Let $\mu$ and $\mu'$ be probability measures on $\lbrace0,1\rbrace^\Lambda,\:\: \Lambda:= {\lbrace 0,1,\ldots,n\rbrace}$. Assume that

$\mu(X_i=1|X = \zeta \text{ on } \Lambda \setminus \lbrace i\rbrace) \leq \mu'(X_i=1|X = \eta \text{ on } \Lambda \setminus \lbrace i\rbrace),$

for any $i\in \lbrace 0,1,\ldots,n\rbrace$ and $\zeta \leq \eta$ (which is understood componentwise). Then by the Holley theorem (under some additional minor assumptions) we have

$\mu' \text{ stochastically dominates } \mu, \text{ i.e. } \mu'(g)\geq \mu(g),$

for any increasing function $g$. My question is: Assume that we have "a good upper-estimate" on:

$ \frac{\mu'(X_i=1|X = \eta \text{ on } \Lambda \setminus \lbrace i\rbrace)}{\mu(X_i=1|X = \zeta \text{ on } \Lambda \setminus \lbrace i\rbrace)}.$

could we then obtain some "reverse dominance", e.g.:

$\frac{\mu'(g)}{\mu(g)}\leq 1+\epsilon,$

for an increasing function $g$ (some condition may be required on $g$). So the task would be to determine what "good upper-estmimate means and what may be additional conditions on g.

Here (page 21) is a precise formulation of the Holley theorem (along with a proof).

  • $\begingroup$ Consider i.i.d. product measures with marginals $p,p'$. If $p'$ is close to $p$ you have an estimate as given, but the resulting measures are very different. (A concrete test function: $g(X) = 1_{\sum X_i > qn}$ for some $p<q<p'$.) This suggests that any bound along the you ask for will be very week. $\endgroup$ – Omer Jun 4 '11 at 16:41
  • $\begingroup$ @Omer, you are right. In the problem, from which my question originated, I was finally able to prove "some global bound", more or less a bound for $\sum_{i\in\Lambda} \frac{\mu'(X_i=1|X = \eta \text{ on } \Lambda \setminus \lbrace i\rbrace)}{\mu(X_i=1|X = \zeta \text{ on } \Lambda \setminus \lbrace i\rbrace)}$ which is obviously much stronger assumption and gives hope to prove something :) $\endgroup$ – Piotr Miłoś Jun 7 '11 at 11:56

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