[proofplan] Project the multivariate observations onto each fixed contrast direction $a_j$, reducing the problem to scalar normal samples. For every nonzero contrast, the usual centered and studentized sample mean has a Student $t$ distribution with $n-1$ degrees of freedom, so the chosen Bonferroni quantile gives marginal failure probability $\alpha/k$. Zero contrast vectors are harmless because their target and interval both collapse to $0$. Finally, the union bound converts the $k$ marginal failure bounds into the stated simultaneous coverage bound. [/proofplan]

[step:Project each fixed contrast to a scalar normal sample]For each $j \in \{1,\dots,k\}$ and $i \in \{1,\dots,n\}$, define the real-valued random variable \begin{align*} Y_{ij}: (\Omega,\mathcal{F}) &\to (\mathbb{R},\mathcal{B}(\mathbb{R})) \\ \omega &\mapsto a_j^\top X_i(\omega). \end{align*} Since $X_1,\dots,X_n$ are independent and identically distributed as $\mathcal{N}_p(\mu,\Sigma)$ and $a_j$ is deterministic, the random variables $Y_{1j},\dots,Y_{nj}$ are independent and identically distributed with univariate normal distribution \begin{align*} Y_{ij} \sim \mathcal{N}(a_j^\top \mu,\, a_j^\top \Sigma a_j). \end{align*} Define the scalar sample mean and scalar sample variance for the $j$th projected sample by \begin{align*} \bar Y_j := \frac{1}{n}\sum_{i=1}^n Y_{ij}, \qquad V_j := \frac{1}{n-1}\sum_{i=1}^n (Y_{ij}-\bar Y_j)^2 . \end{align*} By the definitions of $\bar X$ and $S$, \begin{align*} \bar Y_j = a_j^\top \bar X \end{align*} and \begin{align*} V_j &= \frac{1}{n-1}\sum_{i=1}^n \left(a_j^\top X_i-a_j^\top \bar X\right)^2 \\ &= \frac{1}{n-1}\sum_{i=1}^n a_j^\top (X_i-\bar X)(X_i-\bar X)^\top a_j \\ &= a_j^\top S a_j . \end{align*}[/step]

[guided]The goal is to prove a statement about $a_j^\top \bar X$ and $a_j^\top S a_j$. For a fixed $j$, these are exactly the ordinary sample mean and sample variance of the scalar observations obtained by applying the fixed linear functional $x \mapsto a_j^\top x$ to each vector observation. For each $j \in \{1,\dots,k\}$ and $i \in \{1,\dots,n\}$, define \begin{align*} Y_{ij}: (\Omega,\mathcal{F}) &\to (\mathbb{R},\mathcal{B}(\mathbb{R})) \\ \omega &\mapsto a_j^\top X_i(\omega). \end{align*} The map $x \mapsto a_j^\top x$ is deterministic and linear. Therefore applying it to independent random vectors preserves independence, and since the $X_i$ have the same distribution, the $Y_{ij}$ also have the same distribution for fixed $j$. A linear image of a multivariate normal random vector is univariate normal, with mean and variance computed by the usual linear transformation formulas: \begin{align*} Y_{ij} \sim \mathcal{N}(a_j^\top \mu,\, a_j^\top \Sigma a_j). \end{align*} Now define the scalar sample mean and scalar sample variance by \begin{align*} \bar Y_j := \frac{1}{n}\sum_{i=1}^n Y_{ij}, \qquad V_j := \frac{1}{n-1}\sum_{i=1}^n (Y_{ij}-\bar Y_j)^2 . \end{align*} The sample mean is \begin{align*} \bar Y_j = \frac{1}{n}\sum_{i=1}^n a_j^\top X_i = a_j^\top \left(\frac{1}{n}\sum_{i=1}^n X_i\right) = a_j^\top \bar X . \end{align*} For the variance, substitute $Y_{ij}=a_j^\top X_i$ and $\bar Y_j=a_j^\top \bar X$: \begin{align*} V_j &= \frac{1}{n-1}\sum_{i=1}^n \left(a_j^\top X_i-a_j^\top \bar X\right)^2 \\ &= \frac{1}{n-1}\sum_{i=1}^n \left(a_j^\top (X_i-\bar X)\right)^2 \\ &= \frac{1}{n-1}\sum_{i=1}^n a_j^\top (X_i-\bar X)(X_i-\bar X)^\top a_j \\ &= a_j^\top \left[ \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)(X_i-\bar X)^\top \right] a_j \\ &= a_j^\top S a_j . \end{align*} Thus the displayed interval in the theorem is precisely the usual scalar $t$ interval for the projected observations $Y_{1j},\dots,Y_{nj}$.[/guided]

