Understand better stochastic integral through a.s. convergence

Question

I know that $$\int_0^T f(B_t, t)dB_t=\lim_{n\to \infty }\sum_{i=1}^n f(B_{t_i^{(n)}},t_i^{(n)})(B_{t_{i+1}^{(n)}}-B_{t_i^{(n)}}),\quad \text{in }L^2,$$ where $\{t_i^{(n)}\}_{i=1}^n$ is a sequence of partition of $[0,T]$ s.t. $$\sup|t_{i+1}^{(n)}-t_i^{(n)}|\to 0,\quad \text{when }n\to \infty .$$

Now, there is a subsequence $n_k$ s.t. $$\lim_{k\to \infty }\sum_{i=1}^{n_k}f(B_{t_i^{(n_k)}},t_i^{(n_k)})(B_{t_{i+1}^{(n_k)}}-B_{{t_i}^{(n_k)}})=\int_0^T f(B_t,t)dB_t\quad a.s.$$

So, up to a subsequence, can the stochastic integral be seen as a Stiljes-Riemann integral ? A sort of week Stiljes-Riemann integral in the sense that indeed $$\lim_{n\to \infty }\sum_{i=1}^{n}f(B_{t_i^{(n)}},t_i^{(n)})(B_{t_{i+1}^{(n)}}-B_{{t_i}^{(n)}}),$$ may not exist, but if it exist, then it converges to $\int_0^T f(B_t,t)dB_t$ a.s. Does this makes sense ?

Answer 1

The convergence of the subsequence holds almost surely, i.e. there is an exceptional null set where convergence fails to hold. This null set depends on the partitioning sequence $(t^{(n)})_{n \geq 1}$. Since there are uncountably many sequences, this is pretty bad - the union of the null exceptional sets is going to be quite huge, in general, and therefore we cannot expect to have a "uniform" exceptional null set for all sequences $(t^{(n)})_{n \in \mathbb{N}}$ with mesh size converging to zero. In fact, that's exactly the reason why the Itô integral is defined as an $L^2$-limit of the Riemann sums and not as a pointwise limit.

The phenomena which you are observing is a very general one: If a sequence of random variables $Y_n$ converges to a random variable $Y$ in probability (or in $L^2$), then we can choose a subsquence which converges almost surely to $Y$. Nevertheless, convergence in probability is a much weaker notion of convergence than pointwise convergence; this means, in particular, that the pointwise convergence of a subsequence of Riemann sums is far from giving a notion for a pointwise integration: If we want to get a pointwise notion for a stochastic integral, then we would like to fix $\omega \in \Omega$ and then define the stochastic integral, say, as a pointwise limit of Riemann sums along a suitable partition. That's, however, not what happens if we use the subsequence procedure from your question. If we take a sequence $(t^{(n)})_n$ then we get pointwise convergence with probability $1$ but we have no control about the null set. In particular, we don't have a clue how to choose a sequence $(t^{(n)})_n$ such that the Riemann sums convergence for our fixed $\omega$.

Let me give one further remark. In order to get the convergence

$$\int_0^T f(s) \, dB_s = \lim_{n \to \infty} \sum_{i=1}^n f(t_i^{(n)}) (B_{t_{i+1}^{(n)}}-B_{t_i^{(n)}}) \quad \text{in $L^2$}$$

(and hence the pointwise convergence of the subsequence of Riemann sums) you will typically need some continuity assumptions on $f$. If $f$ is a general progressively measurable function with $\mathbb{E}\int_0^t f(s)^2 \, ds < \infty$ for all $T>0$, then there exists some sequence of approximating simple functions $(f_n)_{n \in \mathbb{N}}$, i.e. a sequence of functions such that$$\mathbb{E}\int_0^t |f_n(s)-f(s)|^2 \ ds \to 0 \quad \text{and} \quad \int_0^t f(s) \, dB_s = L^2-\lim_{n \to \infty} \int_0^t f_n(s) \, dB_s;$$

however, the approximating functions $f_n$ will be, in general, not of the form

$$f_n(s) := \sum_{i=1}^n f(t_i^{(n)}) 1_{[t_i^{(n)},t_{i+1}^{(n)})}(s)$$

(which would give rise to the Riemann sums you are stating at the very beginning of the question); see e.g. Proposition 15.16 and Lemma 15.19/Theorem 15.20 in the book by Schilling & Partzsch (2nd edition) for more information.

Answer 2

The short answer is no and the subsequence here is important but there is more to this answer as there exists an a.s. theory of stochastic integration but the price to pay is that you have to give up something for this. I give below motivated references on both of those subjects.

So first why is stochastic integration impossible(*) ?

The proof of this lies on the Banach-Steinhaus Theroem which allows us to prove that :

*Th. 56 of Protter's book : *

If the sum $S_n=\sum_{i=1}^{n} f(t^{(n)}_i).(B_{t^{(n)}_{i+1}} - B_{t^n_i})$ converges to a limit for every continuous function $f$ then $B$ is of finite variations.

(*)You can take a look at the argument in Philip E. Protter's book "Stochastic Integration and Differential Equations" at Chapter 1 Section 8.

So this is why to circumvent this limitation it is necessary to have recourse to weaker modes of convergence like $L^2$ but in fact u.c.p. (uniform convergence in probability)

Now the second point "Riemann sum stochastic integration and a.s. convergence"

Since the paper "On pathwise stochastic integration" by Rajeeva L. Karandikar in 1994 a lot of refrence are available on the subjet the absract of this paper goes as follows (I htink there is also a prior paper on this subject but this is already quite "old"):

Abstract
In this article, we construct a mapping $\mathcal{J} : D[0, \infty) \times D[0, \infty)\to , D[0, \infty) $ such that if $(X_t)$ is a semimartingale on a probability space ($\Omega, \mathcal{F},\mathbb{P}$) with respect to a filtration ($\mathcal{F}_t$) and if ($f_t$) is an càdlàg $\mathcal{F}$$_t$-adapted process, then $\mathcal{J}(f_.(\omega), X_.(\omega)) = \int_0^. f_-dX(\omega) ~a.s.$ This is of significance when using stochastic integrals in statistical inference >problems. Similar results on solutions to SDEs are also given.

A more recent and one of the most general account on this subject is the article by M. Nutz "Pathwise Construction of Stochastic Integrals"

Abstract We propose a method to construct the stochastic integral simultaneously under a non-dominated family of probability measures. Path-by-path, and without referring to a probability measure, we construct a sequence of Lebesgue-Stieltjes integrals whose medial limit coincides with the usual stochastic integral under essentially any probability measure such that the integrator is a semimartingale. This method applies to any predictable integrand

Answer 3

You are right in noting that \begin{align} \lim_{k \to \infty} \sum_{i=1}^{n_k} f(B_{t^{(n_k)}_i},t^{(n_k)}_i)(B_{t^{(n_k)}_{i+1}}-B_{t^{(n_k)}_i}) = \int_0^T f(B_t,t) dB_t \quad a.s. \end{align} However, the almost sure limit in the equation above is still the Itô-integral. In particular, the integral in the limit is not defined pathwise as a Riemann-Stieltjes integral. So we do not have \begin{align*} \bigg(\int_0^T f(B_t,t) dB_t\bigg)(\omega) = \int_0^T f(B_t(\omega),t) d\big(B_t(\omega)\big) \end{align*} for almost all $\omega$.

I hope that helps.