A criterion for independence based on Characteristic function
As user75064 already pointed out, the answer is "no". However, there is the following result:
Let $X,Y$ be $\mathbb{R}^d$-valued random variables. Then the following statements are equivalent.
- $X,Y$ are independent
- $\forall \eta,\xi \in \mathbb{R}^d: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot \mathbb{E}e^{\imath \, Y \cdot \eta}$
i.e. if the characteristic function of the random vector $(X,Y)$ equals the product of the characteristic function of $X$ and $Y$, then $X$ and $Y$ are independent (proof).
No. The property in your post is called subindependence, and it is strictly weaker than independence. (Note that some people use the term "subindependent" as a synonym for "uncorrelated".) In addition to the references given in Wikipedia, you can find an example in this short note. Unfortunately it's behind a paywall. The example consists of two random variables with joint pdf $$h(x,y)=f(x)f(y)(1+\cos x\cos 3y)$$ where $$f(x)=C\left(\int_0^{1/2} \exp(1/(4s^2-1))\cos (sx)\,ds\right)^2$$ with appropriate normalizing constant $C$.