Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I found the following theorem without a proof:

Let $X$ be a r.v. with respect to $\mathcal{F}$ and $\mathcal{G}\subset\mathcal{F}$ a sub $\sigma$-algebra. If for ever $c\in\mathbb{R}$ we have $$E[e^{cX}|\mathcal{G}]=E[e^{cX}]$$ then $X$ and $\mathcal{G}$ are independent.

I was wondering how to prove this. To show independence I have to prove:

$$P[A\cap B]=P[A]P[B]$$

for $B\in\mathcal{G}$ and $A\in\sigma (X)$. But it seems the wrong approach, since I can not use the above relation. Maybe someone could explain why the above theorem is true, or give me a reference.

share|improve this question
1  
Must assume every exponential moment of X is finite. –  Did Feb 9 '13 at 10:42
1  
@Did Ok, I think we can also assume this. –  user20869 Feb 9 '13 at 10:57
1  
I wonder why the answer below was downvoted. –  Did Feb 9 '13 at 11:05
    
@Did so do I. I was not the "down-voter" –  user20869 Feb 9 '13 at 12:00

1 Answer 1

up vote 0 down vote accepted

Sketch of proof:

  • deduce that $E[X^p\mid \mathcal G]=E(X^p)$ for each integer $p$;
  • this gives that $E[e^{itX}\mid\mathcal G]=E[e^{itX}]$ for all real number $t$;
  • conclude using characteristic functions. We have that $E[e^{itX}Y]=E[e^{itX}]\cdot E[Y]$ for all $Y$ which is $\mathcal G$-measurable.
share|improve this answer
    
For the first bullet: Can I use the property in this way: $E[X^p|\mathcal{G}]=E[e^{p\log{X}}|\mathcal{G}]$. Can I now apply my result also to $\log{X}$ instead of $X$? If so, why? I just know that it is true for $X$ not for $\log{X}$. How else should I prove your first bullet? –  user20869 Mar 21 '13 at 16:45
    
You don't know a priori that $X$ is positive. But you can identify the terms in the power series expansion. –  Davide Giraudo Mar 21 '13 at 16:53
    
thanks for the fast response. I do not see at the moment what you exactly mean, but I will try to figure it out. However at the end you want to conclude with characteristic function. I know the result: $X$ and $Y$ are independent iff $\mu_{X+Y}=\mu_Y\mu_X$, where $\mu_X:=E[e^{itX}]$. But here I want to show the independence of sub-$\sigma$-algebra. How can I use this result? –  user20869 Mar 21 '13 at 17:00
    
@hulik I've added details. –  Davide Giraudo Mar 25 '13 at 12:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.