Proof of Uncorrelated Jointly Gaussian Blocks are Independent

Let $m_{X}=\mathbb{E}X$, $m_{Y}=\mathbb{E}Y$, $\Sigma_{X}=\operatorname{Cov}(X)$ and $\Sigma_{Y}=\operatorname{Cov}(Y)$, and let $\iota$ be the concatenation map of Euclidean Space and Real Matrices: Standing Notation (viii), so that the first $d$ entries of $\iota(u,v)$ are those of $u$ and the last $q$ those of $v$.

Step 1 (moments of a concatenated pair). Let $U$ and $V$ be random vectors in $\mathbb{R}^{d}$ and $\mathbb{R}^{q}$ such that $W=(U,V)$ is square-integrable; then so are $U$ and $V$, whose components are components of $W$ (Square-Integrable Vector-Valued Functions: Components, Dot Products and Almost-Everywhere Equality (i)). Expectations of random vectors are taken entrywise (Expectation (iii)), so $\mathbb{E}W=\iota(\mathbb{E}U,\mathbb{E}V)$. Entrywise, $\operatorname{Cov}(W)_{ij}=\operatorname{Cov}(W_{i},W_{j})$ and $\operatorname{Cov}(U,V)_{ij}=\operatorname{Cov}(U_{i},V_{j})$ (Variance and Covariance (iv), Variance and Covariance (v)), and $\operatorname{Cov}(W)$ is symmetric (Rules for Expectations, Variances and Covariances (xii)); hence

$$\operatorname{Cov}(W)=\begin{pmatrix}\operatorname{Cov}(U)&\operatorname{Cov}(U,V)\\\operatorname{Cov}(U,V)^{\top}&\operatorname{Cov}(V)\end{pmatrix}.$$

For the Gaussian, hence square-integrable (Square-Integrability, Mean and Covariance Matrix of a Gaussian Random Vector (i)), pair $(X,Y)$ this says, as $\operatorname{Cov}(X,Y)=0_{d\times q}$,

$$\mathbb{E}(X,Y)=\iota(m_{X},m_{Y}),\qquad\operatorname{Cov}\bigl((X,Y)\bigr)=\begin{pmatrix}\Sigma_{X}&0\\0&\Sigma_{Y}\end{pmatrix}.$$

Step 2 (an independent pair with the same marginals). On the probability space $(\Omega',\mathcal{F}',P')=(\mathbb{R}^{d+q},\mathcal{B}(\mathbb{R}^{d+q}),\gamma_{d+q})$ the identity $Z$ is a standard normal vector (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (ii)), so it is Gaussian (Affine Transformations of Gaussian Random Vectors are Gaussian (iii)), square-integrable, with $\mathbb{E}Z=0$ and $\operatorname{Cov}(Z)=I_{d+q}$ (Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (iv)). Write $Z=(Z^{1},Z^{2})$ with $Z^{1}$ and $Z^{2}$ its first $d$ and last $q$ coordinates. They are Gaussian (Affine Transformations of Gaussian Random Vectors are Gaussian (ii)) and independent by Independent Random Elements: Preimages, Functions and Blocks (v), since the coordinates of $Z$ are independent by Standard Normal Vectors and the Standard Normal Distribution on ℝᵏ (iii); and reading off the blocks of $\mathbb{E}Z$ and $\operatorname{Cov}(Z)=I_{d+q}$ by Step 1,

$$\mathbb{E}Z^{1}=0,\quad\mathbb{E}Z^{2}=0,\qquad\operatorname{Cov}(Z^{1})=I_{d},\quad\operatorname{Cov}(Z^{2})=I_{q},\quad\operatorname{Cov}(Z^{1},Z^{2})=0_{d\times q}.$$

