most recent 30 from http://mathoverflow.net 2013-05-23T11:27:57Z http://mathoverflow.net/feeds/question/20839 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20839/entropy-of-markov-processes Alekk 2010-04-09T13:28:14Z 2010-04-09T14:43:43Z

<p>Consider a Markov process $X_t$ with generator $L$ and invariant distribution $\pi$, whose distribution at time $t$ is given by $\pi(t,dx)=\phi(t,x) \pi(dx)$ - in other word, $\phi(t,x)$ is the density of $\pi(t, dx)$ wiht respect to the invariant distribution $\pi$. Define the (relative) entropy $$ S(t) = -\int \phi(t,x) \ln \phi(t,x) \pi(dx) \leq 0.$$</p> <p>One can expect (Boltzman H-theorem) the entropy $S$ to increase over time, and eventually to converge to $0$.</p> <p><strong>question:</strong> what conditions should be imposed in order for such a result to be true ?</p> <p>Fokker-Planck equation shows that for any test function $f$, $$ \int f(x) \partial_t \phi(t,x) \pi(dx) = \int (Lf)(x) \phi(t,x) \pi(dx)$$ so that $$S'(t) = -\int L (\ln \circ \phi)(x,t) \phi(x,t) \pi(dx), $$ but I still do not see why this quantity should be non-negative.</p>

http://mathoverflow.net/questions/20839/entropy-of-markov-processes/20845#20845 Steve Huntsman 2010-04-09T14:43:43Z 2010-04-09T14:43:43Z

<p>You can find the relevant calculation <a href="http://books.google.com/books?id=ZKPGzjS0IhcC&amp;pg=PA210#v=onepage&amp;q&amp;f=false" rel="nofollow">here</a>.</p>