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assume $\mathcal{A}$ is a $\sigma$-algebra and $\xi$ is a r.v.  and $\forall x\in \mathbb R$  $$\mathbb{E}\left[e^{ix\xi} \mid \mathcal{A}\right] = \mathbb{E} \left[ e^{ix\xi} \right]\tag{*}$$

try to prove: $\xi$ is independent of $\mathcal{A}$

what I have tried:

according to the def of conditional expectation, we get from (*): $\forall A\in\mathcal A$ $$\mathbb E\left[e^{ix\xi}1_A\right] = \mathbb{E}\left[e^{ix\xi}\right] \mathbb{E}\left[1_A\right] \tag{1}$$

then I tried to prove: $$\mathbb E\left[e^{ix\xi}e^{iy1_A}\right] = \mathbb{E}\left[e^{ix\xi}\right] \mathbb{E}\left[e^{iy1_A}\right]\tag{2}$$

if (2) holds, then $\xi$ and $1_A$ is independent, then $\xi$ is independent of $\mathcal{A}$.

Lookout
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    Where you say $\operatorname{E} \left[e^{ix\xi} \mid \mathcal{A}\right] =\operatorname{E} \left[e^{ix\xi} \right]$, do you mean this holds for ALL values of $x$? $\qquad$ – Michael Hardy Mar 31 '16 at 03:22
  • @MichaelHardy yes, I have added it to my question. – Lookout Mar 31 '16 at 04:36
  • @user166445 How does the computation in the accepted answer prove the independence of $\xi$ and $A$ (I mean, if the first identity in your post does not suffice to prove the independence of $\xi$ and $\mathcal A$)? The independence of $\xi$ and $A$ is not defined by this identity but by the fact that $P(X\in B,A)=P(X\in B)P(A)$ for every Borel set $B$, right? – Did Mar 31 '16 at 17:43
  • @Did the lhs in (2) is the characteristic function of $(X,1_A)$, according to http://math.stackexchange.com/q/287138/161406, they are independent. Is there something wrong in Marcel 's answer? – Lookout Apr 03 '16 at 01:19
  • The trouble lies at least as much in your question, where one does not really know what is known, what is assumed and what is to prove. For example, I fail to see how anybody seeing as obvious that the definition of conditional expectation implies (1) (since you state this without justification) can need a proof for the implication (1) $\implies$ (2). Re the answer, note that it basically comes back to the definition of conditional expectation, which you took for granted, hence how did the answer bring anything new to you, one wonders. Finally, if really the question is to prove (2) ... – Did Apr 03 '16 at 07:49
  • ... starting from (1), then the answer is to note that $$e^{iy\mathbf 1_A}=1+(e^{iy}-1)\mathbf 1_A$$ and to apply this identity plus the linearity of the expectations to the identity (1). To sum up, yes, every identity on the page holds, including those in the answer, but the whole does not fit together logically. – Did Apr 03 '16 at 07:51
  • @Did I updated my question. – Lookout Apr 03 '16 at 08:12
  • Yes, and this puts even more clearly the accepted answer offtopic (unfortunately for its well-meaning author) and suggests even more strongly the $e^{iy\mathbf 1_A}=1+(e^{iy}-1)\mathbf 1_A$ approach. – Did Apr 03 '16 at 08:15

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Actually, you do not need to go back to the definition of conditional expectation but use the tower property instead: $$\mathbb E[e^{ix\xi}e^{iy1_A}] = \mathbb E[\mathbb E[e^{ix\xi}e^{iy1_A} \mid \mathcal A]] = \mathbb E[\mathbb E[e^{ix\xi} \mid \mathcal A]e^{iy1_A}] = \mathbb E[\mathbb E[e^{ix\xi}]e^{iy1_A}] = \mathbb E[e^{ix\xi}]\mathbb E[ e^{iy1_A}]$$

Michael Hardy
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Marcel
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