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8
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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,
$$
rift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } \mbox{ } \lbrace ( \omega^1, \omega^2 ) \in X \times X \ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$
Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.
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7
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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,
$$
rift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) rbrace \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$
Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.
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6
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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,
$$
rift(crift(C) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$
Definition: the coupling is not necessarily the product between the two measures. Let's consider two measures $\mu^1$ and $\mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.
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5
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Let $X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $C$ be a coupling between the two measures and let's define,
$$
rift(c) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that, if A is a cylinder subset of $X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$
P.S.
Definition: the coupling is not necessarily the product between the two measures. The coupling $C$ between Let's consider two measures $\mu^1$ and $\mu^2$ \mu^2$, acting each one on the $\sigma$-algebra of subsets of $X$. The coupling $C$ is any measure $C$ acting on the $\sigma$-algebra of subsets of $X^1 \times X^2$, with $X^i = X$, which has as marginals $\mu^1$ and $\mu^2$, i.e. $\mu^1 = C \circ \pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : X^1 \times X^2 \longrightarrow X^i$.
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4
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Let $\Omega X = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $c$ C$ be a coupling between the two measures and let's define,
$$
rift(c) = sup_{s \in S}\lbrace \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in X\ Omega\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that, if A is a cylinder subset of $\Omega$, X$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures as before, $$|\mu_1(A) - \mu_2(A)| \leq r \mbox{ } dist(\mu_1, \mu_2)$$
Definition of
P.S. the coupling : a is not necessarily the product between the two measures. The coupling $c$ C$ between the two measures $\mu_1$ \mu^1$ and $\mu_2$, acting both on the $\sigma$-algebra of subsets of $\Omega$, \mu^2$ is any measure $c$ defined C$ acting on the $\sigma$-algebra of subsets of $\Omega_1 X^1 \times \Omega_2$, X^2$, with $\Omega_i X^i = \Omega$, X$, which has as marginals $\mu_1$ \mu^1$ and $\mu_2$ as marginals\mu^2$, i.e. This means that $\mu_i \mu^1 = c C \circ \pi_i^{-1}$, pi^{-1}_1$, $\mu^2 = C \circ \pi^{-1}_2$, where $\pi_i$ is the projection $\pi_i : \Omega X^1 \times \Omega X^2 \longrightarrow \Omega_i$. The product between two measures is just one of the possible couplings between the two measuresX^i$.
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3
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Let $\Omega = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $c$ be a coupling between the two measures and let's define,
$$
dist(crift(c) = sup_{s \in S}\lbrace c \mbox{ } C \mbox{ } ( \mbox{ } \lbrace \omega^1, \omega^2 \in \Omega\ \text{ s.t. } \omega^1_s \neq \omega^2_s \mbox{ } \rbrace) \mbox{ } \mbox{ } \rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity.
How can I prove that:
If C , if A is a cylinder subset of $\Omega$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures defined on the product spaceas before, $$|\mu_1(C) $|\mu_1(A) - \mu_2(C)| mu_2(A)| \leq r \, mbox{ } dist(\mu_1, \mu_2)$$
Definition of coupling: a coupling $c$ between the two measures $\mu_1$ and $\mu_2$, acting both on the $\sigma$-algebra of subsets of $\Omega$, is any measure $c$ defined on the $\sigma$-algebra of subsets of $\Omega_1 \times \Omega_2$, with $\Omega_i =\Omega$, which has $\mu_1$ and $\mu_2$ as marginals. This means that $\mu_i = c \circ \pi_i^{-1}$, where $\pi_i$ is the projection $\pi_i : \Omega \times \Omega \longrightarrow \Omega_i$. The product between two measures is just one of the possible couplings between the two measures.
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2
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Let $\Omega = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $c$ be a coupling between the two measures and let's define,
$$
rift(cdist(c) = sup_{s \in S}{ \, \, \, S}\lbrace c \, ( \, { lbrace \omega^1, \omega^2 \in \Omega,\, Omega\ \, text{ s.t. \, \, } \omega^1_s \neq \omega^2_s\, } omega^2_s \, rbrace) \, }.$$rbrace.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity. How can I prove that:
If C is a cylinder subset of $\Omega$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures defined on the product space, $$|\mu_1(C) - \mu_2(C)| \leq r \, dist(\mu_1, \mu_2)$$
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1
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inequality for coupling of measures
Let $\Omega = \prod _{s \in S} \Omega_s$, with $\Omega_s$ finite and all the same, $S$ countable. Let $\mu_1$ and $\mu_2$ be two probability measures on the product space (not necessarily the product measure). Let $c$ be a coupling between the two measures and let's define,
$$
rift(c) = sup_{s \in S}{ \, \, \, c \, \, { \omega^1, \omega^2 \in \Omega,\, \, s.t. \, \, \omega^1_s \neq \omega^2_s\, } \, \, }.$$
Then we call distance between the two measures $dist(\mu_1, \mu_2)$ the infimum over all the possible couplings $c$ of the previous quantity. How can I prove that:
If C is a cylinder subset of $\Omega$, specified by $r$ components $\omega_i$, then $\forall \mu_1, \mu_2$ probability measures defined on the product space, $$|\mu_1(C) - \mu_2(C)| \leq r \, dist(\mu_1, \mu_2)$$
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