# Show that r.v-s X and Y are independent via $\mathbb{E}[e^{iuX}p(Y)]=\mathbb{E}[e^{iuX}]\;\mathbb{E}[p(Y)]$

Let X and Y be two random variables on some $(\Omega, \mathcal{F}, \mathbb{P}).$ We can assume that Y is bounded. It is clear that X and Y are independent iff:

$$\mathbb{E}[e^{iuX}e^{ivY}]=\mathbb{E}[e^{iuX}]\;\mathbb{E}[e^{ivY}], \ \forall u,v \in \mathbb{R}.$$

My question is, can we show that X and Y are independent iff

$$\mathbb{E}[e^{iuX}p(Y)]=\mathbb{E}[e^{iuX}]\;\mathbb{E}[p(Y)]$$ $\forall u \in \mathbb{R}$ and for all polynomial functions $p(\cdot)$ ?

Thank you!

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Yes. If $|Y| \le B$, we can approximate $\exp(ivy)$ uniformly by polynomials $p(y)$ on $[-B,B]$, and since $|e^{iuX}| = 1$ this approximates ${\mathbb E}[e^{iuX} e^{ivY}]$ and ${\mathbb E}[e^{ivY}]$ by ${\mathbb E}[e^{iuX} p(Y)]$ and ${\mathbb E}[p(Y)]$ respectively.