# Time Series Multiple Choice

1) Consider a standard AR(2) process. When is the maximum likelihood estimator identical to the OLS estimator? (a) when $\varepsilon$~ (N 0,$\Sigma^2) (b) always? I'm thinking (a), but that I also need to add in large samples this would be correct. • Indeed, OLS estimator is the same as ML estimator if you assume error terms are a Gaussian white noise. No large samples required off the top of my head. Mar 19 '18 at 14:57 • Actually, I think if the errors are correlated then your estimator is biased but consistent. Mar 19 '18 at 15:17 • @phdstudent thanks for the additional info. This does ring a bell now that you mention it. Mar 19 '18 at 15:39 ## 1 Answer In an simpler AR(1) case (you can generalize to AR(2)) we have that: \begin{equation*} y_{t}=\beta y_{t-1}+\epsilon _{t}, \end{equation*} Even under the assumption$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/ 