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Please explain why the OLS estimator is consistent and unbiased withregard to the equation below

\begin{equation} \label{eq:1} C_{t} = \beta_{1} + \lambda C_{t-1} + \epsilon_{t} \end{equation}


1 Answer 1


We can write $\varepsilon_t=\sum_{s=0}^\infty \rho^su_{t-s}$.

Thus, $Cov(\varepsilon_t, C_{t-1}) = Cov(\sum_{s=0}^\infty \rho^su_{t-s}, C_{t-1})$.

We are given that $E[u_t|C_{t-1},\varepsilon_{t-1}]=0$. Thus, $Cov(u_t, C_{t-1})=0$. To conclude there is consistency also requires that $Cov(u_{t-s},C_{t-1})=0$ for all $s>0$.

OLS is definitely biased. For unbiasedness, we need $E[u_t|C]=0$ where $C$ is a vector of $C_t$ at all time periods. This is impossible because $u_t$ is definitely correlated with $C_t$ (at the same time period).

  • $\begingroup$ and one last follow up: $\endgroup$ Apr 25, 2022 at 10:16
  • $\begingroup$ Consider the following equation * \begin{equation} \label{eq:1} C_{t} = \beta_{1} + \lambda Y_{t} + \epsilon_{t} \end{equation} where, \begin{equation} \label{eq:2} E(\epsilon_{t}\mid Y_{t}) = 0 \end{equation} \begin{equation} \label{eq:3} \epsilon_{t} = \rho\epsilon_{t-1} + u_{t} \end{equation} and the error component Ut, is iid with mean 0, constant variance, and\begin{equation} \label{eq:4} E(u_{t}\mid Y_{t},\epsilon_{t-1}) = 0 \end{equation} questions: (i) Is the OLS estimator of the coefficients in (*) unbiased and consistent? Explain. $\endgroup$ Apr 25, 2022 at 10:17
  • $\begingroup$ I would assume the ols estimator is unbiased and consistent because Yt covers all time periods unlike c(t-1). $\endgroup$ Apr 25, 2022 at 10:18
  • $\begingroup$ Also how would you conduct hypothesis tests in this case? Why? $\endgroup$ Apr 25, 2022 at 10:50
  • $\begingroup$ It's still possible that $Cov(u_t, Y_s)\ne 0$ for some $t\ne s$, which would cause bias. For inference, you can still do a standard t-test. $\endgroup$ Apr 25, 2022 at 16:14

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