1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.