$$ \Omega=\{a,b,c\}, \mathcal{F}=2^{\Omega}\\ P(\{a\})=P(\{b\})=P(\{c\})=\frac{1}{3}\\ X(a)=1, X(b)=2, X(c)=15 $$
$$\mathcal{G}=\sigma (\{a\})=\{\phi, \{a\}, \{a,b,c\}, \{b,c\}\}\}$$
I have to find $E(X|\mathcal{G})$.
I have a solution:
$$E(X|\mathcal{G})(a)=1\\
E(X|\mathcal{G})(b)=E(X|\mathcal{G})(c)=\frac{17}{2}
$$
But I don't know exactly how the solution was found - could somebody give me the formula used here?
The formula that you have to use is the following: $$\mathbb{E}[X \mid \sigma (B)]=\begin{cases} \mathbb{E}[X \mid B] & \mbox{if } \omega\in B \\ \mathbb{E}[X \mid B^c] & \mbox{if } \omega \in B^c \\ \end{cases}.$$
In this particular case we have that: $$\mathbb{E}[X \mid \mathcal{G}]=\mathbb{E}[X \mid \sigma ( \{ a \})]=\begin{cases} \mathbb{E}[X \mid \{ a \}] & \mbox{if } \omega=a \\ \mathbb{E}[X \mid \{b,c\}] & \mbox{if } \omega \neq a \\ \end{cases}$$ So using the formula of mean value respect a specific event we have: $$\mathbb{E}[X \mid \{a\}]=\frac{1}{\mathbb{P}(\{a \})}(X(a)\mathbb{P}(\{a \}))=1,$$ and $$\mathbb{E}[X \mid \{b,c\}]=\frac{1}{\mathbb{P}(\{b,c \})}(X(b)\mathbb{P}(\{b \})+X(c)\mathbb{P}(\{c \})=\frac{3}{2}(2\cdot\frac{1}3+15\cdot\frac13)=\frac{17}{2}.$$ In conclusion $$\mathbb{E}[X \mid \mathcal{G}]=\begin{cases} 1 & \mbox{if } \omega=a \\ \frac{17}{2} & \mbox{if } \omega \neq a \\ \end{cases}$$
