Why isn't the Ito integral just the Riemann-Stieltjes integral?
What I mean is, given a continuous function $f$, some path of standard brownian motion $B$, and the integral:
$$\int_0^Tf(t)\;dB(t).$$
So what if we can't apply the change of variables formula to make sense of
$$\int_0^Tf(t)B'(t)\;dt,$$
the Riemann-Stieltjes integral never required differentiability of the integrator anyways.
Is there a reason to distinguish the Ito integral from the Riemann-Stieltjes integral above and beyond the need to develop a theory (Ito Calculus) to get around all the problems caused by the failure of change of variables?
Best Answer
First of all, Brownian motion is almost surely nondifferentiable meaning that you can't just apply the rule you cite to Stieltjes integrals.
Second, in essence the Ito integral is Riemann Stieltjes integration when you observe the path of Brownian motion, but not entirely. You can think of it as Stieltjes integration where the "integrator" as you call it has an extra variable of dependence, in this case on the probability space of realizable Brownian paths. Specifically $B_t$ is a random variable. The definition is then essentially the same as its a limit over partitions. However Brownian motion is not of bounded variation so you cannot apply the usual definitions of Riemann Stieltjes to evaluate the integrals along some realized path. In particular you get the integral evaluated as a limit of sums over partitions converges in probability.