CALCULATING THE COEFFICIENTS (SLOPE AND INTERCEPT)

OF THE WEIGHTED LEAST SQUARES LINE

The equation of the Weighted Least Squares line is :

y* = a + b.x

with "b" the slope, and "a" the intercept.

The residual in i is :

y_i* - y_i  =  (a + b.x_i ) - y_i  

WLS wants to minimize the sum of the weighted squares of the residuals :

SSR_w = _i w_i.(y_i^* - y_i)² = _i w_i.(a + b.x_i - y_i)²

The coefficients "a" and "b" are obtained by setting to "0" the partial derivatives of SSR_w with respect to "a" and to "b", that is :

            *  SSR_w /a = 2.(a._i w_i + b._i w_i.x_i  - _i w_i.y_i ) = 0

            *  SSR_w /b = 2.(a._i w_i.x_i   + b._i w_i.x_i² - _i w_i.x_i.y_i ) = 0

This linear system is easily solved to yield the solutions :

 These equations are to be used as such only when each weight w_i can be evaluated individually as the reciprocal of the (supposedely known) local variance of y.

Note :

    1) Set all w_i to the same value w (homoscedasticity), and these equations reduce to the ordinary normal equations of Simple Linear Regression :

b = _i(x_i -  ).(y_i -  ) /_i(x_i -  )²

a = - b.

    2) If var(y) is assumed to be inversely proportional to x, these equations reduce to :

 with the following notations :

    * I = _i (1/x_i) ,

    * J =  _i (y_i /x_i).

    * S_x = _i x_i 

    * S_y = _i y_i 

I and J are infinitely large if one of the x_i, say x₀, is 0. But you can easily find the limits of "a" and of "b" when w₀ .