Thread: Martingales with brownian motion

Hi,

I've to prove that a couple of equations are martingales. yet i don't really know where to start (well i know that if i can show them as driftless or as an itos integral that would answer the questions.)

I was hoping someone could give me suggestions on where to start - rather than simply answer the questions (i do want to learn it)

note i've used | to mean integral from 0 to t
and X(t) is the Brownian Motion

the equations are:

Y(t) = [1 - X(t)|sin(kX(t)).dk ] * e^(t/2)

Y(t) = sqrt(t)X(t) - | X(s)/2sqrt(s).ds

Y(t) = wX₁(t) + sqrt(1-w^2)*X₂(t)

I think I have to integrate equations 1 and 2 (in respect to k and s) but i'm getting into a muddle. Any suggestions?

with respect to the first equation i'm thinking the following:
focusing on X int Sin (kX) .dk
as the integral is in respect to k, simple integration gives
X *[1/X* -cos(kX)] = -cos(kX) as over 0_1
-cos(1X) + cos(0X) = 1 - cos(X)
Plugging in to formula
Y(t) = [1 - 1 - cos(X) ] * e^(t/2)
gives
Y(t) = e^(t/2)cos(X)
as there is no drift term (.... .dt) this must mean its a martingale.
but i don't think i can perform simple integration on a function containing Brownian motion.

The second equation i'm thinking of using a change of measure.
so z= sqrt(s)
dz = 1/2sqrt(s)

simplifies formula to Y(t) = sqrt(t)X(t) - int X(z^2).dz
integrating Y(t) = sqrt(t)X(t) - (z^2)X(z^2) int X(z^2).dz
or
Y(t) = sqrt(t)X(t) – sX(s)[0_t] - integral Sqrt(s).dX
so
Y(t) = (sqrt(t) - t)X(t) - integral Sqrt(s).dX
as integral is an ito must mean martingale?