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Feb 7, 2014 at 15:24 vote accept Tom LaGatta
Feb 7, 2014 at 3:22 comment added Tom LaGatta @Miguel: I do like the idea of bootstrapping off the classical Cameron-Martin theorem for Gaussians, if they are available.
Feb 7, 2014 at 3:10 comment added Tom LaGatta Furthermore, there may not be a Gaussian measure corresponding to covariance k. For example, the identity operator on an infinite-dimensional Hilbert space.
Feb 7, 2014 at 1:16 comment added Nate Eldredge @Miguel: No, for example $\mathbb{P}$ could be a point mass, or supported on two points.
Feb 7, 2014 at 1:14 answer added Nate Eldredge timeline score: 4
Feb 6, 2014 at 23:31 comment added Miguel 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 $k$?
Feb 6, 2014 at 23:09 history asked Tom LaGatta CC BY-SA 3.0