MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\Phi$?

share|cite|improve this question
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 $k$? – Miguel Feb 6 '14 at 23:31
@Miguel: No, for example $\mathbb{P}$ could be a point mass, or supported on two points. – Nate Eldredge Feb 7 '14 at 1:16
Furthermore, there may not be a Gaussian measure corresponding to covariance k. For example, the identity operator on an infinite-dimensional Hilbert space. – Tom LaGatta Feb 7 '14 at 3:10
@Miguel: I do like the idea of bootstrapping off the classical Cameron-Martin theorem for Gaussians, if they are available. – Tom LaGatta Feb 7 '14 at 3:22
up vote 2 down vote accepted

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.

share|cite|improve this answer
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? – Tom LaGatta Feb 7 '14 at 3:21
@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. – Nate Eldredge Feb 7 '14 at 4:00
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. – Tom LaGatta Feb 7 '14 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.