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 $\mathbb{E} \left [\mathbb{E} \left [ X|F \right ]|G \right ]$...

• What have you tried and where do you encounter problems? (Please visit stats.stackexchange.com/tags/self-study/info for more information.)
– whuber
Commented Feb 28, 2015 at 20:25
• The outcome $\omega \in \Omega$ is not set valued.... Commented Mar 1, 2015 at 20:38
• Perhaps the related thread at stats.stackexchange.com/questions/74332, on computing conditional expectations in a similar circumstance, will be of some help here. You will want to pay particular attention to measurability of the random variables: $F$-measurable variables are not necessarily $G$-measurable and vice versa.
– whuber
Commented Mar 1, 2015 at 20:54
• If the formal abstract definition of conditional expectation is opaque, then first try calculating the "naive" conditional expectation $E[X|A] = \sum_x x P(X=x|A)$ (for an event $A$) Commented Mar 1, 2015 at 21:12

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 $\omega =\{a, b\}$, $\mathbb{E} \left[Y \middle\vert G\right] = \frac{1}{2} \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = \frac{1}{2} \middle\vert \omega =\{a, b\} \right) + 1 \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = 1\middle\vert \omega =\{a, b\} \right) = \frac{1}{2} \bullet \frac{\mathbb{Q}(a)}{\mathbb{Q}(a)+\mathbb{Q}(b)} + 1 \bullet \frac{\mathbb{Q}(b)}{\mathbb{Q}(a)+\mathbb{Q}(b)} = \frac{2}{3} \, .$

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}