PREVIOUS SECTION: Properties of UI

1.5  Coupling inequality. Maximal coupling. Dobrushin's theorem

In this section, we assume that random variables are not necessary real-valued, but may take values in some measurable space eq176 that is assumed to be complete separable metric space.

Coupling inequality

Let eq177  be two eq178 -valued r.v.'s.

Put  eq179

Then, for eq180,  

 
 

 

Therefore, for any eq180, that is

Maximal coupling

Let's reformulate the statement. Note that l.h.s. of inequality (*) depends on "marginal" distributions eq185 and eq186 only and does not depend on the joint distribution of eq039 and eq041. Therefore, we get the following:

for given eq185 and eq186 and for any their coupling (*) takes place. Or, equivalently,

eq188

(?) May be, in eq188 is equality?

(??) If "yes", then does there exists such a coupling that

eq189

Both answers are positive! And this is the statement of Dobrushin's theorem.

Proof.  eq190 is a signed measure. Therefore, Banach theorem states that

there exists a subset eq191:

(a) eq192  eq193 ;

(b) eq194 eq195.

Note:

1) if eq196, then eq197 and the coupling is obvious;

2) eq198.

Assume eq199. Introduce 4 distributions (probability measures):

eq200

 eq201

Similarly,

eq202  

eq203

Then, define 5 independent r.v.'s:  eq204 and

 eq205 
   eq206   eq207   eq208 

 

Now we can "construct" eq039 and eq041:  

eq209  

eq210

Simple calculations show that eq211eq212.

Problem No.3. "Indeed, . . ."

Then,  

eq214

So,  

eq215

and the proof is completed.

QDE

NEXT SECTION: Probabilistic Metrics


File translated from TEX by TTH, version 1.58, Htex, and with Natalia Chernova as the TeX2gif-convertor.