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 :
yi* - yi = (a + b.xi ) - yi
WLS wants to minimize the sum of the weighted squares of the residuals :
SSRw = i wi.(yi* - yi)² = i wi.(a + b.xi - yi)²
The coefficients "a" and "b" are obtained by setting to "0" the partial derivatives of SSRw with respect to "a" and to "b", that is :
* SSRw /a = 2.(a.i wi + b.i wi.xi - i wi.yi ) = 0
* SSRw /b = 2.(a.i wi.xi + b.i wi.xi² - i wi.xi.yi ) = 0
This linear system is easily solved to yield the solutions :
b = [(iwi ).(iwi.xi.yi) - (iwi.xi).(i wi.yi)] / [(i wi).(iwi.xi²) - (iwi.x)²] Slope
a = [(i wi.xi²).(iwi.yi) - (iwi.xi).(iwi.xi.yi)] / [(i wi).(iwi.xi²) - (iwi.x)²] Intercept
These equations are to be used as such only when each weight wi can be evaluated individually as the reciprocal of the (supposedely known) local variance of y.
1) Set all wi to the same value w (homoscedasticity), and these equations reduce to the ordinary normal equations of Simple Linear Regression :
b = i(xi - ).(yi - ) /i(xi - )²
a = - b.
2) If var(y) is assumed to be inversely proportional to x, these equations reduce to :
b = (I.Sy - J.n) / (I.Sx - n²) Slope
a = (J.Sx - n.Sy ) / (I.Sx - n²) Intercept
with the following notations :
* I = i (1/xi) ,
* J = i (yi /xi).
* Sx = i xi
* Sy = i yi
I and J are infinitely large if one of the xi, say x0, is 0. But you can easily find the limits of "a" and of "b" when w0 .