As $\Sigma_{X}$ and $\Sigma_{Y}$ are positive semidefinite (Rules for Expectations, Variances and Covariances (xii)), their square roots $\Sigma_{X}^{1/2}$ and $\Sigma_{Y}^{1/2}$ are symmetric with $\Sigma_{X}^{1/2}\Sigma_{X}^{1/2}=\Sigma_{X}$ and $\Sigma_{Y}^{1/2}\Sigma_{Y}^{1/2}=\Sigma_{Y}$. Let

$$X'=m_{X}+\Sigma_{X}^{1/2}Z^{1},\qquad Y'=m_{Y}+\Sigma_{Y}^{1/2}Z^{2}.$$

As Borel functions of $Z^{1}$ and $Z^{2}$, $X'$ and $Y'$ are independent (an affine map $x\mapsto Ax+b$ from $\mathbb{R}^{k}$ to $\mathbb{R}^{p}$ is Borel, being a Gaussian random vector on $(\mathbb{R}^{k},\mathcal{B}(\mathbb{R}^{k}),\gamma_{k})$ as the affine image of the identity, which is standard normal and hence Gaussian by Standard Normal Vectors: Density, Moments, Orthogonal Invariance and Concatenation (ii), Affine Transformations of Gaussian Random Vectors are Gaussian (iii) and Affine Transformations of Gaussian Random Vectors are Gaussian (i), and Independent Random Elements: Preimages, Functions and Blocks (iv) applies). They are Gaussian (Affine Transformations of Gaussian Random Vectors are Gaussian (i)), and by Rules for Expectations, Variances and Covariances (vi) and Rules for Expectations, Variances and Covariances (xiii), as $\Sigma_{X}^{1/2}$ and $\Sigma_{Y}^{1/2}$ are symmetric,

$$\mathbb{E}X'=m_{X},\quad\operatorname{Cov}(X')=\Sigma_{X}^{1/2}I_{d}\,\Sigma_{X}^{1/2}=\Sigma_{X},\qquad\mathbb{E}Y'=m_{Y},\quad\operatorname{Cov}(Y')=\Sigma_{Y}^{1/2}I_{q}\,\Sigma_{Y}^{1/2}=\Sigma_{Y}.$$

So $X'$ and $X$ are Gaussian with the same mean and covariance matrix, and so are $Y'$ and $Y$; by The Law of a Gaussian Random Vector Is Determined by Its Mean and Covariance Matrix (ii), $X$ and $X'$ are identically distributed, and so are $Y$ and $Y'$.

Step 3 (the pairs have the same law). Let $D$ be the block matrix

$$D=\begin{pmatrix}\Sigma_{X}^{1/2}&0\\0&\Sigma_{Y}^{1/2}\end{pmatrix}.$$

Computing $DZ$ blockwise gives $DZ=\iota(\Sigma_{X}^{1/2}Z^{1},\Sigma_{Y}^{1/2}Z^{2})$, and by the linearity of concatenation (Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (iv))

$$(X',Y')=\iota(m_{X},m_{Y})+DZ .$$

So $(X',Y')$ is Gaussian (Affine Transformations of Gaussian Random Vectors are Gaussian (i)) and square-integrable (Rules for Expectations, Variances and Covariances (xiii)). By Step 1 and Step 2 its mean is $\iota(m_{X},m_{Y})$ and its covariance matrix has the diagonal blocks $\Sigma_{X}$ and $\Sigma_{Y}$ and the off-diagonal block

$$\operatorname{Cov}(X',Y')=\Sigma_{X}^{1/2}\operatorname{Cov}(Z^{1},Z^{2})\,\Sigma_{Y}^{1/2}=0_{d\times q}=\operatorname{Cov}(X,Y)$$

(Rules for Expectations, Variances and Covariances (xiii)) and its transpose. Thus $(X,Y)$ and $(X',Y')$ have the same mean and the same covariance matrix, and by The Law of a Gaussian Random Vector Is Determined by Its Mean and Covariance Matrix (ii) they are identically distributed.

Step 4 (the joint law is the product law). Let $C\in\mathcal{B}(\mathbb{R}^{d})\otimes\mathcal{B}(\mathbb{R}^{q})$. Then $\iota[C]$ is Borel and $\{\omega:(X(\omega),Y(\omega))\in C\}=\{(X,Y)\in\iota[C]\}$, and likewise for $(X',Y')$ (Borel Sets of ℝ^{m+n} and the Product σ-Algebra of ℝ^m and ℝ^n (ii), Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space (iii)); by Steps 2 and 3 and Independent Random Elements: Product Laws and Iterated Expectations (i) applied to the independent pair $X',Y'$,

$$P\bigl((X,Y)\in\iota[C]\bigr)=P'\bigl((X',Y')\in\iota[C]\bigr)=(P'_{X'}\otimes P'_{Y'})(C)=(P_{X}\otimes P_{Y})(C).$$

By Independent Random Elements: Product Laws and Iterated Expectations (i), $X$ and $Y$ are independent.