Mixing
On showing the existence of a Markov chain's steady state distribution
I want to show that the Markov chain with such transition matrices written below has a unique stationary distribution $mu$.
For a space dimension of 6 :
$$begin{equation}
Pi^{(6)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & b_2\
0 & 0 & a_1 & a_2 & a_3 & b_3\
0 & 0 & 0 & a_1 & a_2 & b_4\
end{pmatrix}
end{equation}$$
or for a space dimension equal to 7 :
$$begin{equation}
Pi^{(7)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & a_5 & b_2 \
0 & 0 & a_1 & a_2 & a_3 & a_4 & b_3 \
0 & 0 & 0 & a_1 & a_2 & a_3 & b_4 \
0 & 0 & 0 & 0 & a_1 & a_2 & b_5 \
end{pmatrix}
end{equation}$$
with $forall i in mathbb{N} 0 < a_i < 1$ and $forall i in mathbb{N} 0 < b_i < 1$
I know I can solve the linear system $Pi mu = mu$ in order to solve this problem.
However I would like to know if from the specific "structure" of this matrix, one can say if there is or not a steady distribution ? Is there a direct way to see that this Markov Chain is aperiodic and irreductible ?
Indeed, I would like to extend this to a greater space dimension Markov process with a similar transition matrix and therefore solving $Pi mu = mu$ would lead to very long computations.
Thank you.
markov-chains steady-state
asked Nov 20 '18 at 14:21
AlexC75
215
add a comment |
I want to show that the Markov chain with such transition matrices written below has a unique stationary distribution $mu$.
For a space dimension of 6 :
$$begin{equation}
Pi^{(6)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & b_2\
0 & 0 & a_1 & a_2 & a_3 & b_3\
0 & 0 & 0 & a_1 & a_2 & b_4\
end{pmatrix}
end{equation}$$
or for a space dimension equal to 7 :
$$begin{equation}
Pi^{(7)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & a_5 & b_2 \
0 & 0 & a_1 & a_2 & a_3 & a_4 & b_3 \
0 & 0 & 0 & a_1 & a_2 & a_3 & b_4 \
0 & 0 & 0 & 0 & a_1 & a_2 & b_5 \
end{pmatrix}
end{equation}$$
with $forall i in mathbb{N} 0 < a_i < 1$ and $forall i in mathbb{N} 0 < b_i < 1$
I know I can solve the linear system $Pi mu = mu$ in order to solve this problem.
However I would like to know if from the specific "structure" of this matrix, one can say if there is or not a steady distribution ? Is there a direct way to see that this Markov Chain is aperiodic and irreductible ?
Indeed, I would like to extend this to a greater space dimension Markov process with a similar transition matrix and therefore solving $Pi mu = mu$ would lead to very long computations.
Thank you.
markov-chains steady-state
asked Nov 20 '18 at 14:21
AlexC75
215
add a comment |
I want to show that the Markov chain with such transition matrices written below has a unique stationary distribution $mu$.
For a space dimension of 6 :
$$begin{equation}
Pi^{(6)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & b_2\
0 & 0 & a_1 & a_2 & a_3 & b_3\
0 & 0 & 0 & a_1 & a_2 & b_4\
end{pmatrix}
end{equation}$$
or for a space dimension equal to 7 :
$$begin{equation}
Pi^{(7)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & a_5 & b_2 \
0 & 0 & a_1 & a_2 & a_3 & a_4 & b_3 \
0 & 0 & 0 & a_1 & a_2 & a_3 & b_4 \
0 & 0 & 0 & 0 & a_1 & a_2 & b_5 \
end{pmatrix}
end{equation}$$
with $forall i in mathbb{N} 0 < a_i < 1$ and $forall i in mathbb{N} 0 < b_i < 1$
I know I can solve the linear system $Pi mu = mu$ in order to solve this problem.
However I would like to know if from the specific "structure" of this matrix, one can say if there is or not a steady distribution ? Is there a direct way to see that this Markov Chain is aperiodic and irreductible ?
Indeed, I would like to extend this to a greater space dimension Markov process with a similar transition matrix and therefore solving $Pi mu = mu$ would lead to very long computations.
