I am right now studying stochastic integral, and facing the following dilemma! I just want to check whether my understanding is right!
The stochastic integral is defined by following:
$I(t) =\int_0^t\Delta(t)dW(t)$
Now assuming $\Delta(t)$ a simple process, my understanding on the Brownian motion is that $\omega$ is fixed and then I am getting the value from this path with $t$.
Now when the author is trying to prove Ito Isometry, which is the following:
$E(I^2(t)) = \int_0^t\Delta^2(u)du$
He is using the fact that, $E[(W(t_{j+1}) - W(t_j))\cdot W(t_{k+1}) - W(t_k))|F(t_k)]=0$
where $t_k \lt t_j$. Though we already know the path in advance (we know $\omega)$, we are marching with time $t$, and get to know the value of the path as we are doing the integration. By marching I mean, I am splitting the time scale $(0,t)=[1, t_1, t_2,\cdots,t]$ and then we first go to $t_1$, then $t_2$, and so on. This way we are getting the value from the Brownian motion.
My main question, is the machinery that I described above, how stochastic integral of the form described in this question works. That is we fix $\omega$, and the pick the values with time $t$, and we don't know the values of the path for any $t_n \gt t$, even though we know the entire Brownian path in advance! To explain more, if we reach $t_1$, we don't know the value of it at $t_2$, even though we know the entire path in advance.