[step:Obtain the marginal coverage probability for every contrast]Define the Bonferroni critical value \begin{align*} c := t_{n-1;1-\alpha/(2k)} . \end{align*} For each $j \in \{1,\dots,k\}$, define the coverage event \begin{align*} A_j := \left\{ \omega \in \Omega : |a_j^\top \bar X(\omega)-a_j^\top \mu| \leq c\sqrt{\frac{a_j^\top S(\omega)a_j}{n}} \right\}. \end{align*} If $a_j=0$, then $a_j^\top \bar X=0$, $a_j^\top \mu=0$, and $a_j^\top S a_j=0$, so $A_j=\Omega$ and hence \begin{align*} \mathbb{P}(A_j^c)=0 \leq \frac{\alpha}{k}. \end{align*} Now suppose $a_j \neq 0$. Since $\Sigma$ is positive definite, \begin{align*} \sigma_j^2 := a_j^\top \Sigma a_j > 0. \end{align*} By the standard univariate normal theory fact that the studentized mean of an independent normal sample has a Student $t$ distribution with $n-1$ degrees of freedom (citing a result not yet in the wiki: Student $t$ statistic for a normal sample), the statistic \begin{align*} T_j := \frac{\bar Y_j-a_j^\top \mu}{\sqrt{V_j/n}} \end{align*} has the Student $t$ distribution with $n-1$ degrees of freedom. Using $\bar Y_j=a_j^\top \bar X$ and $V_j=a_j^\top S a_j$, the event $A_j$ is exactly $\{|T_j|\leq c\}$. Therefore, by the definition of $c$ as the two-sided $1-\alpha/k$ Student $t$ critical value, \begin{align*} \mathbb{P}(A_j^c) = \mathbb{P}(|T_j|>c) = \frac{\alpha}{k}. \end{align*} Thus for every $j \in \{1,\dots,k\}$, \begin{align*} \mathbb{P}(A_j^c) \leq \frac{\alpha}{k}. \end{align*}[/step]

