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In finite dimensions, if $T$ is a linear operator and $x$ is a (centered) Gaussian random variable, then $Tx$ is again a (centered) Gaussian random variable.

Now suppose that $x$ is a (say, centered) random variable on $\mathbb{R}^\mathbb{N}$.

Lemma 2.2.8 together with Example 2.3.5 in Bogachev's monograph on Gaussian measures seem to demonstrate that $Y =\sum_{j=1}^\infty f_j x_j$ is (distributed with) a centered Gaussian measure on the real line with variance $\|f\|_{\ell^2}^2$, when $f$ is in $\ell^2$.

This is interesting because even though $x$ lies in $\ell^2$ with probability 0, one can nonetheless define the random variable $\sum_{j=1}^\infty f_j x_j$---looking through the proof it seems to be because if we define $f^{(n)} \colon x \mapsto \sum_{j \leq n} f_j x_j$, then $\|f^{(n)} - f\|_{L^2(\gamma)}^2 = \sum_{j>n} f_j^2 \to 0$, demonstrating that $f$ is the limit of elements in $(\mathbb{R}^\mathbb{N})^\ast \cong c_{00}$, the space of sequences with finitely many nonzero. One can then use this to analyze the characteristic functional of $Y$.

Question: Now suppose that $T \colon \ell^2 \to \ell^2$ is a bounded linear operator and as above that $x$ is a random variable drawn according to the Gaussian law on $\mathbb{R}^\mathbb{N}$. What additional assumptions are needed so that $Z = Tx$ is a centered Gaussian random variable?

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  • $\begingroup$ Reference: Bogachev, Vladimir I. Gaussian measures. Mathematical Surveys and Monographs, 62. American Mathematical Society, Providence, RI, 1998. xii+433 pp. ISBN: 0-8218-1054-5 $\endgroup$
    – Drew Brady
    Commented Jan 26, 2023 at 0:08
  • $\begingroup$ I wouldn’t expect any additional assumptions to be needed $\endgroup$
    – user44143
    Commented Jan 26, 2023 at 2:39
  • $\begingroup$ @MattF., I tried to find a proof of this claim, and I couldn't locate it. Do you know of one? $\endgroup$
    – Drew Brady
    Commented Jan 26, 2023 at 3:25
  • $\begingroup$ A precise statement is that one can find a Hilbert space $H$ containing $\ell^2$ such that $T$ can be extended (uniquely up to measure zero modifications) to a linear map from a measurable linear subspace of $\mathbb{R}^{\mathbb{N}}$ to $H$. In particular, the law of $Z$ is a Gaussian measure on $H$. See for example Prop. 3.46 and Thm 3.47 in my lecture notes on SPDEs at hairer.org/notes/SPDEs.pdf. $\endgroup$
    – Martin Hairer
    Commented Jan 26, 2023 at 7:32
  • $\begingroup$ Since the vector $x$ is assumed to be Gaussian and centered, the components $x_i$ must be orthonormal therefore i.i.d. if you want the map $f \mapsto \sum_i f_ix_i$ from $\ell^2$ to $L^2(P)$ to be an isometry. $\endgroup$
    – Christophe Leuridan
    Commented Jan 26, 2023 at 15:53

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