Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a Borel space and let us consider a Markov Chain $(\Phi_n)_{n\geq 0}$ on this space given by the stochastic kernel
$$ P(x,\mathrm dy) = p(x,y)\mu(\mathrm dy) $$ where the density $p$ is continuous and the measure $\mu$ is bounded. For any set $A$ we define $$ \tau_A = \inf(n\geq 0:\Phi_n\in A) $$ to be the first hitting time of $A$ and let $f_n(x,A) = \mathsf P_x(\tau_A>n)$ be the survival probability of $\tau_A$. Clearly, for any fixed $x$ the sequence $(f_n(x,A))_{n\geq 0}$ is in $[0,1]$ and non-increasing. Let us define $$ m(A) = \inf\left(n\geq 0:\sup_x f_n(x,A)<1\right). $$ The set $A$ is called uniformly transient (UT) whenever $m(A)<\infty$. In such case functions $f_n(x,A)$ uniformly and exponentially fast converge to $0$. Suppose, that although I don't know functions $f_n$ explicitly, I can compute functions them with any given precision and I am interested in checking whether a given set $A$ is UT or not.

In case it appears that at some moment $f_n(x,A)\leq 1-\delta$ for all $x$, then $m(A)\leq n<\infty$. However, suppose that $A$ is not UT (which we never know in advance) - then computing $f_n$ I'll never meet a condition $f_n\leq 1-\delta$.

Q: I wonder, if it is possible to provide some sort of a threshold condition, i.e.

if $\sup\limits_x f_{n'}(x,A) \geq 1-\delta'$ then $m(A) = \infty$

Why do I have a hope for this? Well, in case $X$ is a finite space we can take $\mu$ to be a counting measure, so $p(x,y)$ are just transition probabilities. The correspondent adjacency graph is $(V,E)$ where $V = X$ and $E_{ij} = 1$ if $p(i,j)>0$ and is zero otherwise. Then $$ m(A) = \max_{x\in A}m_x(A) $$ where $m_x(A)$ is the shortest path from $x$ to $X\setminus A$ in the graph $(V,E)$. Clearly, if $m_x(A)<\infty$ then it contains at most as many elements as there are in $A$ so that $$ m(A)\leq \mu(A). $$ Thus if $\sup\limits_xf_{\mu(A)+1}(x,A) = 1$ then $m(A) = \infty$, which is an example of the threshold condition I'm interested in.

It is known, that when $p$ is continuous and $\mu$ is nice enough, Markov Chains on the topological state spaces behave very close to those over discrete state spaces, so I tried to look for the similar results, but didn't find anything in particular so far. I was trying to use some minorization conditions like $$ P(x,\mathrm dy)\geq \beta \cdot 1_{A_k}(x)\nu(\mathrm dy) $$ which tell intuitively that points in $A_k$ make transitions "uniformly", here $A_k$ is an element of a partition of $A$. However, no interesting results I have found. Any help is highly appreciated.

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.