Fix a Gaussian representation (m,(ΞΌkβ),(akjβ),(Zjβ)) of (X1β,β¦,Xdβ), so that the events Ekβ={Xkβ=ΞΌkβ+βj=1mβakjβZjβ} satisfy P(Ekβ)=1 for 1β€kβ€d; when m=0 all sums over j below are empty and equal to 0, in accordance with Gaussian Random Vectors and Jointly Gaussian Random Variables. Each Yiβ is a random variable by the closure of random variables under sums and scalar multiples recorded in the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product.
Step 1: A representation for (Y1β,β¦,Ypβ). Let Ξ©0β=βk=1dβEkβ. Its complement is the finite union of the complements Ξ©βEkβ, each of probability 0, so by the monotonicity and countable additivity of the probability measure P (finite subadditivity being the special case with cofinitely many empty sets), P(Ξ©βΞ©0β)=0, i.e., P(Ξ©0β)=1. For every ΟβΞ©0β and 1β€iβ€p, substituting and rearranging the finite sums,
Yiβ(Ο)=ciβ+k=1βdβMikβ(ΞΌkβ+j=1βmβakjβZjβ(Ο))=Ξ½iβ+j=1βmβbijβZjβ(Ο),
where
Ξ½iβ=ciβ+k=1βdβMikβΞΌkβ,bijβ=k=1βdβMikβakjβ.
Hence P(Yiβ=Ξ½iβ+βjβbijβZjβ)β₯P(Ξ©0β)=1 for each i, so (m,(Ξ½iβ),(bijβ),(Zjβ)) is a Gaussian representation of (Y1β,β¦,Ypβ), and (Y1β,β¦,Ypβ) is a Gaussian random vector.
Step 2: Mean vector and covariances. Apply Claim 2 of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector twice: to the representation of Step 1, giving E[Yiβ]=Ξ½iβ and Cov(Yiβ,Ylβ)=βjβbijβbljβ, and to the original representation, giving E[Xkβ]=ΞΌkβ and Cov(Xkβ,Xkβ²β)=βjβakjβakβ²jβ. Then
E[Yiβ]=Ξ½iβ=ciβ+k=1βdβMikβE[Xkβ],
and, exchanging the order of the finite sums,
Cov(Yiβ,Ylβ)=j=1βmβbijβbljβ=j=1βmβ(k=1βdβMikβakjβ)(kβ²=1βdβMlkβ²βakβ²jβ)=k=1βdβkβ²=1βdβMikβMlkβ²βj=1βmβakjβakβ²jβ=k=1βdβkβ²=1βdβMikβMlkβ²βCov(Xkβ,Xkβ²β).
Step 3: Particular cases. The linear-combination claim is the case p=1 with c1β=c and M1kβ=ckβ. The subfamily claim is the case ciβ=0 and Mikβ=1 if k=iiβ and Mikβ=0 otherwise, where i1β<β―<ipβ are the selected indices, so that Yiβ=Xiiββ everywhere on Ξ©. Sums and differences are the case p=1 with coefficients in {1,β1} and c=0. β