2
$\begingroup$

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 $A,B.$

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $f_p$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.