Multidimensional CLT
Gaussian Vectors
An [math]\R^n[/math]-valued r.v. [math]X=(X_1,...,X_n)[/math] is called a gaussian random vector if every linear combination [math]\sum_{j=1}^n\lambda_jX_j[/math], with [math]\lambda_j\in\R[/math], is a gaussian r.v. (Possibly degenerated [math]\mathcal{N}(\mu,0)=\mu[/math] a.s.).
[math]X[/math] is an [math]\R^n[/math]-valued gaussian r.v. if and only if its characteristic function has the form
Suppose that [math](*)[/math] holds. Let [math]Y=\sum_{j=1}^na_jX_j=\langle a,X\rangle[/math]. For [math]v\in\R[/math], [math]\varphi_Y(v)=\varphi_X(va)=\exp\left(iv\langle a,\mu\rangle-\frac{v^2}{2}\langle a,Qa\rangle\right)\Longrightarrow Y\sim \mathcal{N}\left(\langle a,\mu\rangle,\langle a,Qa\rangle\right)\Longrightarrow X[/math] is a gaussian vector. Conversely assume that [math]X[/math] is a gaussian vector and let [math]Y=\sum_{j=1}^na_jX_j=\langle a,X\rangle[/math]. Let [math]\omega=Cov(X)[/math] and note that [math]\E[Y]=\langle \sigma,\mu\rangle[/math] and [math]Var(Y)=\sigma^2(Y)=\langle a,Qa\rangle[/math]. Since [math]Y[/math] is a gaussian r.v.
Notation: We write [math]X\sim \mathcal{N}(\mu,Q)[/math].
Example
Let [math]X_1,...,X_n[/math] be independent gaussian r.v.'s with [math]X_k\sim\mathcal{N}(\mu_k,\sigma_k^2)[/math]. Then [math]X=(X_1,...,X_n)[/math] is a gaussian vector. Indeed, we have
where [math]\mu=(\mu_1,...,\mu_n)[/math], [math]Q=\begin{pmatrix}\sigma_1^2&\dotsm &0\\ \vdots&\ddots&\vdots\\0&\dotsm&\sigma_n^2\end{pmatrix}[/math].
Let [math]X[/math] be an [math]\R^n[/math]-valued gaussian vector. The components [math]X_j[/math] of [math]X[/math] are independent if and only if [math]Q[/math] is a diagonal matrix.
Suppose [math]Q=\begin{pmatrix}\sigma_1^2&\dotsm &0\\ \vdots&\ddots&\vdots\\0&\dotsm&\sigma_n^2\end{pmatrix}[/math], then [math](*)[/math] shows that
Let [math]X[/math] be an [math]\R^n[/math]-valued gaussian vector with mean [math]\mu[/math]. Then there exists independent gaussian r.v.'s [math]Y_1,..,Y_n[/math] with
It is possible that [math]\lambda_j=0[/math]. In that case, it is also possible to get [math]Y_j=0[/math] a.s.
There is an [math]A\in O(\R)[/math], such that [math]Q=A\Lambda A^*[/math], with [math]\Lambda=\begin{pmatrix}\lambda_1&\dotsm &0\\ \vdots&\ddots&\vdots\\0&\dotsm&\lambda_n\end{pmatrix}[/math], [math]\lambda_j\leq 0[/math]. Set [math]Y=A^*(X-\mu)[/math]. Then one can check that [math]Y[/math] is gaussian. So we get that [math]Cov(Y)=A^*QA=\Lambda[/math], which implies that [math]Y_1,...,Y_n[/math] are independent because [math]Cov(Y)[/math] is diagonal.
An [math]\R^n[/math]-valued gaussian vector [math]X[/math] has density on [math]\R^n[/math] if and only if [math]\det(Q)\not=0[/math].
If [math]\det(Q)\not=0[/math], then [math]f_X(x)=\frac{1}{\sqrt{2\pi}\sqrt{\det(Q)}}e^{-\frac{1}{2}\langle x-\mu,Q^{-1}(\lambda-\mu)\rangle}[/math].
Let [math]X[/math] be an [math]\R^n[/math]-valued gaussian r.v. and let [math]Y[/math] be an [math]\R^m[/math]-valued gaussian r.v. If [math]X[/math] and [math]Y[/math] are independent, then [math]Z=(X,Y)[/math] is an [math]\R^{n+m}[/math]-valued gaussian vector.
Let [math]u=(w,v)[/math], [math]w\in\R^n[/math] and [math]v\in\R^m[/math]. Take [math]Q=\begin{pmatrix}Q^X&0\\ 0&Q^Y\end{pmatrix}[/math]. Now we get
Let [math]X[/math] be an [math]\R^n[/math]-valued gaussian vector. Two components [math]X_j[/math] and [math]X_k[/math] of [math]X[/math] are independent if and only if [math]Cov(X_j,X_k)=0[/math].
Consider [math]Y=(Y_1,Y_2)[/math], with [math]Y_1=X_j[/math] and [math]Y_2=X_k[/math]. If [math]Y[/math] is a gaussian vector, then [math]Cov(Y_1,Y_2)=0[/math], which implies that [math]Y_1[/math] and [math]Y_2[/math] are independent.
Warning!: Let [math]Y\sim\mathcal{N}(0,1)[/math] and for [math]a \gt 0[/math] fix [math]Z=Y\one_{\vert Y\vert\leq a}-Y\one_{\vert Y\vert \gt a}[/math]. Then [math]Z\sim\mathcal{N}(0,1)[/math]. But [math]Y+Z=2Y\one_{\vert Y\vert \leq a}[/math] is not gaussian because it is a bounded r.v. and it is not constant. Therefore [math](Y,Z)[/math] is not a gaussian vector.
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].