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I can give you the answer for a certain class of Markov Chains, i.e. for reversible, irreducible and aperiodic Markov Chains on a finite state space.

Then the question is related to the spectral gap of the transition matrix $P:\Omega\times\Omega\to [0,1]$ of your chain. The transition matrix entry $P(x,y)$ is the probability of going from state $x$ to $y$. Thus one has $\sum_{y\in \Omega} P(x,y)=1$ If your chain is irreducible ($\forall x,y\in \Omega \exists n\in \mathbb{N}: P^n(x,y)>0$) and aperiodic. Aperiodic means (means that $\forall x\in \Omega$ the period defined by the greatest common divisor of the set $T(x)=\{n\in \mathbb{N}: P^n(x,x)>0\}$ is $1$. Then) then one has the following Theorem:

If $\Omega$ is finite and the chain defined by the transition matrix $P$ is irreducible and aperiodic than there exists a unique stationary distribution $\pi$ such that $\pi P =\pi$.

A chain satisfying a detailed balance relation $\pi(x) P(x,y) =\pi(y) P(y,x)$ is reversible.

Under these conditions the eigenvalues of $P$ are bounded in modulus by $1$ and the largest is $1$, the corresponding left eigenvector is $\pi$. Then the spectral gap is defined by $\gamma=1-\max_{\lambda\ne 1} |\lambda| $. Note that $0<\gamma<1$. Then it yields the following estimate for the variance of a test function $f:\Omega\to\mathbb{R}$ with respect to the stationary measure $\pi$

${\mathrm{Var}}_\pi(P^n f) \leq (1-\gamma)^{2t} {\mathrm{Var}}_\pi(f)$

I can give you the answer for a certain class of Markov Chains, i.e. for reversible, irreducible and aperiodic Markov Chains on a finite state space.

Then the question is related to the spectral gap of the transition matrix $P:\Omega\times\Omega\to [0,1]$ of your chain. The transition matrix entry $P(x,y)$ is the probability of going from state $x$ to $y$. Thus one has $\sum_{y\in \Omega} P(x,y)=1$ If your chain is irreducible ($\forall x,y\in \Omega \exists n\in \mathbb{N}: P^n(x,y)>0$) and aperiodic. Aperiodic means that $\forall x\in \Omega$ the period defined by the greatest common divisor of the set $T(x)=\{n\in \mathbb{N}: P^n(x,x)>0\}$ is $1$. Then one has the following Theorem:

If $\Omega$ is finite and the chain defined by the transition matrix $P$ is irreducible and aperiodic than there exists a unique stationary distribution $\pi$ such that $\pi P =\pi$.

A chain satisfying a detailed balance relation $\pi(x) P(x,y) =\pi(y) P(y,x)$ is reversible.

Under these conditions the eigenvalues of $P$ are bounded in modulus by $1$ and the largest is $1$, the corresponding left eigenvector is $\pi$. Then the spectral gap is defined by $\gamma=1-\max_{\lambda\ne 1} |\lambda| $. Note that $0<\gamma<1$. Then it yields the following estimate for the variance of a test function $f:\Omega\to\mathbb{R}$ with respect to the stationary measure $\pi$

${\mathrm{Var}}_\pi(P^n f) \leq (1-\gamma)^{2t} {\mathrm{Var}}_\pi(f)$

I can give you the answer for a certain class of Markov Chains, i.e. for reversible, irreducible and aperiodic Markov Chains on a finite state space.

Then the question is related to the spectral gap of the transition matrix $P:\Omega\times\Omega\to [0,1]$ of your chain. The transition matrix entry $P(x,y)$ is the probability of going from state $x$ to $y$. Thus one has $\sum_{y\in \Omega} P(x,y)=1$ If your chain is irreducible ($\forall x,y\in \Omega \exists n\in \mathbb{N}: P^n(x,y)>0$) and aperiodic (means that $\forall x\in \Omega$ the period defined by the greatest common divisor of the set $T(x)=\{n\in \mathbb{N}: P^n(x,x)>0\}$ is $1$) then one has the following Theorem:

If $\Omega$ is finite and the chain defined by the transition matrix $P$ is irreducible and aperiodic than there exists a unique stationary distribution $\pi$ such that $\pi P =\pi$.

A chain satisfying a detailed balance relation $\pi(x) P(x,y) =\pi(y) P(y,x)$ is reversible.

Under these conditions the eigenvalues of $P$ are bounded in modulus by $1$ and the largest is $1$, the corresponding left eigenvector is $\pi$. Then the spectral gap is defined by $\gamma=1-\max_{\lambda\ne 1} |\lambda| $. Note that $0<\gamma<1$. Then it yields the following estimate for the variance of a test function $f:\Omega\to\mathbb{R}$ with respect to the stationary measure $\pi$

${\mathrm{Var}}_\pi(P^n f) \leq (1-\gamma)^{2t} {\mathrm{Var}}_\pi(f)$

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I can give you the answer for a certain class of Markov Chains, i.e. for reversible, irreducible and aperiodic Markov Chains on a finite state space.

Then the question is related to the spectral gap of the transition matrix $P:\Omega\times\Omega\to [0,1]$ of your chain. The transition matrix entry $P(x,y)$ is the probability of going from state $x$ to $y$. Thus one has $\sum_{y\in \Omega} P(x,y)=1$ If your chain is irreducible ($\forall x,y\in \Omega \exists n\in \mathbb{N}: P^n(x,y)>0$) and aperiodic. Aperiodic means that $\forall x\in \Omega$ the period defined by the greatest common divisor of the set $T(x)=\{n\in \mathbb{N}: P^n(x,x)>0\}$ is $1$. Then one has the following Theorem:

If $\Omega$ is finite and the chain defined by the transition matrix $P$ is irreducible and aperiodic than there exists a unique stationary distribution $\pi$ such that $\pi P =\pi$.

A chain satisfying a detailed balance relation $\pi(x) P(x,y) =\pi(y) P(y,x)$ is reversible.

Under these conditions the eigenvalues of $P$ are bounded in modulus by $1$ and the largest is $1$, the corresponding left eigenvector is $\pi$. Then the spectral gap is defined by $\gamma=1-\max_{\lambda\ne 1} |\lambda| $. Note that $0<\gamma<1$. Then it yields the following estimate for the variance of a test function $f:\Omega\to\mathbb{R}$ with respect to the stationary measure $\pi$

${\mathrm{Var}}_\pi(P^n f) \leq (1-\gamma)^{2t} {\mathrm{Var}}_\pi(f)$