# How is Gaussian assumption used in proving Gaussian comparison inequality

I'm studying the HDS book by Martin Wainwright. In section 5.4.1, there is a Theorem 5.25, which is a general Gaussian comparison inequality, reproduced as follows.

Let $$(X_1,X_2,\dots,X_N)$$ and $$(Y_1,Y_2,\dots,Y_N)$$ be a pair of centered Gaussian random vectors, and suppose that there exist disjoint subsets $$A$$ and $$B$$ of $$[N]\times[N]$$ such that $$E(X_iX_j)\leq E(Y_iY_j),\quad\forall(i,j)\in A,$$ $$E(X_iX_j)\geq E(Y_iY_j),\quad\forall(i,j)\in B,$$ $$E(X_iX_j)=E(Y_iY_j),\quad\forall(i,j)\not\in A\cup B.$$ Let $$F:\mathbb{R}^N\to\mathbb{R}$$ be a twice-differentiable function, and suppose that $$\dfrac{\partial^2F}{\partial u_i\partial u_j}(u)\geq0,\quad\forall(i,j)\in A,$$ $$\dfrac{\partial^2F}{\partial u_i\partial u_j}(u)\leq0,\quad\forall(i,j)\in B.$$ Then we are guaranteed that $$E(F(X))\leq E(F(Y)).$$ In the proof, the author lets $$Z(t)=\sqrt{1-t}X+\sqrt{t}Y$$ for each $$t\in[0,1]$$, and calculates $$E(Z_i(t)Z_j'(t))=\dfrac{1}{2}(E(Y_iY_j)-E(X_iX_j)),$$ where $$Z_j'(t)$$ is the derivative of $$Z_j(t)$$. Here comes the part I couldn't understand: the author claims that we can write $$Z_i(t)=\alpha_{ij}Z_j'(t)+W_{ij}$$, where $$\alpha_{ij}\geq0$$ for $$(i,j)\in A$$, and similarly for other cases. Also, $$W_j=(W_{1j},\dots,W_{Nj})$$ is independent of $$Z_j'(t)$$ due to Gaussianity.

My question is: how is Gaussianity used to write down the $$W_{ij}$$'s, and why are they independent of $$Z_j'(t)$$. The book doesn't provide more information. This proof in the book was adapted from Theorem 3.11 in Probability in Banach Spaces (Ledoux, Telegrand, 1991), but the proof is even more succinct and harder to understand. Any comments are appreciated!

I might have figured it out. Since $$Z_i(t)$$ and $$Z_j'(t)$$ are zero-mean, $$E(Z_i(t)Z_j'(t))=\text{Cov}(Z_i(t),Z_j'(t))$$. For $$(i,j)\in A$$, this quantity is non-negative. Thinking of covariance as an inner product, we see that $$Z_i(t)$$ can be written as a non-negative multiple of $$Z_j'(t)$$ plus an orthogonal component $$W_{ij}$$. In addition, for Gaussian vectors, zero correlation means independence, so $$Z_j'(t)$$ and $$W_{ij}$$ are independent.