Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stochastic Caluclus: dt^2=0, dW*dt = 0?

  1. Oct 27, 2009 #1
    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

    [tex]X(t) = X(0) + \int_0^tfdW + \int_0^tgds[/tex]

    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?
  2. jcsd
  3. Jan 6, 2010 #2

    \text dW dT = 0


    rigorously means

    [tex] \int_{t_{0}}^{t}G(t) dW dt = 0 [/tex]

    for a non anticipating function G(t). And is the same for the others.
    Last edited: Jan 6, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook