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Well, we cannot, see for example https://en.wikipedia.org/wiki/Subindependence for an interesting counterexample. But the real question is: Is there some way to strengthen the condition so that independence follows? For example, is there some set of functions $g_1, \dotsc, g_n$ so that if $\E g_i(X) g_j(Y) =\E g_i(X) \E g_j(Y)$ for all $i,j$ then independence follows? And, how large must such a set of function be, infinite?

And, in addition, is there some good reference that treats this question?

kjetil b halvorsen
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  • have you had any luck with this? I'd love to see if there is a finite set of functions that works for any pair of RVs, and especially the justification is something other than CDF factorization – jld Sep 21 '17 at 22:53
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    I will look into it! I doubt there are in general a finite set, but any set which is a basis of a linear set of functions should do (so for instance, if $X, Y$ both have values in $0,1,2,\dotsc,n$ then a set of $n+1$linearly independent polynomials (or other) functions should do. – kjetil b halvorsen Sep 21 '17 at 22:57
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    https://stats.stackexchange.com/questions/503835/zero-correlation-of-all-functions-of-random-variables-implying-independence – kjetil b halvorsen Jan 09 '21 at 11:53
  • Thanks for the link. Good discussion there – jld Jan 10 '21 at 03:03

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Let $(\Omega, \mathscr F, P)$ be a probability space. By definition two random variables $X, Y :\Omega \to \mathbb R$ are independent if their $\sigma$-algebras $S_X := \sigma(X)$ and $S_Y := \sigma(Y)$ are independent, i.e. $\forall A \in S_X, B \in S_Y$ we have $P(A \cap B) = P(A)P(B)$.

Let $g_a(x) = I(x \leq a)$ and take $G = \{g_a : a \in \mathbb Q\}$ (thanks to @grand_chat for pointing out that $\mathbb Q$ suffices). Then we have $$ E\left(g_a(X)g_b(Y)\right) = E(I(X \leq a)I(Y \leq b)) = E(I(X \leq a, Y \leq b)) = P(X \leq a \cap Y \leq b) $$ and $$ E(g_a(X))E(g_b(Y)) = P(X \leq a)P(Y \leq b). $$

If we assume that $\forall a, b \in \mathbb Q$ $$ P(X \leq a \cap Y \leq b) = P(X \leq a)P(Y \leq b) $$ then we can appeal to the $\pi-\lambda$ theorem to show that $$ P(A \cap B) = P(A)P(B) \hspace{5mm} \forall A \in S_X, B \in S_Y $$ i.e. $X \perp Y$.

So unless I've made a mistake, we've at least got a countable collection of such functions and this applies to any pair of random variables defined over a common probability space.

jld
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    What have you shown, actually? Although you have defined an uncountable collection of functions, where have you demonstrated they are all needed? It's hard to imagine such a quantity of functions would be necessary when $X$ and $Y$ each have finite sets of possible values, for instance. – whuber Sep 19 '17 at 18:35
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    @whuber i was attempting to answer the question as to whether or not there exists any such collection of functions at all. I agree that the more interesting aspect is to find a minimal such set (which i'm still working on) – jld Sep 19 '17 at 18:36
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    You can reduce $G$ to a countable set by considering just rational $a$. – grand_chat Sep 19 '17 at 19:10
  • @grand_chat great point, i've updated – jld Sep 19 '17 at 19:18