Finding Stationary Distribution for a Stochastic Process

In summary, the conversation discusses the process of finding the stationary distribution of a Markov chain with the given properties. There is some confusion about the correct method, but it is determined that the chain is aperiodic and the stationary distribution is N(0,1/1-rho^2). There is also discussion about the definition of aperiodicity for a continuous state space.
  • #1
Ardla
5
0
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 I am very sure I am not doing it right... Can someone please help me out?
 
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  • #2
the first line of your post is garbled. You need to fix it to get a response.
 
  • #3
mathman said:
the first line of your post is garbled. You need to fix it to get a response.


Sorry i didn't realize, I fixed it now, can you please help?
 
  • #4
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
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
Ardla said:
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
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 I am not sure about how to show it
 
  • #8
Ardla said:
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 I am 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
hmm! well i think that the stationary distribution is right..

umm yeahh I am 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!
 

1. What is a stationary distribution?

A stationary distribution, also known as an equilibrium distribution or steady-state distribution, is a probability distribution that remains unchanged over time in a stochastic process. In other words, it is a distribution that remains constant regardless of the number of iterations or time steps in a particular system.

2. How is a stationary distribution different from a transient distribution?

A transient distribution is a probability distribution that changes over time in a stochastic process. It represents the probability of being in a particular state at a specific time step. On the other hand, a stationary distribution remains unchanged over time and represents the long-term probabilities of being in each state in a stochastic process.

3. What types of systems use stationary distributions?

Stationary distributions are commonly used in various fields such as physics, biology, economics, and computer science. They are often applied in systems that involve random or unpredictable events, such as stock market fluctuations, population dynamics, and chemical reactions.

4. How is a stationary distribution calculated?

The calculation of a stationary distribution depends on the specific system and its underlying mathematical model. In general, it involves setting up a system of equations to represent the probabilities of transitioning between different states over time. These equations are then solved to determine the long-term stationary probabilities for each state in the system.

5. What is the significance of a stationary distribution?

Stationary distributions are important in understanding the long-term behavior of a system. They provide insight into the steady-state probabilities of being in different states and can help predict the overall behavior of a system over time. They are also essential in various statistical and mathematical models used in a wide range of scientific fields.

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