most recent 30 from http://mathoverflow.net 2013-06-20T08:17:47Z http://mathoverflow.net/feeds/question/88129 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88129/when-is-mathbfexfx-t-a-continuous-function-of-x ShawnD 2012-02-10T19:01:19Z 2013-03-22T04:22:00Z

<p>Let $E$ be a locally compact Hausdorff space with countable base and $X_t$ be a stochastic process taking values in the one-point compactification of $E$ (with the Borel $\sigma$-algebra). Let $f$ be a continuous function vanishing at infinity. I'm wondering under what conditions it is true that $x \mapsto \mathbf{E}^{x}[f(X_t)]$ is a continuous function?</p> <p>If $X_t$ is a Brownian motion on $\mathbb{R}$, it is straightforward to verify this is true and, in fact, it is true whenever $X_t$ is an Ito diffusion.</p> <p>An example where this fails is if we let $E=\mathbf{R}$ and let $X_t$ be a reflected Brownian motion on ${ x: x\geq 0 }$ and be the negative of the absolute value of a Brownian motion in $\mathbb{R}^3$ on ${x : x&lt;0 }$.</p> <p>More broadly, I'm curious as to what kind of conditions on the sample paths of a Markov process $X$ with continuous paths force it to be a Feller process. The condition I'm asking about seems to be the one that does not come for free if you start with such an $X$.</p>

http://mathoverflow.net/questions/88129/when-is-mathbfexfx-t-a-continuous-function-of-x/90190#90190 Pascal Maillard 2012-03-04T10:19:25Z 2012-03-05T15:05:04Z

<p>UPDATE: This is at best a partial answer.</p> <p>If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion it be <em>strong</em> Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0). In that case, however, it is still not necessarily Feller (this was stated previously).</p> <p>If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost, see <a href="http://mathoverflow.net/questions/67067/extending-state-space-to-make-a-process-feller" rel="nofollow">this thread</a>) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)</p> <p>If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.</p>