Let

{B_(t)}_(t)>=0

be Brownian motion started at 0 and

F_(t)^(B)=\\\\sigma ({B_(s):0

. Are the\ following

x_(t)(F_(t)^(B))

-martingale? (Explain the reason as well.)\

x_(t):=B_(t)^(2)-t

\

x_(t):=t^(2)B_(t)-\\\\int_0^t sB_(s)ds

\

x_(t):=e^(B_(t)-(t)/(2))

\

{x_(t)}_(t)>=0

is called the It process if there exist

\\\\sigma inL^(2)

and

\\\\mu inL^(1)

such that\

x_(t)(\\\\omega )=x_(0)(\\\\omega )+\\\\int_0^t \\\\sigma (s,\\\\omega )dB_(s)(\\\\omega )+\\\\int_0^t \\\\mu (s,\\\\omega )ds,t>=0.

\ Here

{B_(t)(\\\\omega )}_(t)>=0

is

(F_(t))

-Brownian motion.

1. Let {Bt}t0 be Brownian motion started at 0 and FtB=({Bs:0st})N. Are the following Xt(FtB)-martingale? (Explain the reason as well.) 1) Xt:=Bt2t 2) Xt:=t2Bt0tsBsds 3) Xt:=eBtt/2 - {Xt}t0 is called the It process if there exist L2 and L1 such that Xt()=X0()+0t(s,)dBs()+0t(s,)ds,t0. Here {Bt()}t0 is (Ft)-Brownian motion