Thank you.
markov-chains steady-state
asked Nov 20 '18 at 14:21
AlexC75
215
I want to show that the Markov chain with such transition matrices written below has a unique stationary distribution $mu$.
For a space dimension of 6 :
$$begin{equation}
Pi^{(6)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & b_2\
0 & 0 & a_1 & a_2 & a_3 & b_3\
0 & 0 & 0 & a_1 & a_2 & b_4\
end{pmatrix}
end{equation}$$
or for a space dimension equal to 7 :
$$begin{equation}
Pi^{(7)} = begin{pmatrix}
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & b_1 \
0 & a_1 & a_2 & a_3 & a_4 & a_5 & b_2 \
0 & 0 & a_1 & a_2 & a_3 & a_4 & b_3 \
0 & 0 & 0 & a_1 & a_2 & a_3 & b_4 \
0 & 0 & 0 & 0 & a_1 & a_2 & b_5 \
end{pmatrix}
end{equation}$$
with $forall i in mathbb{N} 0 < a_i < 1$ and $forall i in mathbb{N} 0 < b_i < 1$
I know I can solve the linear system $Pi mu = mu$ in order to solve this problem.
However I would like to know if from the specific "structure" of this matrix, one can say if there is or not a steady distribution ? Is there a direct way to see that this Markov Chain is aperiodic and irreductible ?
Indeed, I would like to extend this to a greater space dimension Markov process with a similar transition matrix and therefore solving $Pi mu = mu$ would lead to very long computations.
Thank you.
markov-chains steady-state
markov-chains steady-state
asked Nov 20 '18 at 14:21
AlexC75
215
asked Nov 20 '18 at 14:21
AlexC75
215
edited Nov 20 '18 at 16:04
asked Nov 20 '18 at 14:21
AlexC75
215
asked Nov 20 '18 at 14:21
AlexC75
215
asked Nov 20 '18 at 14:21
AlexC75
215
215
add a comment |
add a comment |
1 Answer
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Yes, you can view the transition matrix as a graph and analyze its connectivity.
For a transition matrix $Pi in mathbb{R}^{n times n}$, let $G=(V,E)$ be the directed graph where $V = [n]$ and $E = {(i,j) in [n]^2 : Pi(i,j) > 0}$.
The Markov chain is irreducible if and only if $G$ has one strongly connected component. Again, there are linear-time algorithms for checking this condition. More simply for the matrix above, we can see that all nodes $i > 1$ can directly reach node $i-1$. Moreover, node $1$ can transition to any node. Therefore, the graph has one strongly connected component and thus is irreducible.
An irreducible Markov chain is aperiodic if and only if $G$ has a node with period 1. In this case, it is even easier because every node has a self-loop since $Pi(i,i) > 0$ for all $i in [n]$.
It follows from the fundamental theorem of Markov chains that $Pi$ has a unique stationary distribution.
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
|
show 3 more comments
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1 Answer
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Yes, you can view the transition matrix as a graph and analyze its connectivity.
For a transition matrix $Pi in mathbb{R}^{n times n}$, let $G=(V,E)$ be the directed graph where $V = [n]$ and $E = {(i,j) in [n]^2 : Pi(i,j) > 0}$.
The Markov chain is irreducible if and only if $G$ has one strongly connected component. Again, there are linear-time algorithms for checking this condition. More simply for the matrix above, we can see that all nodes $i > 1$ can directly reach node $i-1$. Moreover, node $1$ can transition to any node. Therefore, the graph has one strongly connected component and thus is irreducible.
An irreducible Markov chain is aperiodic if and only if $G$ has a node with period 1. In this case, it is even easier because every node has a self-loop since $Pi(i,i) > 0$ for all $i in [n]$.
It follows from the fundamental theorem of Markov chains that $Pi$ has a unique stationary distribution.
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
|
show 3 more comments
Yes, you can view the transition matrix as a graph and analyze its connectivity.
