Combining Couplings of Random Variables

Question

Given $n$, let $(A_i){i\le n}, (B_i)_{i\le n}$ be sequences of distinct random variables, where $(B_i)_i$ is an independent process. For each $i$, I have a coupling of $A_i,B_i$ with some desired feature (e.g., $A_i$ divides $B_i$).

How can I construct a probability space that contains all of these couplings? That is, how can I construct a probability space so that $A_i$ divides $B_i$ for all $i\le n$?

One idea is to consider the product measure, but how can we be certain that the desired features (division) will carry over into the product space?

Answer 1

Answer to the 1st version of the question

In general, this may not be possible.

Say, $n=2$.

• Let $A_1$ be a $(\frac{1}{2},\frac{1}{2})$ Bernoulli variables over $\{-1,1\}$ and $A_2:=A_1$.
• Let $B_1$ be a $(\frac{1}{2},\frac{1}{2})$ Bernoulli variables over $\{-1,1\}$ and $B_2:=-B_1$.

Then there is an obvious coupling $(A'_1,B'_1)$ of $A_1$ and $B_1$ in which $A'_1=B'_1$, and likewise there is a coupling $(A'_2,B'_2)$ of $A_2$ and $B_2$ in which $A'_2=B'_2$. But clearly, there is no coupling $((A''_1,A''_2),(B''_1,B''_2))$ of $(A_1,A_2)$ and $(B_1,B_2)$ such that $A''_1=B''_1$ and $A''_2=B''_2$.

If you are particularly interested in the property "$A_i$ divides $B_i$", still this may not be possible.

• Let $A_1$ be a $(\frac{1}{2},\frac{1}{2})$ Bernoulli variables over $\{1,2\}$ and $A_2:=A_1$.
• Let $B_1$ be a $(\frac{1}{2},\frac{1}{2})$ Bernoulli variables over $\{1,2\}$ and $B_2:=3-B_1$.

Note that in a coupling $(A'_1,B'_1)$ of $A_1$ and $B_1$, we have $A'_1|B'_1$ almost surely if and only if $A'_1=B'_1$ almost surely. Now, there is a coupling $(A'_1,B'_1)$ of $A_1$ and $B_1$ in which $A'_1|B'_1$ (choose $A'_1=B'_1$), and there is a coupling $(A'_2,B'_2)$ of $A_2$ and $B_2$ in which $A'_2|B'_2$ (again, choose $A'_2=B'_2$), but there is no coupling $((A''_1,A''_2),(B''_1,B''_2))$ of $(A_1,A_2)$ and $(B_1,B_2)$ such that $A''_1|B''_1$ and $A''_2|B''_2$.

Answer to the 2nd version of the question

The answer to your modified answer is still "no". The "equalities" in my example above are meant to exaggerate what can go wrong. The same problem persists. Namely, suppose $A_1,A_2$ are "strongly dependent", for instance, $A_1=A_2$ with probability close to $1$.

• Let $(A'_1,B'_1)$ be a coupling of $A_1$ and $B_1$ such that $A_1=B_1$ with probability close to $1$,
• Let $(A'_2,B'_2)$ be a coupling of $A_2$ and $B_2$ such that $A_2=B_2$ with probability close to $1$,

Then, if you were able to construct a coupling $((A''_1,A''_2),(B''_1,B''_2))$ of $(A_1,A_2)$ and $(B_1,B_2)$ such that $(A''_1,B''_1)$ and $(A''_2,B''_2)$ have the same properties (i.e., $A''_1=B''_1$ with probability close to $1$ and $A''_2=B''_2$ with probability close to $1$), then you would necessarily get $B''_1=B''_2$ with probability close to $1$, contradicting with the independence of $B_1$ and $B_2$.