Consider the sample space $\Omega = \{\omega_{i}\}_{i=1}^{3}$ with $ω_{i}$ are independent identically distributed according to,
$ω_{i} = 1$ with probability $0.5$
$ω_{i} = −1$ with probability $0.5$
Furthermore, define $X_{n}(\omega) = \sum_{i=1}^{n}{\omega_{i}}$ and let $\mathcal{F_{1}} = \sigma(X_{1}), \mathcal{F_{12}} = σ({\omega_{1}, \omega_{2}})$ (the sigma algebras generated by the outcome of the first flip and of the first two flips respectively). Compute
a) $\mathbb{E}[X_{3}|X_{1}]$
b) $\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{1}}]|\mathcal{F_{12}}]$
c) $\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{12}}]|\mathcal{F_{1}}]$
d) Check that the answer to the above two points is the same.
$\textbf{My Attempt (Edited)}$
a) $\mathbb{E}[X_{3}|X_{1}] = \mathbb{E}[X_{1}+X_{2}+X_{3}|X_{1}] = X_{1} \mathbb{E}[X_{2}+X_{3}|X_{1}] = X_{1}\mathbb{E}[X_{2}+X_{3}] = 0$
b) $\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{1}}]|\mathcal{F_{12}}]$
Let $Z=\mathbb{E}[X_{3}|\mathcal{F_{1}}]$. We know that,
$\mathbb{E}[X_{3}|\omega_{1}=1] = \mathbb{E}[\omega_{1} + \omega_{2} + \omega_{3}|\omega_{1}=1] = 1 + \mathbb{E}[\omega_{2} + \omega_{3}|\omega_{1}=1]= 1$
$\mathbb{E}[X_{3}|\omega_{1}= -1] = \mathbb{E}[\omega_{1} + \omega_{2} + \omega_{3}|\omega_{1}=-1] = -1 + \mathbb{E}[\omega_{2} + \omega_{3}|\omega_{1}=-1]= -1$
So, Z = $X_{1}$ and $\mathbb{E}[Z|\mathcal{F_{12}}]= X_{1}$ since $X_{1}$ is $\mathcal{F_{12}}$ measurable.
c) $\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{12}}]|\mathcal{F_{1}}]$
We know that, $\mathcal{F_{1}} \subset \mathcal{F_{12}}$, so by the tower property,
$\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{12}}]|\mathcal{F_{1}}] = \mathbb{E}[X_{3}|\mathcal{F_{1}}] = Z = X_{1}$.
Any help will be appreciated.
According what I think I make a little change to clear the question.
$$\Omega=\{ (1,1,1),(1,1,-1),(1,-1,-1),(1,-1,1), (-1,1,1),(-1,1,-1),(-1,-1,-1),(-1,-1,1) \}$$ when $(1,1,1)$ means $(\omega_1=1,\omega_2=1,\omega_3=1)$ .
so
\begin{eqnarray} \begin{array}{c|ccc} &(1,1,1)&(1,1,-1)&(1,-1,-1)&(1,-1,1)& (-1,1,1)&(-1,1,-1)&(-1,-1,-1)& (-1,-1,1) \\ \hline X_1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 \\ X_2 & 2 & 2 & 0 & 0 & 0 & 0 & -2 & -2 \\ X_3 & 3 & 1 & -1 & 1 & 1 & -1 &-3 & -1 \\ \end{array} \end{eqnarray}
Part a)
$$E(X_3|X_1)= E(\omega_1 + \omega_2 + \omega_3|\omega_1) =\omega_1 + E( \omega_2 + \omega_3|\omega_1)$$ $$=\omega_1 + E( \omega_2 + \omega_3)=\omega_1 + 0=X_1$$
In part $b$ and $c$ i use Tower property lets $F_1$ and $F_2$ are two sigma field such that $F_1\subset F_2$ so $$E(E(X|F_1)|F_2)=E(E(X|F_2)|F_1)=E(X|F_1)$$
part b)
$$\mathbb{E}[\mathbb{E}[X_{3}|\mathcal{F_{1}}]|\mathcal{F_{12}}]= E\bigg(E\color{red}{\big(}X_3|\sigma(\omega_1)\color{red}{\big)}|\sigma(\omega_1,\omega_2)\bigg) \overset{Tower.p}{=}E\color{red}{\big(}X_3|\sigma(\omega_1)\color{red}{\big)}=\omega_1$$ in $\overset{Tower.p}{=}$ equation since $\sigma(\omega_1)\subset \sigma(\omega_1,\omega_2)$ i used Tower property. part $c$ is similar.
we can handle it without Tower property. for example part $b$
$$E(X_3|\sigma(\omega_1))=E(X_3|\{\omega_1=1\}) 1_{\{\omega_1=1\}} + E(X_3|\{\omega_1=-1\}) 1_{\{\omega_1=-1\}} $$ $$=(1) 1_{\{\omega_1=1\}} + (-1) 1_{\{\omega_1=-1\}}= 1_{\{\omega_1=1\}} -1_{\{\omega_1=-1\}}=w_1 $$
so
$$E\bigg(E\color{red}{\big(}X_3|\sigma(\omega_1)\color{red}{\big)}|\sigma(\omega_1,\omega_2)\bigg) =E\bigg(w_1|\sigma(\omega_1,\omega_2)\bigg)=w_1$$