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Feb 7, 2014 at 15:24 comment added Tom LaGatta 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.
Feb 7, 2014 at 15:24 vote accept Tom LaGatta
Feb 7, 2014 at 4:00 comment added Nate Eldredge @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.
Feb 7, 2014 at 3:21 comment added Tom LaGatta 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?
Feb 7, 2014 at 1:14 history answered Nate Eldredge CC BY-SA 3.0