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Proof. By Corollary 2.1.10 in Christensen, we know that $\lambda^T\beta$ is estimable if and only if $\lambda^T\beta = E [c^TY]$ for any $\beta$. We can expand the RHS of the equation to find \begin{align*} \lambda^T\beta = E [c^TY]\\ = E [c^TY] \\ = c^TE [Y] \\ =c^T X\beta \implies \lambda^T = c^TX \end{align*} Note that since $X$ has full rank, then $\mathcal{R} (X) = \mathbb{R}^p$ . Hence, any $\lambda^T \in \mathbb{R}^p$ can be expressed as a linear combination of some row vector $c^T \in \mathbb{R}^n$ and $X$. This implies that $\lambda^T \in \mathcal{R} (X)$. The function $\lambda^T \beta$ is said to be estimable if and only if $\lambda^T \in \mathcal{R} (X)$ $\lambda^T\beta$ is an estimable function for any $\lambda^T \in \mathbb{R}^p$. $\Box $
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Fri, 24 Jan 2025 19:25 GMT