[proofplan] We use the [one-sample Hotelling confidence ellipsoid](/theorems/4026) for $\mu$ and show that every linear functional $m \mapsto a^\top m$ is bounded on that ellipsoid by an explicit support-function calculation. On the Hotelling event, the displacement $\mu-\bar X$ has controlled length in the $S^{-1}$-inner product. A finite-dimensional Cauchy-Schwarz argument in this inner product gives the bound for each fixed $a$, and because the same Hotelling event is used for all $a \in \mathbb{R}^p$, the resulting intervals are simultaneous. [/proofplan]

[step:Define the Hotelling ellipsoid event with its exact coverage]Define the critical constant $c_\alpha \in (0,\infty)$ by \begin{align*} c_\alpha &:= \frac{p(n-1)}{n-p}F_{p,n-p;1-\alpha}. \end{align*} Let $E \in \mathcal{F}$ be the event \begin{align*} E := \left\{ S \text{ is positive definite and } n(\bar X-\mu)^\top S^{-1}(\bar X-\mu) \le c_\alpha \right\}. \end{align*} By the one-sample Hotelling confidence ellipsoid theorem (citing a result not yet in the wiki: One-Sample Hotelling Confidence Ellipsoid Theorem), applied to the independent sample $X_1,\dots,X_n$ from $\mathcal{N}_p(\mu,\Sigma)$ with $n>p$ and positive definite $\Sigma$, we have \begin{align*} \mathbb{P}(E)=1-\alpha. \end{align*}[/step]

[guided]The role of Hotelling's theorem is to produce one event, not one event for each direction $a$. Define \begin{align*} c_\alpha &:= \frac{p(n-1)}{n-p}F_{p,n-p;1-\alpha}. \end{align*} The Hotelling confidence ellipsoid event is \begin{align*} E := \left\{ S \text{ is positive definite and } n(\bar X-\mu)^\top S^{-1}(\bar X-\mu) \le c_\alpha \right\}. \end{align*} The hypotheses needed for the one-sample Hotelling confidence ellipsoid theorem are exactly the hypotheses in the statement: $X_1,\dots,X_n$ are independent and identically distributed as $\mathcal{N}_p(\mu,\Sigma)$, the covariance matrix $\Sigma$ is positive definite, and $n>p$. Therefore that theorem gives \begin{align*} \mathbb{P}(E)=1-\alpha. \end{align*} Everything that follows is deterministic on this single event $E$. This is what makes the final confidence intervals simultaneous in $a$.[/guided]

[step:Control every linear contrast on the ellipsoid]Fix an outcome $\omega \in E$. Define $h \in \mathbb{R}^p$ by \begin{align*} h := \mu-\bar X(\omega). \end{align*} Since $S(\omega)$ is positive definite, define the bilinear map \begin{align*} (\cdot,\cdot)_{S(\omega)} : \mathbb{R}^p \times \mathbb{R}^p &\to \mathbb{R} \\ (u,v) &\mapsto u^\top S(\omega)^{-1}v . \end{align*} This is an inner product on $\mathbb{R}^p$. We record the needed Cauchy-Schwarz estimate in this inner product. For $u,v \in \mathbb{R}^p$, the non-negativity of \begin{align*} (u-tu,u-tu)_{S(\omega)} \end{align*} is not the useful quadratic; instead take $u-tv$ with $t \in \mathbb{R}$. Then \begin{align*} 0 &\le (u-tv,u-tv)_{S(\omega)} \\ &= (u,u)_{S(\omega)} -2t(u,v)_{S(\omega)} + t^2(v,v)_{S(\omega)}. \end{align*} If $v \ne 0$, this quadratic in $t$ has non-positive discriminant, so \begin{align*} |(u,v)_{S(\omega)}|^2 \le (u,u)_{S(\omega)}(v,v)_{S(\omega)}. \end{align*} The case $v=0$ is immediate. Now fix $a \in \mathbb{R}^p$. Since $S(\omega)$ is symmetric, we have \begin{align*} a^\top h &= h^\top a \\ &= h^\top S(\omega)^{-1}S(\omega)a \\ &= (h,S(\omega)a)_{S(\omega)}. \end{align*} Applying the estimate above with $u=h$ and $v=S(\omega)a$ gives \begin{align*} |a^\top h|^2 &\le (h,h)_{S(\omega)}(S(\omega)a,S(\omega)a)_{S(\omega)} \\ &= h^\top S(\omega)^{-1}h \, a^\top S(\omega)a. \end{align*} Because $\omega \in E$, \begin{align*} h^\top S(\omega)^{-1}h \le \frac{c_\alpha}{n}. \end{align*} Therefore, for every $a \in \mathbb{R}^p$, \begin{align*} |a^\top(\mu-\bar X(\omega))| \le \sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a}. \end{align*}[/step]

