@FedorPetrov, in your example, one row will have all zeros, so that one of the $r_i$s is zero, and thus violates the assumption that all $r_i, s_j$ are positive.

@darijgrinberg How are only finite many of the $r_i$ nonzero? In my case, each $s_j$ is $1/n$ and the rows are indexed by integers $i$ which have no prime factor larger than the largest prime $\le n$ (so $m$ is unbounded but doesn't take on every integer). For such values if $m=i$, I have $$r_i=\frac{\prod_{p\le n, p\text{ prime}}\frac{p-1}{p}}{i}$$, so that $r_i$ is never zero and their sum is 1.

@FedorPetrov Yes.

@ darij grinberg, it means that $A$ has $0$'s at entries $(i,j) $for which $W$ has a zero, and $A$ has a positive number at $(i,j)$ in locations for which $W$ has a $1$. @PeterTaylor I closed the other one.

@RonP, the entries in the joint mass distribution are $P(X=i, Y=j)=P(X=i)P(Y=j \vert X=i).$ Is this how the conditional probability describes the coupling, or do you mean something else?

To obtain a new coupling of $X,Y$ from a given one. Will edit.

What if $(B_p)$ is an independent process while $(A_p) $ is a dependent process? The dependence is not major since in my case $(A_p)_{p \le n} $ converges to $(B_p)_{p\le n} $ in distribution (as $n \to \infty$).

@michael, thanks. No, I'm not trying to prove that in the general case. But for a specific example, I'd like some ideas on how I can attempt to combine these couplings to a larger space. Should the marginals be $A_p's, B_p's$ or random vectors $(A_p, B_p) $?