Consider the following regression equation * \begin{equation} \label{eq:1} C_{t} = \beta_{1} + \lambda C_{t-1} + \epsilon_{t} \end{equation} and let \begin{equation} \label{eq:3} \epsilon_{t} = \rho\epsilon_{t-1} + u_{t} \end{equation} where the error component \begin{equation} u_{t}\end{equation} is iid with mean 0 and constant variance, and \begin{equation} \label{eq:4} E(u_{t}\mid C_{t-1},\epsilon_{t-1}) = 0 \end{equation} Is the OLS estimator of the coefficients in (*) unbiased and consistent under this error specification? Why?
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1 Answer
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By the definition of OLS estimator we can rewrite the $\hat\lambda$ as
$\hat\lambda=argmin \sum_{i=1}^{n}\epsilon_t^2$
From first order condition, we get $\hat\lambda=\frac{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})(C_t-\bar{C_t} )}{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})^2} $
You can get more simple expression as $\hat\lambda=\lambda+\frac{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})(\epsilon_t-\bar{\epsilon_t})}{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})^2}$
Since $E[(C_{t-1}-\bar{C}_{t-1})(\epsilon_t-\bar{\epsilon_t})]=0$, $E(\hat\lambda)=\lambda$.(unbiased)
Now apply Law of Large Numbers to show consistency.
$\hat\lambda =\frac{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})(C_t-\bar{C_t} )}{\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})^2}=\frac{1/n\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})(C_t-\bar{C_t} )}{1/n\sum_{i}^{n}(C_{t-1}-\bar{C}_{t-1})^2}$
By LLN, $\hat\lambda =\frac{Cov(C_t, C_t-1)}{Var(C_t-1)}=\frac{Cov(\beta_1 +\lambda C_t-1 +\epsilon_t, C_t-1)}{Var(C_t-1)}=\frac{\lambda Var(C_t-1)+Cov(\epsilon_t, C_t-1)}{Var(C_t-1)}=\lambda $.(consistent)
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$\begingroup$ could you explain why its unbiased and consistent with more words? $\endgroup$ Apr 22, 2022 at 4:47
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$\begingroup$ @JohnWilliams I edited my solution. You need to know the concept of LLN to prove consistency. $\endgroup$– guestApr 22, 2022 at 5:14
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$\begingroup$ The AR(1) is not unbiased for the lagged coefficient even in the case where the error term is purely iid. See this link for a couple of different and insightful explanations. The link doesn't cover whether it's consistent for the AR(1) error term case but I don't think it is. stats.stackexchange.com/questions/240383/… $\endgroup$ Apr 25, 2022 at 15:53