Weighted Least Squares Without Intercept

Question

I am studying the WLS model for $y=\beta x+\epsilon$, where $\beta$ and $x$ are vectors and $\epsilon$ is the error term. This is a multiple regression model without an intercept. How would I find the WLS estimator $b$ for $\beta$ and how would I find the variance of this estimator? I was thinking of taking $b=\frac{\sum{w_ix_iy_i}}{\sum{w_ix_i^2}}$, where the $w_i$ are the weights. Then $Var(b)$ would be $\frac{\sigma^2}{\sum{w_ix_i^2}}$. Is this correct? I'm afraid of the lack of intercept will throw it off.

Answer 1

$\mathbb{E}(\epsilon_i|x_i)=0$ and $\operatorname{Var}(\epsilon_i|x_i)=\sigma_i^2 $, thus WLS is $$ \hat{\beta} = (X'WX)^{-1}X'WY. $$ Without the intercept $X$ is $n\times 1$ vector, $ W = diag (w_1,.., w_n)$, hence $$ (X'WX)^{-1} = (\sum_i x_i^2w_i)^{-1}=1/\sum_i x_i^2w_i . $$ Where $X'WY = \sum x_iw_iy_i $, thus
$$ \hat{\beta} = \frac{\sum x_i w_iy_i}{\sum x_i^2w_i}. $$ As $w_i =1/\sigma_i^2$ $$ \operatorname{Var}(\hat{\beta}|X)=\frac{\sum x_i^2 \sigma^4_i /\sigma_i^2}{(\sum x_i^2/\sigma_i^2 )^2 } = \frac{1}{\sum x_i^2/\sigma_i^2}, $$ or in matrix form, denote $\mathbb{\operatorname{Var}}[\mathbf{\vec{\epsilon}}|X] = \Sigma$ so $W = \Sigma^{-1}$, thus \begin{align} \operatorname{Var}(\hat{\beta}|X) & = (X'WX)^{-1}X'W \operatorname{Var}(Y) WX(X'WX)^{-1} \\ &= (X'\Sigma^{-1}X)^{-1} X'\Sigma^{-1} \Sigma \Sigma^{-1}X (X'\Sigma^{-1}X)^{-1}\\ & = (X'\Sigma^{-1}X)^{-1}\\ & =(\sum x_i /\sigma_i^2)^{-1}. \end{align}