Given a fixed positive integer $n$, I have two random variables $$A(n)=2^{A_2}\cdots p^{A_{p_n}}, B(n)=2^{B_2}\cdots p^{B_{p_n}},$$ where $p_n$ is the largest prime number not exceeding $n$, $(A_p)_{p\le n}$ and $(B_p)_{p\le n}$ are two random processes, and the index $p$ ranges over prime numbers. I am searching for a coupling of $A(n), B(n)$ such that $A\vert B$, and this holds if and only if we have the vector inequality $(A_2,A_3\ldots, A_{p_n})\le (B_2,B_3\ldots, B_{p_n})$.

Thus, I seek a probability space in which $A_p\le B_p$ for each $p\le n$.

For each $p$, I have a coupling of $A_p$ and $B_p$ such that $A_p\le B_p$. I.e., for each $p$ I have a joint distribution $(A_p',B_p')$ with marginals $A_p,B_p$ such that $\mathbb{P}(A_p=a, B_p=b)=0$ when $a > b$.

Is there any literature/technique on how to combine these couplings into a new probability space in which $A_p\le B_p$ for *all* $p\le n$?

My idea was to consider the joint distribution of the $n$ coupled random vectors $(A_p,B_p)$. Then the $p$th marginal is $(A_p, B_p)$ which only has positive probability when $A_p \le B_p$. However, I am not sure if this is the desired probability space since the marginals are not the $A_p$'s or $B_p$'s but are instead random vectors. I am unsure if this joint distribution gives the desired coupling of $A,B.$