1
$\begingroup$

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $\mathcal G\subset \mathcal F$ a subfield of $\mathcal F$. Let $X\in L^1(\Omega )$. Then, there is a r.v. $Z\in L^1(\Omega )$ s.t. $Z$ is $\mathcal G$-measurable and $\mathbb E[ZU]=\mathbb E[XU]$ for all $U$ bounded and $\mathcal G-$measurable.

Prove that $Z$ exist and is unique.

Ques 1 : First of all, is it equivalent to consider a function $U$ which is $\mathcal G-$measurable and bounded and $\boldsymbol 1_G$ where $G\in \mathcal G$ ? And if yes, why ? Because in proof in my course, my teacher always consider $\boldsymbol 1_G$ for $G\in \mathcal g$ despite of $U$ a function $\mathcal G-$measurable and bounded.

My attempts

For the existence, the thing would be to find a $Z\neq X$ s.t. $\mathbb E[(Z-X)U]=0$ for all $U$ $\mathcal G-$measurable and bounded, but I don't see how.

Quest 2 : Any idea ?

For the uniqueness, let $Z_1$ and $Z_2$ s.t. $\mathbb E[Z_1U]=\mathbb E[Z_2 U]=\mathbb E[XU]$ for all $U$ bounded and $\mathcal G-$measurable. In particular $\mathbb E[(Z_1-Z_2)U]=0$.

Quest 3 : How can I conclude that $\mathbb P\{Z_1=Z_2\}=1$ from here ?

  • 0
    The existence and uniqueness follow from the Lebesgue-Nikodym Theorem. – 2017-01-03
  • 0
    What is the motivation of the question? What is the context? The statement is very standard in any measure theoretical probability textbook which should contain at least a sketch of a proof and I doubt that it would be an exercise. – 2017-01-03
  • 0
    @Jack: It's a theorem of my course that has no proof. Could you please at least answer to the question 1 ? I will check the Lebesgue-Nikodym theorem for the rest – 2017-01-03
  • 0
    I don't quite understand Question 1: "is it equivalent to consider..." Do you mean replace the conclusion in the theorem by only indicator functions of $\mathcal{G}$-measurable sets? (Presumably there is a textbook for the class?) – 2017-01-03
  • 0
    For example, to prove that $Z=\mathbb E[X\mid \mathcal G]$, in my course, my teacher take $U$ a v.a. $\mathcal G$-measurable and bounded, and he prove that $\mathbb E[ZU]=\mathbb E[XU]$, but in the exercise, he take $G\in \mathcal G$ and prove that $\mathbb E[Z\boldsymbol 1_G]=\mathbb E[X\boldsymbol 1_G]$, but I don't understand why these two method are equivalent (if they are), and if they are not, why can we do both ? @Jack – 2017-01-03
  • 0
    There seems to be confusion in your question. To clarify, there are two things: (1)The theorem you quoted in the beginning which allows one to *define* the condition expectation given a sub $\sigma$-field; (2) If one wants to *prove* $Z=E(X|\mathcal{G})$, then by the *definition* of conditional expectations, one needs to prove two things (i) $Z$ is $\mathcal{G}$-measurable; (ii) $E(Z1_G)=E(X1_G)$ for every $G\in\mathcal{G}$. – 2017-01-03
  • 0
    @Jack: Thank you. Does it come frome the fact that simple function are dense in $L^1$ ? – 2017-01-03
  • 1
    Well, for the proof of $Z=E(X|\mathcal{G})$, I would say that it comes from the *definition* of conditional expectations. You might want to check your notes to see how it is defined. – 2017-01-03
  • 0
    What I wanted to say (sorry if it's ambiguous), the fact that to prove that $Z=\mathbb E[X\mid \mathcal G]$ it's enough to prove that $\mathbb E[Z\boldsymbol 1_G]=\mathbb E[X\boldsymbol 1_G]$ for all $G\in \mathcal G$ come from the fact that simple function are dense in $L^1$, right ? (since every $\mathcal G-$measurable function that is bounded is in $L^1$) – 2017-01-03
  • 1
    Again, do you know what is the definition of $E[X|\mathcal{G}]$? – 2017-01-03
  • 0
    It's written in my original post... $Z=\mathbb E[X\mid \mathcal G]$ if $\mathbb E|Z|<\infty $, $Z$ is $\mathcal G$ measurable and $\mathbb E[ZU]=\mathbb E[XU]$ for all $U$ bounded and $\mathcal G-$measurable. @Jack – 2017-01-03
  • 0
    Oh, that is an unusual version. You might want to read [this one](https://en.wikipedia.org/wiki/Conditional_expectation#Conditional_expectation_with_respect_to_a_.CF.83-algebra). Yes, simple functions are dense in $L^1$. – 2017-01-03

0 Answers 0