Is this a linear estimator? 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.
 A: 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 $y$.
A: To check the definition, you want to bring $\widehat β_1$ to the form $$\widehat β_1=c_1y_1+\dots+c_ny_n$$ So, to simplify notation, let $d_i:=\dfrac{1}{n}\dfrac{1}{x_i-\bar x}$ for $i=1,2,\dots,n$ and write \begin{align}\hat β_1&=\frac{1}{n}\sum_{i=1}^n\frac{y_i-\bar y}{x_i-\bar x}=\sum_{i=1}^nd_i(y_i-\bar y)=\frac1n\sum_{i=1}^n\left(d_iy_i-\frac{d_i}{n}\sum_{j=1}^ny_j\right)\\[0.2cm]&=d_1y_1-\frac{d_1}{n}(y_1+\dots+y_n)+d_2y_1-\frac{d_2}{n}(y_1+\dots+y_n)+\dots\\[0.2cm]&=\left(d_1y_1-\frac{d_1}{n}y_1-\frac{d_2}{n}y_1-\dots\right)+\left(d_2y_2-\frac{d_1}{n}y_2-\frac{d_2}{n}y_2-\dots\right)+\dots\\[0.2cm]&=\sum_{i=1}^n \left(d_i\left(1-\frac1n\right)-\frac1n\sum_{j=1}^nd_j\right) y_i=:\sum_{i=1}^nc_iy_i\end{align} which shows that the definition is satisfied with $$c_i:=d_i\left(1-\frac1n\right)-\frac1n\sum_{j=1}^nd_j$$ for $i=1,2,\dots,n$.
