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I am wondering whether a non-degenerate continuous Gaussian process is equivalent in distribution to a linear transformation of itself.

More specifically, let $T$ be a separable, complete and compact metric space, $C(T,\mathbb{R})$ the set of continuous, real-valued functions on $T$ and $\mathcal{B}$ the Borel-$\sigma$-algebra on $C(T,\mathbb{R})$. Let $$X = (X_t)_{t\in T} \colon (\Omega,\mathcal{A},\mathbb{P}) \to (C(T,\mathbb{R}),\mathcal{B})$$ be a Gaussian process with continuous mean function $m\colon T\to \mathbb{R}$ and continuous, positive definite covariance function $K\colon T\times T\to \mathbb{R}$. Then the distribution $\mathbb{P}^X = \mathbb{P}(X^{-1}(\cdot))$ of $X$ is a probability measure on $\mathcal{B}$.

Now let $a\in \mathbb{R}\setminus\{0\}$ and $f\in C(T,\mathbb{R})$. Then $aX + f = (aX_t + f(t))_{t\in T}$ again defines a Gaussian process with mean function $am + f$ and covariance function $a^2K$. My question now is whether $aX + f$ is equivalent to $X$ in distribution, that is, if $$\mathbb{P}^X \stackrel{?}{\sim} \mathbb{P}^{aX + f},$$ where by equivalence of two measures I mean absolute continuity with respect to each other. Clearly, it is necessary for $X$ to be non-degenerate (i.e. for $K$ to be positive definite), but in that case it seems like an intuitive statement to me. If it is at all true, I could also imagine it only holds if $f$ is a "typical" path of $X$ in some sense.

If I am not mistaken, it follows from Theorem 1 in "Equivalence and perpendicularity of Gaussian processes" by J. Feldman (https://projecteuclid.org/euclid.pjm/1103039696) that the distributions of two Gaussian processes in $C(T,\mathbb{R})$ can only be equivalent or orthogonal, so it would suffice to show that the distributions of $X$ and $aX + f$ are not orthogonal.

Edit: An approach suggested by E-A in the comments is to use the fact that X (as well as its translate) assigns positive probabilities to open neighbourhoods of continuous functions (see f.ex. https://arxiv.org/abs/math/0702686, Theorem 4). Some more thought would have to go into this, however, since in general, two Borel probability measures that both assign positive probability to open balls can still be orthogonal (f.ex. any probability measure with positive weights on $\mathbb{Q}$ and its translate by $\pi$).

Peter Koepernik
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  • And by equivalent you mean absolutely continuous with respect to each other? – WoolierThanThou Sep 02 '20 at 13:51
  • Yes, I will add it to the question to clarify. – Peter Koepernik Sep 02 '20 at 14:33
  • @Peter Does every continuous path not have positive probability? I am assuming when you made $K$ positive definite, you also ended up ruling out processes like a Brownian bridge, and processes like X_t = cX. I think if you can show that for Gaussian processes with PD $K$, all epsilon neighbourhoods of continous functions have positive probability, then the result you want will follow, since then you can freely shift by $f$ – E-A Sep 06 '20 at 19:35
  • @E-A I do believe that holds (a similar result can be found in https://arxiv.org/abs/math/0702686, Theorem 4). Could you please elaborate on how this would imply equivalence? In general, two Borel measures that both assign positive probability to every open ball can still be orthogonal, for example $\mu(\cdot) = |\cdot \cap \mathbb{Q}|$ and $\nu(\cdot) = |\cdot \cap (\mathbb{Q} + \pi)|$ as measures on $\mathbb{R}$. – Peter Koepernik Sep 07 '20 at 07:31
  • Thanks for the example; I think my suggestion was too naive! I mean, the measures you picked are not probability measures as written (assuming $|\cdot|$), but there exists a probability measure on $\mathbb{Q}$, so you can adjust your example as such so it is a good counterexample. While you should really not be having anything so pathological given that you have a gaussian process with full support, I guess I will have to think about it more; good question! – E-A Sep 07 '20 at 08:04
  • Thanks! I will Include the idea as an edit in the question. – Peter Koepernik Sep 07 '20 at 08:24
  • My guess would be that this only holds for $f$ in the Reproducing Kernel Hilbert Space associated with $K$. It cannot hold for "typical" f: for $K$ the kernel of Brownian motion and $f$ a typical path, the typical path in $X+f$ has twice the quadratic variation. – Bananach Sep 07 '20 at 08:37
  • @Bananach Good point, thanks. Do you have an idea how one could prove the statement in that case (when $f$ is in the RKHS of $K$)? – Peter Koepernik Sep 07 '20 at 11:13

1 Answers1

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If $T$ is infinite, then the necessary and sufficient conditions for the distribution of $aX + f$ to be equivalent to that of $X$ are:

  1. $|a| = 1$
  2. $(a-1)m + f$ is in the Cameron-Martin space (RKHS) of $X$

For more details, see any book on Gaussian measures e.g. Kuo or Bogachev.

zhoraster
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  • Thanks! This does answer the question, but I am not very familiar with Cameron-Martin spaces. How "rich" is it? Is it, for example, dense in $C(T,\mathbb{R})$ under some assumptions on $T$ and $X$? – Peter Koepernik Sep 07 '20 at 13:19
  • @Peter, the completion of the Cameron-Martin space is the support of the distribution of $X$. And the support is full iff the covariance operator is injective.

    Say, for $X$ in $C(T,\mathbb R)$, the Cameron-Martin space is dense iff there exists no non-zero finite signed measure $\mu$ on $T$ such that $\langle X,\mu\rangle = 0$ a.s.

     – zhoraster Sep 07 '20 at 13:39
  • Thank you very much! – Peter Koepernik Sep 07 '20 at 15:20