QR decomposition gives exact t and F pivots in the fixed Gaussian linear model

By full column rank, the columns of $X$ are linearly independent. Apply [F2] to obtain orthonormal columns of $Q\in {\mathbb{R}}^{n\times p}$ with the same successive spans. Expressing each column of $X$ in this basis gives $X=QR$, where $R$ is upper triangular; its diagonal entries are the nonzero Gram–Schmidt residual norms, so $R$ is invertible. Let $W=\mathrm{col}X$. The columns of $X$ form a basis of $W$, so $\dim W=p$. By [F3], ${\mathbb{R}}^n=W\oplus W^{\perp }$. The subspace $W^{\perp }$ is finite-dimensional by [F10], and direct-sum dimension additivity in [F10] gives $n=p+\dim W^{\perp }$, hence $\dim W^{\perp }=n-p$. By [F3], choose an orthonormal basis of this complement with exactly $n-p$ vectors as the columns of $Q_0$; together with $Q$ it forms an orthonormal basis $[Q{Q}_0]$ of ${\mathbb{R}}^n$.

Set $U=[Q{Q}_0]$. Then $U$ is orthogonal. Standardizing the independent errors gives $\eta =\epsilon \mathrm{/}\sigma$ with independent standard-normal coordinates, so [F4] makes $U^T\eta =(Z_Q,Z_0)$ a standard-normal product vector with independent coordinate blocks. Since $X=QR$ and $X^{T}X=R^{T}R$, the normal-equation solution in [F1] is $\widehat{\beta }=R^{-1}{Q}^{T}Y=\beta +\sigma R^{-1}{Z}_Q$. Substituting $X=QR$ and $(R^{T}R{)}^{-1}=R^{-1}{R}^{-T}$ gives $P_X=QR(R^{T}R{)}^{-1}{R}^T{Q}^T=Q[R{R}^{-1}{R}^{-T}{R}^T]Q^T=Q{Q}^T$, because $R{R}^{-1}=I_p$ and $R^{-T}{R}^T=I_p$. Since $U$ is orthogonal, $I_n-P_X=Q_0{Q}_0^T$. Thus $(I_n-P_X)Y=\sigma Q_0{Z}_0$, so $RS{S}_X\mathrm{/}{\sigma }^2=\mathrm{\parallel }{Z}_0{\mathrm{\parallel }}^2\sim {\chi }_{n-p}^2$ by [F5]. By [A2] and [F7], the linear maps defining $\widehat{\beta }$ and $Q_0{Z}_0$, and the squared-norm map defining $RS{S}_X$, are Borel; hence these quantities are measurable functions of the independent coordinate blocks and $\widehat{\beta }$ is independent of $RS{S}_X$ by [F4]. The map defining $\widehat{\beta }$ is linear and Borel, so it is a statistic for the stated experiment. The residual map and squared norm are continuous, so $RS{S}_X$ and $S$ are continuous. The piecewise t version in the Statement is Borel: its quotient is continuous on the Borel set $S>0$ and its value on $S=0$ is constant. Its denominator has positive degree $n-p$, hence $S>0$ almost surely by [F5].

For $c\mathrm{\ne }0$, put $v=R^{-T}c$. Invertibility of $R$ gives $v\mathrm{\ne }0$. Complete $v\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }$ to an orthonormal basis of ${\mathbb{R}}^p$ by [F3]; applying [F4] to the resulting orthogonal matrix shows $G_c=v^T{Z}_Q\mathrm{/}\mathrm{\parallel }v\mathrm{\parallel }\sim N(0,1)$. Under [A2], the continuous linear map defining $G_c$ is Borel. It is independent of $Q_0:=\mathrm{\parallel }{Z}_0{\mathrm{\parallel }}^2\sim {\chi }_{n-p}^2$, because $Z_Q$ and $Z_0$ are independent blocks and $G_c$ and the squared-norm map are measurable functions of those blocks. Since $X^{T}X=R^{T}R$, $\mathrm{\parallel }v{\mathrm{\parallel }}^2=c^T{R}^{-1}{R}^{-T}c=c^T(X^{T}X{)}^{-1}c$. Thus on $Q_0>0$ the displayed contrast ratio is $G_c\mathrm{/}\sqrt{Q_0\mathrm{/}(n-p)}$; on $Q_0=0$ its version ${\tilde{T}}_c$ is zero. This is exactly the extension in the definition of $t_{n-p}$, so [F5]–[F6] give ${\tilde{T}}_c\sim t_{n-p}$.

For the nested models, let $W_D=W_U\cap W_R^{\perp }$. For $u\in W_U$, its orthogonal decomposition relative to $W_R$ has a $W_R$ component and a remaining component in $W_U\cap W_R^{\perp }=W_D$, because $W_R\subseteq W_U$. Thus [F3] gives $W_U=W_R\oplus W_D$ and ${\mathbb{R}}^n=W_U\oplus W_U^{\perp }$. All these subspaces are finite-dimensional by [F10]. Applying direct-sum dimension additivity in [F10] to the two decompositions gives $p=(p-q)+\dim W_D$ and $n=p+\dim W_U^{\perp }$, hence $\dim W_D=q$ and $\dim W_U^{\perp }=n-p$. Combining the orthogonal decompositions gives ${\mathbb{R}}^n=W_R\oplus W_D\oplus W_U^{\perp }$, including the zero-dimensional $W_R$ case $q=p$. Choose orthonormal bases in these three spaces and concatenate them into an orthogonal matrix $V$. By [F4], the coordinates of $V^T\epsilon \mathrm{/}\sigma$ in the three blocks are independent standard normals. Under the null, $P_R(X\beta )=P_U(X\beta )=X\beta$, so $(I-P_R)Y=(I-P_R)\epsilon =P_{W_D}\epsilon +P_{W_U^{\perp }}\epsilon$ and $(I-P_U)Y=(I-P_U)\epsilon =P_{W_U^{\perp }}\epsilon$: here [F8] identifies $I-P_R=P_{W_R^{\perp }}$ and $I-P_U=P_{W_U^{\perp }}$, while the direct sum in [F3] splits the first projection into the two orthogonal components. The coordinate-block maps and the linear projections are continuous, and squared norms are continuous, so [A2] and [F7] make all these random vectors and real variables measurable. Orthogonality and [F9] give $RS{S}_R-RS{S}_U={\sigma }^2{\sum }_{j=1}^q{G}_j^2$ and $RS{S}_U={\sigma }^2{\sum }_{j=1}^{n-p}{H}_j^2$, where the $G$ and $H$ coordinate blocks are independent standard-normal blocks. By [F5] these sums are independent ${\sigma }^2{\chi }_q^2$ and ${\sigma }^2{\chi }_{n-p}^2$ variables; both degrees are positive. The piecewise ratio $\tilde{F}$ is Borel, since it is continuous on the Borel set $RS{S}_U>0$ and constant on $RS{S}_U=0$. Its value is the scaled ratio on $Q_H:={\sum }_{j=1}^{n-p}{H}_j^2>0$ and $0$ when $Q_H=0$; by [F6] this is exactly the defined $F_{q,n-p}$ version. The F denominator is positive almost surely by [F5], so $\tilde{F}\sim F_{q,n-p}$.