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 $A,B.$

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    $\begingroup$ Is your conjecture " if you can couple each marginal so that $A_p < B_p$ then you can couple the processes so that the same holds "? The answer to that is no, take for example, B to be fixed, say $B_p = 2 $ for all p with probability .5 or $B_p = 4 $ for all p with probability .5 , and $A_p$ to be iid values 1 and 3 with prob .5 . $\endgroup$ – user83457 Oct 3 '17 at 10:48
  • $\begingroup$ @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) $? $\endgroup$ – The Substitute Oct 3 '17 at 20:32

One way to achieve your goal is as follows. You have nonnegative numbers $f_p(a_p,b_p)=P(A_p=a_p,B_p=b_p)$ for $p=2,3,5,\dots$, $a_p=0,1,\dots$, $b_p=0,1,\dots$ such that for each $p$ you have $\sum_{a_p,b_p}f_p(a_p,b_p)=1$ and $f_p(a_p,b_p)=0$ when $a_p>b_p$. Consider the corresponding conditional probabilities $f_p(a_p|b_p):=f_p(a_p,b_p)/g_p(b_p)$, where $g_p(b):=\sum_{a_p}f_p(a_p,b)=P(B_p=b)$, your marginal probabilities for $B_p$ -- assuming $g_p(b)>0$ for all relevant $b$. You can use these couplings of $A_p$ and $B_p$ by requiring that the sequence $(A_2,A_3,A_5,\dots)$ be conditionally independent given $(B_2,B_3,B_5,\dots)$ and that each $A_p$ depend on the sequence $(B_2,B_3,B_5,\dots)$ only through $B_p$: \begin{equation} P(A_2=a_2,A_3=a_3,A_5=a_5,\dots|B_2=b_2,B_3=b_3,B_5=b_5,\dots) =f_2(a_2|b_2)f_3(a_3|b_3)f_5(a_5|b_5)\dots \end{equation} for all $a_p$'s and $b_p$'s. For the joint distribution of $(B_2,B_3,B_5,\dots)$ you can take any distribution with the marginals $(g_2,g_3,g_5,\dots)$. It is easy to check that thus you will have a joint distribution of $(A_2,A_3,A_5,\dots,B_2,B_3,B_5,\dots)$ with the prescribed two-dimensional marginals $f_p$: $P(A_p=a_p,B_p=b_p)=f_p(a_p,b_p)$ for all $p=2,3,5,\dots$, $a_p=0,1,\dots$, $b_p=0,1,\dots$.

In particular, letting $B_2,B_3,B_5,\dots$ be independent, you will just get independent pairs $(A_p,B_p)$ with the prescribed two-dimensional marginals $f_p$.

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  • $\begingroup$ 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 $n \to \infty$). $\endgroup$ – The Substitute Oct 3 '17 at 20:37
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    $\begingroup$ That is also possible. However, you may want to ask this additional question as a separate one, where you can state all the additional conditions you want the joint distribution of the $A_p$'s and $B_p$'s to satisfy. $\endgroup$ – Iosif Pinelis Oct 3 '17 at 21:28

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