An example s.t. $E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)$

Question

If $\Omega=\{a,b,c\}$ give an example that $E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)$

If I choose the $\sigma$-algebras disjoint (or better said: one is not fully contained in the other) is the inequality always satisfied ?

Answer 1

Let $$\mathcal F_1=\{\emptyset,\{a\},\{b,c\},\{a,b,c\}\}$$ and $$\mathcal F_2=\{\emptyset,\{b\},\{a,c\},\{a,b,c\}\}$$ and let $$P(\{a\})=P(\{b\})=P(\{c\})=\frac13.$$ Finally, let

$$X(\omega)=\begin{cases}1,&\text{ if } \,\omega\in\{a,b\}\\0,&\text{ if }\,\omega=c.\end{cases}$$

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1. $$Y(\omega):=E[X\mid\mathcal F_1]= \begin{cases} P(X=1\mid \omega=a),&\text{ if }\,\omega=a\\ P(X=1\mid \omega\in\{b,c\}),&\text{ if }\,\omega\in\{b,c\} \end{cases}=$$ $$ =\begin{cases} 1,&\text{ if }\,\omega=a\\ \frac12,&\text{ if }\,\omega\in\{b,c\} \end{cases} $$

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2. $$Z(\omega):=E[X\mid\mathcal F_2]= \begin{cases} P(X=1\mid \omega=b),&\text{ if }\,\omega=b\\ P(X=1\mid \omega\in\{a,c\}),&\text{ if }\,\omega\in\{a,c\} \end{cases}=$$ $$ =\begin{cases} 1,&\text{ if }\,\omega=b\\ \frac12,&\text{ if }\,\omega\in\{a,c\} \end{cases} $$

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3. $$E[E[X\mid\mathcal F_1]\mid\mathcal F_2]=E[Y\mid\mathcal F_2]=$$ $$=\begin{cases} P(Y=1\mid \omega=b)+\frac12P(Y=\frac12\mid \omega=b),&\text{ if }\,\omega=b\\ P(Y=1\mid \omega\in\{a,c\})+\frac12P(Y=\frac12\mid \omega\in\{a,c\}),&\text{ if }\,\omega\in\{a,c\} \end{cases}=$$ $$ =\begin{cases} \frac12,&\text{ if }\,\omega=b\\ \frac34,&\text{ if }\,\omega\in\{a,c\} \end{cases} $$

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4. $$E[E[X\mid\mathcal F_2]\mid\mathcal F_1]=E[Z\mid\mathcal F_1]=$$ $$=\begin{cases} P(Z=1\mid \omega=a)+\frac12P(Z=\frac12\mid \omega=a),&\text{ if }\,\omega=a\\ P(Z=1\mid \omega\in\{b,c\})+\frac12P(Z=\frac12\mid \omega\in\{b,c\}),&\text{ if }\,\omega\in\{b,c\} \end{cases}=$$ $$ =\begin{cases} \frac12,&\text{ if }\,\omega=a\\ \frac34,&\text{ if }\,\omega\in\{b,c\} \end{cases} $$

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So

$$E[E[X\mid\mathcal F_1]\mid\mathcal F_2]\not=E[E[X\mid\mathcal F_2]\mid\mathcal F_1].$$