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Are all invariant measures to a Markov Process found by following the process?
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Some (hand-wavy) context: Suppose that we have found the invariant measures to a Markov Process, $(X_t)_t=1^infty$, on a compact set $S$ by the following the process itself, i.e. looking at the weak-$ast$ limit points of empirical measures $frac1tsum_s=0^t-1 delta_X(t)(cdot )$.
Question: Are we guaranteed that there are no other invariant measures to the process $(X(t))_t=1^infty$? As in, we can follow the process (say with any initial condition $X(0) = x$) and find the invariant measures to this process by following the process itself, but are we missing any others? Let's say $Omega$ is actually just $mathbbR^n$.
Attempt: I tried to do something in a more general format. I took an identity map between $imath: (Omega, phi, tilde mu) to (Omega, phi, mu) $, where $tilde mu$ and $mu$ are both invariant measures to the process $X(t)$, and $phi$ is a measure-preserving map for both spaces. Then I used Birkhoff's Ergodic Theorem to say that:
$mathbbE(mathbb1_A | tilde I) = lim_ttoinftyfrac1tsum_s=0^t-1 mathbb1_A(phi^s(x)) = frac1tsum_s=0^t-1 mathbb1_A(imath( phi^s(x))) = mathbbE(mathbb1_A | I)$,
where $I$ and $tilde I$ are the invariant sets of $mu$ and $tilde mu$, respectively. Then if we took the expectation of both sides of the equation, then we'd get $mu(A) = tilde mu(A)$ for all $A$ measurable. In particular, In particular if $tilde A in tilde I$ was an invariant set of $tilde mu$, then $mu(tilde A) = tilde mu(imath^-1(tilde A)) = tilde mu((phi^-1 imath^-1(tilde A))) = tilde mu((phi^-1(tilde A)) = mu(phi^-1(tilde A))$, so that $tilde A$ is an invariant set of $I$ as well. But something seems a bit iffy about what I'm doing and I wonder if there's just an easy answer.
dynamical-systems markov-process ergodic-theory
asked Jul 18 at 22:23
willscue
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Some (hand-wavy) context: Suppose that we have found the invariant measures to a Markov Process, $(X_t)_t=1^infty$, on a compact set $S$ by the following the process itself, i.e. looking at the weak-$ast$ limit points of empirical measures $frac1tsum_s=0^t-1 delta_X(t)(cdot )$.
Question: Are we guaranteed that there are no other invariant measures to the process $(X(t))_t=1^infty$? As in, we can follow the process (say with any initial condition $X(0) = x$) and find the invariant measures to this process by following the process itself, but are we missing any others? Let's say $Omega$ is actually just $mathbbR^n$.
Attempt: I tried to do something in a more general format. I took an identity map between $imath: (Omega, phi, tilde mu) to (Omega, phi, mu) $, where $tilde mu$ and $mu$ are both invariant measures to the process $X(t)$, and $phi$ is a measure-preserving map for both spaces. Then I used Birkhoff's Ergodic Theorem to say that:
$mathbbE(mathbb1_A | tilde I) = lim_ttoinftyfrac1tsum_s=0^t-1 mathbb1_A(phi^s(x)) = frac1tsum_s=0^t-1 mathbb1_A(imath( phi^s(x))) = mathbbE(mathbb1_A | I)$,
where $I$ and $tilde I$ are the invariant sets of $mu$ and $tilde mu$, respectively. Then if we took the expectation of both sides of the equation, then we'd get $mu(A) = tilde mu(A)$ for all $A$ measurable. In particular, In particular if $tilde A in tilde I$ was an invariant set of $tilde mu$, then $mu(tilde A) = tilde mu(imath^-1(tilde A)) = tilde mu((phi^-1 imath^-1(tilde A))) = tilde mu((phi^-1(tilde A)) = mu(phi^-1(tilde A))$, so that $tilde A$ is an invariant set of $I$ as well. But something seems a bit iffy about what I'm doing and I wonder if there's just an easy answer.
dynamical-systems markov-process ergodic-theory
asked Jul 18 at 22:23
willscue
1
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Some (hand-wavy) context: Suppose that we have found the invariant measures to a Markov Process, $(X_t)_t=1^infty$, on a compact set $S$ by the following the process itself, i.e. looking at the weak-$ast$ limit points of empirical measures $frac1tsum_s=0^t-1 delta_X(t)(cdot )$.
Question: Are we guaranteed that there are no other invariant measures to the process $(X(t))_t=1^infty$? As in, we can follow the process (say with any initial condition $X(0) = x$) and find the invariant measures to this process by following the process itself, but are we missing any others? Let's say $Omega$ is actually just $mathbbR^n$.
Attempt: I tried to do something in a more general format. I took an identity map between $imath: (Omega, phi, tilde mu) to (Omega, phi, mu) $, where $tilde mu$ and $mu$ are both invariant measures to the process $X(t)$, and $phi$ is a measure-preserving map for both spaces. Then I used Birkhoff's Ergodic Theorem to say that:
$mathbbE(mathbb1_A | tilde I) = lim_ttoinftyfrac1tsum_s=0^t-1 mathbb1_A(phi^s(x)) = frac1tsum_s=0^t-1 mathbb1_A(imath( phi^s(x))) = mathbbE(mathbb1_A | I)$,
where $I$ and $tilde I$ are the invariant sets of $mu$ and $tilde mu$, respectively. Then if we took the expectation of both sides of the equation, then we'd get $mu(A) = tilde mu(A)$ for all $A$ measurable. In particular, In particular if $tilde A in tilde I$ was an invariant set of $tilde mu$, then $mu(tilde A) = tilde mu(imath^-1(tilde A)) = tilde mu((phi^-1 imath^-1(tilde A))) = tilde mu((phi^-1(tilde A)) = mu(phi^-1(tilde A))$, so that $tilde A$ is an invariant set of $I$ as well. But something seems a bit iffy about what I'm doing and I wonder if there's just an easy answer.
dynamical-systems markov-process ergodic-theory
asked Jul 18 at 22:23
willscue
1
Some (hand-wavy) context: Suppose that we have found the invariant measures to a Markov Process, $(X_t)_t=1^infty$, on a compact set $S$ by the following the process itself, i.e. looking at the weak-$ast$ limit points of empirical measures $frac1tsum_s=0^t-1 delta_X(t)(cdot )$.
