Consider an at most countable set $S$ and the corresponding bit space $\{0, 1\}^S$ that is often considered in percolation, interacting particle systems, and other lattice models. Suppose that $\le$ denotes stochastic domination, and random variables are assumed to be valued in $\{0, 1\}^S$, and are identified with their distributions. Let $Z$ have iid components as $s\in S$ varies with its one-dimensional marginals $Z_s$ being bernoulli(p) with $p\in (0, 1)$. It is easy to see that if $X\le Y$, then $Z'X\le Z''Y$ where $Z', Z''$ are identically distributed, $\{Z', X\}$ are independent (defined on the space probability space), and as are $\{Z'', Y\}$.
What is less clear to me is if the converse holds. I see right away that the question is reduced to the case where $S$ is finite. I believe the answer is probably going to be that the converse fails because I am reading a proof of a result of Liggett Schonmann and Stacey that takes special care to dilute both fields before claiming stochastic domination. But as far as my own efforts, I have tried to prove that the converse holds by extracting the information from a subspace. I try to look at the "subspace of the probability space where Z'=1 at every $s\in S$". The plan was to find a coupling of $Z'X$ with $Z''Y$ where pointwise $\le$ holds, and then to restrict that information to just where $Z'=1$. But the trouble is that there is no $Z', Z''$, much less a notion of comparison between the two. Once I couple the laws of $Z'X$ and $Z''Y$ there is no reason to expect to be able to recover $Z', Z''$.
An example where $S$ has two sites is as follows. Consider the random variable $X$ which has probability $1/4$ to be $(1, 1)$, $1/6$ to be $(1, 0)$, $1/6$ to be $(0, 1)$ and $5/12$ to be $(0, 0)$. Let $Y$ have $1/2$ probability to be $(1, 1)$ and $1/2$ to be $(0, 0)$. Consider dilution by $1/2.$ Then $X$ is not stochastically dominated by $Y$, but the dilution of $X$ is dominated by the dilution of $Y$.