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

1. Oct 27, 2009

### logarithmic

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?

2. Jan 6, 2010

### maxplankj

$$\text dW dT = 0$$

rigorously means

$$\int_{t_{0}}^{t}G(t) dW dt = 0$$

for a non anticipating function G(t). And is the same for the others.

Last edited: Jan 6, 2010