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given two shift invariant measures in the Bernoulli space $\{0,1\}^{\mathbb{N}}$, is there a way to construct joinings of them? It's very diffcult, in general, to find exactly the minimal joining i.e, the joining which achieves the d-bar distance of the measures. I would like a way to construct joinings efficiently... The shift here is the usual full shift and the d-bar distance is

$d(\eta,\nu) = \inf_{\mu \in J}\mu([0] \times [1]) + \mu([1] \times [0])$

where $J$ is the space of joinings of $\mu$ and $\nu$. A Joining is a invariant measure in $\{0,1\}^{\mathbb{N}} \times \{0,1\}^{\mathbb{N}}$.which projects $\eta$ in the first coordinate and $\nu$ in the second... (Anthony Quas has defined this distance above, equal as my definition)

thanks for your attention

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  • $\begingroup$ I've modified my text, professor. Thank you :) $\endgroup$ – Bruno Brogni Uggioni Aug 18 '15 at 20:32
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I assume that the context you're talking about is ergodic measures on $A^{\mathbb Z}$ for some finite set $A$. One definition of the $\bar d$ distance is $$ \bar d(\mu,\nu)=\inf_{\lambda\in J(\mu,\nu)}\int \mathbb 1_{x_0\ne y_0} \,d\lambda(x,y), $$ where $J(\mu,\nu)$ is the (weak$^*$-compact) collection of joinings of $\mu$ and $\nu$, that is the collection of shift-invariant measures, $\lambda$, on $A^{\mathbb Z}\times A^{\mathbb Z}$ such that $\lambda(B\times A^{\mathbb Z}=\mu(B)$ and $\lambda(A^{\mathbb Z}\times B)=\nu(B)$.

I have a paper that estimates $\bar d$ distance for infinite order Markov processes with Zaqueu Coelho. Xavier Bressaud, Roberto Fernandez and Antonio Galves have a paper dealing with a similar situation.

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    $\begingroup$ I don't think ergodicity plays a direct role; but in fact the processes studied in these papers have very strong ergodicity properties. In general, it can be shown if $\mu$ and $\nu$ are ergodic, that the extreme points of $J(\mu,\nu)$ are also ergodic, so that the $\bar d$ distance is attained by an ergodic joining. $\endgroup$ – Anthony Quas Aug 19 '15 at 0:55
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    $\begingroup$ So this is the Wasserstein-Monge-Kantorovich-Rubinshtein-transportation-and-other-names distance. Some keywords for you @BrunoBrogniUggioni. $\endgroup$ – Stéphane Laurent Sep 3 '15 at 9:29
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    $\begingroup$ @StéphaneLaurent: A joining is actually quite a lot stronger: you require the coupling to be shift-invariant. $\endgroup$ – Anthony Quas Sep 3 '15 at 15:09
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    $\begingroup$ Ok! You're right, I missed this point. $\endgroup$ – Stéphane Laurent Sep 3 '15 at 16:50

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