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Give an example on $\Omega = \{a, b, c\}$ in which

$E(E(X|F_{1})|F_{2}) \neq E(E(X|F_{2})|F_{1})$

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Obviously X is a random variable and $F_{1}$ and $F_{2}$ are sigma-algebras... but I'm not even sure how to get started on the actual example. Any help is appreciated.

Thanks.

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    Let $X$ be the random variable defined by $X: \Omega \to \mathbb{R}$ as $X(a) = 1$, $X(b) = 2$ and $X(c) = 3$. Let $F_1 = \{\{a\}, \{b,c \}, \emptyset, \{a,b,c \} \}$ and $F_2 = \{\{a,b,c\}, \emptyset \}$. Perhaps something can be gleaned from this.2011-03-28
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    Can you write down all possible $\sigma$-algebras?2011-03-28
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    @PEV You are getting close, but the proposed counterexample cannot work with your $F_2$ (the trivial $\sigma$-field). The OP should consider "Why not?".2011-03-28
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    @Byron I understand you wish the OP to construct an example by himself and concur, so will not give one.2011-03-28
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    Or perhaps let $F_2 = \{ \{b \}, \{a,c \}, \emptyset, \{a,b,c \} \}$.2011-03-28
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    Ok... so am I on the right track thinking that the example has something to do with $F_{1}$ and $F_{2}$ having subsets that don't "complement" each other?2011-03-28
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    @tsiki My advice is to take PEV's example and start calculating.2011-03-28

1 Answers 1

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Suppose, without loss of generality, that $\Omega = \{1,2,3\}$. Recall that ${\rm E}[X|Y]$ stands for ${\rm E}[X|\sigma(Y)]$, where $\sigma(Y)$ is the $\sigma$-algebra generated by $Y$. Let ${\rm P}$ be the uniform probability measure on $\Omega$, that is, ${\rm P}(\{\omega\})=1/3$, $\forall \omega \in \Omega$. Now, define random variables $X$, $Y$, and $Z$ as follows: $$ X(\omega)=\omega,\;\; \forall \omega \in \Omega , $$ $$ Y(1)=1, Y(\omega)=2.5, \omega \in \{2,3\}, $$ and $$ Z(\omega)=1.5, \omega \in \{1,2\}, Z(3)=3. $$ Then, $$ {\rm E}[X|Y] = Y $$ and $$ {\rm E}[X|Z] = Z. $$ It thus suffices to show that $$ {\rm E}[Y|Z] \ne {\rm E}[Z|Y]. $$ Indeed, ${\rm E}[Y|Z]$ takes the values $1.75$ and $2.5$ with probabilities $2/3$ and $1/3$, respectively, whereas ${\rm E}[Z|Y]$ takes the values $2.25$ and $1.5$ with probabilities $2/3$ and $1/3$, respectively. So, these random variables are never equal.

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    (This is a friendly comment.) In relation with @Byron's comment to the OP, you might wish to meditate on Alex Bartel's advice given on meta http://meta.math.stackexchange.com/questions/1254, and also on the thread http://meta.math.stackexchange.com/questions/1277, called *A suggestion for dealing with users who post homework problems*. As I said, these are on the meta site, so you could have missed them. Please let me know what you think of it.2011-03-30
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    @Didier: I agree I should have given a less detailed example. On the other hand, I left out many details (relative to a beginner), and the OP might need to work hard in order to work out a complete answer (and understand it). Anyway, I have found a general solution for this problem: give hints, and promise to elaborate in the future (say two weeks after the question was posted). In general, however, detailed answers may be valuable for the general SE community.2011-03-30