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Given the following model:

$$ y_t = \alpha x_t + \epsilon_t~~~~,~~~\epsilon_t \sim NID(0,\sigma^2) ~\text{and}~~x_t = y_t + z_t$$

Given that $z_t$ is a non covariate variable with $\epsilon_t$, how is it possible to derive the following unbiased OLS estimator of $\alpha$:

$$ 1 - \dfrac{\sum_{t=1}^T z_t^2}{\sum_{t=1}^T z_t y_t + \sum_{t=1}^T z_t^2} $$

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Substitute $x_t$ in the model to obtain $y_t=\alpha y_t + \alpha z_t + \varepsilon_t$. Manipulating, $y_t = \frac{\alpha}{1-\alpha}z_t + \frac{1}{1-\alpha}\varepsilon_t$. Rename the errors $\delta_t=\frac{1}{1-\alpha}\varepsilon_t$, then $\delta_t\sim NID(0,\tau^2)$, where $\tau^2=\frac{\sigma^2}{(1-\alpha)^2}$, so homoscedasticidy still holds. Rename $\beta=\frac{\alpha}{1-\alpha}$. You now have the model $y_t=\beta z_t+\delta_t$, with the usual assumptions. The OLS estimator of $\beta$ is $$\hat{\beta}=\frac{\sum_{t=1}^T z_t y_t}{\sum_{t=1}^T z_t^2}.$$ Hence, $$\frac{\hat{\alpha}}{1-\hat{\alpha}}=\frac{\sum_{t=1}^T z_t y_t}{\sum_{t=1}^T z_t^2}.$$ Solving for $\hat{\alpha}$, $$\hat{\alpha}=\frac{\sum_{t=1}^T z_t y_t}{\sum_{t=1}^T z_t y_t+\sum_{t=1}^T z_t^2}=\frac{\sum_{t=1}^T z_t y_t + \sum_{t=1}^T z_t^2 - \sum_{t=1}^T z_t^2}{\sum_{t=1}^T z_t y_t+\sum_{t=1}^T z_t^2},$$

so $$\hat{\alpha}=1-\frac{\sum_{t=1}^T z_t^2}{\sum_{t=1}^T z_t y_t+\sum_{t=1}^T z_t^2}.$$

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  • $\begingroup$ Thanks a lot. The last thing, how you pass from the expression above "Solving for a" to the expression below "Solving for a"? $\endgroup$
    – Archimede
    Commented Mar 14, 2017 at 8:58
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    $\begingroup$ Multiply both sides by $1-\hat{\alpha}$ and by $\sum_{t=1}^T z_t^2$ and rearrange terms. And sorry for mixing the $\hat{\alpha}$ notation and the $a$ notation, let me correct it. $\endgroup$
    – Anna SdTC
    Commented Mar 14, 2017 at 9:02
  • $\begingroup$ This doesn't actually answer the question of whether or not the estimator is unbiased ... it just defines the estimator. $\endgroup$
    – Igor
    Commented Mar 14, 2017 at 9:16
  • $\begingroup$ No the question is how to derive it. $\endgroup$
    – Archimede
    Commented Mar 14, 2017 at 9:26
  • $\begingroup$ @AnnaSdTC please if you can answer to a similar question stats.stackexchange.com/questions/267149/… $\endgroup$
    – Archimede
    Commented Mar 14, 2017 at 9:31

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