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Convergence of ergodic averages

I think I've found an invariant distribution for a transient discrete Markov chain - Where is my mistake?Uniform integrability of functional of an ergodic Markov Chain?Computational methods for the limiting distribution of a finite ergodic Markov chainConvergence of empirical average of Markov chain from transient classThis is a Markov Chain?Determining the Stationary Distribution of a Homogeneous Markov ChainCan a Markov process be ergodic but not stationary?How can we measure how good a Metropolis-Hastings estimator of an integral is?Show that i.i.d. process is ergodicFind a modified coupling $((X_n,tilde Y_n))_{n∈ℕ_0}$ with the same coupling time $τ$ and $tilde Y_n=X_n$ for $n≥τ$ in the coupling lemma

4

$begingroup$

Let $(X_n)_{ninmathbb N_0}$ be a time-homogeneous Markov chain on a probability space $(Omega,mathcal A,operatorname P)$ with transition kernel $pi$, invariant measure $mu$ and initial distribution $nu$. Let $finmathcal L^1(mu)$ and $$A_{b,:n}f:=frac1nsum_{i=b}^{b+n-1}f(X_i).$$

Assuming that the total variation distance $|mu-nupi^n|$ tends to $0$ as $ntoinfty$, it is claimed here in Theorem 3.18 that $$A_{b,:n}fxrightarrow{ntoinfty}int f:{rm d}mu.$$ In the proof, the author is basically reducing the problem to the case $nu=mu$ (i.e. the chain is started in stationarity). However, even in that case, we should need that $operatorname P_nu:=nupi$ (composition of transition kernels) is ergodic with respect to the shift $$tau:mathbb R^{mathbb N_0}tomathbb R^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}.$$ What am I missing?

probability-theory stochastic-processes dynamical-systems markov-chains ergodic-theory

share|cite|improve this question

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

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This question had a bounty worth +50
reputation from 0xbadf00d that ended 22 hours ago. Grace period ends in 1 hour

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you write explicitly what $nu pi$ is? I don't see why it's a measure on $mathbb R^{mathbb N_0}$.
  $endgroup$
  – Roberto Rastapopoulos
  Mar 2 at 17:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could also add that the convergence seeked is in the almost sure sense. For the constant Markov chain ($X_{n+1} = X_{n})$, all measures are invariant and the statement obviously does not hold, so I agree that an additional ergodicity condition should be assumed.
  $endgroup$
  – Roberto Rastapopoulos
  Mar 2 at 17:08
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Let $(X_n)_{ninmathbb N_0}$ be a time-homogeneous Markov chain on a probability space $(Omega,mathcal A,operatorname P)$ with transition kernel $pi$, invariant measure $mu$ and initial distribution $nu$. Let $finmathcal L^1(mu)$ and $$A_{b,:n}f:=frac1nsum_{i=b}^{b+n-1}f(X_i).$$

Assuming that the total variation distance $|mu-nupi^n|$ tends to $0$ as $ntoinfty$, it is claimed here in Theorem 3.18 that $$A_{b,:n}fxrightarrow{ntoinfty}int f:{rm d}mu.$$ In the proof, the author is basically reducing the problem to the case $nu=mu$ (i.e. the chain is started in stationarity). However, even in that case, we should need that $operatorname P_nu:=nupi$ (composition of transition kernels) is ergodic with respect to the shift $$tau:mathbb R^{mathbb N_0}tomathbb R^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}.$$ What am I missing?

probability-theory stochastic-processes dynamical-systems markov-chains ergodic-theory

share|cite|improve this question

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

$endgroup$

This question had a bounty worth +50
reputation from 0xbadf00d that ended 22 hours ago. Grace period ends in 1 hour

