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limit of (2) state of Markov chain, probability of being in state $A$ or $B$ in the long term

Steady State Markov ChainMarkov chain probability that a state changesLong term probability in Markov Chainsmarkov chain: 2 state chainFind steady state of continuous time Markov chainExpected payoff of a 2-State Markov ChainUnderstanding the “first step analysis” of absorbing Markov chainsA question on Random Walks and Markov ChainProbability $mathbb{P}[X_N = 1 | X_0 = 1]$ for a Markov Chain over $mathbb{X} = {1,2,3}$How to find probability transition matrix for continuous time markov chain?

0

0

$begingroup$

Let there be a frog jumping on two spots $A$ and $B$ such that
$$mathbb P[X_n=Bmid X_{n-1}=A]=alpha=:p_{AB}\
mathbb P[X_n=Amid X_{n-1}=B]=beta=:p_{BA}$$
and so $p_{AA}=1-alpha, p_{BB}=1-beta$, where $X_n$ is the position of the frog at time $ninBbb N$

The transition matrix is $P=begin{pmatrix}1-alpha & alpha\beta & 1-beta end{pmatrix}$

I found the eigenvalues $1$ and $1-alpha-beta$ and diagonalized the matrix to get the transition matrix for $n$ steps $P^n=begin{pmatrix}frac{beta}{alpha+beta}+frac{alpha}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{alpha}{alpha+beta}(1-alpha-beta)^n\
frac{beta}{alpha+beta}+frac{beta}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{beta}{alpha+beta}(1-alpha-beta)^n
end{pmatrix}$ which clearly converges to

$$P^infty:=begin{pmatrix}frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}\
frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}end{pmatrix}$$ as long as $0<alpha+beta<2$

How do you understand what $P^infty$ is? Can we say that after an eternity the frog is on spot $A$ with probability $frac{beta}{alpha+beta}$ and on spot $B$ with probability $frac{alpha}{alpha+beta}$? Why?

stochastic-processes markov-chains conditional-probability

share|cite|improve this question

edited 2 days ago

John Cataldo

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let for a state $i$, let $V_i(n)=sum_{k=0}^{n-1} mathsf 1_{{X_k=i}}$. The ergodic theorem states that $$mathbb P(lim_{ntoinfty} V_i(n)/n = pi_i)=1, $$ where $pi$ is the (unique) stationary distribution of the Markov chain. In other words, the fraction of time spent in state $i$ converges to $pi_i$ almost surely.
  $endgroup$
  – Math1000
  yesterday
  

  
  
  
  

add a comment | 

0

0

$begingroup$

Let there be a frog jumping on two spots $A$ and $B$ such that
$$mathbb P[X_n=Bmid X_{n-1}=A]=alpha=:p_{AB}\
mathbb P[X_n=Amid X_{n-1}=B]=beta=:p_{BA}$$
and so $p_{AA}=1-alpha, p_{BB}=1-beta$, where $X_n$ is the position of the frog at time $ninBbb N$

The transition matrix is $P=begin{pmatrix}1-alpha & alpha\beta & 1-beta end{pmatrix}$

I found the eigenvalues $1$ and $1-alpha-beta$ and diagonalized the matrix to get the transition matrix for $n$ steps $P^n=begin{pmatrix}frac{beta}{alpha+beta}+frac{alpha}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{alpha}{alpha+beta}(1-alpha-beta)^n\
frac{beta}{alpha+beta}+frac{beta}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{beta}{alpha+beta}(1-alpha-beta)^n
end{pmatrix}$ which clearly converges to

$$P^infty:=begin{pmatrix}frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}\
frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}end{pmatrix}$$ as long as $0<alpha+beta<2$

How do you understand what $P^infty$ is? Can we say that after an eternity the frog is on spot $A$ with probability $frac{beta}{alpha+beta}$ and on spot $B$ with probability $frac{alpha}{alpha+beta}$? Why?

stochastic-processes markov-chains conditional-probability

share|cite|improve this question

edited 2 days ago

John Cataldo

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let for a state $i$, let $V_i(n)=sum_{k=0}^{n-1} mathsf 1_{{X_k=i}}$. The ergodic theorem states that $$mathbb P(lim_{ntoinfty} V_i(n)/n = pi_i)=1, $$ where $pi$ is the (unique) stationary distribution of the Markov chain. In other words, the fraction of time spent in state $i$ converges to $pi_i$ almost surely.
  $endgroup$
  – Math1000
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Let there be a frog jumping on two spots $A$ and $B$ such that
$$mathbb P[X_n=Bmid X_{n-1}=A]=alpha=:p_{AB}\
mathbb P[X_n=Amid X_{n-1}=B]=beta=:p_{BA}$$
and so $p_{AA}=1-alpha, p_{BB}=1-beta$, where $X_n$ is the position of the frog at time $ninBbb N$

The transition matrix is $P=begin{pmatrix}1-alpha & alpha\beta & 1-beta end{pmatrix}$

I found the eigenvalues $1$ and $1-alpha-beta$ and diagonalized the matrix to get the transition matrix for $n$ steps $P^n=begin{pmatrix}frac{beta}{alpha+beta}+frac{alpha}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{alpha}{alpha+beta}(1-alpha-beta)^n\
frac{beta}{alpha+beta}+frac{beta}{alpha+beta}(1-alpha-beta)^n & frac{alpha}{alpha+beta}-frac{beta}{alpha+beta}(1-alpha-beta)^n
end{pmatrix}$ which clearly converges to

$$P^infty:=begin{pmatrix}frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}\
frac{beta}{alpha+beta} & frac{alpha}{alpha+beta}end{pmatrix}$$ as long as $0<alpha+beta<2$

How do you understand what $P^infty$ is? Can we say that after an eternity the frog is on spot $A$ with probability $frac{beta}{alpha+beta}$ and on spot $B$ with probability $frac{alpha}{alpha+beta}$? Why?

stochastic-processes markov-chains conditional-probability

share|cite|improve this question

edited 2 days ago

John Cataldo

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

$endgroup$

share|cite|improve this question

edited 2 days ago

John Cataldo

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

share|cite|improve this question

edited 2 days ago

John Cataldo

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

asked 2 days ago

John CataldoJohn Cataldo

1,1881316

asked 2 days ago

John CataldoJohn Cataldo

1,1881316