Solved: Weighted Least Squares. Suppose that we are
Chapter , Problem 114MEE(choose chapter or problem)
Weighted Least Squares. Suppose that we are fitting the line \(Y=\beta_{0}+\beta_{1} x+\epsilon\), but the variance of \(Y \) depends on the level of \(x); that is,
\(V\left(Y_{i} \mid x_{i}\right)=\sigma_{i}^{2}=\frac{\sigma^{2}}{w_{i}} \quad i=1,2, \ldots, n\)
where the \(w_{i}\) are constants, often called weights. Show that for an objective function in which each squared residual is multiplied by the reciprocal of the variance of the corresponding observation, the resulting weighted least squares normal equations are
\(\widehat{\beta}_{0} \sum_{i=1}^{n} w_{i}+\widehat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i}=\sum_{i=1}^{n} w_{i} y_{i}\)
\(\widehat{\beta}_{0} \sum_{i=1}^{n} w_{i} x_{i}+\widehat{\beta}_{1} \sum_{i=1}^{n} w_{i} x_{i}^{2}=\sum_{i=1}^{n} w_{i} x_{i} y_{i}\)
Find the solution to these normal equations. The solutions are weighted least squares estimators of \(\beta_{0} \text { and } \beta_{1}\)
Equation Transcription:
Text Transcription:
Y=\beta_0+\beta_1 x+\epsilon
Y
x
V(Y_i \mid x_i\right)=\sigma_i^2=\sigma^2 w_i\quad i=1,2, \ldots, n
w_i
\widehat\beta_0\sum_i=1^n w_i+\widehat\beta_1\sum_i=1^n w_i x_i=\sum_i=1^n w_i y_i
\widehat\beta}_0 \sum_i=1^n w_i x_i+\widehat\beta_1 \sum_i=1^n w_i x_i^2=\sum_i=1^n w_i x_i y_i
\beta_0 and \beta_1
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer