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      A stochastic process is a collection of random variables usually indexed by a totally ordered set.

      3
      votes
      There is no need to talk about "some probability space" on which your random variable is defined (as probabilists like to do). Everything is determined just by the uniform (aka Lebesgue) measure on $[ …
      answered Apr 13 '14 by R W
      2
      votes
      Unfortunately, the probabilistic argot can be sometimes not so easy to understand for outsiders. Also, I think that your notation is unnecessarily complicated. So, what do we have here? The state spa …
      answered Jul 12 '16 by R W
      2
      votes
      I am not exactly sure about what is meant by "probabilistic", but there is a simple argument based on the Cauchy inequality (no spectral theory involved) which provides a so-called "ratio limit theore …
      answered Apr 26 '11 by R W
      2
      votes
      There is a well-known characterization of weakly mixing dynamical systems as the ones whose product by any ergodic system is also ergodic. In your case $Y(t)$ is weakly mixing, because it is Bernoulli …
      answered Dec 2 '17 by R W
      8
      votes
      One can see it much easier by using the subadditive ergodic theorem, which is a standard tool for dealing with the law of large numbers. Since the second moment of $W_t$ is obviously finite (this is …
      answered Sep 26 '13 by R W
      3
      votes
      You don't actually formulate your question, but I presume that you want to know about the probability $p=p(a,b,m)$ (strictly speaking, one has to specify the measure $P$ on the path space, but in fact …
      answered Jul 4 '16 by R W
      2
      votes
      It's much easier to understand and answer this question by using the (equivalent) language of families of transition probabilities corresponding to stochastic kernels. So, let $(\pi_x)$ (resp., $(\til …
      answered Jul 26 '12 by R W
      2
      votes
      Since you don't want to exclude the case of infinite stationary measures, it is very easy to produce a counterexample. The simple random walk on $\mathbb Z$ has a unique stationary measure (which coin …
      answered Jun 16 '16 by R W
      1
      vote
      One way to do it is by first proving it for an appropriate "comparison" chain just on $\mathbb R_+$, and then showing that there is a measure preserving map $\pi$ from the path space of the compariso …
      answered Jul 26 '16 by R W
      1
      vote
      I do not think that one can expect any general results in the case of ininite graphs. Take just the integer line $\mathbb Z$, for which all bounded harmonic functions are constants, so that the limit …
      answered Apr 20 '18 by R W
      7
      votes
      This is a particular case of random walks with internal degrees of freedom (aka semi-Markov random walks or covering Markov chains). In the case when the translation group is just $\mathbb Z$ (like in …
      answered May 29 '13 by R W
      1
      vote
      No. The "theory" which you refer to is actually wrong. The mistake is that you confuse several conditions on a sequence of invertible matrices $B_n$. In the order of strengthening these conditions are …
      answered Mar 11 '15 by R W
      2
      votes
      The identity you are referring to can be interpreted as saying that if you run your Markov chain on $\mathbb Z$ from the initial distribution $\pi$, then the probability of crossing the edge $[k,k+1]$ …
      answered Nov 8 '10 by R W
      0
      votes
      The point $x=(1,1,\dots,1)$ is an absorbing state, and $p(y,x)$ is bounded away from 0 for any other state $y$. The only stationary measure of any countable Markov chain with this property is $\delta_ …
      answered Mar 27 '13 by R W
      1
      vote
      There is a general formula for the infinitesimal generator of the Brownian bridge on an arbitrary Riemannian manifold, and I do not see how it would become simpler in your particular case. For instanc …
      answered Sep 5 '15 by R W

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