I'm having trouble understanding the following I read from a book:
If $m < n$, then $E\left( E [ Y| F_n ] | F_m \right) = E [ Y| F_m ]$.
My arguments are as follows: say $Z_m = E [ Y| F_m ]$ and $Z_n = E [ Y| F_n ]$. $Z_m$ is $m$-measurable. Therefore, I would say that $E[Z_m | F_n ]$ is the same as $Z_m$, since we are given more information than needed, and thus $Z_m$ is known provided the information $F_n$. Question 1: is this correct?
However, the book says that $E[Z_n | F_m] = Z_m$, and my best guess is that the book is right. Question 2: why is this true?
Q1: For an intuition, you might need the property of the conditional expectation as a projection (of course, when these make sense).
Q2: So we need to show that $\mathbb{E}(Z_n|F_m)=Z_m$, when $m<n$. We need to show that $\mathbb{E}[Z_n1_A]=\mathbb{E}[Z_m1_A]$ for every $A \in F_m$.
$\mathbb{E}[Y|F_n]1_{A}=\mathbb{E}[Y1_A|F_n]$ almost surely since $A \in F_m \subseteq F_n$. Hence, $\mathbb{E}[Z_n1_A]=\mathbb{E}[Y1_{A}]$ for every $A \in \mathbb{F_m}$, by total expectation. This means that $\mathbb{E}[Y|F_m]=\mathbb{E}[Z_n|F_m]$ , which finishes the proof. There are still some gaps, but you can fill them in!