<p>By the definition of OLS estimator we can rewrite the <span class="math-container">$\hat\lambda$</span> as<br />
<span class="math-container">$\hat\lambda=argmin \sum_{i=1}^{n}\epsilon_t^2$</span><br />
From first order condition, we get <span class="math-container">$\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} $</span><br />
You can get more simple expression as <span class="math-container">$\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}$</span><br />
Since <span class="math-container">$E[(C_{t-1}-\bar{C}_{t-1})(\epsilon_t-\bar{\epsilon_t})]=0$</span>, <span class="math-container">$E(\hat\lambda)=\lambda$</span>.(unbiased)</p>
<p>Now apply Law of Large Numbers to show consistency.</p>
<p><span class="math-container">$\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}$</span><br />
By LLN, <span class="math-container">$\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 $</span>.(consistent)</p>

CC BY-SA 4.0
2022-04-22T04:29:41.180Z
what does lln mean?
John Williams
CC BY-SA 4.0
2022-04-22T04:37:26.143Z
@JohnWilliams LLN means Law of Large Numbers
guest
CC BY-SA 4.0
2022-04-22T04:47:13.737Z
could you explain why its unbiased and consistent with more words?
John Williams
CC BY-SA 4.0
2022-04-22T05:14:22.130Z
@JohnWilliams I edited my solution. You need to know the concept of LLN to prove consistency.
guest
CC BY-SA 4.0
2022-04-25T15:53:05.713Z
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. https://stats.stackexchange.com/questions/240383/why-is-ols-estimator-of-ar1-coefficient-biased
mark leeds
CC BY-SA 4.0
2022-04-22T04:14:13.873Z
2024-11-06T00:34:57.430Z
2024-11-06T00:34:57.430Z
guest
<p>Consider the following regression equation * <span class="math-container">\begin{equation} \label{eq:1}
C_{t} = \beta_{1} + \lambda   C_{t-1} + \epsilon_{t}
\end{equation}</span>
and let <span class="math-container">\begin{equation} \label{eq:3}
\epsilon_{t} = \rho\epsilon_{t-1} + u_{t}
\end{equation}</span> where the error component <span class="math-container">\begin{equation} u_{t}\end{equation}</span> is iid with mean 0 and constant
variance, and <span class="math-container">\begin{equation} \label{eq:4}
E(u_{t}\mid   C_{t-1},\epsilon_{t-1}) = 0
\end{equation}</span>
Is the OLS estimator of the coefficients in (*)
unbiased and consistent under this error specification? Why?</p>

<econometrics><regression><statistics>
error specification of OLS regression models
John Williams
# error specification of OLS regression models

(Asked by: username_0 on 2022-04-22)

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?

**Tags:** econometrics, regression, statistics


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## 1 answers

### Answer by username_1 on 2022-04-22. Score: 3

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)

> **Comments:**
> **username_0**: what does lln mean?  
> **username_1**: @JohnWilliams LLN means Law of Large Numbers  
> **username_0**: could you explain why its unbiased and consistent with more words?  
> **username_1**: @JohnWilliams I edited my solution. You need to know the concept of LLN to prove consistency.  
> **username_2**: 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. https://stats.stackexchange.com/questions/240383/why-is-ols-estimator-of-ar1-coefficient-biased  

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