Convergence in sup norm of elementary integrals to the Itô integral process

Question

Let $W$ be a standard one dimensional Brownian motion, and $X$ a continuous process adapted to $W$ such that $\int_0^T X^2 \, ds < \infty$ almost surely for some $T > 0$.

Define for any sequence of partitions $\mathcal P_n = \{t_1^n, \dotsc t_{k_n} ^n\}$ of $[0, T]$, the elementary integral process $Y^n$ associated to $\mathcal P_n$ by

$$Y_t^n := \sum_{i = 1}^{k_n - 1} X_{t^n_i} (W_{t^n_{i+i} \wedge t} - W_{t^n_{i} \wedge t} \,).$$

Question: Is it true that for all partitions $\mathcal P_n$ of $[0, T]$ with mesh going to $0$, we have

$$\sup_{0 \leq t \leq T} \left\lvert\int_0^t X_s \, dW_s - Y_t^n\right\rvert \to 0$$

in probability?

Answer 1

Let \begin{equation*} X^{(n)}_t=\sum_{i=1}^{k_n-1} X_{t_i^n}1_{(t_i^n,t_{i+1}^n]}(t). \end{equation*} Since $X=\{X_t,0\le t\le T\}$ is a continuous adapted process, then \begin{gather*} \lim_{n\to\infty} \sup_{0\le t\le T}|X^{(n)}_t - X_t |=0, \qquad \text{a.s.}, \\ \lim_{n\to\infty} \int_{0}^{T} (X^{(n)}_t - X_t )^2\,\mathrm{d}t=0, \qquad \text{a.s.}, \end{gather*} and \begin{equation*} \lim_{n\to\infty}\mathsf{P}\Big(\Big|\int_{0}^{T} (X^{(n)}_t - X_t )^2\, \mathrm{d}t \Big| >d\Big)=0, \quad \forall d>0. \tag{1} \end{equation*} Due to $(\int_0^t (X^{n}_s-X_s)\mathrm{d}W_s)^2$ is L-dominated by \begin{equation*} \langle \int_0^{\cdot} (X^{(n)}_s-X_s)\mathrm{d}W_s \rangle_t = \int_{0}^{t} (X^{(n)}_s - X_s )^2\,\mathrm{d}s. \end{equation*} Using Lenglart's inequality(cf. S. W. He et al., Semimartingale Theory and Stochastic Calculus, Sci. Press and CRC(1992). Th9.26, pp240, or J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, Lemma I.3.30, pp35.), we get \begin{align*} &\mathsf{P}\Big(\sup_{0\le t\le T}\Big[\int_0^t (X^{(n)}_s-X_s) \mathrm{d}W_s \Big]^2 \ge\epsilon \Big)\\ &\quad\le \frac{d}{\epsilon}+\mathsf{P}\Big[\int_{0}^{T} (X^{(n)}_t - X_t )^2\,\mathrm{d}t \ge d\Big]. \tag{2} \end{align*} Now, by (1), letting $n\to\infty$ and $d\downarrow 0$ consecutively in (2), we obtain immediately the following assertion, \begin{equation*} \lim_{n\to\infty}\mathsf{P}\Big(\sup_{0\le t\le T}\Big|\int_0^t (X^{(n)}_s-X_s) \mathrm{d}W_s \Big|\ge\epsilon \Big)=0, \qquad \forall \epsilon>0. \end{equation*}

Answer 2

The difference of two martingales is still a martingale by linearity of conditional expectation and so

$$Z^n_{t}:=\int_{0}^{t}X_{s}dW_{s}-\sum_{i = 1}^{k_n - 1} X_{t^n_i} (W_{t^n_{i+i} \wedge t} - W_{t^n_{i} \wedge t} \,)$$

is a martingale for each $n$. Indeed, the first term is a martingale by the properties of stochastic integration, while the latter is a martingale transform of a martingale.

Thus we can apply Doob's inequality for $p=2$:

$$\text{E}[| Z^n_T |^p] \leq \text{E}\left[\sup_{0 \leq s \leq T} |Z^n_s|^p\right] \leq \left(\frac{p}{p-1}\right)^p\text{E}[|Z^n_T|^p]$$

to get $L^{2}$-convergence of the supremum as well. For probability-convergence we can use the weaker version

$$\text{P}[\sup_t|Z_t^{n}|\geq C]\leq \frac{\text{E}[|Z_T^{n}|^p]}{C^p}.$$

In terms of the $L_{2}$ convergence one needs a bit more structure. There are many references eg. see here Theorem 5.3.

Now we verify the assumption for the L2 convergence of the elementary processes

$$\phi_{n}(t):=\sum_{i = 1}^{k_n - 1} X_{t^n_i} 1_{[t^n_i,t^n_{i+1}]}$$

to $X_{t}$. For reference of the last one, also see Oksendal SDEs on pags 26-28. When $X_{t}$ is bounded one can use the above simple-functions.

and when $X_{t}$ is unbounded, one can use the following approximation from his step 3

$$\psi_{n}(t):=1_{X_{t}\in [-n,n]}\sum_{i = 1}^{k_n - 1} X_{t^n_i} 1_{[t^n_i,t^n_{i+1}]}-n1_{X_{t}\leq -n}+n1_{X_{t}\geq n}.$$

So if one wants to just have $\phi_{n}$ as the approximation, they would need the following terms to go to zero

$$\int_{0}^{T} E[|\phi_{n}(t)-n|^{2}1_{X(t)>n}]+E[|\phi_{n}(t)+n|^{2}1_{X(t)<-n}]dt.$$

I will try to check if this can be obtained given the above particular assumptions on $X$ or whether there are counterexamples. Other answers are welcome.