# Definition of $\rho$-mixing and its relation to strong mixing?

From Wikipedia:

Suppose $\{X_t\}$ is a stationary Markov process, with stationary distribution $Q$. Denote $L²(Q)$ the space of Borel-measurable functions that are square-integrable with respect to measure $Q$. Also let $ℰ_tϕ(x) = E[ϕ(X_t) | X_0 = x]$ denote the conditional expectation operator on $L²(Q)$. Finally, let $Z = \{ϕ∈L²(Q): ∫ ϕdQ = 0\}$ denote the space of square-integrable functions with mean zero.

The $ρ$-mixing coefficients of the process $\{x_t\}$ are $$\rho_t = \sup_{\phi\in Z:\,\|\phi\|_2=1} \| \mathcal{E}_t\phi \|_2.$$ The process is called $ρ$-mixing if these coefficients converge to zero as $t → ∞$.

I was wondering

1. How $\rho_t$ can represent the "difference" between the stationary distribution $Q$ and the conditional distribution of $X_t$ given $X_0 = x$?

2. why restricting focus on $Z = \{ϕ∈L²(Q): ∫ ϕdQ = 0\}$?

3. How is $\rho$-mixing related to strong mixing defined in an earlier part of the same Wikipedia article? $\rho$-mixing seems to mean convergence of measure of $X_t$ to the limiting distribution, while strong mixing seems to mean $X_t$ become more and more independent from $X_0$, again from Wikipedia:

implies that for any two possible states of the system (realizations of the random variable), when given a sufficient amount of time between the two states, the occurrence of the states is independent.

What kind of mixing is for Markov chain mixing time?

4. Whenever talking about the stationary distribution, is the underlying Markov process discrete-time? If not, how is the stationary distribution for a continuous time Markov process defined?

Thanks!

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