question - integrating brownian motion

misiti3780 · May 9th, 2008, 12:39 am

Why is the following integral equal to zerointegral(0,T) { W^3(t) dW(t) }where W(t) is Brownian motion and W^3(t) is Brownian motion cubed.Thanks

Zedr0n · May 9th, 2008, 3:31 am

hm, only the expectation of this integral is equal to zero, I think. It would have been strange if this whole integral was equal to zero(with probability 1 anyway?).

misiti3780 · May 9th, 2008, 3:56 am

ok, i get what you are saying.so the expectation is zero becuase W(t) is continuous, and when you turn the integral into a summation, and let the the number of partitions (n) go to infinity,the max [W(t_j+1) - W(t_j)] for 0 < k < n-1 goes to zero, making the entire summation go to zero also ??

mjy · May 9th, 2008, 4:10 am

QuoteOriginally posted by: misiti3780ok, i get what you are saying.so the expectation is zero becuase W(t) is continuous, and when you turn the integral into a summation, and let the the number of partitions (n) go to infinity,the max [W(t_j+1) - W(t_j)] for 0 < k < n-1 goes to zero, making the entire summation go to zero also ??no. the increment of a Wiener process can be anywhere in R, i.e. W_t is not differentiable wrt t. the expectation is zero because Wiener integral is a stochastic process that originates from zero AND on the other hand it is a martingale.