Stationary distribution

  • Thread starter Ardla
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  • #1
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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?
 
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Answers and Replies

  • #2
mathman
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the first line of your post is garbled. You need to fix it to get a response.
 
  • #3
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the first line of your post is garbled. You need to fix it to get a response.


Sorry i didn't realise, I fixed it now, can you please help?
 
  • #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)
 
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  • #5
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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)?
 
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  • #6
mathman
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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)?
I need to know what your terms mean. "MC aperiodic " - what is MC?, "d(i)" - what is d and what is i?
 
  • #7
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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
 
  • #8
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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

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.
 
  • #9
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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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