The standard Wishart inverse quadratic form has an exact Fisher law

By [F1], realize $W=G{G}^T$ with $G=[g_1\mathrm{\mid }\cdots \mathrm{\mid }{g}_m]$ and independent $g_j\sim N_p(0,I_p)$. For any square matrix $A$, extend inversion by setting $J(A)=A^{-1}$ when $\det A\mathrm{\ne }0$ and $J(A)=0$ otherwise. This is Borel: the invertible matrices form an open set, inversion is continuous there, and the complementary branch is constant; [F11] supplies Euclidean Borel measurability. Thus $J(G{G}^T)$ is a Borel matrix-valued random element and agrees with $W^{-1}$ whenever $G{G}^T$ is invertible. Since $m\ge p$, [F2] makes $G$ full row rank almost surely; hence for nonzero $v$, $v^{T}Wv=\mathrm{\parallel }{G}^{T}v{\mathrm{\parallel }}^2>0$ and $W$ is invertible almost surely.

For $p\ge 2$, write the first row of $G$ as $g$ and the remaining $p-1$ rows as $G_2$. The event $E_2=\{\det (G_2{G}_2^T)\mathrm{\ne }0\}$ is Borel. Define $C^{†}=(G_2{G}_2^T{)}^{-1}$ on $E_2$ and $C^{†}=0$ on $E_2^c$, and put $P=G_2^T{C}^{†}{G}_2$; this is Borel by the same piecewise-inversion argument as in step 1.1. By [F2], $E_2$ has probability one. On $E_2$, direct multiplication gives $P^T=P$ and $P^2=P$, and its image is the row space of $G_2$, of dimension $p-1$; [F9] identifies it as an orthogonal projection. For $p=1$, use the empty-row convention $G_2=0$, $P=0$. In either case define the Borel scalar $Q=g(I_m-P)g^T$ on the whole matrix space. On the full-probability event that $G$ has full row rank, the block matrix $W=G{G}^T=\left(\begin{array}{cc}g{g}^T& g{G}_2^T\\ G_2{g}^T& C\end{array}\right)$ has invertible lower block $C$ when $p\ge 2$, and solving $Wx=e_1$ gives $e_1^T{W}^{-1}{e}_1=(g{g}^T-g{G}_2^T{C}^{-1}{G}_2{g}^T{)}^{-1}$. Thus for $p\ge 2$ this $Q$ equals $(e_1^T{W}^{-1}{e}_1{)}^{-1}$ on the full-probability event. For $p=1$, $W=g{g}^T$ and the same identity reads $Q=g{g}^T=(e_1^T{W}^{-1}{e}_1{)}^{-1}$.

The entries of $G$ are independent standard normals, so its first row $g$ is independent of the remaining rows $G_2$ by [F7]. For each fixed full-row-rank value of $G_2$, $I_m-P$ is an orthogonal projection of rank $m-p+1$. Therefore [F4] gives the section law $Q\sim {\chi }_{m-p+1}^2$ for every such $G_2$; the $p=1$ case follows from [F5]. The matrix rank event is Borel because it is determined by continuous minors, and the extended map $P(G_2)$ is Borel by step 2.1. The exceptional set has probability zero by [F2]. Applying the product-law identity [F8] to these Borel sections shows that the unconditional law of the Borel extension $Q=g(I_m-P)g^T$ is also ${\chi }_{m-p+1}^2$.

For any fixed unit vector $u\in {\mathbb{R}}^p$, [F10] supplies an orthogonal matrix $U$ with $Uu=e_1$. By [F3] and [F7], applying $U$ to each independent column preserves their joint standard-normal product law; hence $UW{U}^T$ has the same law as $W$. Let $h(t)=t^{-1}$ for $t>0$ and $h(t)=0$ otherwise. Since $J(W)=W^{-1}$ almost surely by step 1.1 and $W\succ 0$ almost surely, $h(u^{T}J(W)u)$ has the same law as $Q$ in step 3.1 for every unit $u$, using the orthogonal invariance just proved. Set $U_Z=Z\mathrm{/}\mathrm{\parallel }Z\mathrm{\parallel }$ when $Z\mathrm{\ne }0$ and let $U_Z=e_1$ when $Z=0$. The map $z\mapsto U_z$ is Borel, and $Q_Z:=h(U_Z^{T}J(W)U_Z)$ is Borel because $J$ and $h$ are Borel; $Z=0$ is null by [F5]–[F6]. For Borel sets $A,B$, independence and the product law give $\mathbb{P}(\mathrm{\parallel }Z{\mathrm{\parallel }}^2\in A,Q_Z\in B)={\int }_{\{z:\mathrm{\parallel }z{\mathrm{\parallel }}^2\in A\}}{\mathbb{P}}_W(h(u_z^{T}J(W)u_z)\in B)d{\mathbb{P}}_Z(z),$ where $u_z=z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$ for $z\mathrm{\ne }0$ and $u_0=e_1$. The inner probability is the same for every unit $u_z$, so this factors as $\mathbb{P}(\mathrm{\parallel }Z{\mathrm{\parallel }}^2\in A)\mathbb{P}(Q\in B)$. Thus $Q_Z$ is independent of $\mathrm{\parallel }Z{\mathrm{\parallel }}^2$ and has the same ${\chi }_{m-p+1}^2$ law as $Q$ in step 3.1.

By [F5], $\mathrm{\parallel }Z{\mathrm{\parallel }}^2\sim {\chi }_p^2$ and it is independent of $Q_Z\sim {\chi }_{m-p+1}^2$ by step 4.1. Therefore $\frac{m-p+1}{p}{Z}^T{W}^{-1}Z=\frac{\mathrm{\parallel }Z{\mathrm{\parallel }}^2\mathrm{/}p}{Q_Z\mathrm{/}(m-p+1)}$ has law $F_{p,m-p+1}$ by the defining Fisher ratio; its denominator is positive almost surely by [F6].