stochastic calculus problem

**johnbarkwith** · January 14th 2009, 09:05 AM

I have the equation:

F(B,S)=F(0,0) + IFs(B,S) dS + IFb(B,S) dB,

where I denotes the integral between 0 and t, B is a standard brownian motion, and S is the maximum of the brownian motion between 0 and t, Fs and Fb are the derivatives of F wrt s and b.

I have to show that if Fs=0 for B=S, then F is a continuous local martingale.

any idea how to show this?

**Laurent** · January 14th 2009, 10:36 AM

Originally Posted by johnbarkwith

I have the equation:

F(B,S)=F(0,0) + IFs(B,S) dS + IFb(B,S) dB,

where I denotes the integral between 0 and t, B is a standard brownian motion, and S is the maximum of the brownian motion between 0 and t, Fs and Fb are the derivatives of F wrt s and b.

I have to show that if Fs=0 for B=S, then F is a continuous local martingale.

any idea how to show this?

Consider the Stieltjes measure $dS$ : it is intuitive, and you can prove, that this measure is supported by the set of the real numbers $t\geq0$ such that $S_t=B_t$ . (For $t$ to be a growth point of $S$ , $B_t$ has to be a maximum)

So if your condition is fulfilled, $\partial_s F(B_s,S_s)$ is thus zero on the support of $dS$ . This is why $\int_0^t \partial_s F(B_s,S_s) dS_s=0$ . And this gives the result.

**johnbarkwith** · January 14th 2009, 02:18 PM

thanks so much for your help!!!

in basic terms is this argument that dS only has positive values when B=S because the brownian motion is at its maximum there, which implies S could be changing.

So if dS only takes non-zero values when B=S and Fs equals zero at exactly these points then the integrand with the dS term equals zero.

So the only term left is a continuous local martingale which can be shown to be a martingale....

i think i understand..

thanks alot

**Laurent** · January 14th 2009, 11:12 PM

Originally Posted by johnbarkwith

thanks so much for your help!!!

in basic terms is this argument that dS only has positive values when B=S because the brownian motion is at its maximum there, which implies S could be changing.

So if dS only takes non-zero values when B=S and Fs equals zero at exactly these points then the integrand with the dS term equals zero.

That's almost it, but $dS$ is a measure, so it is not "zero" or "positive" at $t$ . However, there are intervals with zero or positive measure, and these correspond to what you say: if $B_s<S_s$ on an interval, then $S$ is constant on this interval, and the $dS$ -measure of this interval is 0 ; with some more work, you can show what I wrote about the support of $dS$ .

So the only term left is a continuous local martingale which can be shown to be a martingale....

The only term left is a continuous local martingale, but I think it is not always a martingale (that was not your question, btw).

**johnbarkwith** · January 15th 2009, 03:00 PM

thanks for your help. i understand it now... the other thing i was stuck on was computing E[I B^2 dB] and E[(I B dB)^2), the expectation value where B is standard brownian motion and I is the integral between 0 and t.

Have no idea how to go about calculating these and as you are obviously good at stochastic calculus thought you might be able to help.

**Laurent** · January 16th 2009, 06:41 AM

Originally Posted by johnbarkwith

thanks for your help. i understand it now... the other thing i was stuck on was computing E[I B^2 dB] and E[(I B dB)^2), the expectation value where B is standard brownian motion and I is the integral between 0 and t.

Have no idea how to go about calculating these and as you are obviously good at stochastic calculus thought you might be able to help.

I haven't done stochastic calculus for a while, so I may be missing a straightforward argument here.
I would say that $M_t=\int_0^t B^2_s dB_s$ is a martingale, so that $E[M_t]=0$ for any $t$ . What is clear is that it is a local martingale. In order to show it is a martingale, you may for instance use the following result if you know it: given a local martingale $M$ , we have that $M$ is a square integrable martingale if, and only if $E[\langle M\rangle_t]<\infty$ for all $t$ . In addition, you probably know that $\langle \int_0^t B_s^2 dB_s\rangle_t = \int_0^t B_s^4 ds$ and by Fubini you can see that the expectation of this integral is finite (switch the expectation and the integral sign). Perhaps you know another theorem for the specific case of Brownian motion.

Another way would be Ito's formula: it gives $\frac{1}{3}B_t^3=\int_0^t B^2_s dB_s+\int_0^t B_s ds$ , and you integrate both sides. By symmetry, it is clear that $E[B_t^3]=0$ and $E[\int_0^t B_s ds]=\int_0^t E[B_s]ds=0$ .

As for $E[\left(\int_0^t B_s dB_s\right)^2]$ , again I give you two ways. First, the process $N_t=\int_0^t B_s dB_s$ is again seen to be a square-integrable martingale (using the same result), so that $N_t^2-\langle N\rangle_t$ is a martingale (this comes from the previous result as well). As a consequence, its expectation is 0 for all $t$ , which implies $E[N_t^2]=E[\langle N\rangle_t]=E[\int_0^t B_s^2 ds]$ and you know how to compute that.

Or you can use Ito's formula: it gives $B_t^2-t=2\int_0^t B_s dB_s$ , so that $E[\left(\int_0^t B_s dB_s\right)^2]=\frac{1}{4}E[(B_t^2-t)^2]$ , you expand and compute (you'll need to find the fourth moment of a Gaussian random variable).

NB: If you click on the formulas, you'll see what I typed to get them. There's a section in the forum about the way of writing formulas, look for "LaTeX".

**johnbarkwith** · January 16th 2009, 08:39 AM

thanks so much for your help. I understand everything apart from why E $[B_t^3]=0$ and $E[B_s^4]$ is finite???

I am studying stochastic calculus at masters level at university and was having problems with this, so you obviously have a great deal of mathematics knowledge.

**Laurent** · January 16th 2009, 09:51 AM

Originally Posted by johnbarkwith

thanks so much for your help. I understand everything apart from why E $[B_t^3]=0$ and $E[B_s^4]$ is finite???

Remember the random variable $X=B_t$ is just a Gaussian random variable with mean 0 and variance $t$ . In particular, it is well known that all the moments of $X$ are finite. In addition, because the Gaussian distribution is symmetric, $E[X^k]=0$ for every odd $k$ . The Brownian motion must have confused you, I'm sure you knew this

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