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0

$begingroup$

I have two different invariant measure/probability measure say $m_1^*, m_2^*$ and I am given $P_1m_1^*=m_1^*$ and $P_2m_2^*=m_2^*$ where $P_1, P_2$ are two Markov operators on the space of finite measure or say probability measure. all are in same, metric space say $M$. Now, what would be the best approach to find $|m_1^*-m_2^*|$ in total variation norm? My set up $X_n+1=f_omega_n(X_n)$ with $(f_k,p_k)_k=1^m, p_k=prob(w_i=k)$ and I defined operators as $P_1f(x)=sum_k=1^m p_kf(f_k(x))$ and its adjoint as $P_1^*m_n(A)=sum_k=1^m p_k m_n(f^-1(A)), P_1m_n-1=m_n$. Thank you.

markov-chains iterated-function-system

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edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

$endgroup$

add a comment | 

0

$begingroup$

I have two different invariant measure/probability measure say $m_1^*, m_2^*$ and I am given $P_1m_1^*=m_1^*$ and $P_2m_2^*=m_2^*$ where $P_1, P_2$ are two Markov operators on the space of finite measure or say probability measure. all are in same, metric space say $M$. Now, what would be the best approach to find $|m_1^*-m_2^*|$ in total variation norm? My set up $X_n+1=f_omega_n(X_n)$ with $(f_k,p_k)_k=1^m, p_k=prob(w_i=k)$ and I defined operators as $P_1f(x)=sum_k=1^m p_kf(f_k(x))$ and its adjoint as $P_1^*m_n(A)=sum_k=1^m p_k m_n(f^-1(A)), P_1m_n-1=m_n$. Thank you.

markov-chains iterated-function-system

share|cite|improve this question

edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

$endgroup$

add a comment | 

$begingroup$

I have two different invariant measure/probability measure say $m_1^*, m_2^*$ and I am given $P_1m_1^*=m_1^*$ and $P_2m_2^*=m_2^*$ where $P_1, P_2$ are two Markov operators on the space of finite measure or say probability measure. all are in same, metric space say $M$. Now, what would be the best approach to find $|m_1^*-m_2^*|$ in total variation norm? My set up $X_n+1=f_omega_n(X_n)$ with $(f_k,p_k)_k=1^m, p_k=prob(w_i=k)$ and I defined operators as $P_1f(x)=sum_k=1^m p_kf(f_k(x))$ and its adjoint as $P_1^*m_n(A)=sum_k=1^m p_k m_n(f^-1(A)), P_1m_n-1=m_n$. Thank you.

markov-chains iterated-function-system

share|cite|improve this question

edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

$endgroup$

share|cite|improve this question

edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

share|cite|improve this question

edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

edited Mar 30 at 2:16

Saad

20.4k92352

edited Mar 30 at 2:16

Saad

20.4k92352

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184

asked Mar 29 at 11:28

Ding DongDing Dong

17.4k1060184