February 18th, 2009, 2:48 pm
If I miss something in the questions raised about dW and dt, please enlighten me.But according to the part 3 of the definition of Brownian motion:W_(s+t)-W_s is normal N(0, t), and is independent of filtration up to sThis means dW is normal N(0, dt). Right?As for defining dW (or dt for that matter), you can define it as the differential form such that its integration is W (or t), bypassing all the infinitesimal nonsense. You can prove that the integral of dW*dW is t, hence dW*dW = dt.