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: $$\mathbb{E} \left [ X|F \right ] = \begin{cases} 0.5\cdot 1=0.5 & \text{ if } \omega = \left \{ a \right \}\\ 2\cdot 0.25+2\cdot0.25=1& \text{ if } \omega = \left \{ b,c \right \}\\ \end{cases}$$


$$\mathbb{E} \left [ X|G \right ] = \begin{cases} 0.5\cdot 1+0.25\cdot 2=1 & \text{ if } \omega = \left \{ a,b \right \}\\ 2\cdot 0.25=0.5& \text{ if } \omega = \left \{ c \right \}\\ \end{cases}$$

However, I do not know how to calculate $\mathbb{E} \left [\mathbb{E} \left [ X|F \right ]|G \right ]$...

  • 1
    $\begingroup$ What have you tried and where do you encounter problems? (Please visit stats.stackexchange.com/tags/self-study/info for more information.) $\endgroup$ – whuber Feb 28 '15 at 20:25
  • $\begingroup$ The outcome $\omega \in \Omega$ is not set valued.... $\endgroup$ – P.Windridge Mar 1 '15 at 20:38
  • 2
    $\begingroup$ 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. $\endgroup$ – whuber Mar 1 '15 at 20:54
  • $\begingroup$ 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 $A$) $\endgroup$ – P.Windridge Mar 1 '15 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 $\omega =\{a, b\}$, $\mathbb{E} \left[Y \middle\vert G\right] = \frac{1}{2} \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = \frac{1}{2} \middle\vert \omega =\{a, b\} \right) + 1 \bullet \Pr \left( \mathbb{E} \left[X \middle\vert F\right] = 1\middle\vert \omega =\{a, b\} \right) = \frac{1}{2} \bullet \frac{\mathbb{Q}(a)}{\mathbb{Q}(a)+\mathbb{Q}(b)} + 1 \bullet \frac{\mathbb{Q}(b)}{\mathbb{Q}(a)+\mathbb{Q}(b)} = \frac{2}{3} \, .$


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