Question: Are we guaranteed that there are no other invariant measures to the process $(X(t))_t=1^infty$? As in, we can follow the process (say with any initial condition $X(0) = x$) and find the invariant measures to this process by following the process itself, but are we missing any others? Let's say $Omega$ is actually just $mathbbR^n$.
Attempt: I tried to do something in a more general format. I took an identity map between $imath: (Omega, phi, tilde mu) to (Omega, phi, mu) $, where $tilde mu$ and $mu$ are both invariant measures to the process $X(t)$, and $phi$ is a measure-preserving map for both spaces. Then I used Birkhoff's Ergodic Theorem to say that:
$mathbbE(mathbb1_A | tilde I) = lim_ttoinftyfrac1tsum_s=0^t-1 mathbb1_A(phi^s(x)) = frac1tsum_s=0^t-1 mathbb1_A(imath( phi^s(x))) = mathbbE(mathbb1_A | I)$,
where $I$ and $tilde I$ are the invariant sets of $mu$ and $tilde mu$, respectively. Then if we took the expectation of both sides of the equation, then we'd get $mu(A) = tilde mu(A)$ for all $A$ measurable. In particular, In particular if $tilde A in tilde I$ was an invariant set of $tilde mu$, then $mu(tilde A) = tilde mu(imath^-1(tilde A)) = tilde mu((phi^-1 imath^-1(tilde A))) = tilde mu((phi^-1(tilde A)) = mu(phi^-1(tilde A))$, so that $tilde A$ is an invariant set of $I$ as well. But something seems a bit iffy about what I'm doing and I wonder if there's just an easy answer.
dynamical-systems markov-process ergodic-theory
asked Jul 18 at 22:23
willscue
1
asked Jul 18 at 22:23
willscue
1
asked Jul 18 at 22:23
willscue
1
asked Jul 18 at 22:23
willscue
1
1
add a comment |Â
add a comment |Â
1 Answer
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Take a Markov chain with states $(a,b,c)$ and connections $arightarrow b$, $arightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $mu_1=(0,1,0)$, $mu_2=(0,0,1)$ and $mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $mu_1$ and $mu_2$. Yet, $mu_3$ is not achievable by starting from any state in $iin(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.
answered Jul 18 at 23:09
Alex R.
23.7k12352
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
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1 Answer
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Take a Markov chain with states $(a,b,c)$ and connections $arightarrow b$, $arightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $mu_1=(0,1,0)$, $mu_2=(0,0,1)$ and $mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $mu_1$ and $mu_2$. Yet, $mu_3$ is not achievable by starting from any state in $iin(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.
answered Jul 18 at 23:09
Alex R.
23.7k12352
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
add a comment |Â
up vote
0
down vote
Take a Markov chain with states $(a,b,c)$ and connections $arightarrow b$, $arightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $mu_1=(0,1,0)$, $mu_2=(0,0,1)$ and $mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $mu_1$ and $mu_2$. Yet, $mu_3$ is not achievable by starting from any state in $iin(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.
answered Jul 18 at 23:09
Alex R.
23.7k12352
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take a Markov chain with states $(a,b,c)$ and connections $arightarrow b$, $arightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $mu_1=(0,1,0)$, $mu_2=(0,0,1)$ and $mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $mu_1$ and $mu_2$. Yet, $mu_3$ is not achievable by starting from any state in $iin(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.
answered Jul 18 at 23:09
Alex R.
23.7k12352
Take a Markov chain with states $(a,b,c)$ and connections $arightarrow b$, $arightarrow c$, with $P(b|a)=P(c|a)=1/2$, i.e. $b,c$ are fixed points and $a$ is transient. Then there are at least three invariant measures $mu_1=(0,1,0)$, $mu_2=(0,0,1)$ and $mu_3=(0,1/2,1/2)$. In fact you can form infintely many more by taking convex combinations of $mu_1$ and $mu_2$. Yet, $mu_3$ is not achievable by starting from any state in $iin(a,b,c)$.
Section 2 of this reference, along with Theorem 1.7 which says that any invariant measure of a markov chain must be a convex combination of the ergodic measures of that markov chain. It follows that if you've found all ergodic measures, then you've also found all invariant measures.
answered Jul 18 at 23:09
Alex R.
23.7k12352
edited Jul 18 at 23:26
answered Jul 18 at 23:09
Alex R.
23.7k12352
answered Jul 18 at 23:09
Alex R.
23.7k12352
answered Jul 18 at 23:09
Alex R.
23.7k12352
23.7k12352
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
add a comment |Â
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
Thank you for the example! I think I'm a little confused. If I have the vector [1,0,0] have a left action on the transition matrix, then I will get (0,1/2,1/2), which is the invariant measure that you've mentioned. Also, would one be able to find all the ergodic measures then by following the process itself (starting from different initial conditions)?
– willscue
Jul 19 at 18:11
add a comment |Â
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