logarithmic
Oct27-09, 03:45 AM
Can someone explain to me the rigorous meaning of statements like:
dt^2 = 0
dW*dt = 0
dW^2 = dt
Here W = W(t) is standard Brownian motion.
I know that a SDE such as
dX = f dW + g dt
rigorously means
X(t) = X(0) + \int_0^tfdW + \int_0^tgds
But what does dt^2 mean? And why is it equal to 0. Same with the other statements. Is the above definition useful for this?
dt^2 = 0
dW*dt = 0
dW^2 = dt
Here W = W(t) is standard Brownian motion.
I know that a SDE such as
dX = f dW + g dt
rigorously means
X(t) = X(0) + \int_0^tfdW + \int_0^tgds
But what does dt^2 mean? And why is it equal to 0. Same with the other statements. Is the above definition useful for this?