Can we conclude from $\DeclareMathOperator{\E}{\mathbb{E}}\E g(X)h(Y)=\E g(X) \E h(Y)$ that $X,Y$ are independent?

Question

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?

Answer 1

This is almost Proposition 7.1.3(i) from Athreya & Lahiri 2006 sans some minor differences in formatting:

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $\{X_1, \ldots, X_k \}$, $2 \leq k < \infty$ be a collection of random variables on $(\Omega, \mathcal{F}, P)$.

(i) Then $\{X_1, \ldots, X_k \}$ is independent iff $$\mathbb{E} \left[ \prod_{i=1}^k h_i(X_i) \right] = \prod_{i=1}^k \mathbb{E} [h_i(X_i)]$$ for all bounded Borel measurable functions $h_i: \mathbb{R} \mapsto \mathbb{R}$, $i=1,\ldots,k$.

Proof:

If (i) holds, then taking $h_i = I_{B_i}$ with $B \in \mathcal{B}(\mathbb{R})$, $i=1,2,\ldots,k$ yields the independence of $\{X_1, \ldots, X_k \}$. Conversely, if $\{X_1, \ldots, X_k \}$ are independent, then (i) holds for $h_i = I_{B_i}$ for $B_i \in \mathcal{B}(\mathbb{R})$, $i=1,2,\ldots,k$ and hence for simple functions $\{h_1, h_2, \ldots, h_k \}$. Now (i) follows from the BCT.

BCT refers to the bounded convergence theorem which is given as corollary 2.3.13 of the same reference:

Let $\mu(\Omega) < \infty$. If there exists a $0 < k < \infty$ such that $|f_n| \leq k$ a.e. $(\mu)$ and $f_n \rightarrow f$ a.e. $(\mu)$, then $$\lim_{n \rightarrow \infty} \int f_n\ d\mu = \int f\ d\mu$$ and $$\lim_{n \rightarrow \infty} \int |f_n - f|\ d\mu = 0.$$

Proof: Take $g(\omega) \equiv k$ for all $\omega \in \Omega$ in the previous corollary.

The previous corollary (Corollary 2.3.12) is Lebesgue's dominated convergence theorem:

If $|f_n| \leq g$ a.e. $(\mu)$ for all $n \geq 1$, $\int g\ d\mu < \infty$ and $f_n \rightarrow f$ a.e. $(\mu)$, then $f \in L^1 (\Omega, \mathcal{F}, \mu)$, $$\lim_{n \rightarrow \infty} \int f_n\ d\mu = \int f\ d\mu$$ and $$\lim_{n \rightarrow \infty} \int |f_n - f|\ d\mu = 0.$$

Answer 2

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.