Question

Help on estimators please and thank you.

Eaniv hat Kese eshi arr equiml the hote averp art eau

Answer 1

$a.~\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})=\sum_{i=1}^nx_iy_i-n\bar{x}\bar{y}=\sum_{i=1}^nx_iy_i-n\left(\frac{\sum_{i=1}^nx_i}{n} \right )\left(\frac{\sum_{i=1}^ny_i}{n} \right )\\ =\sum_{i=1}^nx_iy_i-\frac{1}{n}\left(\sum_{i=1}^nx_i \right )\left(\sum_{i=1}^ny_i \right )..........(1)\\ \sum_{i=1}^n(x_i-\bar{x})^2=\sum_{i=1}^nx_i^2-n\bar{x}^2=\sum_{i=1}^nx_i^2-\frac{1}{n}\left(\sum_{i=1}^nx_i \right )^2.............(2)\\ Then~from~(1)~and~(2),\\ b_1=\frac{\sum_{i=1}^nx_iy_i-\frac{1}{n}\left(\sum_{i=1}^nx_i \right )\left(\sum_{i=1}^ny_i \right )}{\sum_{i=1}^nx_i^2-\frac{1}{n}\left(\sum_{i=1}^nx_i \right )^2}................(a)$

$b_0=\bar{y}-b_1\bar{x}=\frac{1}{n}\sum_{i=1}^ny_i-b_1\left(\frac{1}{n}\sum_{i=1}^nx_i \right )..........(b)\\ (b)~\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})=\sum_{i=1}^n(x_i-\bar{x})y_i,~~since~\bar{y}\sum_{i=1}^n(x_i-\bar{x})=0\\ Hence~b_1=\frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2}=\frac{\sum_{i=1}^n(x_i-\bar{x})y_i}{\sum_{i=1}^n(x_i-\bar{x})^2}\\ b_0=\bar{y}-b_1\bar{x}.\\ From~(a),~(b),~we~conclude~that~these~are~equivalent~estimators~(1)~and~(2)\\ of~part~(a).$

$(c)~ b_1=\frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2}=\frac{\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2}=\frac{Cov(x,y)}{Var(x)}\\ =\frac{rSD(x)SD(y)}{SD^2(x)}=r\times\frac{SD~of~y}{SD~of~x}$