MathB.in
New
Demo
Tutorial
About
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 some** $c^T\in \mathbb{R}^p$. The linearity of expectation and $\mathbf{E}[Y]=X\beta$ give $\mathbf{E}[a^TY]=a^TX\beta\,$ for any $a^T\in\mathbb{R}$. Consequently $\lambda^T\beta$ is estimable if and only if $\lambda^T\beta=c^TX\beta$ for some $c^T\in \mathbb{R}^p$. It follows that if $\lambda^T=c^TX\,$ for some $c^T\in\mathbb{R}^p$, then $\lambda^T\beta$ is estimable. Since $X$ is full rank, $R(X)=\mathbb{R}^p$. Therefore any vector in $\mathbb{R}^p$ may be express uniquely as a linear combination of the rows of $X$. In particular, there exists a unique $c^T$ such that $\lambda^T=c^TX$. Therefore $\lambda^T\beta$ is estimable for all $\lambda^T\in \mathbb{R}^p$. $\Box$
ERROR: JavaScript must be enabled to render input!
Fri, 24 Jan 2025 21:27 GMT