Context
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Let $t\_1, .., t\_n$ be jointly independent boolean random variables.


Let $X, Y$ be monotone functions (i.e. $\forall i: t\_i \geq t'\_i$ implies $X(t\_1, ... t\_i ..., t\_n) \geq X(t'\_1, ... t'\_i, ..., t'\_n)$.


Then, $E(XY) \geq E(X)E(Y)$


This is easily proved by induction on $n$:


* noramlize so that $E(X| t\_n = 0) = E(Y| t\_n=0) = 0$, expand


Question
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Do we know a continuous version of this? I.e. is there a way to prove this if the $t\_i$ are continuous rather than boolean? If not, what is the simplest counter example?



Continuous variant of KFG Inequality
There are many variants of the FKG inequality. For example, consider ${\mathbb R}^n$ as a lattice with partial order $x \le y$ if $x\_i \le y\_i$ for all $i$, and let $X\_i$ be independent random variables. Then for any increasing functions $f$, $g$ on ${\mathbb R}^n$, 
$E[f(X\_1,\ldots,X\_n) g(X\_1,\ldots,X\_n)] \ge E[f(X\_1,\ldots,X\_n)] E[g(X\_1,\ldots,X\_n)]$
provided all the expected values exist.


See J. Kemperman, "On the FKG inequality for measures on a partially ordered space", Indag. Math. 39 (1977) 313-331 for more.


