[proofplan] We first compute the distribution of the sample mean directly from the characteristic function of a multivariate normal vector. For the covariance estimator, we temporarily replace the empirical centering by the true centering and apply the [multivariate central limit theorem](/theorems/1854) to the vector of products $Y_iY_i^\top-\Sigma$, where $Y_i=X_i-\mu$. The Gaussian fourth moment formula gives the limiting covariance, and the correction from replacing $\mu$ by $\widehat{\mu}_n$ is negligible after multiplication by $\sqrt n$. Finally, an [orthogonal decomposition](/theorems/436) of the Gaussian sample shows that the sample mean and the centred residual covariance are independent, so the joint limit has independent blocks. [/proofplan] [step:Compute the exact distribution of the scaled sample mean] Let $(\Omega, \mathcal{F}, \mathbb{P})$ denote the probability space on which the sequence $(X_i)_{i \in \mathbb{N}}$ is defined. Let $\mathcal{B}(\mathbb{R}^p)$ denote the Borel $\sigma$-algebra on $\mathbb{R}^p$, and for each $i \in \mathbb{N}$ regard $X_i$ as the measurable random vector \begin{align*} X_i:(\Omega, \mathcal{F}) &\to (\mathbb{R}^p, \mathcal{B}(\mathbb{R}^p)). \end{align*} For each $i \in \{1,\dots,n\}$, define the centred random vector \begin{align*} Y_i:\Omega &\to \mathbb{R}^p, \\ \omega &\mapsto X_i(\omega)-\mu . \end{align*} Then $Y_1,\dots,Y_n$ are independent and identically distributed as $\mathcal{N}_p(0,\Sigma)$. Define \begin{align*} Z_n:\Omega &\to \mathbb{R}^p, \\ \omega &\mapsto \sqrt{n}(\widehat{\mu}_n(\omega)-\mu) = \frac{1}{\sqrt n}\sum_{i=1}^{n}Y_i(\omega). \end{align*} For $t \in \mathbb{R}^p$, the characteristic function of $Y_i$ is \begin{align*} \mathbb{E}\left[e^{i t^\top Y_i}\right] = \exp\left(-\frac{1}{2}t^\top \Sigma t\right). \end{align*} Using independence, \begin{align*} \mathbb{E}\left[e^{i t^\top Z_n}\right] &= \prod_{i=1}^{n}\mathbb{E}\left[e^{i t^\top Y_i/\sqrt n}\right] \\ &= \left[ \exp\left(-\frac{1}{2n}t^\top \Sigma t\right) \right]^n \\ &= \exp\left(-\frac{1}{2}t^\top \Sigma t\right). \end{align*} This is the characteristic function of $\mathcal{N}_p(0,\Sigma)$, so $Z_n \sim \mathcal{N}_p(0,\Sigma)$ for every $n$. Hence \begin{align*} \sqrt n(\widehat{\mu}_n-\mu)\xrightarrow{d}\mathcal{N}_p(0,\Sigma). \end{align*} [/step] [step:Apply the multivariate central limit theorem to the true-centred covariance] Define the true-centred covariance fluctuation \begin{align*} A_n:\Omega &\to \mathbb{R}^{p \times p}, \\ \omega &\mapsto \frac{1}{n}\sum_{i=1}^{n}Y_i(\omega)Y_i(\omega)^\top-\Sigma . \end{align*} Also define the product vector \begin{align*} V_i:\Omega &\to \mathbb{R}^{p^2}, \\ \omega &\mapsto \operatorname{vec}\left(Y_i(\omega)Y_i(\omega)^\top-\Sigma\right). \end{align*} Then $V_1,V_2,\dots$ are independent and identically distributed, $\mathbb{E}[V_i]=0$, and every second moment is finite because Gaussian random variables have finite moments of all orders. Define the covariance matrix $\Gamma \in \mathbb{R}^{p^2 \times p^2}$ by \begin{align*} \Gamma := \operatorname{Cov}(V_1). \end{align*} The hypotheses of the multivariate [central limit theorem](/theorems/1848) are therefore satisfied: the summands are independent and identically distributed in $\mathbb{R}^{p^2}$, have mean zero, and have finite second moments. The multivariate [central limit theorem](/theorems/521) gives \begin{align*} \sqrt n\,\operatorname{vec}(A_n) = \frac{1}{\sqrt n}\sum_{i=1}^{n}V_i \xrightarrow{d} \mathcal{N}_{p^2}(0,\Gamma). \end{align*} In coordinates, this definition of $\Gamma$ means \begin{align*} \Gamma_{(j,k),(l,m)} = \operatorname{Cov}(Y_{1,j}Y_{1,k},Y_{1,l}Y_{1,m}). \end{align*} It remains to compute this covariance. Since $Y_1$ is centred Gaussian, its moment generating function \begin{align*} M:\mathbb{R}^p &\to \mathbb{R}, \\ t &\mapsto \mathbb{E}\left[e^{t^\top Y_1}\right] = \exp\left(\frac{1}{2}t^\top\Sigma t\right) \end{align*} is smooth. Differentiating four times at $t=0$ gives the Gaussian fourth moment identity \begin{align*} \mathbb{E}[Y_{1,j}Y_{1,k}Y_{1,l}Y_{1,m}] = \Sigma_{jk}\Sigma_{lm} + \Sigma_{jl}\Sigma_{km} + \Sigma_{jm}\Sigma_{kl}. \end{align*} Therefore \begin{align*} \Gamma_{(j,k),(l,m)} &= \mathbb{E}[Y_{1,j}Y_{1,k}Y_{1,l}Y_{1,m}] - \mathbb{E}[Y_{1,j}Y_{1,k}] \mathbb{E}[Y_{1,l}Y_{1,m}] \\ &= \left( \Sigma_{jk}\Sigma_{lm} + \Sigma_{jl}\Sigma_{km} + \Sigma_{jm}\Sigma_{kl} \right) - \Sigma_{jk}\Sigma_{lm} \\ &= \Sigma_{jl}\Sigma_{km}+\Sigma_{jm}\Sigma_{kl}. \end{align*} [/step] [step:Show that empirical centering changes the covariance estimator by a negligible term] Define \begin{align*} \delta_n:\Omega &\to \mathbb{R}^p, \\ \omega &\mapsto \widehat{\mu}_n(\omega)-\mu . \end{align*} For every $i \in \{1,\dots,n\}$, \begin{align*} X_i-\widehat{\mu}_n=Y_i-\delta_n. \end{align*} Expanding the outer product and using $\frac{1}{n}\sum_{i=1}^{n}Y_i=\delta_n$, \begin{align*} \widehat{\Sigma}_n &= \frac{1}{n}\sum_{i=1}^{n}(Y_i-\delta_n)(Y_i-\delta_n)^\top \\ &= \frac{1}{n}\sum_{i=1}^{n}Y_iY_i^\top - \delta_n\delta_n^\top . \end{align*} Thus \begin{align*} \widehat{\Sigma}_n-\Sigma=A_n-\delta_n\delta_n^\top. \end{align*} Using the Frobenius norm $|\cdot|_F$ on $\mathbb{R}^{p\times p}$, \begin{align*} \left|\sqrt n\,\operatorname{vec}(\delta_n\delta_n^\top)\right| = \sqrt n\,|\delta_n\delta_n^\top|_F = \sqrt n\,|\delta_n|^2 = \frac{1}{\sqrt n}\left|\sqrt n\,\delta_n\right|^2. \end{align*} The sequence $\sqrt n\,\delta_n$ is tight because it has the fixed distribution $\mathcal{N}_p(0,\Sigma)$ by the first step. Hence \begin{align*} \sqrt n\,\operatorname{vec}(\delta_n\delta_n^\top)\xrightarrow{\mathbb{P}}0. \end{align*} The preceding convergence in probability and the distributional convergence of $\sqrt n\,\operatorname{vec}(A_n)$ satisfy the hypotheses of Slutsky's theorem. Hence \begin{align*} \sqrt n\,\operatorname{vec}(\widehat{\Sigma}_n-\Sigma) = \sqrt n\,\operatorname{vec}(A_n) - \sqrt n\,\operatorname{vec}(\delta_n\delta_n^\top) \xrightarrow{d} \mathcal{N}_{p^2}(0,\Gamma). \end{align*} [/step] [step:Establish independence of the limiting mean and covariance components] Let $\Sigma^{1/2}\in \mathbb{R}^{p\times p}$ denote the symmetric positive definite square root of $\Sigma$, and define \begin{align*} R_i:\Omega &\to \mathbb{R}^p, \\ \omega &\mapsto \Sigma^{-1/2}(X_i(\omega)-\mu). \end{align*} Then $R_1,\dots,R_n$ are independent standard normal random vectors in $\mathbb{R}^p$. Choose an orthogonal matrix $Q_n\in\mathbb{R}^{n\times n}$ whose first row is \begin{align*} \left(\frac{1}{\sqrt n},\dots,\frac{1}{\sqrt n}\right). \end{align*} For $a\in\{1,\dots,n\}$, define \begin{align*} W_a:\Omega &\to \mathbb{R}^p, \\ \omega &\mapsto \sum_{i=1}^{n}(Q_n)_{ai}R_i(\omega). \end{align*} Because the joint vector $(R_1,\dots,R_n)$ is centred Gaussian and the transformation induced by $Q_n$ is orthogonal, $W_1,\dots,W_n$ are independent standard normal random vectors in $\mathbb{R}^p$. The first row of $Q_n$ gives \begin{align*} \sqrt n(\widehat{\mu}_n-\mu)=\Sigma^{1/2}W_1. \end{align*} The remaining rows span the orthogonal complement of the constant vector in $\mathbb{R}^n$, so the residual sum of squares satisfies \begin{align*} \widehat{\Sigma}_n = \Sigma^{1/2} \left( \frac{1}{n}\sum_{a=2}^{n}W_aW_a^\top \right) \Sigma^{1/2}. \end{align*} Thus $\sqrt n(\widehat{\mu}_n-\mu)$ is a measurable function of $W_1$, while $\widehat{\Sigma}_n$ is a measurable function of $(W_2,\dots,W_n)$. Since $W_1$ is independent of $(W_2,\dots,W_n)$, the random vectors $\sqrt n(\widehat{\mu}_n-\mu)$ and $\sqrt n\,\operatorname{vec}(\widehat{\Sigma}_n-\Sigma)$ are independent for each $n$. We now pass from marginal convergence to joint convergence using characteristic functions. For $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^{p^2}$, define \begin{align*} \phi_n(u,v) := \mathbb{E}\left[ \exp\left( i u^\top \sqrt n(\widehat{\mu}_n-\mu) + i v^\top \sqrt n\operatorname{vec}(\widehat{\Sigma}_n-\Sigma) \right) \right]. \end{align*} The finite-sample independence just proved gives the factorisation \begin{align*} \phi_n(u,v) &= \mathbb{E}\left[\exp\left(i u^\top \sqrt n(\widehat{\mu}_n-\mu)\right)\right] \mathbb{E}\left[\exp\left(i v^\top \sqrt n\operatorname{vec}(\widehat{\Sigma}_n-\Sigma)\right)\right]. \end{align*} The first factor converges to $\exp(-u^\top\Sigma u/2)$ by the exact normal distribution of the scaled sample mean. The second factor converges to $\exp(-v^\top\Gamma v/2)$ by the covariance asymptotic normality proved above. Therefore \begin{align*} \phi_n(u,v) \to \exp\left(-\frac{1}{2}u^\top\Sigma u\right) \exp\left(-\frac{1}{2}v^\top\Gamma v\right) = \exp\left( -\frac{1}{2} \begin{pmatrix} u \\ v \end{pmatrix}^\top \begin{pmatrix} \Sigma & 0 \\ 0 & \Gamma \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} \right). \end{align*} This is the characteristic function of the centred normal vector in $\mathbb{R}^{p+p^2}$ with block diagonal covariance matrix. Hence \begin{align*} \left(\sqrt{n}(\widehat{\mu}_n-\mu),\sqrt{n}\operatorname{vec}(\widehat{\Sigma}_n-\Sigma)\right) \xrightarrow{d} \mathcal{N}_{p+p^2}\left(0, \begin{pmatrix} \Sigma & 0 \\ 0 & \Gamma \end{pmatrix} \right). \end{align*} The factorisation of the limiting characteristic function into a function of $u$ times a function of $v$ shows that the limiting mean component and limiting covariance component are independent. [/step]