7
$\begingroup$

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

$\endgroup$
4
  • 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
    Commented Feb 28, 2015 at 20:25
  • 1
    $\begingroup$ The outcome $\omega \in \Omega$ is not set valued.... $\endgroup$ Commented Mar 1, 2015 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
    Commented Mar 1, 2015 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 $A$) $\endgroup$ Commented Mar 1, 2015 at 21:12

2 Answers 2

1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $ $$\begin{aligned} \mathbb{E} \left [\mathbb{E} \left [ X|\mathcal{F} \right ]|\mathcal{G} \right ] & = \mathbb{E} \left [ X|\mathcal{G} \right ] \\ & = \begin{cases} 4/3 , & w = \{ a,b \}, P(Y = 4/3) = 3/4 \\ 2 , & w = \{ c \}, P(Y = 2) = 1/4 \\ \end{cases} \\ & := Y \\ \end{aligned}$$

$$\begin{aligned} \mathbb{E} \left [\mathbb{E} \left [ X|\mathcal{G} \right ]|\mathcal{F} \right ] & = \mathbb{E} \left [ Y|\mathcal{F} \right ] \\ & = \begin{cases} 4/3 , & w = \{ a \}, P(Y = 4/3) = 1/2 \\ 5/3 , & w = \{ b,c \}, P(Y = 5/3) = 1/2 \\ \end{cases} \\ \end{aligned}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.