TheoremBase

Proof

All covariances below are defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector.

Reduction to a full factorization. We first show that it suffices to prove: for all Borel sets C1,…,CpC_1,\dots,C_p,

P(⋂k=1p{Xk∈Ck})=∏k=1pP(Xk∈Ck),P\Bigl(\bigcap_{k=1}^{p}\{X_k\in C_k\}\Bigr)=\prod_{k=1}^{p}P(X_k\in C_k),

with the finite product notation. Indeed, let Borel sets B1,…,BpB_1,\dots,B_p and a nonempty subset S⊆{1,…,p}S\subseteq\{1,\dots,p\} be given, as in Independence of Events and of Random Variables. Apply the full factorization with Ci=BiC_i=B_i for i∈Si\in S and Ci=RC_i=\mathbb{R} for i∉Si\notin S. Since {Xi∈R}=Ω\{X_i\in\mathbb{R}\}=\Omega, the intersection on the left reduces to ⋂i∈S{Xi∈Bi}\bigcap_{i\in S}\{X_i\in B_i\}, and since P(Ω)=1P(\Omega)=1, the factors with i∉Si\notin S equal 11; hence

P(⋂i∈S{Xi∈Bi})=∏i∈SP(Xi∈Bi).P\Bigl(\bigcap_{i\in S}\{X_i\in B_i\}\Bigr)=\prod_{i\in S}P(X_i\in B_i).

As SS and the Borel sets were arbitrary, the events {X1∈B1},…,{Xp∈Bp}\{X_1\in B_1\},\dots,\{X_p\in B_p\} are independent for every Borel choice, i.e., X1,…,XpX_1,\dots,X_p are independent random variables.

Induction. Fix Borel sets C1,…,CpC_1,\dots,C_p. We prove by induction on k∈{1,…,p}k\in\{1,\dots,p\} that

P(⋂l=1k{Xl∈Cl})=∏l=1kP(Xl∈Cl).P\Bigl(\bigcap_{l=1}^{k}\{X_l\in C_l\}\Bigr)=\prod_{l=1}^{k}P(X_l\in C_l).

For k=1k=1 both sides equal P(X1∈C1)P(X_1\in C_1) and there is nothing to prove.

Let 1≤k<p1\le k<p and assume the identity holds for kk. The initial segment (X1,…,Xk,Xk+1)(X_1,\dots,X_k,X_{k+1}) is a Gaussian random vector, being a subfamily of the Gaussian random vector (X1,…,Xp)(X_1,\dots,X_p), by Affine Transformations of Gaussian Random Vectors are Gaussian. By hypothesis its cross-covariances with the last component vanish:

Cov⁡(Xi,Xk+1)=0(1≤i≤k).\operatorname{Cov}(X_i,X_{k+1})=0\qquad(1\le i\le k).

By Uncorrelated Jointly Gaussian Blocks are Independent, applied with the blocks (X1,…,Xk)(X_1,\dots,X_k) and (Xk+1)(X_{k+1}) (that is, with d=kd=k and q=1q=1), the σ\sigma-algebras σ(X1,…,Xk)\sigma(X_1,\dots,X_k) and σ(Xk+1)\sigma(X_{k+1}) are independent.

The event E=⋂l=1k{Xl∈Cl}E=\bigcap_{l=1}^{k}\{X_l\in C_l\} belongs to σ(X1,…,Xk)\sigma(X_1,\dots,X_k): each {Xl∈Cl}=Xl−1(Cl)\{X_l\in C_l\}=X_l^{-1}(C_l) is one of the generating sets of σ(X1,…,Xk)\sigma(X_1,\dots,X_k) exhibited in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, and a σ\sigma-algebra contains every finite intersection of its members, a finite intersection being the complement of the finite union of the complements. Likewise {Xk+1∈Ck+1}∈σ(Xk+1)\{X_{k+1}\in C_{k+1}\}\in\sigma(X_{k+1}). Independence of the two σ\sigma-algebras therefore gives

P(E∩{Xk+1∈Ck+1})=P(E) P(Xk+1∈Ck+1),P\bigl(E\cap\{X_{k+1}\in C_{k+1}\}\bigr)=P(E)\,P(X_{k+1}\in C_{k+1}),

and inserting the inductive hypothesis for P(E)P(E) yields the identity for k+1k+1. Taking k=pk=p gives the full factorization, which by the reduction completes the proof. ■\blacksquare

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