Mixing
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Derivation of a stationary distribution of a Markov chain
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Consider the following markov chain in index notation
$$P_j(t+1)=sum_i=1^nM_ijP_i(t)$$
where $M_ij$ is a column transition matrix .
I would like to develop the steady state given by $pi M=pi $ in an index notation for the Column transition matrix.
This is what I have so far
$$P_j(t+1)=sum_i=1^n M_ijP_i(t)=sum_i=1^nleft(M_ijP_i(t)+M_iiP_i(t)right)=$$
$$=sum_i=1^nM_ijP_i(t)+sum_i=1^nM_iiP_i(t)=sum_i=1,ineq j^nM_ijP_i(t)+left(1-sum_j=1,ineq j^nM_jiright)P_j(t)=$$
$$=sum_i=1,ineq j^nM_ijP_i(t)+P_j(t)-sum_j=1,ineq j^nM_jiP_j(t)=$$
$$longrightarrow$$
$$P_j(t+1)-P_j(t)=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)$$
$$=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)+sum_i=1^nM_ii P_i(t)-sum_j=1^nM_jj P_j(t)$$
$$=sum_i=1^nM_ijP_i(t)-sum_j=1^nM_jiP_j(t)$$
At steady state
$$sum_i=1^nM_ijP_i(t)=sum_j=1^nM_jiP_j(t)$$
Which is a trivial solution, I don't understand how to obtain the steady state solution $pi=Mpi$ in matrix form the equation above.
Please advise, note that it is important for me, if possible, to remain in mattrix form for a column transition matrix.
linear-algebra markov-chains
asked Jul 25 at 10:05
jarhead
1148
add a comment |Â
up vote
0
down vote
favorite
Consider the following markov chain in index notation
$$P_j(t+1)=sum_i=1^nM_ijP_i(t)$$
where $M_ij$ is a column transition matrix .
I would like to develop the steady state given by $pi M=pi $ in an index notation for the Column transition matrix.
This is what I have so far
$$P_j(t+1)=sum_i=1^n M_ijP_i(t)=sum_i=1^nleft(M_ijP_i(t)+M_iiP_i(t)right)=$$
$$=sum_i=1^nM_ijP_i(t)+sum_i=1^nM_iiP_i(t)=sum_i=1,ineq j^nM_ijP_i(t)+left(1-sum_j=1,ineq j^nM_jiright)P_j(t)=$$
$$=sum_i=1,ineq j^nM_ijP_i(t)+P_j(t)-sum_j=1,ineq j^nM_jiP_j(t)=$$
$$longrightarrow$$
$$P_j(t+1)-P_j(t)=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)$$
$$=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)+sum_i=1^nM_ii P_i(t)-sum_j=1^nM_jj P_j(t)$$
$$=sum_i=1^nM_ijP_i(t)-sum_j=1^nM_jiP_j(t)$$
At steady state
$$sum_i=1^nM_ijP_i(t)=sum_j=1^nM_jiP_j(t)$$
Which is a trivial solution, I don't understand how to obtain the steady state solution $pi=Mpi$ in matrix form the equation above.
Please advise, note that it is important for me, if possible, to remain in mattrix form for a column transition matrix.
linear-algebra markov-chains
asked Jul 25 at 10:05
jarhead
1148
I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the following markov chain in index notation
$$P_j(t+1)=sum_i=1^nM_ijP_i(t)$$
where $M_ij$ is a column transition matrix .
I would like to develop the steady state given by $pi M=pi $ in an index notation for the Column transition matrix.
This is what I have so far
$$P_j(t+1)=sum_i=1^n M_ijP_i(t)=sum_i=1^nleft(M_ijP_i(t)+M_iiP_i(t)right)=$$
$$=sum_i=1^nM_ijP_i(t)+sum_i=1^nM_iiP_i(t)=sum_i=1,ineq j^nM_ijP_i(t)+left(1-sum_j=1,ineq j^nM_jiright)P_j(t)=$$
$$=sum_i=1,ineq j^nM_ijP_i(t)+P_j(t)-sum_j=1,ineq j^nM_jiP_j(t)=$$
$$longrightarrow$$
$$P_j(t+1)-P_j(t)=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)$$
$$=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)+sum_i=1^nM_ii P_i(t)-sum_j=1^nM_jj P_j(t)$$
$$=sum_i=1^nM_ijP_i(t)-sum_j=1^nM_jiP_j(t)$$
At steady state
$$sum_i=1^nM_ijP_i(t)=sum_j=1^nM_jiP_j(t)$$
Which is a trivial solution, I don't understand how to obtain the steady state solution $pi=Mpi$ in matrix form the equation above.
Please advise, note that it is important for me, if possible, to remain in mattrix form for a column transition matrix.
linear-algebra markov-chains
asked Jul 25 at 10:05
jarhead
1148
Consider the following markov chain in index notation
$$P_j(t+1)=sum_i=1^nM_ijP_i(t)$$
where $M_ij$ is a column transition matrix .
I would like to develop the steady state given by $pi M=pi $ in an index notation for the Column transition matrix.
This is what I have so far
$$P_j(t+1)=sum_i=1^n M_ijP_i(t)=sum_i=1^nleft(M_ijP_i(t)+M_iiP_i(t)right)=$$
$$=sum_i=1^nM_ijP_i(t)+sum_i=1^nM_iiP_i(t)=sum_i=1,ineq j^nM_ijP_i(t)+left(1-sum_j=1,ineq j^nM_jiright)P_j(t)=$$
$$=sum_i=1,ineq j^nM_ijP_i(t)+P_j(t)-sum_j=1,ineq j^nM_jiP_j(t)=$$
$$longrightarrow$$
$$P_j(t+1)-P_j(t)=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)$$
$$=sum_i=1,ineq j^nM_ijP_i(t)-sum_j=1,ineq j^nM_jiP_j(t)+sum_i=1^nM_ii P_i(t)-sum_j=1^nM_jj P_j(t)$$
$$=sum_i=1^nM_ijP_i(t)-sum_j=1^nM_jiP_j(t)$$
At steady state
$$sum_i=1^nM_ijP_i(t)=sum_j=1^nM_jiP_j(t)$$
Which is a trivial solution, I don't understand how to obtain the steady state solution $pi=Mpi$ in matrix form the equation above.
Please advise, note that it is important for me, if possible, to remain in mattrix form for a column transition matrix.
linear-algebra markov-chains
asked Jul 25 at 10:05
jarhead
1148
edited Jul 25 at 10:31
asked Jul 25 at 10:05
jarhead
1148
asked Jul 25 at 10:05
jarhead
1148
asked Jul 25 at 10:05
jarhead
1148
1148
I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14
add a comment |Â
I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14
I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14
add a comment |Â
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I'm not really sure what you mean by "develop the steady state...in an index notation." Note that the $n$-dimensional vector whose entries are identically $1$ is always a solution to $pi=Mpi$.
– Math1000
Jul 25 at 11:56
$mathbf1$ will always work as pointed out above. More generally $pi$ can be any eigenvector of $M$ with eigenvalue 1. Basically you just need to find the nullspace of $M-I$
– Casey
Jul 25 at 20:14