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So what I did was use the fact that $\lambda^T\beta$ is estimable iff $\lambda^T \beta = \mathbb{E} [c^Ty]$ for any $\beta$. After some steps, I showed that $\lambda^T = c^TX$. Since $X$ has full rank, then $\mathcal{R} (X) = \mathbb{R}^p$. Therefore, for 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 $\lambda^T \in \mathcal{R}(X)$, and we know that $\lambda^T\beta$ is estimable $\iff$ $\lambda^T \in \mathcal{R}(X)$. Hence, $\lambda^T\beta$ is estimable for any $\lambda^T \in \mathbb{R}^p$. Is this a valid proof?
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Fri, 24 Jan 2025 19:21 GMT