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I currently reading Whitney Newey and Kenneth West's paper "a simple, positive semidefinite, heteroskedasticity and autocorrelation consistent covariance matrix"

For a multi-regression linear model: \begin{equation} y_t = X_t \beta_t + e_t, \qquad t=1,\cdots, T \end{equation} where $y_t$ and $X_t$ are given, $\beta_t$ is a non-random but unobservable vector, $e_t$ is the noise term.

According to the paper, the noise covariance of $e_t$, $S_T$ can be estimated using sample errors $\hat{e}_t := y_t - X_t \hat{\beta}_t$, such that \begin{equation} \hat{S}_T = \hat{\Omega}_0 + \sum_{j=1}^m w_j\left[\hat{\Omega}_j \hat{\Omega}_j^\top\right], \end{equation} where \begin{equation} \hat{\Omega}_j := \frac{1}{T}\sum_{t=j+1}^T\hat{e}_t\hat{e}_{t-j}^\top. \end{equation} Can anyone explain to me the derivation of $\hat{S}_T$? Why $\hat{S}_T$ can be expressed as sum of the covariance matrices with different lags?

my first guess is that $e_t$ might follow MA(m) process, such that \begin{equation} e_t:= \hat{e}_t + \phi_1\hat{e}_{t-1} + \cdots + \phi_m\hat{e}_{t-m} \end{equation} but traditionally for MA(m) process, we assume $\mathbb{E}\{e_{t} e_{t-j}^\top\} = 0$ for nonzero lag $j$.

If we assume that $\mathbb{E}\{\hat{e}_{t} \hat{e}_{t-j}^\top\} \neq 0$ for all $j$, then $e_t$ more like a AR(m) process, hence the covariance matrix is given by \begin{equation} \mathbb{E}\{e_te_t^\top\} = \begin{bmatrix} 1 & \phi_1 & \cdots & \phi_m\end{bmatrix} \begin{bmatrix} \hat{\Omega}_0 & \hat{\Omega}_1 & \cdots & \hat{\Omega}_m \\ \hat{\Omega}_1 & \hat{\Omega}_0 & \cdots & \hat{\Omega}_{m-1}\\ \vdots & \vdots & \ddots & \vdots \\ \hat{\Omega}_m & \hat{\Omega}_{m-1} & \cdots & \hat{\Omega}_0 \end{bmatrix}\begin{bmatrix} 1 \\ \phi_1 \\ \vdots \\ \phi_m\end{bmatrix} \end{equation}

Stephen Ge
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  • Related: https://stats.stackexchange.com/questions/153444/what-is-the-long-run-variance/153543#153543 – Christoph Hanck Mar 15 '23 at 10:41
  • Thank you very much Christoph! it explains a lot. I have one more question, for the sample mean $\bar{Y}_T$, as T approches infinity, does $\bar{Y}_T$ equals expectation of $Y_t$ equals $\mu$? – Stephen Ge Mar 16 '23 at 03:38
  • Yes, provided what is known in time series analysis as (mean) ergodicity holds, so basically a law of large numbers. – Christoph Hanck Mar 16 '23 at 05:48
  • Thank you, but if $\bar{Y}T \rightarrow \mu$ as $T\rightarrow \infty$, why $\lim{T\rightarrow \infty} {T\mathbb{E}(\bar{Y}_T - \mu)}$ not equals to zero? – Stephen Ge Mar 16 '23 at 06:28
  • you are overlooking a square – Christoph Hanck Mar 16 '23 at 07:47
  • Sorry, I forgot to put square sign, but the point I confused is that as $T\rightarrow \infty$, should $\bar{Y}_T - \mu = 0$? – Stephen Ge Mar 16 '23 at 10:40
  • Yes, the sample mean converges to the expected value, but this is not what my link discusses (although that result over there implies consistency of the sample mean). – Christoph Hanck Mar 16 '23 at 12:09

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