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Q's: I suspect this is true: if two states in a markov chain communicate and one is recurrent, then the other is recurrent.

My approach is, lets say i and j are two states that communicate and as i is recurrent, this means you are guaranteed to revisit i everytime you reach it, so you have infinite points in discrete time which you reach i and as i communicates with j, this implies j is recurrent as we are always guaranteed to revisit j after reaching it.

My problem with this proof is it involves alot of words, so an alt.proof would be great. Thanks

Raul
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  • your argument seems sound to me. no need to overdo the symbolic reasoning. perhaps, if you wish to be thorough, you can state the definition of "communicate" in terms of a conditional probability, and state the definition of "recurrent" in symbolic terms, and then deduce the required result without a great deal of fuss – David Holden Dec 13 '13 at 15:16
  • @DavidHolden well my definition for communicating is there is exist a natural number k s.t. Pi,j(k)>0, and my definition for recurrence is Pi(Ti<infinity)=1, where Ti is the first passage time to i. I can't seem to get Pj(Tj<infinity)=1 without mentioning the words above, is that ok? – Raul Dec 13 '13 at 15:25
  • it looks like you could do with some help learning to write mathjax so your symbols will be more legible, and you might then be less dependent on verbal explanations. if you look at any post on MSE and right-click on a formula a menu will come up.press "show math as" then click "LaTex" and the formula will appear as the way it has to be written. it may sound comple3x but in fact it is very logical, and an absolutely wonderful tool. i've only been using it for about a month and it has improved my math no end. so i urge you to do this a.s.a.p. – David Holden Dec 13 '13 at 19:01
  • for example, where you have written Pj(Tj)<infinity)=1, if you change that to P_j(T_j) \lt \infty =1 and then enclose that between two dollar signs it will come out like this: $P_j(T_j) \lt \infty =1$. so try it now - just right-click on the correct version – David Holden Dec 13 '13 at 19:03
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    the good news is that your understanding of the question you raised is completely sound. i've noticed quite a few people here get a bit bogged down with symbols, and often miss a more concise solution. i admire the tenacity of some people to bash their way through tricky manipulations, but i am error-prone, which, actually, serves as a stimulus to seek an elegant approach, if one exists. math is very much an art, as well as a science. – David Holden Dec 13 '13 at 19:06

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You are decribing what's called a Regular Markov Chain. Every location has visitors and from every location you can visit another location. No visitor can leave the Chain and no outsider cen enter. This means that the transition matrix (there are different names for such a matrix) has columns that add up to 1. The fact is that the dominant eigenvalue of such a matrix is 1. That ought to be the first part of your proof. Then any particular distribution vector can be expressed as a Linear Combination of the eigenvectors of the matrix. In the long run (i.e. taking high powers of the matrix), Linear Algebra slhows us that the dominant eigenvalue prevails as the others "die out" for high powers. Thus, the entries of the corresponding eigenvector gives the ratio of the equilibrium. This is the proof in "words"

imranfat
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