1楼: Originally posted by jollage at 2011-11-03 03:15:56:
如果{Sn}是martingale adapted to the filtration {Fn}，
那我们有E[S(n+1) | Fn]=Sn, 因为根据Fn时的信息S(n+1)的期望和Sn相等，
那E[S(n+1) | F(n+m)]=E[S(n+1) | Fn]吗？就是说S(n+1)的期望在Fn和F(n+m)的不 ...

4楼: Originally posted by randallxu868 at 2011-11-04 21:09:55:
应该是E[S_{n+1}|F_{n+m}]=S_{n+1}, if m>=1. 因此二者不等。