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.

  • $\begingroup$ Thanks a lot! This is a very nice proof. $\endgroup$ – Sam Feb 15 at 6:20

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