[guided]We now prove the marginal confidence statement for one contrast at a time. Define \begin{align*} c := t_{n-1;1-\alpha/(2k)} . \end{align*} For each $j \in \{1,\dots,k\}$, define the event \begin{align*} A_j := \left\{ \omega \in \Omega : |a_j^\top \bar X(\omega)-a_j^\top \mu| \leq c\sqrt{\frac{a_j^\top S(\omega)a_j}{n}} \right\}. \end{align*} This is the event that the $j$th displayed interval contains its target value $a_j^\top \mu$. There is a small degeneracy to handle first. If $a_j=0$, then the contrast is identically zero: \begin{align*} a_j^\top \bar X=0, \qquad a_j^\top \mu=0, \qquad a_j^\top S a_j=0. \end{align*} Hence the interval is $[0,0]$, which contains $0$ for every outcome $\omega \in \Omega$. Thus $A_j=\Omega$, and \begin{align*} \mathbb{P}(A_j^c)=0 \leq \frac{\alpha}{k}. \end{align*} Assume now that $a_j \neq 0$. Since $\Sigma$ is positive definite, the projected variance is strictly positive: \begin{align*} \sigma_j^2 := a_j^\top \Sigma a_j > 0. \end{align*} From the previous step, $Y_{1j},\dots,Y_{nj}$ are independent and identically distributed as $\mathcal{N}(a_j^\top \mu,\sigma_j^2)$. Therefore the usual studentized mean statistic \begin{align*} T_j := \frac{\bar Y_j-a_j^\top \mu}{\sqrt{V_j/n}} \end{align*} has the Student $t$ distribution with $n-1$ degrees of freedom by the standard univariate normal theory result for the Student statistic (citing a result not yet in the wiki: Student $t$ statistic for a normal sample). The hypotheses of that result are satisfied here: the observations are independent, normally distributed, have common finite mean $a_j^\top \mu$, have common positive variance $\sigma_j^2$, and $n\geq 2$. Because $\bar Y_j=a_j^\top \bar X$ and $V_j=a_j^\top S a_j$, the event $A_j$ can be rewritten as \begin{align*} A_j = \{|T_j|\leq c\}. \end{align*} The Student $t$ distribution is symmetric about $0$, and $c=t_{n-1;1-\alpha/(2k)}$ is chosen so that the upper tail probability is $\alpha/(2k)$. Hence the two tails together have probability $\alpha/k$: \begin{align*} \mathbb{P}(A_j^c) = \mathbb{P}(|T_j|>c) = \frac{\alpha}{2k}+\frac{\alpha}{2k} = \frac{\alpha}{k}. \end{align*} Combining this with the zero-contrast case gives, for every $j \in \{1,\dots,k\}$, \begin{align*} \mathbb{P}(A_j^c) \leq \frac{\alpha}{k}. \end{align*}[/guided]

[step:Apply the union bound to pass from marginal to simultaneous coverage]The event that all intervals cover their corresponding targets is \begin{align*} \bigcap_{j=1}^k A_j. \end{align*} Its complement is the event that at least one interval fails: \begin{align*} \left(\bigcap_{j=1}^k A_j\right)^c = \bigcup_{j=1}^k A_j^c. \end{align*} By finite subadditivity of probability, \begin{align*} \mathbb{P}\left(\bigcup_{j=1}^k A_j^c\right) \leq \sum_{j=1}^k \mathbb{P}(A_j^c) \leq \sum_{j=1}^k \frac{\alpha}{k} = \alpha . \end{align*} Therefore \begin{align*} \mathbb{P}\left(\bigcap_{j=1}^k A_j\right) = 1- \mathbb{P}\left(\left(\bigcap_{j=1}^k A_j\right)^c\right) \geq 1-\alpha . \end{align*} This is exactly the asserted simultaneous coverage probability.[/step]

[guided]The marginal estimates do not require the events $A_1,\dots,A_k$ to be independent. This is precisely why the Bonferroni argument is useful: it only uses finite subadditivity of probability. The event of simultaneous success is \begin{align*} \bigcap_{j=1}^k A_j, \end{align*} because this says that every one of the $k$ intervals contains its corresponding target. Its complement is the event that at least one interval fails. By De Morgan's law, \begin{align*} \left(\bigcap_{j=1}^k A_j\right)^c = \bigcup_{j=1}^k A_j^c. \end{align*} Now apply finite subadditivity of probability to the events $A_1^c,\dots,A_k^c$: \begin{align*} \mathbb{P}\left(\bigcup_{j=1}^k A_j^c\right) \leq \sum_{j=1}^k \mathbb{P}(A_j^c). \end{align*} From the previous step, each failure probability is at most $\alpha/k$, so \begin{align*} \mathbb{P}\left(\bigcup_{j=1}^k A_j^c\right) \leq \sum_{j=1}^k \frac{\alpha}{k} = \alpha . \end{align*} Taking complements gives \begin{align*} \mathbb{P}\left(\bigcap_{j=1}^k A_j\right) = 1- \mathbb{P}\left(\left(\bigcap_{j=1}^k A_j\right)^c\right) \geq 1-\alpha . \end{align*} Since $A_j$ was defined exactly as the event that the $j$th displayed interval contains $a_j^\top \mu$, this proves the claimed simultaneous coverage statement.[/guided]