Throughout, expectations, variances, and covariances of the Gaussian random variables involved are defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector. Write ΞΌiβ=E[Xiβ] and Οi2β=Var(Xiβ) for 1β€iβ€p, with the variance of square-integrable random variables, and let
J={iβ{1,β¦,p}Β :Β Οi2β>0},
let m be the number of elements of J, and enumerate J={i1β,β¦,imβ} with i1β<β―<imβ; the case m=0 (empty J) is allowed. For iβ/J we have Οi2β=0: by the moment formulas (claim 2) of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to any Gaussian representation of the one-term tuple (Xiβ), the variance is a sum of squares, hence nonnegative.
Step 1: Standardized variables. Suppose mβ₯1 and fix lβ{1,β¦,m}. Let Οilββ be the positive square root of Οilβ2β and define
Zlβ=ΟilββXilβββΞΌilβββ.
By the standardization claim (claim 2) of Standardization and Cumulative Distribution Function of a Gaussian Random Variable, Zlβ is a standard normal random variable.
Step 2: Z1β,β¦,Zmβ are independent. Fix lβ{1,β¦,m} and define glβ:RβR by glβ(x)=(xβΞΌilββ)/Οilββ, so that Zlβ=glβ(Xilββ) pointwise on Ξ©. For every real a, since Οilββ>0,
{xβR:glβ(x)>a}=(ΞΌilββ+Οilββa,Β β),
an open interval and hence a Borel set; by the generator criterion of Measurable Function and Real-Valued Measurable Function, glβ is measurable from R with the Borel Ο-algebra to itself. Hence for every Borel set B the set glβ1β(B) is Borel, and
{ZlββB}={Xilβββglβ1β(B)}.
Now fix Borel sets B1β,β¦,Bmβ and put Clβ=glβ1β(Blβ) for 1β€lβ€m. Since X1β,β¦,Xpβ are independent, the events {X1ββE1β},β¦,{XpββEpβ} are independent for the Borel choice Eilββ=Clβ (1β€lβ€m) and Eiβ=R for iβ/J. Independence of events requires the product identity for every nonempty subfamily, in particular for every nonempty subfamily of the events {XilβββClβ}={ZlββBlβ} (1β€lβ€m). Hence for every nonempty Sβ{1,β¦,m},
P(lβSββ{ZlββBlβ})=lβSββP(ZlββBlβ),
and, the Borel sets B1β,β¦,Bmβ being arbitrary, the random variables Z1β,β¦,Zmβ are independent.
Step 3: A Gaussian representation. Define real numbers aijβ for 1β€iβ€p, 1β€jβ€m by: aijβ=Οilββ if i=ilβ and j=l for some lβ{1,β¦,m}, and aijβ=0 otherwise.
If i=ilββJ, then pointwise on Ξ©,
ΞΌiβ+j=1βmβaijβZjβ=ΞΌilββ+ΟilββZlβ=Xilββ,
directly from the definition of Zlβ in Step 1; in particular the equality holds with probability one. If iβ/J, then Οi2β=0, so P(Xiβ=ΞΌiβ)=1 by the degenerate case (claim 1) of Standardization and Cumulative Distribution Function of a Gaussian Random Variable, while the i-th row of (aijβ) is zero, so that
ΞΌiβ+j=1βmβaijβZjβ=ΞΌiβ.
Hence P(Xiβ=ΞΌiβ+βj=1mβaijβZjβ)=1 for every iβ{1,β¦,p}, and by Steps 1 and 2 the random variables Z1β,β¦,Zmβ are independent standard normal. Therefore (m,(ΞΌiβ),(aijβ),(Zjβ)) is a Gaussian representation of (X1β,β¦,Xpβ); the case m=0, in which every Xiβ is almost surely constant, is expressly allowed by Gaussian Random Vectors and Jointly Gaussian Random Variables. Thus (X1β,β¦,Xpβ) is a Gaussian random vector.
Step 4: Distinct components are uncorrelated. By the moment formulas (claim 2) of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to the representation of Step 3, for 1β€i<kβ€p,
Cov(Xiβ,Xkβ)=j=1βmβaijβakjβ.
Fix j=lβ{1,β¦,m}. By construction ailβξ =0 only if i=ilβ, and aklβξ =0 only if k=ilβ; since iξ =k, at least one of the two factors vanishes, so every summand is 0. Hence Cov(Xiβ,Xkβ)=0. β