My workbook gives the following property for Conditional Expectation and martingale:
Measurability:
If $Y_n$ is "$F_n$ measurable", then:
$E[Y_n~|~F_n] = Y_n$
Let's say I have a Stochastic Process $\{A_n, n\ge 0\}$ that is "$F_{n-1}$" measurable? (My rule is written for "$F_n$ measurable", so I'm wondering how to handle this case when $A_n$ is $F_{n-1}$ measurable.)
For Example, i'm assuming the following relationship holds if $A_n$ is "$F_{n-1}$" measurable? (but, no idea if its correct):
$$E[A_n | F_{n-1}] = A_n???$$
And, let $n = m+1~~~ \Rightarrow ~~~n-1 = m$
$$E[A_{m+1} | F_m] = A_{m+1}???$$
Does this make any sense?
Here's where I need know how the "$F_{n-1}$ measurability" works:
Doob's decomposition: Let $X = \{X_n, n \ge 0\}$ be a sub-martingale with respect to $F_n$. Then, there exists a martingale $M = \{M_n, n \ge 0\}$ and a process $A = \{A_n, n \ge 0\}$ such that:
- M is a martingale with respect to $F_n$
- A is an increasing process $A_{n+1} \ge A_n$
- $A_n$ is "$F_{n-1}$ measurable" for all n. $\Leftarrow$
- $X_n = M_n + A_n$
Note, condition 3 says $F_{n-1}$ measurable instead of $F_n$ measurable.
A process $A_n$ such that $A_n$ is $F_{n-1}$ measurable for each $n$ is often said to be predictable. You are quite correct that if $A_n$ is a predictable process then $E[A_n \mid F_{n-1}] = A_n$; I don't see why you would doubt it? And of course it is equally true that $E[A_{m+1} \mid F_m] = A_{m+1}$.
Note that martingales typically do not have this property; indeed, you can show that if $A_n$ is a martingale which is predictable then it is constant with respect to time, e.g. $A_0 = A_1 = A_2 = \dots$ almost surely.