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Stationary distribution

  1. Sep 29, 2009 #1
    Hi, can someone please provide some guidance on how i should go about finding the stationary distribution of:

    [tex]X_t = [/tex] [tex] \rho X_{t-1} + \epsilon_t[/tex], [tex]X_0 = 0 [/tex]and [tex]|\rho|<1[/tex]
    where [tex]\epsilon_1, \epsilon_2, \cdots[/tex] are all independent N(0,1)..

    i have no idea what to do, so here's my attempt which i know to be completely wrong:
    suppose,
    [tex]Var(X_1) = \rho \sigma^2 < \infty [/tex]
    [tex] Var(X_2) = \rho\sigma^2 + 1 [/tex]
    [tex] \vdots [/tex]
    [tex]Var(X_{n+1}) = \rho\sigma^2 + t [/tex]
    As [tex] t \rightarrow \infty, Var(X_{n+1} = \rho \sigma^2 + t [/tex] ????????

    yeah im very sure im not doing it right... Can someone please help me out?
     
    Last edited: Sep 30, 2009
  2. jcsd
  3. Sep 29, 2009 #2

    mathman

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    the first line of your post is garbled. You need to fix it to get a response.
     
  4. Sep 30, 2009 #3

    Sorry i didn't realise, I fixed it now, can you please help?
     
  5. Sep 30, 2009 #4

    mathman

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    Since X0=0, V(X1)=1.
    Next V(X2)=r2+1.
    V(X3)=r2(r2+1)+1.
    etc. (r=rho).

    The limit as n->oo of V(Xn)=1/(1-r2)
     
    Last edited: Sep 30, 2009
  6. Sep 30, 2009 #5
    THank you!!

    Sorry can i ask another question in this same post? Well is that MC aperiodic too because [tex]V(X_1) = 1[/tex]? Is that the value of d(i)?
     
    Last edited: Sep 30, 2009
  7. Oct 1, 2009 #6

    mathman

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    I need to know what your terms mean. "MC aperiodic " - what is MC?, "d(i)" - what is d and what is i?
     
  8. Oct 1, 2009 #7
    Sorry I meant Markov Chain. Um the d(i) is the period defined as:
    [tex] d(i) = gcd[/tex]{[tex]n:p_{ii}(n)>0[/tex]}
    i.e.the greatest common divisor
    and something is aperiodic if d(i) = 1.

    Coz like in my notes it says Markov is aperiodic if it can access all states. But im not sure about how to show it
     
  9. Oct 2, 2009 #8
    Hi,
    A Markov chain is aperiodic if all the states are in one class (as periodicity is a class property and the chain itself is called aperiodic in your case) and starting from state i, there is a non-zero probability of transition to state i (this is of course given by your definition of d(i)). I think if it can access all states the chain would be called irreducible (because it is a single communicating class and there are no other states so that it is also closed).

    About your first question, I just want to confirm whether you are also getting the stationary distribution as N(0,1/1-rho^2).

    Now as your state space is continuous I am not really very sure about the definition of periodicity. I guess that a reasonable defintion of aperiodicity would be when P(X(1) \in A | X(0) \in A) > 0 for A \in the state space. (Also there would be some restriction on what A could be). Because epsilon is Gaussian I think that the chain is aperiodic.
     
  10. Oct 2, 2009 #9
    hmm! well i think that the stationary distribution is right..

    umm yeahh im not sure about the aperiodicity too, but i agree i think that its aperiodic. I think that i'll just say coz it can reach any subspace in one step? its ok, i think that i'll get the answer from my lecturer when uni starts again..

    but thank you guys!
     
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