The motivation for developing a theory of stochastic integration is that many important processes — such as standard Brownian motion — have sample paths which are extraordinarily badly behaved. With probability one, the path of a Brownian motion is nowhere differentiable and has infinite variation over all nonempty time intervals. This rules out the application of the techniques of ordinary calculus. In particular, the Stieltjes integral can be applied with respect to integrators of finite variation, but fails to give a well-defined integral with respect to Brownian motion. The Ito stochastic integral was developed to overcome this difficulty, at the cost both of restricting the integrand to be an adapted process, and the loss of pathwise convergence in the dominated convergence theorem (convergence in probability holds intead).
However, as I demonstrate in this post, the stochastic integral represents a strict generalization of the pathwise Lebesgue-Stieltjes integral even for processes of finite variation. That is, if V has finite variation, then there can still be predictable integrands such that the integral is undefined as a Lebesgue-Stieltjes integral on the sample paths, but is well-defined in the Ito sense. The relevant result from the notes is the following. Recall that an FV process is a cadlag adapted process of finite variation over all bounded time intervals.
Lemma 1 Every FV process V is a semimartingale. Furthermore, let be a predictable process satisfying
almost surely, for each . Then, and the stochastic integral agrees with the Lebesgue-Stieltjes integral, with probability one.
The condition that is finite is equivalent to being V-integrable in the pathwise Lebesgue-Stieltjes sense. So Lemma 1 tells us that being pathwise V-integrable implies integrability in the Ito sense, and that the two definitions of the integral are in agreement. It is natural to ask about the converse: is being integrable in the stochastic sense sufficient to guarantee integrability as a pathwise Lebesgue-Stieltjes integral? The answer is, in general, no it isn’t. I give an example below of an FV process V and predictable such that is almost-surely infinite, so the pathwise integral does not exist, but for which is V-integrable in the stochastic sense. The resulting integral is then not an FV process. So, this example also demonstrates that stochastic integration does not preserve FV processes.
Before going ahead and constructing the example, it is worth noting that V must be a noncontinuous process. In fact, it can be shown that if V is a continuous FV process, then a predictable process is V-integrable if and only if it is almost-surely Lebesgue-Stieltjes integrable. This simple result will be covered later in my stochastic calculus notes but, for now, I just state it here without any proof.
Let us now construct to be a finite variation martingale. It will then have quadratic variation
and, a sufficient condition for a predictable process to be V-integrable is that
has finite expectation for each time t. Choose any infinite sequence of distinct times in the unit interval [0,1]. For example, as in Figure 1. Also, choose a sequence of independent random variables defined on a (complete) probability space , each with . The process V is then defined by the following absolutely convergent sum,
This is a cadlag pure jump process with and variation . Let be its natural filtration. That is, is the sigma-algebra generated by the random variables and the -null sets. Then, is a complete filtered probability space with respect to which V is an FV process. It is furthermore the case that V is a square integrable martingale. In fact, it has the independent increments property, so that is independent of with zero mean, for .
Next, let be the process
This is measurable and deterministic, so is certainly predictable. It will also be V-integrable since
is bounded. By dominated convergence, its integral with respect to V is the sum over n of the integrals of , with convergence in probability.
This expresses as a sum of the independent variables with zero mean, and with bounded sum of squares . So, by martingale convergence for -bounded martingales, this sum converges almost surely. The jumps of W are
This gives the variation of W as
So, W is not an FV process. Similarly,
so is not pathwise V-integrable.
In this example, only the values of at the times mattered. In the case where we could have instead used the piecewise continuous integrand which still satisfies as above.
One comment about the example given above is that, although the Lebesgue-Stieltjes integral over the unit interval is undefined on the sample paths, the improper integral does still exist. At least, this is true for the choice . Then, is bounded on for any time so is well defined on the individual sample paths, and
with almost-sure convergence. However, other choices of can be used so that is infinite for all , so the Lebesgue-Stieltjes integral is not well defined in the neighbourhood of any point in the unit interval.
For example, suppose that have uniform density on the unit interval,
as for all . This happens if we let be the fractional part of for any irrational number , by Weyl’s equidistribution theorem. Then, satisfies as and
Integration by parts can be applied to this integral,
The final integral on the right hand side has integrand converging asymptotically to as , so the integral diverges at rate . This shows that is infinite on all nonempty subintervals , so the pathwise integral is undefined, even as an improper integral. An example sample path with is shown in Figure 2.