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you write explicitly what $nu pi$ is? I don't see why it's a measure on $mathbb R^{mathbb N_0}$.
  $endgroup$
  – Roberto Rastapopoulos
  Mar 2 at 17:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could also add that the convergence seeked is in the almost sure sense. For the constant Markov chain ($X_{n+1} = X_{n})$, all measures are invariant and the statement obviously does not hold, so I agree that an additional ergodicity condition should be assumed.
  $endgroup$
  – Roberto Rastapopoulos
  Mar 2 at 17:08
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $(X_n)_{ninmathbb N_0}$ be a time-homogeneous Markov chain on a probability space $(Omega,mathcal A,operatorname P)$ with transition kernel $pi$, invariant measure $mu$ and initial distribution $nu$. Let $finmathcal L^1(mu)$ and $$A_{b,:n}f:=frac1nsum_{i=b}^{b+n-1}f(X_i).$$

Assuming that the total variation distance $|mu-nupi^n|$ tends to $0$ as $ntoinfty$, it is claimed here in Theorem 3.18 that $$A_{b,:n}fxrightarrow{ntoinfty}int f:{rm d}mu.$$ In the proof, the author is basically reducing the problem to the case $nu=mu$ (i.e. the chain is started in stationarity). However, even in that case, we should need that $operatorname P_nu:=nupi$ (composition of transition kernels) is ergodic with respect to the shift $$tau:mathbb R^{mathbb N_0}tomathbb R^{mathbb N_0};,;;;(x_n)_{ninmathbb N_0}mapsto(x_{n+1})_{ninmathbb N_0}.$$ What am I missing?

probability-theory stochastic-processes dynamical-systems markov-chains ergodic-theory

share|cite|improve this question

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

$endgroup$

share|cite|improve this question

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

share|cite|improve this question

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

asked Feb 26 at 11:54

0xbadf00d0xbadf00d

1,91441532

1 Answer
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active

oldest

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2

$begingroup$

[There was a mistake in my initial reply; I've corrected it now.]

I think you're right:

If $mu$ is only invariant and not ergodic,$^ast$ if we take the simple case $nu=mu$, the ergodic averages $A_{b,n}f$ converge almost surely to the random variable $mathbb{E}_mu[f|mathcal{I}](X_k)$ (which may alternatively be written as $mathbb{E}_mathrm{P}[f(X_k)|X_k^{-1}mathcal{I}]$), where $k$ may be any natural number and $mathcal{I}$ is the $sigma$-algebra of measurable sets $A$ satisfying $pi(x,A)=1$ for $mu$-almost all $x in A$. This limiting random variable is only equal (mod null sets) to the constant $mu(f)$ if $mu$ is ergodic.

So I think an additional condition is meant to be included in the theorem, namely that $mu$ is ergodic.$^ast$ Indeed, Corollary 2.15 -- which is probably the "stationary case" of the CLT referred to several times in Section 3.4.3 (including the statement of Theorem 3.18) -- explicitly assumes ergodicity.

[By the way, $mathrm{P}_nu$ is not $nupi$ -- which would just be the same as $nu$ if $nu=mu$ -- nor is it any other kind of "composition" of $nu$ with $pi$. It is a measure on the sequence space, just as you have written, defined as the law of a homogeneous Markov process with transition kernel $pi$ and initial distribution $nu$.]

$^ast$Ergodicity of $mu$ (with respect to $pi$) is defined as meaning that the equivalent conditions in Theorem 2.1 are satisfied.

share|cite|improve this answer

edited yesterday

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

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add a comment | 

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2

$begingroup$

[There was a mistake in my initial reply; I've corrected it now.]

I think you're right:

If $mu$ is only invariant and not ergodic,$^ast$ if we take the simple case $nu=mu$, the ergodic averages $A_{b,n}f$ converge almost surely to the random variable $mathbb{E}_mu[f|mathcal{I}](X_k)$ (which may alternatively be written as $mathbb{E}_mathrm{P}[f(X_k)|X_k^{-1}mathcal{I}]$), where $k$ may be any natural number and $mathcal{I}$ is the $sigma$-algebra of measurable sets $A$ satisfying $pi(x,A)=1$ for $mu$-almost all $x in A$. This limiting random variable is only equal (mod null sets) to the constant $mu(f)$ if $mu$ is ergodic.

So I think an additional condition is meant to be included in the theorem, namely that $mu$ is ergodic.$^ast$ Indeed, Corollary 2.15 -- which is probably the "stationary case" of the CLT referred to several times in Section 3.4.3 (including the statement of Theorem 3.18) -- explicitly assumes ergodicity.

