# Stochastics problem: give an example

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. –  PEV Mar 28 '11 at 21:08
Can you write down all possible $\sigma$-algebras? –  Gerben Mar 28 '11 at 21:10
@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?". –  Byron Schmuland Mar 28 '11 at 21:13
@Byron I understand you wish the OP to construct an example by himself and concur, so will not give one. –  Did Mar 28 '11 at 21:17
Or perhaps let $F_2 = \{ \{b \}, \{a,c \}, \emptyset, \{a,b,c \} \}$. –  PEV Mar 28 '11 at 21:20

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.