Part a) Check whether it satisfies the following 3 conditions.
(i) E|M_t| < \infty, this follows from the assumption that EY^2<\infty.
(ii) M_t is adapted to F_t by the definition of M_t and property of conditional expectation.
(iii) E(M_t|F_s) = M_s, if s<t
To show (iii), you need apply the theorem below.
If F1  F2 then (i) E(E(X|F1)|F2) = E(X|F1)
(ii) E(E(X|F2)|F1) = E(X|F1).

Part b). E(exp(\sigmaB(T)|Ft)=exp(\sigma B(t)) exp(B(T)-B(t)|F_t)=exp(\sigma B(t)) exp(B(T)-B(t)) since the second part is independent of F_t.
Use moment generating function of the normal distribution to get the solution to the second part. Then apply Ito's formula.

第一题已经解出来了 第二题能说的详细点吗 谢谢啦 大侠 这个题目要计分的 2# ash2010

2# ash2010

求求你啦

鞅很重要