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

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?

• is the concern the supremum? Because L2 convergence implies in probability. Commented Mar 6, 2023 at 19:56
• math.stackexchange.com/questions/2056322/… Commented Mar 6, 2023 at 19:56
• Yes, the usual result says that the term inside the sup converges to $0$ in probability, but I don’t see that it implies the sup converges to $0$ in probability. Commented Mar 6, 2023 at 19:59
• Ohh, that’s a good idea. The term in the sup (without the absolute values) is a martingale no? Being the difference of two martingales. Commented Mar 6, 2023 at 20:01
• Is $(X_s)$ appropriately adapted? Commented Mar 6, 2023 at 20:34

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, $$\begin{equation*} \lim_{n\to\infty}\mathsf{P}\Big(\sup_{0\le t\le T}\Big|\int_0^t (X^{(n)}_s-X_s) \mathrm{d}W_s \Big|\ge\epsilon \Big)=0, \qquad \forall \epsilon>0. \end{equation*}$$

• Thank you for your answer! Commented Aug 6, 2023 at 11:36

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.

• Wait, I think there is a problem. How do we know that $Z^n_t \to 0$ in $L^2$? The usual result only says we have convergence in probability. Commented Mar 6, 2023 at 20:17
• Hm, I don’t see a proof of $L^2$ convergence there… it is claimed in the linked post but I believe the claim is false - the usual $L^2$ construction of elementary integrals does not use $X_{t^n_i}$ as the elementary integrands. Commented Mar 6, 2023 at 20:43
• Thanks for the updated answer. However, the hypotheses of Theorem 5.3, namely that the elementary processes converge in $L^2$, has not been verified in our case. Commented Mar 6, 2023 at 21:14
• Thanks again. But I think there is still a problem - Lemma 5.1 and Theorem 5.2 say that there exists a sequence of elementary processes approximating $X$, however nowhere is it said that these processes are given by our particular choice of elementary integrands. Commented Mar 6, 2023 at 21:22
• Updated answer. I agree with you; it is tricker than it looks. So Oksendal has a slightly different approximation that also includes the unbounded case. Commented Mar 6, 2023 at 23:01