Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Multiple linear regression ▷ Weighted least squares

Theorem: Given a linear regression model with correlated observations

$\label{eq:MLR} y = X\beta + \varepsilon, \; \varepsilon \sim \mathcal{N}(0, \sigma^2 V) \; ,$

the parameters minimizing the weighted residual sum of squares are given by

$\label{eq:WLS} \hat{\beta} = (X^\mathrm{T} V^{-1} X)^{-1} X^\mathrm{T} V^{-1} y \; .$

Proof: Let there be an $n \times n$ square matrix $W$, such that

$\label{eq:W-def} W V W^\mathrm{T} = I_n \; .$

Since $V$ is a covariance matrix and thus symmetric, $W$ is also symmetric and can be expressed the matrix square root of the inverse of $V$:

$\label{eq:W-V} W V W = I_n \quad \Leftrightarrow \quad V = W^{-1} W^{-1} \quad \Leftrightarrow \quad V^{-1} = W W \quad \Leftrightarrow \quad W = V^{-1/2} \; .$

Left-multiplying the linear regression equation \eqref{eq:MLR} with $W$, the linear transformation theorem implies that

$\label{eq:MLR-W} Wy = WX\beta + W\varepsilon, \; W\varepsilon \sim \mathcal{N}(0, \sigma^2 W V W^T) \; .$

Applying \eqref{eq:W-def}, we see that \eqref{eq:MLR-W} is actually a linear regression model with independent observations

$\label{eq:MLR-W-dev} Wy = WX\beta + W\varepsilon, \; W\varepsilon \sim \mathcal{N}(0, \sigma^2 I_n) \; .$

With this, we can express the weighted residual sum of squares as

$\label{eq:wRSS} \mathrm{wRSS}(\beta) = \sum_{i=1}^n (W \varepsilon)_i = (W \varepsilon)^\mathrm{T} (W \varepsilon) = (Wy-WX\beta)^\mathrm{T} (Wy-WX\beta)$

which can be developed into

$\label{eq:wRSS-dev} \begin{split} \mathrm{wRSS}(\beta) &= y^\mathrm{T} W^\mathrm{T} W y - y^\mathrm{T} W^\mathrm{T} W X \beta - \beta^\mathrm{T} X^\mathrm{T} W^\mathrm{T} W y + \beta^\mathrm{T} X^\mathrm{T} W^\mathrm{T} W X \beta \\ &= y^\mathrm{T} W W y - 2 \beta^\mathrm{T} X^\mathrm{T} W W y + \beta^\mathrm{T} X^\mathrm{T} W W X \beta \\ &\overset{\eqref{eq:W-V}}{=} y^\mathrm{T} V^{-1} y - 2 \beta^\mathrm{T} X^\mathrm{T} V^{-1} y + \beta^\mathrm{T} X^\mathrm{T} V^{-1} X \beta \; . \end{split}$

The derivative of $\mathrm{wRSS}(\beta)$ with respect to $\beta$ is

$\label{eq:wRSS-der} \frac{\mathrm{d}\mathrm{wRSS}(\beta)}{\mathrm{d}\beta} = - 2 X^\mathrm{T} V^{-1} y + 2 X^\mathrm{T} V^{-1} X \beta$

and setting this deriative to zero, we obtain:

$\label{eq:WLS-qed} \begin{split} \frac{\mathrm{d}\mathrm{wRSS}(\hat{\beta})}{\mathrm{d}\beta} &= 0 \\ 0 &= - 2 X^\mathrm{T} V^{-1} y + 2 X^\mathrm{T} V^{-1} X \hat{\beta} \\ X^\mathrm{T} V^{-1} X \hat{\beta} &= X^\mathrm{T} V^{-1} y \\ \hat{\beta} &= (X^\mathrm{T} V^{-1} X)^{-1} X^\mathrm{T} V^{-1} y \; . \end{split}$

Since the quadratic form $y^\mathrm{T} V^{-1} y$ in \eqref{eq:wRSS-dev} is positive, $\hat{\beta}$ minimizes $\mathrm{wRSS}(\beta)$.

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Metadata: ID: P136 | shortcut: mlr-wls2 | author: JoramSoch | date: 2020-07-22, 06:48.