Why dZ(t) x dZ(t) = dt ??

[Arthur Kade]

Originally Posted by david849

would that imply (dZ(t))^3 = ±sqrt(dt)*dt?

Yes, except that you don't keep higher order terms.

[jas66Kent]

You obviously have little understanding of Calc in general. But hey, what else is new.

In case you didn't know, the stochastic PDE can be re-written in terms of integration.

I suggest you look into that before acting like an ass.

[truonda]

For the purpose of this exam you do not need to know why dZ(t) x dZ(t) = dt; just accept it as a fact.

[zywen]

it is because the quadratic variation of brownian motion ----> t

[Numerical Neuroticism]

Originally Posted by david849

would that imply (dZ(t))^3 = ±sqrt(dt)*dt?

I think (dZ(t))^3 = (dZ(t))^2 * dZ(t) = dt * dZ(t) = 0

[zywen]

Originally Posted by Numerical Neuroticism

I think (dZ(t))^3 = (dZ(t))^2 * dZ(t) = dt * dZ(t) = 0

correct!

[jas66Kent]

Originally Posted by AAABBBCCC

yes

I suspect that people actually dont understand what dz(t)^2=dt

"means".

You can also use the cross product to prove that dz(t)*dz(t) = dt

[AAABBBCCC]

almost surely. Doesnt work along any partition.

[AAABBBCCC]

You have to think of dz(t)^2 as something you can integrate against, but it is a stochastic integral. Basically says

as a stochastic integral, with g anything reasonable and perhaps stochastic. So when you "prove" Ito's Lemma, you can use a taylor series expansion and get rid of any higher powers of dz(t) beyond 2 and convert dz(t)^2 into dt. IT explains why you have the Ito correction term 1/2 g''(X(t))dt in Ito's Lemma.

[algebrat]