Let $\Omega=\{\omega_1,\omega_2,\omega_3,\omega_4\}$ and $X=\{X_t\}_{t=0,1,2}$, where $$X(0)=3, X(1,\omega)=\begin{cases}5, \omega=\omega_1,\omega_2 \\ 2, \omega=\omega_3,\omega_4 \end{cases}, X(2,\omega)=\begin{cases}6, \omega=\omega_1,\omega_2 \\ 3, \omega=\omega_3 \\ 3, \omega=\omega_4 \end{cases}.$$
- Find the natural filtration of $X$, i.e., find $\mathcal{F}_0^x=\sigma(X(0))$, $\mathcal{F}_1^x=\sigma(X(0),X(1))$, and $\mathcal{F}_2^x=\sigma(X(0),X(1),X(2))$.
- Find the conditional expectation $E[X(2)\mid \mathcal{F}_1^x]$.
Attempt:
- We have to find $\sigma(X(0))$,$\sigma(X(1))$, and $\sigma(X(2))$.
First of all, I don't know what the meaning of $X(0)=3$ is. Does the value of $X(0)$ matter? I think that it doesn't matter, since at time $t=0$, we are told nothing about the events, and so $\sigma(X(0))=\{\emptyset, \Omega\}$.
Furthermore,
$\sigma(X(1))=\{\emptyset, \Omega, \{\omega_1,\omega_2\}, \{\omega_3,\omega_4\}\}$,
$\sigma(X(2))=\{\emptyset, \Omega, \{\omega_1,\omega_2\}, \{\omega_3,\omega_4\}\}$.
Hence,
$\mathcal{F}_0^x=\{\emptyset, \Omega\}$,
$\mathcal{F}_1^x= \sigma(X(1))$,
$\mathcal{F}_2^x= \sigma(X(1))$.
- Let $A_1=\{\omega_1,\omega_2\}$ and $A_2=\{\omega_3,\omega_4\}$ be the set of events in $X(2)$. Then, $$E[X(2) \mid A_1]= \frac{6 \cdot \frac14+6\cdot \frac14}{\frac{1}{4}+\frac{1}{4}}= 6$$ and $$E[X(2) \mid A_2]= \frac{3 \cdot \frac14+3\cdot \frac14}{\frac{1}{4}+\frac{1}{4}}= 3$$, so $$E[X(2) \mid \mathcal{F}_1^x]= \begin{cases} 6, \omega=\omega_1,\omega_2 \\ 3, \omega=\omega_3,\omega_4 \end{cases}.$$
Am I correct? Any help please? Many thanks in advanced.