[By the way, $mathrm{P}_nu$ is not $nupi$ -- which would just be the same as $nu$ if $nu=mu$ -- nor is it any other kind of "composition" of $nu$ with $pi$. It is a measure on the sequence space, just as you have written, defined as the law of a homogeneous Markov process with transition kernel $pi$ and initial distribution $nu$.]

$^ast$Ergodicity of $mu$ (with respect to $pi$) is defined as meaning that the equivalent conditions in Theorem 2.1 are satisfied.

share|cite|improve this answer

edited yesterday

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

$endgroup$

add a comment | 

2

$begingroup$

[There was a mistake in my initial reply; I've corrected it now.]

I think you're right:

If $mu$ is only invariant and not ergodic,$^ast$ if we take the simple case $nu=mu$, the ergodic averages $A_{b,n}f$ converge almost surely to the random variable $mathbb{E}_mu[f|mathcal{I}](X_k)$ (which may alternatively be written as $mathbb{E}_mathrm{P}[f(X_k)|X_k^{-1}mathcal{I}]$), where $k$ may be any natural number and $mathcal{I}$ is the $sigma$-algebra of measurable sets $A$ satisfying $pi(x,A)=1$ for $mu$-almost all $x in A$. This limiting random variable is only equal (mod null sets) to the constant $mu(f)$ if $mu$ is ergodic.

So I think an additional condition is meant to be included in the theorem, namely that $mu$ is ergodic.$^ast$ Indeed, Corollary 2.15 -- which is probably the "stationary case" of the CLT referred to several times in Section 3.4.3 (including the statement of Theorem 3.18) -- explicitly assumes ergodicity.

[By the way, $mathrm{P}_nu$ is not $nupi$ -- which would just be the same as $nu$ if $nu=mu$ -- nor is it any other kind of "composition" of $nu$ with $pi$. It is a measure on the sequence space, just as you have written, defined as the law of a homogeneous Markov process with transition kernel $pi$ and initial distribution $nu$.]

$^ast$Ergodicity of $mu$ (with respect to $pi$) is defined as meaning that the equivalent conditions in Theorem 2.1 are satisfied.

share|cite|improve this answer

edited yesterday

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

$endgroup$

add a comment | 

$begingroup$

[There was a mistake in my initial reply; I've corrected it now.]

I think you're right:

If $mu$ is only invariant and not ergodic,$^ast$ if we take the simple case $nu=mu$, the ergodic averages $A_{b,n}f$ converge almost surely to the random variable $mathbb{E}_mu[f|mathcal{I}](X_k)$ (which may alternatively be written as $mathbb{E}_mathrm{P}[f(X_k)|X_k^{-1}mathcal{I}]$), where $k$ may be any natural number and $mathcal{I}$ is the $sigma$-algebra of measurable sets $A$ satisfying $pi(x,A)=1$ for $mu$-almost all $x in A$. This limiting random variable is only equal (mod null sets) to the constant $mu(f)$ if $mu$ is ergodic.

So I think an additional condition is meant to be included in the theorem, namely that $mu$ is ergodic.$^ast$ Indeed, Corollary 2.15 -- which is probably the "stationary case" of the CLT referred to several times in Section 3.4.3 (including the statement of Theorem 3.18) -- explicitly assumes ergodicity.

[By the way, $mathrm{P}_nu$ is not $nupi$ -- which would just be the same as $nu$ if $nu=mu$ -- nor is it any other kind of "composition" of $nu$ with $pi$. It is a measure on the sequence space, just as you have written, defined as the law of a homogeneous Markov process with transition kernel $pi$ and initial distribution $nu$.]

$^ast$Ergodicity of $mu$ (with respect to $pi$) is defined as meaning that the equivalent conditions in Theorem 2.1 are satisfied.

share|cite|improve this answer

edited yesterday

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

$endgroup$

share|cite|improve this answer

edited yesterday

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215

answered Mar 1 at 17:51

Julian NewmanJulian Newman

101215