# Independence of a random variable and a sigma algebra

Let $$(\Omega,\mathcal{F},P)$$ be a probability space and $$X$$ be a random variable on it. Consider a sub $$\sigma$$-algebra $$\mathcal{G}$$. $$X$$ is said to be independent of $$\mathcal{G}$$ if $$\sigma(X)$$ and $$\mathcal{G}$$ are independent as $$\sigma$$-algebras.

I already know the fact that independence of $$X$$ and $$\mathcal{G}$$ implies $$\mathbb{E}[X|\mathcal{G}]=\mathbb{E}[X]$$ but not necessarily the other way round. However, if $$X$$ satisfies the equality $$\mathbb{E}[e^{itX}|\mathcal{G}]=\mathbb{E}[e^{itX}]$$, for all $$t\in\mathbb{R}$$, then can we conclude that $$X$$ and $$\mathcal{G}$$ are independent?

Yes. The equality $$\mathbb{E}[e^{itX}|\mathcal{G}]=\mathbb{E}[e^{itX}]$$ means that for any $$A\in\mathcal G$$ and for all $$t$$ $$\mathbb{E}[e^{itX} \mathbb 1_A]=\mathbb{E}[e^{itX}]\cdot\mathbb P(A).$$ And also $$A^c\in\mathcal G$$, so this equality holds for $$A^c$$ as well.
Let us use Kac's theorem $$\mathbb{E}[e^{itX} e^{is\mathbb 1_A}]=\mathbb{E}[e^{itX}\left(e^{is}\mathbb 1_A+\mathbb 1_{A^c}\right)]=e^{is}\mathbb{E}[e^{itX}\mathbb 1_A]+\mathbb{E}[e^{itX}\mathbb 1_{A^c}]$$ $$= e^{is} \mathbb{E}[e^{itX}]\mathbb P(A) + \mathbb{E}[e^{itX}]\mathbb P(A^c) = \mathbb{E}[e^{itX}]\left(e^{is}\mathbb P(A) +\mathbb P(A^c)\right) = \mathbb{E}[e^{itX}]\mathbb{E}[e^{is\mathbb 1_A}].$$ We can see that for all $$t,s\in\mathbb R$$, the joint characteristic function of $$X,\mathbb 1_A$$ is a product of c.f.'s. Kac's theorem implies that $$X$$ and $$\mathbb 1_A$$ are independent. And since it holds for all $$A\in\mathcal G$$, $$X$$ and $$\mathcal G$$ are independent.