I need help understanding a solution, I have two thirds figured out but I cannot understand one part, I explain:
Problem:
Let $X$, $Y$ be two random variables in a probability space ($\Omega, \mathcal{F}, \mathbb{P}$). Define the process $X$ by:
$$X_t = \begin{cases} Z \mbox{ if }t\lt t_0, \quad \\ Y \mbox{ if }t\ge t_0 \end{cases} $$
Where $t_0\gt0$ is a fixed time
Apologies for the format! I copied other people's code for this on stack exchange and it does not seem to work!
Question:
Assume $Z$ and $Y$ are integrable. Calculate $E[X_t|\mathcal{F}_s^X]$ for each $s < t$.
Solution:
For $s<t<t_0$, it is clear we have $E[X_t|\mathcal{F}_s^X] = E[X_s|\mathcal{F}_s^X] = X_s$
We can also view this as $E[X_t|\mathcal{F}_s^X] = E[X_s|\mathcal{F}_s^X] = E[Z|\sigma(Z)] = Z$
For $t_0 \le s < t$, it is clear we have $E[X_t|\mathcal{F}_s^X] = E[X_s|\mathcal{F}_s^X] = X_s$
We can also view this as $E[X_t|\mathcal{F}_s^X] = E[X_s|\mathcal{F}_s^X] = E[Y|\sigma(Z,Y)] = Y$
My Problem:
What I do not know how to describe is for $s<t_0 \le t$, what is $E[X_t|\mathcal{F}_s^X]$?
The solution says, for $s<t_0 \le t$, $E[X_t|\mathcal{F}_s^X] = E[Y|Z]$
And I am confused, how does this follow? Also, is $E[Z|\sigma(Z)] = E[Z|Z]$ an abuse of notation, or are they different things?
ANY help would be greatly appreciated thank you!
Firstly $E[X|Y]$ is probably defined to be $E[X|\sigma(Y)]$ Since you are after all taking conditional expectation w.r.t. a sub $\sigma$-algebra and $\sigma(Y)$ is the only one which makes natural sense. (You can of course consult wikipedia or your textbook in these cases).
For your problem: Clearly $t\geq t_{0}$ implies that $X_{t} = Y$. For $s< t_{0}\leq t$ you know that $X_{s} = Z$. Recall now that the definition of $\mathcal{F}^{X}_{s}$ is the sigma algebra generated by the union of $X_{r}$ for $0\leq r \leq s$. But these are all $Z$, so it is precisely the sigma algebra generated by $Z$, $\sigma(Z)$. Therefore $$E[X_{t}|\mathcal{F}_{s}^{X}]=E[Y|\sigma(Z)]=E[Y|Z].$$