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We know he fitted estimator for $\beta_1$ is $$ \hat{\beta_1} = \frac{\sum_{i=1}^{n}{(x_i-\bar{x})(y_i-\bar{y})}}{\sum_{i=1}^{n}{(x_i-\bar{x})^2}} $$

Now, given the following estimator: $$ \beta_1' = \frac{1}{n}\sum_{i=1}^{n}\frac{y_i-\bar{y}}{x_i-\bar{x}} $$

How can I find out its bias? How can I calculate its $MSE$? I'm not sure how to treat that random variable, what are its properties, etc.

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  • $\begingroup$ As it stands there is no way this question can be answered, as a bias needs to be assessed relative to some process having generated the data, which you do not state. Is it $y_i=\alpha+\beta x_i+u_i$ where something is assumed about the relationship between $u_i$ and $x_i$? $\endgroup$ Commented Feb 27, 2018 at 13:54
  • $\begingroup$ Exactly, I'm assuming simple linear regression such that $\hat{y_i}=\hat{\beta_0}+\hat{\beta_1}x_i + e_i$. where $e_i$ is normally distributed. $\endgroup$
    – galah92
    Commented Feb 27, 2018 at 14:37
  • $\begingroup$ The more interesting point than the distribution for $e$ is what you assume about (in)dependence between $x$ and $e$. $\endgroup$ Commented Feb 27, 2018 at 15:42

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That estimator is not biased.

\begin{eqnarray} E[\beta_1^\prime] &=& \frac{1}{n} \sum_{i=1}^n E \left[ \frac{y_i - \bar{y}}{x_i - \bar{x}}\right]\\ &=& \frac{1}{n} \sum_{i=1}^n \frac{E[y_i] - E[\bar{y}]}{x_i - \bar{x}} \\ &=& \frac{1}{n} \sum_{i=1}^n \frac{\beta_1 x_i - \beta_1 \bar{x}}{x_i - \bar{x}}\\ &=& \beta_1 \end{eqnarray}

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    $\begingroup$ There is a missing bar over x on the last denominator. $\endgroup$ Commented Feb 27, 2018 at 16:10
  • $\begingroup$ Why is the second equality true? $\endgroup$
    – galah92
    Commented Feb 27, 2018 at 16:32
  • $\begingroup$ @galah92 because expectation is linear and the x's are fixed. $\endgroup$
    – AdamO
    Commented Feb 27, 2018 at 17:13
  • $\begingroup$ @AdamO But why the $x$'s are fixed..? And are you relying on the fact that $E[y_i] = \beta_1x_i + \beta_0$? $\endgroup$
    – galah92
    Commented Feb 28, 2018 at 10:02

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