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](https://en.wikipedia.org/wiki/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](http://bit.ly/1e7sZFg) 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 $\Phi$?



Cameron-Martin theorem for non-Gaussian measures
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.


