For $X,Y,$ RVs on $\{0,1\}^V$, does $P(v \in X) \geq P(v \in Y)$ for every $v \in V$ imply that X stochastically dominates Y?

Question

More specifically, for an infinite graph $G = (V,E)$, and random variables $X,Y$ taking values on the subsets of $V$:

Suppose that, for every $v \in V$, $$\mathsf{P}(v \in X) \geq \mathsf{P}(v \in Y).$$

Can we then conclude that for any increasing event $A \subset \{0,1\}^V$, $\mathsf{P}(X \in A)\geq \mathsf{P}(Y \in A)$?

Answer 1

We can't.

Take $V = \{1,\dots,n\}$ for example. Your property holds if $X$ is uniformly chosen to be one of $\{1\}, \{2\}, \dots, \{n\}$, while $Y = \{1,2,\dots,n\}$ with probability $\frac1{2n}$ and $Y = \emptyset$ otherwise.

However, with this example, if $A$ is the event of having size $\ge 2$, then $\Pr[X \in A] = 0$ and $\Pr[Y \in A] = \frac1{2n}$.