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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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    @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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    @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