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Let $X$ be a locally convex topological linear space, and $\mathbb P$ be a probability measure on $X$. Denote the mean vector $m \in X$ and covariance operator $k : X^* \to X$. Let $\tau_u : X \to X$ be the translation-by-$u$ operator, and define the translated measure $\mathbb P_u := (\tau_u)_* \mathbb P := \mathbb P \circ \tau_u^{-1}$.

Let $U \subseteq X$ be the Cameron-Martin space of the measure $\mathbb P$, i.e., the Hilbert-space completion of the set $kX^* \subseteq X$.

If $\mathbb P$ is Gaussian, then the Cameron-Martin theorem states that $\mathbb P_u$ is absolutely continuous with respect to $\mathbb P$ if and only if $u \in U$. In that case, the Radon-Nikodym derivative equals $\exp\!\big( \langle u, x\rangle^\sim -\tfrac 1 2 \|u\|^2 \big)$, where $x \mapsto \langle u, x\rangle^\sim$ is the Paley-Wiener integral of $u$.

Suppose that $\mathbb P$ is non-Gaussian, and let $u \in U$. Is $\mathbb P_u$ absolutely continuous with respect to $\mathbb P$? If so, can we write the Radon-Nikodym derivative as $\exp\!\big( \langle u, x\rangle^\sim -\tfrac 1 2 \|u\|^2 + \Phi(u,x) \big)$, for some well-behaved function $\Phi$?

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  • $\begingroup$ So you're assuming $\mathbb{P}$ is such that any two linear coordinates on $X$ are random variables of finite vocariance given by $k$. Does that not imply that $\mathbb{P}$ is absolutely continuous with respect to the Gaussian measure with covariance $k$? $\endgroup$
    – Miguel
    Commented Feb 6, 2014 at 23:31
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    $\begingroup$ @Miguel: No, for example $\mathbb{P}$ could be a point mass, or supported on two points. $\endgroup$
    – Nate Eldredge
    Commented Feb 7, 2014 at 1:16
  • $\begingroup$ Furthermore, there may not be a Gaussian measure corresponding to covariance k. For example, the identity operator on an infinite-dimensional Hilbert space. $\endgroup$
    – Tom LaGatta
    Commented Feb 7, 2014 at 3:10
  • $\begingroup$ @Miguel: I do like the idea of bootstrapping off the classical Cameron-Martin theorem for Gaussians, if they are available. $\endgroup$
    – Tom LaGatta
    Commented Feb 7, 2014 at 3:22

1 Answer 1

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If $X = \mathbb{R}$ and $\mathbb{P}$ is any measure of finite second moment that is not a point mass, then $k$ is not zero so $k X^* = X$. But clearly we can choose $\mathbb{P}$ such that not all translates are absolutely continuous to it. Take for example $\mathbb{P}$ supported on two points, or uniform measure on an interval, or (for an example with full support) a measure supported on the rationals.

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  • $\begingroup$ These are great examples, and emphasize the importance of the support sets aligning. Suppose we insist that the support sets agree, eg, $supp \mathbb P = supp \mathbb P_u = m + U$, as in the Gaussian case. Then by assumption, the measures satisfy a Cameron-Martin decomposition for some appropriate $\Phi$. Must this function be well-behaved in any meaningful sense? $\endgroup$
    – Tom LaGatta
    Commented Feb 7, 2014 at 3:21
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    $\begingroup$ @TomLaGatta: If as in my last example, $\mathbb{P} = \sum a_n \delta_{q_n}$, where $\{q_n\}$ are the rationals and $a_n$ are chosen so as to give a measure with finite second moment, then for irrational $u$, $\mathbb{P}$ and $\mathbb{P}_u$ both have (topological) support equal to $\mathbb{R}$, and are mutually singular. $\endgroup$
    – Nate Eldredge
    Commented Feb 7, 2014 at 4:00
  • $\begingroup$ very nice counterexample. This means that the sufficient assumption I'm looking for is really the necessary one: that $\mathbb P_u$ be absolutely continuous with respect to $\mathbb P$. Assuming that by fiat, I think I'm now on the right track. $\endgroup$
    – Tom LaGatta
    Commented Feb 7, 2014 at 15:24

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