An example s.t. $E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)$ 
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 ?
 A: 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
$$X(\omega)=\begin{cases}1,&\text{ if } \,\omega\in\{a,b\}\\0,&\text{ if }\,\omega=c.\end{cases}$$



*

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





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





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





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



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