A Markov process which is not a strong markov process? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T05:41:40Z http://mathoverflow.net/feeds/question/43833 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 A Markov process which is not a strong markov process? Simon Lyons 2010-10-27T17:00:37Z 2011-02-22T12:14:11Z <p>Can anyone give an example of a Markov process which is not a strong Markov process? The Markov property and strong Markov property are typically introduced as distinct concepts (for example in Oksendal's book on stochastic analysis), but I've never seen a process which satisfies one but not the other.</p> <p>Many thanks -Simon</p> http://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/43838#43838 Answer by Byron Schmuland for A Markov process which is not a strong markov process? Byron Schmuland 2010-10-27T17:17:16Z 2010-10-27T17:17:16Z <p>A standard example is Exercise 6.17 in Sharpe's book <em>The general theory of Markov processes</em>. The process stays at zero for an exponential amount of time, then moves to the right at a uniform speed. </p> http://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/43840#43840 Answer by Andrey Rekalo for A Markov process which is not a strong markov process? Andrey Rekalo 2010-10-27T17:20:50Z 2010-10-27T17:20:50Z <p>Let $X(t) = f(W(t) + \pi)$, where $W(t)$ is a standard Wiener process and $$f(x) = \begin{cases} (x,0), &amp; x\leq 0 \\ \\ (\sin x,1-\cos x), &amp; 0 &lt; x &lt; 2\pi \\ \\ (x-2\pi,0), &amp; x\geq 2\pi \end{cases}$$ is a map from $\mathbb R$ to $\mathbb R^2$. $X(t)$ is an $\mathbb R^2$-valued Markov process on $\mathbb R_+$ which is not strongly Markovian. See <a href="http://books.google.co.uk/books?id=5D5O8xyM-kMC&amp;printsec=frontcover&amp;dq=Fristedt+and+Gray&amp;source=bl&amp;ots=7P5pY-tJce&amp;sig=67tNH1Hy40vtSXCyLbaYIY_g-Cg&amp;hl=en&amp;ei=UlvITN-9AsTFswaa1Ij4DQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CBUQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow"><em>"A Modern Approach to Probability Theory"</em></a> by Fristedt and Gray (1997, pp. 626–627).</p> <p>If the time set is discrete, the ordinary Markov property implies the strong Markov property.</p> http://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/43841#43841 Answer by George Lowther for A Markov process which is not a strong markov process? George Lowther 2010-10-27T17:24:15Z 2010-10-27T17:24:15Z <p>Consider the following continuous Markov process X, starting from position x</p> <ol> <li>if x = 0 then X<sub>t</sub>&nbsp;=&nbsp;0 for all times.</li> <li>if x &ne; 0 then X is a standard Brownian motion starting from x.</li> </ol> <p>This is not strong Markov (look at times at which it hits zero).</p> http://mathoverflow.net/questions/43833/a-markov-process-which-is-not-a-strong-markov-process/56272#56272 Answer by vinay for A Markov process which is not a strong markov process? 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>