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Consider the integrable random variable $X$ on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and the sub sigma-algebra $\mathcal{G}\subset \mathcal{F}$.

I want to show that

$$ \text{$X$ is independent of $\mathcal{G}$} \Leftrightarrow \text{ $\mathbb{E} \left[ e^{i u X } \mid \mathcal{G} \right] = \mathbb{E}\left[ e^{i u X } \right]$ } $$

The necessary part seems to be trivial, since if $X$ is independent of $\mathcal{G}$ then so is $e^{i u X }$ as it is a continuous and bounded function of $X$, and the result follows.

For the sufficient part, I must show that the sigma algebras $\sigma(e^{i u X })$ and $\mathcal{G}$ are independent. The hypothesis and definition of conditional expectation yields

$$\mathbb{E}\left[ e^{i u X } \mathbb{1}_G \right] = \mathbb{P}\left[ G \right] \mathbb{E}\left[ e^{i u X } \ \right], ~\forall G \in \mathcal{G}$$

I am wondering what more needs to be shown to justify this part.

shilov
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  • Let $F$ be the set of all $f$ such that $\mathbb E[f(X)1_G] = \mathbb P(G)\mathbb E[f(X)]$ for all $G \in \mathscr G$. Then $F$ is closed under some operations such as addition and some form of limit. Starting with the fact that $e^{iux} \in F$ for all $u$, show that $F$ must contain indicators of all Borel sets. – Sarvesh Ravichandran Iyer Jul 07 '23 at 10:54

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By the Fourier transform, we can write: $I_A(x)=\int\widehat{I_A}(\xi)e^{i2\pi\xi x}d\xi$. $$P_{X/\mathcal{G}}(A)=\int I_A(x) dP_{X/\mathcal{G}}=\int\int\widehat{I_A}(\xi)e^{i2\pi\xi x}d\xi dP_{X/\mathcal{G}}=\int\int\widehat{I_A}(\xi)e^{i2\pi\xi x}dP_{X/\mathcal{G}} d\xi=$$$$=\int\int\widehat{I_A}(\xi)e^{i2\pi\xi x}dP_{X} d\xi=\int\int\widehat{I_A}(\xi)e^{i2\pi\xi x} d\xi dP_{X}=\int I_A(x) dP_{X}=P_{X}(A)$$

Speltzu
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