An 'easy' exercise on conditional expectations and filtrations I am struggling with the following exercise in the context of modeling information structure via filtration to evaluate contingent claims. I hope that someone can explain me how to derive the solution:
Let $ \Omega = ${a,b,c} with $\mathbb{Q}$({a}) = 1/2, $\mathbb{Q}$({b}) = 1/4 and $\mathbb{Q}$({c}) = 1/4 and X a random variable defined by X(a) = 1, X(b) = X(c) = 2. Calculate $\mathbb{E} \left [\mathbb{E} \left [ X|F \right ]|G  \right ]$ and $\mathbb{E} \left [\mathbb{E} \left [ X|G \right ]|F  \right ]$ for $ F:=\left \{  \emptyset   ,\left \{a\right \},\left \{b,c\right \},\Omega\right \}$ and $ G:=\left \{  \emptyset   ,\left \{a,b\right \},\left \{c\right \},\Omega\right \}$
I tried: 
$$\mathbb{E} \left [ X|F \right ] = \begin{cases}
0.5\cdot 1=0.5 & \text{ if } \omega = \left \{ a \right \}\\ 
2\cdot 0.25+2\cdot0.25=1& \text{ if } \omega = \left \{ b,c \right \}\\ 
\end{cases}$$
And: 
$$\mathbb{E} \left [ X|G \right ] = \begin{cases}
0.5\cdot 1+0.25\cdot 2=1 & \text{ if } \omega = \left \{ a,b \right \}\\ 
2\cdot 0.25=0.5& \text{ if } \omega = \left \{ c \right \}\\ 
\end{cases}$$
However, I do not know how to calculate $\mathbb{E} \left [\mathbb{E} \left [ X|F \right ]|G  \right ]$...
 A: Following the hint given by Windridge, let $\mathbb{E} \left[ \mathbb{E} \left[X \middle \vert F\right] \middle\vert G\right] = \mathbb{E} \left[ Y \middle\vert G\right]$, where $Y = \mathbb{E} \left[X \middle\vert F\right]$.
$\mathbb{E} \left[Y \middle\vert G\right] = \sum y \, \Pr\left(Y = y \middle\vert G\right)$.
If $\omega =\{a, b\}$, $\mathbb{E} \left[Y \middle\vert G\right] = \frac{1}{2} \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = \frac{1}{2} \middle\vert \omega =\{a, b\} \right) + 1 \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = 1\middle\vert \omega =\{a, b\} \right) = \frac{1}{2} \bullet \frac{\mathbb{Q}(a)}{\mathbb{Q}(a)+\mathbb{Q}(b)} + 1 \bullet \frac{\mathbb{Q}(b)}{\mathbb{Q}(a)+\mathbb{Q}(b)} = \frac{2}{3} \, .$
A: Let $ \Omega = ${a,b,c} with $\mathbb{P}$({a}) = 1/2, $\mathbb{P}$({b}) = 1/4 and $\mathbb{P}$({c}) = 1/4.
Define a random variable  X
$$ X = 
\begin{cases} 
1 , \quad & w = \{ a \}, P(X = 1) = 1/2\\
2 , \quad & w = \{ b,c \}, P(X = 2) = 1/2\\
\end{cases} \\
$$
Define 2 sigma-algebra:
$ \mathcal{F}:=\left\{  \emptyset   ,\left\{a\right\},\left\{b,c\right\},\Omega\right\} $
$ \mathcal{G}:=\left \{  \emptyset   ,\left \{a,b\right \},\left \{c\right \},\Omega\right \}$
Obviously, $X$ is $\mathcal{F}$-measurable,so $\mathbb{E} \left [ X|\mathcal{F} \right ] = X $
$$\begin{aligned}
\mathbb{E} \left [\mathbb{E} \left [ X|\mathcal{F} \right ]|\mathcal{G}  \right ]
& =
\mathbb{E} \left [ X|\mathcal{G}  \right ] \\ 
& = 
\begin{cases} 
4/3 ,  & w = \{ a,b \}, P(Y = 4/3) = 3/4 \\
2 ,  & w = \{ c \}, P(Y = 2) = 1/4 \\
\end{cases} \\
& := Y \\ 
\end{aligned}$$
$$\begin{aligned}
\mathbb{E} \left [\mathbb{E} \left [ X|\mathcal{G} \right ]|\mathcal{F}  \right ] 
& =
\mathbb{E} \left [ Y|\mathcal{F}  \right ] \\
& = 
\begin{cases} 
4/3 ,  & w = \{ a \}, P(Y = 4/3) = 1/2 \\
5/3 ,  & w = \{ b,c \}, P(Y = 5/3) = 1/2 \\
\end{cases} \\
\end{aligned}$$
