Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $$|\mu_1(C) - \mu_2(C)| \leq r \, dist(\mu_1, \mu_2)$$

share|improve this question
    
I understood all the words except "coupling". –  GEdgar Nov 23 '12 at 16:52
    
A coupling between $\mu_1$ and $\mu_2$ is a measure $c = \mu_1\, \, t \, \, \mu_2$ defined on the space of measures of $\Omega \times \Omega$ and which has as marginals $\mu_1$ and $\mu_2$. –  QuantumLogarithm Nov 23 '12 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.