[guided]Fix one outcome $\omega \in E$. From this point to the end of the step, the argument is deterministic linear algebra. Define the displacement vector $h \in \mathbb{R}^p$ by \begin{align*} h := \mu-\bar X(\omega). \end{align*} Since $\omega \in E$, the matrix $S(\omega)$ is positive definite, so $S(\omega)^{-1}$ exists and is positive definite. Define \begin{align*} (\cdot,\cdot)_{S(\omega)} : \mathbb{R}^p \times \mathbb{R}^p &\to \mathbb{R} \\ (u,v) &\mapsto u^\top S(\omega)^{-1}v . \end{align*} This is an inner product because $S(\omega)^{-1}$ is symmetric positive definite. We need to bound $a^\top h$, where $a \in \mathbb{R}^p$ is arbitrary. The right inner product is the one induced by $S(\omega)^{-1}$ because the Hotelling ellipsoid controls $h^\top S(\omega)^{-1}h$. For completeness, we prove the Cauchy-Schwarz estimate in this inner product. For $u,v \in \mathbb{R}^p$ and $t \in \mathbb{R}$, \begin{align*} 0 &\le (u-tv,u-tv)_{S(\omega)} \\ &= (u,u)_{S(\omega)} -2t(u,v)_{S(\omega)} + t^2(v,v)_{S(\omega)}. \end{align*} If $v \ne 0$, this quadratic polynomial in $t$ is non-negative for every real $t$, so its discriminant is non-positive. Hence \begin{align*} 4(u,v)_{S(\omega)}^2 - 4(u,u)_{S(\omega)}(v,v)_{S(\omega)} \le 0, \end{align*} which gives \begin{align*} |(u,v)_{S(\omega)}|^2 \le (u,u)_{S(\omega)}(v,v)_{S(\omega)}. \end{align*} If $v=0$, the same inequality holds because both sides are zero. Now take an arbitrary $a \in \mathbb{R}^p$. The expression $a^\top h$ can be rewritten as an inner product with first entry $h$: \begin{align*} a^\top h &= h^\top a \\ &= h^\top S(\omega)^{-1}S(\omega)a \\ &= (h,S(\omega)a)_{S(\omega)}. \end{align*} Applying the just-proved Cauchy-Schwarz estimate with $u=h$ and $v=S(\omega)a$ gives \begin{align*} |a^\top h|^2 &\le (h,h)_{S(\omega)}(S(\omega)a,S(\omega)a)_{S(\omega)} \\ &= h^\top S(\omega)^{-1}h \, a^\top S(\omega)S(\omega)^{-1}S(\omega)a \\ &= h^\top S(\omega)^{-1}h \, a^\top S(\omega)a. \end{align*} Because $\omega \in E$, the Hotelling inequality gives \begin{align*} n h^\top S(\omega)^{-1}h \le c_\alpha, \end{align*} and hence \begin{align*} h^\top S(\omega)^{-1}h \le \frac{c_\alpha}{n}. \end{align*} Substituting this into the previous estimate yields \begin{align*} |a^\top(\mu-\bar X(\omega))| \le \sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a}. \end{align*} The direction $a$ was arbitrary, and the outcome $\omega$ was fixed only under the condition $\omega \in E$, so this bound holds for every $a \in \mathbb{R}^p$ on the same event $E$.[/guided]

[step:Translate the contrast bound into the simultaneous interval] For every $\omega \in E$ and every $a \in \mathbb{R}^p$, the previous step gives \begin{align*} -\sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a} \le a^\top\mu-a^\top\bar X(\omega) \le \sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a}. \end{align*} Adding $a^\top\bar X(\omega)$ throughout gives \begin{align*} a^\top\bar X(\omega) - \sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a} \le a^\top\mu \le a^\top\bar X(\omega) + \sqrt{\frac{c_\alpha}{n}\,a^\top S(\omega)a}. \end{align*} Since \begin{align*} \frac{c_\alpha}{n} = \frac{p(n-1)}{n(n-p)}F_{p,n-p;1-\alpha}, \end{align*} this is exactly the stated interval. Because $\mathbb{P}(E)=1-\alpha$ and the assertion holds for every $a \in \mathbb{R}^p$ on $E$, the intervals have simultaneous coverage probability $1-\alpha$. [/step]