1) Consider a standard AR(2) process. When is the maximum likelihood estimator identical to the OLS estimator? (a) when $\varepsilon $~ (N 0,$\Sigma^2) (b) always?
I'm thinking (a), but that I also need to add in large samples this would be correct.
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$\begingroup$ Indeed, OLS estimator is the same as ML estimator if you assume error terms are a Gaussian white noise. No large samples required off the top of my head. $\endgroup$– QuantupleCommented Mar 19, 2018 at 14:57
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$\begingroup$ Actually, I think if the errors are correlated then your estimator is biased but consistent. $\endgroup$– phdstudentCommented Mar 19, 2018 at 15:17
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$\begingroup$ @phdstudent thanks for the additional info. This does ring a bell now that you mention it. $\endgroup$– QuantupleCommented Mar 19, 2018 at 15:39
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1 Answer
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In an simpler AR(1) case (you can generalize to AR(2)) we have that:
\begin{equation*} y_{t}=\beta y_{t-1}+\epsilon _{t}, \end{equation*} Even under the assumption $E(\epsilon_{t}y_{t-1})=0$ we have that \begin{equation*} E(\epsilon_ty_{t})=E(\epsilon_t(\beta y_{t-1}+\epsilon _{t}))=E(\epsilon _{t}^{2})\neq 0. \end{equation*} But, $y_t$ is also a regressor for future values in ain AR model, as $y_{t+1}=\beta y_{t}+\epsilon_{t+1}$.
Or in other words:
$$\hat\beta =\beta + \frac{\sum_{t=2}^Ty_{t-1}\varepsilon_t}{\sum_{t=2}^Ty_{t-1}^2}$$
For unbiasedness we need
$$E\frac{\sum_{t=2}^Ty_{t-1}\varepsilon_t}{\sum_{t=2}^Ty_{t-1}^2}=0.$$
But for that we need that $E(\varepsilon_t|y_{1},...,y_{T-1})=0,$ for each $t$. For AR(1) model this clearly fails, since $\varepsilon_t$ is related to the future values $y_{t},y_{t+1},...,y_{T}$.
Take a look of to this example with gaussian errors: http://www.alexchinco.com/bias-in-time-series-regressions/
answered Mar 19, 2018 at 15:31