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Transience of a MC

  1. Apr 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the following model.

    [itex]X_{n+1}[/itex] given [itex]X_n, X_{n-1},...,X_0[/itex] has a Poisson distribution with mean [itex]\lambda=a+bX_n[/itex] where [itex]a>0,b\geq{0}[/itex]. Show that [itex]X=(X_n)_{n\in\mathrm{N_0}}[/itex] is a transient M.C if [itex]b\geq 1[/itex].


    2. Relevant equations

    How do we approach this question? I was thinking of using the theorem below.

    Let [itex]X[/itex] be an irreducible Markov chain with countable state space [itex]S[/itex]. A necessary and sufficient condition for [itex]X[/itex] to be transient is the existence of a non-constant, non-negative super-harmonic function [itex]\phi[/itex].



    3. The attempt at a solution
    I was thinking of using an exponential function as a superharmonic function, but failed terribly. What superharmonic function can we use to prove transience for [itex]b\geq 1[/itex] Thanks in advance.
     
    Last edited: Apr 19, 2013
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  3. Apr 19, 2013 #2

    Ray Vickson

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    Something is missing: you have included no statement about what happens to/with the super-harmonic function ##\phi## in the context of ##X##.
     
  4. Apr 19, 2013 #3
    I attempted to use functions such as [itex]\phi(x) = e^{bx}, e^{(b-1)x}[/itex], but all of them are not superharmonic w.r.t [itex]X[/itex]. What type of functions should I attempt?
     
  5. Apr 19, 2013 #4

    Ray Vickson

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    You are missing the whole point: WHAT is supposed to happen if I give you a superharmonic function? You quoted only half of a theorem; the other half is vital!
     
  6. Apr 19, 2013 #5
    I am sorry about that! This is the definition of a superharmonic function!

    Let [itex]X[/itex] be a time-homogeneous irreducible Markov chain with countable state space [itex]S[/itex] and one-step transition probability matrix [itex]P(x, y)[/itex]. A function [itex]\phi: S \rightarrow R[/itex] is said to be superharmonic for X at [itex]x \in S[/itex] if [itex]\sum_{y\in S} P(x,y)\phi(y)\leq\phi(x)[/itex]
     
  7. Apr 20, 2013 #6

    Ray Vickson

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    If you don't care about signs, just getting a superharmonic ##\phi## is easy: in this case, ##\phi(x) = -x## is superharmonic if ##a>0,\: b \geq 1##. However, if you want a non-negative ##\phi## it is harder. You can follow the construction in
    http://math.stackexchange.com/questions/165913/markov-chains-recurrence-and-transience
     
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