I would like to prove if
$$\hat{\beta_1} = \frac{1}{n}\sum_{i=0}^n \frac{y_i-\bar{y}}{x_i-\bar{x}}$$
where $y = {\beta_0} + {\beta_1}x+ u$ and $\Bbb E(u\mid x) = 0$, is a linear estimator or not. But I was stuck at not knowing how to deal away $\bar y$.
According to definition from wikipedia
A linear estimator of $\beta_j$ is a linear combination $\widehat{\beta_j} = c_{1j}y_1+\cdots+c_{nj}y_n$ in which the coefficients $c_{ij}$ are not allowed to depend on the underlying coefficients $\beta{j}$.
Could anyone help? Thanks in advance.
Best Answer
If $z=y'+y''$,
$$\frac{1}{n}\sum_{i=0}^n \frac{z_i-\bar{z}}{x_i-\bar{x}}= \frac{1}{n}\sum_{i=0}^n \frac{(y'_i+y''_i)-\overline{y'+y''}}{x_i-\bar{x}}= \hat\beta'_1+\hat\beta''_1 $$
If $z=\lambda y$,
$$\frac{1}{n}\sum_{i=0}^n \frac{z_i-\bar{z}}{x_i-\bar{x}}= \frac{1}{n}\sum_{i=0}^n \frac{\lambda y_i-\lambda \bar{y}}{x_i-\bar{x}}= \lambda\hat{\beta_1} $$
hence $\hat\beta_1$ is clearly linear in $y$.