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 at 14:21
AlexC75
194
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 at 14:21
AlexC75
194
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 at 14:21
AlexC75
194
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 at 14:21
AlexC75
194
asked Nov 20 at 14:21
AlexC75
194
edited Nov 20 at 16:04
asked Nov 20 at 14:21
AlexC75
194
asked Nov 20 at 14:21
AlexC75
194
asked Nov 20 at 14:21
AlexC75
194
194
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 15:38
fahrbach
1,5781915
edited Nov 21 at 22:09
answered Nov 20 at 15:38
fahrbach
1,5781915
answered Nov 20 at 15:38
fahrbach
1,5781915
answered Nov 20 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 21:47
|
show 3 more comments
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Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
