For Brownian motion, we can show that the left endpoint Riemann sum converges to the Ito integral:

$$\sum\limits_{i=0}^{n-1} W_{t_i}(W_{t_{i+1}} - W_{t_i}) \to \int_0^t W_s dW_s$$

However, I am wondering if this is true more generally for any $X$ that is continuous and square-integrable martingale. I first tried to show that

$$X_t^{(n)} := \sum\limits_{i=0}^{n-1} X_{t_i} \mathbb{1}_{(t_i, t_{i+1}]}(t)$$

on $[0,T]$ is Cauchy in $L^2$ under the norm $|| X ||_T := E\int_0^T X_t^2 d\langle X\rangle_t$ for every $T$. Then since $X_t^{(n)} \to X_t$ a.s. as $n\to \infty$ by continuity, we have that $X_t^{(n)} \to X_t$ in $|| \cdot ||_T$ for every T. If we assume that $E\int_0^T X_s^2 d\langle X\rangle_s < \infty$ for every $T$, I believe the stochastic integral $\int_0^TX_s dX_s$ does exist. Then we should get that by definition $\int_0^T X_s^{(n)} dX_s \to \int_0^T X_sdX_s$ in $L^2(P)$. However, I am already stuck showing the first step simply because I cannot use Fubini like in Brownian case where $\langle X \rangle_s = s$ deterministically. Is there a counter-example to my statement or is there a more involved proof?

  • $\begingroup$ The Riemann sum converges certainly in probability... but right now I don't see how to get $L^2$-convergence. $\endgroup$ – saz Apr 5 '18 at 19:24
  • $\begingroup$ Could you give a sketch of how to show convergence in probability? $\endgroup$ – James Yang Apr 5 '18 at 20:53

I believe that, in general, $L^2$ convergence fails to hold. If we pose stronger assumptions on the integrability of $(X_t)_{t \geq 0}$, then $L^2$-convergence holds, but unfortunately I'm not aware of a result on the optimal integrability condition. In my answer I will show that the Riemann sums $$\sum_{i=0}^{n-1} X_{t_i} (X_{t_{i+1}}-X_{t_i})$$ converge in probability.

For $R>0$ define stopping times by

$$\tau_R := \inf\{t \geq 0; |X_t| \geq R\}.$$

Since $(X_t)_{t \geq 0}$ has continuous sample paths, we habe $\tau_R \uparrow \infty$ as $R \to \infty$. For fixed $\epsilon>0$ and $T>0$ we can choose $R>0$ such that $\mathbb{P}(\tau_R \leq T) \leq \epsilon$. If we denote by $M^{(n)}$ the stochastic integral of $X^{(n)}$ with respect to $X$ we find

$$\begin{align*} \mathbb{P} \left( \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) &= \mathbb{P} \left( \tau_R \leq T, \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) \\ &\quad + \mathbb{P} \left( \tau_R>T, \sup_{t \leq T} |M^{(n)}_t-M_t^{(m)}| \geq \delta \right) \\ &\leq \mathbb{P}(\tau_R \leq T) + \mathbb{P} \left( \sup_{t \leq T} |M^{(n)}_{t \wedge \tau_R}-M^{(m)}_{t \wedge \tau_R}| \geq \delta \right) \\ &=: I_1+I_2 \tag{1} \end{align*}$$

By our choice of $R$ we know that $I_1 \leq \epsilon$. For the second term we use Markov's inequality, the maximal inequality and Itô's isometry to conclude that

$$\begin{align*} I_2 &\leq \frac{4}{\delta^2} \mathbb{E} \left( \int_0^T |X_{t \wedge \tau_R}^{(n)}-X^{(m)}_{t \wedge \tau_R}|^2 \, d\langle X \rangle_t \right) \\ &\leq \frac{4}{\delta^2} \mathbb{E} \left( \sup_{t \leq T} |X_{t \wedge \tau_R}^{(n)}-X_{t \wedge \tau_R}^{(m)}|^2 \langle X \rangle_T \right). \tag{2} \end{align*}$$

Since the sample paths of $(X_t)_{t \geq 0}$ are uniformly continuous on compact sets, it is not difficult to see that

$$\sup_{t \leq T} |X_{t \wedge \tau_R}^{(n)}-X_{t \wedge \tau_R}^{(m)}|^2 \to 0$$

as $m,n \to \infty$. On the other hand, the expression is bounded by $4R^2$, and therefore we may apply the dominated convergence theorem in $(2)$ to conlucde that $I_2 \to 0$ as $m,n \to \infty$. Consequently we have shown that $M^{(n)}$ is a ("uniform") Cauchy sequence. On the other hand, $X^{(n)} \to X$, and therefore

$$\int_0^T X_s \, dX_s \tag{3}$$

exists and

$$\int_0^T X_s \, dX_s = \mathbb{P}-\lim_{n \to \infty} M^{(n)}_T.$$

Remark: The existence of the stochastic integral $(3)$ is to be understood in the sense of convergence in probability and not in $L^2$-convergence. In fact, defining stochastic integrals as a limit in probability theory is a standard way to extend stochastic integrals to a larger class of integrators and integrands.

  • $\begingroup$ Thank you for such a complete answer! Just one thing: how do we know $\langle X \rangle_T$ is integrable? In order to use LDCT on $I_2$ I understand that we have the $4R^2$ on the sup term, but we still need integrability of $\langle X \rangle_T$ right? $\endgroup$ – James Yang Apr 8 '18 at 14:47
  • 1
    $\begingroup$ @JamesYang Since $(X_t)_t$ is an $L^2$-martingale we have $\mathbb{E}(\langle X \rangle_T) = \mathbb{E}(X_T^2)<\infty$, i.e. $\langle X \rangle_T \in L^1(\mathbb{P})$. $\endgroup$ – saz Apr 8 '18 at 15:11

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