help with conditional probability with regard to sigma algebra

Question

Suppose we have $\Omega = \{a,b,c\}$ and we know that $\mathbb P(\{a\})=1/2$,$$\mathbb P(\{b\})=1/4,\mathbb P(\{c\})=1/4$$. Also, $$X(a) = 0; X(b) = X(c) = 2$$$$\mathcal F_1=\sigma(\{a\}),F_2=\sigma(\{c\})$$ how to compute $\mathbb E[\mathbb E[X|\mathcal F_1]|\mathcal F_2]$. Can someone explain in details?

Answer 1

We have that

$$\mathbb{E}[X \mid \mathcal{F}_1]=\mathbb{E}[X \mid \sigma(\{ a \})]=\begin{cases} \mathbb{E}[X \mid \{a \}] & \mbox{if } \omega =a \\ \mathbb{E}[X \mid \{b,c \}] & \mbox{if } \omega \neq a \end{cases}$$

Using the formula for mean value condional for a specific event:

$$\mathbb{E}[X \mid \{ a \}]=0 \quad \mbox{and} \quad \mathbb{E}[X \mid \{b,c\}]=\frac{1}{\mathbb{P}(\{b,c\})}(2\mathbb{P}(\{b\})+2\mathbb{P}(\{c\}))=2$$

Let $Z=\mathbb{E}[X \mid \mathcal{F}_1]$. So $Z(a)=0$ and $Z(b)=Z(c)=2$ (i.e. $Z=X$) $$\mathbb{E}[Z \mid \mathcal{F}_2]=\mathbb{E}[Z \mid \sigma(\{ c \})]=\begin{cases} \mathbb{E}[Z \mid \{c \}] & \mbox{if } \omega =c \\ \mathbb{E}[Z \mid \{a,b \}] & \mbox{if } \omega \neq c \end{cases}$$

Similarly: $$\mathbb{E}[Z \mid \{ c \}]=2 \quad \mbox{and} \quad \mathbb{E}[Z \mid \{a,b\}]=\frac{1}{\mathbb{P}(\{a,b\})}(0\cdot\mathbb{P}(\{a\})+2\mathbb{P}(\{b\}))=\frac{2}{3}$$ So the final result is:

$$\mathbb{E}[\mathbb{E}[X\mid \mathcal{F_1}] \mid \mathcal{F}_2]=\begin{cases} 2 & \mbox{if } \omega =c \\ \frac{2}{3} & \mbox{if } \omega \neq c \end{cases}$$