Let $X$ be a homogeneous Markov process in a continuous time with value in the set $E$. Suppose that for some $T>0,x\in E, A\subset E$ we have $$ P_x[X_t\in A] = 0 $$ for all $t\in [0,T]$ but $$ P_x[X_{t'}\in A] >0 $$ for some $t'> T$. I wonder how to find an example of such a process $X$. The simple one $$ dX_t = f(X_t)dt $$ in not interesting because it is purely deterministic.
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2 Answers
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This question was already asked here. Homogenous Markov chains won't work because the probability distributions of $X_t$ and $X_s$ are mutually absolutely continuous for every positive $t$ and $s$. For a homogenous diffusion with nonzero diffusion term, $P[X_{2t}\in A]\ne0$ would require that $P_y[X_t\in A]\ne0$ for $y$ in a set of positive measure with respect to the probability distribution of $X_t$ while $P_x[X_t\in A]=0$. Unless I am missing something, this is impossible.
The idea of the proof for finite Markov chains is as follows. Assume that $P_x[X_t\in A]$ is positive for a given positive $t$. The transitions of the chain can be wriiten as $$ P_y[X_t=z]=(\mathrm{e}^{tQ})(y,z), $$ where $Q$ is the infinitesimal generator of the process. Choose $u$ such that $u\ge -Q(y,y)$ for every state $y$, then $uI+Q$ has nonnegative coefficients. Now, $$ P_x[X_t\in A]=\mathrm{e}^{-ut}(\mathrm{e}^{t(uI+Q)})(x,A) $$ is a linear combination with positive coefficients of nonnegative terms $(uI+Q)^n(x,A)$, hence $(uI+Q)^n(x,y)$ is positive for at least one given $n$.
Now, for every positive time $s$, $P_x[X_s\in A]$ is also a linear combination with nonnegative coefficients of $(uI+Q)^k(x,y)$, in particular $P_x[X_s\in A]$ is at least $\mathrm{e}^{-us}(s^n/n!)(uI+Q)^n(x,A)$, hence is positive.
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$\begingroup$ Sorry, I'm missing a lot in your answer. 1. For the diffusion case you suppose that my hypothesis is true if the coefficient of diffusion term is always non-zero? (It seems to me too) 2. For the continuous Markov chain what did you prove, that I'm right or wrong? $\endgroup$– SBFCommented Jan 5, 2011 at 19:15
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$\begingroup$ The answer gives a proof that for denumerable homogenous continuous time Markov chains, if $P_x[X_t\in A]>0$ for a given positive $t$, then $P_x[X_s\in A]>0$ for every positive $s$. The same result should hold for homogenous diffusions with nonzero diffusion coefficient. $\endgroup$– DidCommented Jan 5, 2011 at 19:27
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$\begingroup$ As I understand you denote $(e^{tQ})(y,z)$ as an element of the matrix on row $y$ and column $z$. Then you are mixing $A$ and $y$ easily - this is a little bit confusing, could you explain what is $(e^{tQ})(y,z)$ and $(e^{tQ})(y,A)$ in your notation? $\endgroup$– SBFCommented Jan 6, 2011 at 16:14
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$\begingroup$ But I like your logic - this is prove based on the generator which I'm missing for the general homogeneous Markov process. If you would explain, I could try to make an analogue for the general case. $\endgroup$– SBFCommented Jan 6, 2011 at 16:22
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$\begingroup$ Both $Q$ and $\mathrm{e}^{tQ}$ are matrices indexed by $E\times E$, where $E$ is the state space of the Markov chain. Since $\mathrm{e}^{tQ}$ is defined by the usual exponential series for matrices, $(\mathrm{e}^{tQ})(y,z)$ is the sum of $(t^n/n!)Q^n(y,z)$ over every $n\ge0$, with the convention that $Q^0$ is the identity matrix. For every such matrix $M$, every $x$ in $E$ and every $A\subset E$, $M(x,A)$ is the sum of $M(x,y)$ over every $y$ in $A$. For example, $(\mathrm{e}^{tQ})(x,E)=\mathrm{e}^{t}$. (What you call "my logic" is explained in standard textbooks on Markov chains.) $\endgroup$– DidCommented Jan 6, 2011 at 17:02
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The porous medium equation provides examples. See, for instance, M. Inoue, A Markov process associated with a porous medium equation, Proc. Japan Acad. 60 (1984), 157-160.
answered Jan 5, 2011 at 10:07
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$\begingroup$ As I understand, it is a non-homogeneous Markov process since it is written as $$ X_t = X_0 + \int\limits_0^t u(s,X_s)dB_s. $$ In the topic I mentioned that I am looking for the homogeneous one. $\endgroup$– SBFCommented Jan 5, 2011 at 10:26