For a transition matrix $Pi in mathbb{R}^{n times n}$, let $G=(V,E)$ be the directed graph where $V = [n]$ and $E = {(i,j) in [n]^2 : Pi(i,j) > 0}$.
The Markov chain is irreducible if and only if $G$ has one strongly connected component. Again, there are linear-time algorithms for checking this condition. More simply for the matrix above, we can see that all nodes $i > 1$ can directly reach node $i-1$. Moreover, node $1$ can transition to any node. Therefore, the graph has one strongly connected component and thus is irreducible.
An irreducible Markov chain is aperiodic if and only if $G$ has a node with period 1. In this case, it is even easier because every node has a self-loop since $Pi(i,i) > 0$ for all $i in [n]$.
It follows from the fundamental theorem of Markov chains that $Pi$ has a unique stationary distribution.
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
|
show 3 more comments
Yes, you can view the transition matrix as a graph and analyze its connectivity.
For a transition matrix $Pi in mathbb{R}^{n times n}$, let $G=(V,E)$ be the directed graph where $V = [n]$ and $E = {(i,j) in [n]^2 : Pi(i,j) > 0}$.
The Markov chain is irreducible if and only if $G$ has one strongly connected component. Again, there are linear-time algorithms for checking this condition. More simply for the matrix above, we can see that all nodes $i > 1$ can directly reach node $i-1$. Moreover, node $1$ can transition to any node. Therefore, the graph has one strongly connected component and thus is irreducible.
An irreducible Markov chain is aperiodic if and only if $G$ has a node with period 1. In this case, it is even easier because every node has a self-loop since $Pi(i,i) > 0$ for all $i in [n]$.
It follows from the fundamental theorem of Markov chains that $Pi$ has a unique stationary distribution.
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
Yes, you can view the transition matrix as a graph and analyze its connectivity.
For a transition matrix $Pi in mathbb{R}^{n times n}$, let $G=(V,E)$ be the directed graph where $V = [n]$ and $E = {(i,j) in [n]^2 : Pi(i,j) > 0}$.
The Markov chain is irreducible if and only if $G$ has one strongly connected component. Again, there are linear-time algorithms for checking this condition. More simply for the matrix above, we can see that all nodes $i > 1$ can directly reach node $i-1$. Moreover, node $1$ can transition to any node. Therefore, the graph has one strongly connected component and thus is irreducible.
An irreducible Markov chain is aperiodic if and only if $G$ has a node with period 1. In this case, it is even easier because every node has a self-loop since $Pi(i,i) > 0$ for all $i in [n]$.
It follows from the fundamental theorem of Markov chains that $Pi$ has a unique stationary distribution.
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
edited Nov 21 '18 at 22:09
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
answered Nov 20 '18 at 15:38
fahrbach
1,5781915
1,5781915
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
|
show 3 more comments
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Thank you very much for your answer. I changed the question in order to explain more on the "structure" of the matrices. In the general case I don't think that the argument to prove the irreductiblity holds. Does it ? Moreover, considering the self-loop argument to prove the aperiodicity, does this means that the period of the matrix is d=1 ? and therefore this is why it is not periodic ?
– AlexC75
Nov 20 '18 at 17:26
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
Yes, one strongly connected component is essentially the definition of irreducibility: It is possible to get to any state from any other state. The period of a matrix is a different property than the period of a state in a Markov chain, so no. To see why a self-loop leads to aperiodicity for irreducible Markov chains, from any starting node you can reach the node with a self-loop, cycle on that node as much as you want (to break the period), and then return to the starting node.
– fahrbach
Nov 20 '18 at 18:55
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I got the argument for the aperiodicity thank you. However in the general case of the matrices I am studying, there are no strongly components : the argument on vertex 3 does not hold I think, does it ?
– AlexC75
Nov 20 '18 at 21:07
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
I misread the matrix at first so you're right about the earlier argument being incorrect. I updated my post with a correct argument about irreducibility.
– fahrbach
Nov 21 '18 at 15:14
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
Thank you very much. What do you mean "the graph is one strongly connected component" ? Did you mean "has" instead of "is" ?
– AlexC75
Nov 21 '18 at 21:47
|
show 3 more comments
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