Stationary probabilities.(Markov chain).

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Discussion Overview

The discussion revolves around characterizing the probability distribution of the number of visits to state 2 in an irreducible and positive recurrent Markov chain after two consecutive visits to state 1. Participants explore the formulation of this probability in the context of stationary probabilities.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant suggests writing out the possible sequences of visits to states 1 and 2 to derive a formula for the probability distribution.
  • Another participant expresses uncertainty about how to formulate the probabilities for the sequences, providing specific examples and calculations for certain sequences.
  • A third participant proposes that the probabilities should be multiplied by the stationary probabilities of states 1 and 2, depending on the transitions, but is unsure about a general equation for all cases.
  • A later reply references a paper that discusses the unconditional distribution of successes in trials, suggesting that a similar approach may be needed for the conditional distribution based on the visits to state 1.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a general formula for the probability distribution, and there are multiple competing approaches and uncertainties expressed throughout the discussion.

Contextual Notes

Participants note the dependence on the definitions of the states and the transitions, as well as the need for a clearer understanding of the conditions under which the probabilities are calculated.

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We are given two states 1,2 in an irreducible and positive recurrent Markov chain, and their stationary probabilities [tex]\pi_1[/tex] and [tex]\pi_2[/tex] respectively, try to characterise in general the probability (distribution) of the number of visits in state 2 after two consecutive visits in state 1.

Any hints?
 
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Write out the ways this can happen, then turn it into a formula:

1,1,2
2,1,1,2
2,2,1,1,2
1,2,1,1,2
...
 
Yes I thought in this direction but not sure how to get to the formula.
I mean for 1,1,2 the probability is: [tex]P_{1,1}\frac{\pi_1}{\pi_1+\pi_2}P_{1,2}[/tex]
2,1,1,2 [tex]\frac{\pi_2}{\pi_1+\pi_2}P_{1,1}P_{2,1}P_{1,2}[/tex]
1,2,1,1,2 [tex]\frac{\pi_1}{\pi_1+\pi_2}P^2_{1,2}P_{1,1}P_{2,1}[/tex]

So my hunch is that if we first get to 1 or 2, we should multiply by pi1/(pi1+pi2) or pi2/(pi1+pi2), and we should always multiply by P1,1, but other than this I don't see a general equation for all cases.
 
I don't have an immediate answer, but you might find the approach in this paper to be helpful:

http://smu.edu/statistics/TechReports/TR211.pdf

The authors derive the unconditional distribution of the number of successes (e.g., state 2) in n+1 trials. It seems to me that you need to derive a similar distribution, conditional on having obtained (exactly? or at least?) two consecutive failures (1,1).
 
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