# Why isn't the Ito integral just the Riemann-Stieltjes integral?

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?

• It seems from your question that you think the only reason why $\int _0^T f(t) dB(t)$ as a Riemann Stieltjes integral doesn't work is because CoV doesn't work. But in fact the integrator in Riemann Stieltjes (the function you use to integrate with) needs to satisfy some nice properties which the Brownian motion do not.
– htd
Jun 22, 2013 at 8:34
• You just mean not differentiable a.e.? Also don't basically all the nice properties of the Riemann Integral which don't work for the Ito integral end up following from CoV in some way or another? I mean it's all the MVT in the end.
– Set
Jun 22, 2013 at 8:36
• Hmm. yea basicly. Usually one says it is not of bounded variation but the two concepts are very much related. See this link for what the integrator must satisfy: encyclopediaofmath.org/index.php/Stieltjes_integral (the brownian motion does not satisfy any of those conditions). I don't know much about regular calculus :p
– htd
Jun 22, 2013 at 8:39

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.

• But generally when we take the Ito Integral we usually imagine this extra variable of dependence as fixed right?
– Set
Jun 22, 2013 at 8:42
• @angry: not quite. You could if its of bounded variation but its not. That's why you need to interpret the limit in terms of convergence in probability. Jun 22, 2013 at 8:53
• So what you're saying is that as $|\Pi|\rightarrow 0$, $\sum_{\Pi}f(x_i)(B(x_i)-B(x_{i-1}))$ will converge to different values depending on the choice of partition sequence $\Pi$ with some positive probability? but that it will weakly converge to the same value no matter $\Pi$?
– Set
Jun 22, 2013 at 8:55

You can't simply integrate with regards to Brownian motion pathwise because it's not of bounded variation (because such functions are differentiable almost everywhere) and functions of bounded variation are precisely those with respect to which you can compute the Riemann–Stieltjes integrals.

• So are you saying that the upper and lower sums will depend on the choice of the sequence of partitions? I was under the impression that that part still worked for Brownian Motion.
– Set
Jun 22, 2013 at 8:46
• If the function is not of bounded variation, there is no easy way to control what is happening with your sums (for all you know, it might diverge, or converge to different things depending on the choice of sequence of partitions). The upshot is that you cannot make sense of the limit those sums pointwise, but only as random variables (they converge in probability). Jun 22, 2013 at 9:00