}e_i \\ \end{matrix}\right] \rightarrow\left[\begin{matrix} E(e_i) \\ E(x_{i}e_i) \\ E(y_{i-1}e_i) \\ \end{matrix}\right]\; \forall i$$
$E(e\mid X) = 0 \Rightarrow E(e_i) = 0$, and also that $E(x_{i}e_i)=0$, for all $i$.
Finally, IF serial correlation has been removed, then $E(y_{i-1}e_i) =0$ also. So this plim goes to zero and therefore
$$\operatorname{plim}\hat \beta = \beta$$ i.e. the OLS estimator is indeed consistent in this case. So the "summary" is correct.
Question 2)
The full sentence from Wooldridge is
"It is also valid to use the SC-robust standard errors in models with lagged dependent variables assuming, of course, that there is good reason for allowing serial correlation in such models".
meaning, when we have good reasons to believe that the inclusion of lagged dependent variables does not fully remove autocorrelation. And it seems we got ourselves a Catch-22: if serial correlation (SC) has been removed, why use SC-robust std errors? And if serial correlation has not been removed, our OLS estimator will be inconsistent, so in such a case is it meaningful/useful/appropriate to use asymptotic inference? Well, it appears that if we do suspect that SC still exists, it is better to try to do something about it, regardless. But your comment has merit, and I would suggest to contact Wooldridge directly on the matter, in order to get an authoritative answer.
Best Answer
Keep in mind what the $t$-ratio is: it's $\hat{\alpha}/S(\hat{\alpha})$, where $\hat{\alpha}$ is the estimated intercept and $S(\hat{\alpha})$ its standard error. What you do by choosing to apply HAC is using a different function $S(\cdot)$.
Assume that $\hat{\alpha}$ is the first regressor in the regressor matrix $X$. Then, while for OLS
$S_{OLS}(\hat{\alpha}) = \sqrt{\text{Var}_{OLS}(\hat{\alpha})_{(1,1)} }= \sqrt{s^2(X'X)^{-1}_{(1,1)}} = \sqrt{(s^2)(1/n)}$,
this is not the case if you use HAC standard errors. They are constructed diffferently (see e.g. here: https://en.wikipedia.org/wiki/Heteroscedasticity-consistent_standard_errors). In particular, for the $i$th estimated regression residual $\hat{u_i} = Y_i - X_i\hat{b}$,
$S_{HAC}(\hat{\alpha}) = \sqrt{\left( (X'X)^{-1}(X'\text{diag}(\hat{u_1}, ... \hat{u_n})X)(X'X)^{-1}\right)_{(1,1)} }$.
Notice that if we replaced $\text{diag}(\hat{u_1}, ... \hat{u_n})$ by $I_n \cdot s^2$, the HAC would become identical to the OLS standard error estimate. But unless $\hat{u_1} = ... = \hat{u_n}$, this will not happen.
This applies to the intercept as it does to all other coefficients. Hence, the denominator of your $t$-ratio changes depending on the choice of OLS vs HAC unless $\hat{u_1} = ... = \hat{u_n}$, which happens with probability $0$.