Brownian motion and martingale

What is F_t?
Is ((B_t^3-3tB_t),(F_t)) a two-dimensional random variable?

Is A_t = 3t(t -2B_t) ?

2009-02-26 23:08:48 補充：
If f(Bt, t) is a martingale, then there is no "dt" term in df(Bt, t),
i.e. df(Bt, t) = G(Bt, t)dBt for some function G(x,t)

Now,
d( Bt ^3 -3tBt)
= {(3Bt ^2 -3t)dBt +(1/2)(6Bt)dt } +(-3Bt)dt
= (3Bt ^2 -3t)dBt
Hence, Bt ^3 -3tBt is a martingale.

On the other hand,
d( Bt ^4 -6tBt)
= { (4Bt ^3 -6t)dBt +(1/2)(12Bt ^2)dt } +(-6Bt)dt
= (4Bt ^3 -6t)dBt
Hence, Bt ^4 -6tBt is a martingale.
i.e. At = Bt ^4 -6tBt

2009-02-27 07:15:19 補充：
Sorry!!
On the other hand,
d( Bt ^4 -6tBt ^2 +3t^2)
= { (4Bt ^3 -12tBt)dBt +(1/2)(12Bt ^2 -12t)dt } +(-6Bt ^2 +6t)dt
= (4Bt ^3 -12tBt)dBt +(6Bt ^2 -6t)dt + (-6Bt ^2 +6t)dt
= (4Bt ^3 -12tBt)dBt
Hence, Bt ^4 -6tBt is a martingale.
i.e. At = Bt ^4 -6tBt ^2 +3t^2