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I am studying probability theory in infinite dimensional spaces and want to know if things that hold in the usual(?) theory holds in a similar manner.

Fix a real separable Hilbert space $H$ with inner product $\langle\cdot,\cdot\rangle$. Let $X$ be a $H$-valued Gaussian random variable on a probability space$(\Omega,\mathscr{F},P)$ such that $E\langle X,h\rangle\langle X,h'\rangle=\langle Qh,h'\rangle$ (e.g. a $Q$-Wiener process at a fixed time) with a trace class covariance operator $Q\colon H\to H$.

The usual probability theory tells us that for a nice function $f\colon\mathbb{R}\to\mathbb{R}$ we have that $f(\langle X,h\rangle)$ and $f(\langle X,h'\rangle)$ are independent. Also that if $f$ is linear (or affine) $f(\langle X,h\rangle)$ is again Gaussian, but not in general, e.g., $f(\cdot)=|\cdot|$.

My question is,

For nice $T\colon H\to H$ not necessarily linear, can we hope $ \langle TX,h\rangle $ and $ \langle TX,h'\rangle $ are independent, or uncorrelated? When $h,h'$ are eigenfunctions of $Q$ such that they are orthogonal, and $T$ is linear diagonalized by $(h)$, then it should be true, but I have no idea how to prove or disprove the above in general. I wanted to go back to the good old definition of independence of random variables (independence of $\sigma$-algebra generated by them), but didn't seem it would help.

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  • $\begingroup$ Is the condition in the first equation as intended? It seems to me that it implies $Q = \lambda I$ for some $\lambda$. $\endgroup$
    – Sangchul Lee
    Feb 2, 2017 at 4:27
  • $\begingroup$ $E\langle X,h\rangle\langle X,h'\rangle=\langle Qh,h'\rangle$? yes, it is as intended. $\endgroup$
    – user412313
    Feb 2, 2017 at 4:38
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    $\begingroup$ What I mean is the condition $$ \langle h, h' \rangle \quad \Rightarrow \quad \langle X, h\rangle \text{ and } \langle X, h' \rangle \text{ are independent}. $$ If $X$ is centered (as it seems tacitly assumed), then this implies that $$ \langle h, h' \rangle = 0 \quad \Rightarrow \quad \langle Qh, h' \rangle = \Bbb{E}[\langle X, h \rangle\langle X, h' \rangle] = 0. $$ This forces that all eigenvalues of $Q$ are identical. $\endgroup$
    – Sangchul Lee
    Feb 2, 2017 at 4:42
  • $\begingroup$ I do not really understand. Let $(e_n)$ be an eigenfunctions of $Q$ that are orthogonal in $H$. Then, if $\langle e_n,e_m\rangle=0$ then $\langle Qe_n,e_m\rangle =\lambda_n \langle e_n,e_m \rangle =0$ regardless of $\lambda_n$? $\endgroup$
    – user412313
    Feb 2, 2017 at 4:55
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    $\begingroup$ Since $Q$ is a compact self-adjoint operator, we can find an orthonormal basis $\{ e_n : n\geq 1\}$ consisting of eigenvectors of $Q$. If the condition above is satisfied, then $h = e_m + e_n$ and $h' = e_m - e_n$ are orthogonal and thus $$ 0 = \langle Q(e_m + e_n), e_m - e_n) = \lambda_m - \lambda_n, $$ where $\lambda_k = \langle Qe_k, e_k\rangle$ is the eigenvalue corresponding to $k$. This forces that $\lambda_k$ are all equal and thus $Q = \lambda_1 I$. Of course, this further forces that $Q = 0$ since $\lambda_n \to 0$ as $n\to\infty$. $\endgroup$
    – Sangchul Lee
    Feb 2, 2017 at 5:01

1 Answer 1

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If

  • $(\Omega,\mathcal A)$, $(\Omega',\mathcal A')$ and $(\Omega'',\mathcal A'')$ are measurable spaces
  • $X:\Omega\to\Omega'$ is measurable with respect to $\mathcal A$-$\mathcal A'$
  • $f:\Omega'\to\Omega''$ is measurable with respect to $\mathcal A'$-$\mathcal A''$

then $f\circ X$ is measurable with respect to $\sigma(X)$-$\mathcal A''$ and hence $$\sigma(f\circ X)\subseteq\sigma (X)\tag1\;.$$ Thus, if $\mathcal F\subseteq\mathcal A$ is a $\sigma$-algebra on $\Omega$ and $X$ is independent of $\mathcal F$, then $f\circ X$ is independent of $\mathcal F$ too.

If

  • $H:=\Omega'$ is a separable $\mathbb R$-Hilbert space, $\mathcal A':=\mathcal B(H)$ and $(e_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$
  • $E$ is a $\mathbb R$-Banach space, $\Omega''=\mathfrak L(H,E)$ and $\mathcal A''$ is the strong Borel $\sigma$-algebra on $\Omega''$

then $$f(X)=\sum_{n\in\mathbb N}\langle X,e_n\rangle_Uf(e_n)$$ and hence (since $f(e_n)$ is measurable with respect to $\mathcal A'$-$\mathcal B(E)$), $f(X)$ is indepdent of $\mathcal F$ as long as $X$ is independent of $\mathcal F$.

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