most recent 30 from http://mathoverflow.net 2013-05-19T12:52:49Z http://mathoverflow.net/feeds/user/13166 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/56272#56272 vinay 2011-02-22T12:14:11Z 2011-02-22T12:14:11Z

<p>I did not quite get the first answer(the one using Brownian Motion). If the process starts at x(not equal to 0), the distribution of X(0) is delta(x) and transition kernels are that of brownian motion and if x = 0 then distribution of x(0) is delta(0) and transition kernels according as a constant stochastic process. How do we mix the 2 processes? Sorry if I am missing something silly.</p>