3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ This probably doesn't help, but this question is related: math.stackexchange.com/questions/166549/…. Note also that Ito stochastic integrals are not "path-by-path", i.e. they are NOT just the Riemann-Stieltjes integral of a given path $W_t(\omega)$ where $\omega$ is an element of Wiener space. My professor went over this a lot, and I think I read it also in Durrett: "Stochastic Calculus: A Practical Introduction" and Revuz, Yor "Brownian Motion and Continuous Martingales", maybe also Protter "Stochastic Calculus". Hope this helps some $\endgroup$ Commented May 22, 2016 at 4:23
  • $\begingroup$ i.e. the Ito integral is a limit IN PROBABILITY of Riemann sums, it is NOT a limit almost surely (which is what I mean in some sense by "path-by-path") -- the almost sure limit/ the limit almost surely/using Riemann-Stieltjes integrals of the random paths/pathwise limit can't even be defined for functions of unbounded variation on any compact interval, like Brownian motion for instance, hence why it has to be a limit in probability and there is no longer any clear correspondence between $\omega$ and $I(\omega)$. $\endgroup$ Commented May 22, 2016 at 4:24

0

You must log in to answer this question.

Browse other questions tagged .