• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $B=(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
  • $\mathcal P$ be a sequence of countable subsets $$\mathcal P^n:=\left\{\cdots<t_i^{(n)}<t_{i+1}^{(n-1)}<\cdots\right\}$$ of $[0,\infty)$ such that
    • $0\in\mathcal P^n\subseteq\mathcal P^{n+1}$
    • $\sup\mathcal P^n=\infty$
    • $\displaystyle|\mathcal P^n|:=\sup_{t\in\mathcal P^n}\min_{s\in\mathcal P^n:s\ne t}|s-t|\stackrel{n\to\infty}{\to}0$
  • $\mathcal P_T^n:=\mathcal P^n\cap [0,T)$, for $T>0$
  • $t':=t_{i+1}^{(n)}\wedge T$, if $t=t_i^{(n)}$

Let $f:\mathbb{R}\to\mathbb{R}$ be continuous and $$I:=\lim_{n\to\infty}\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)\;.$$ How can we show, that $$\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)\stackrel{n\to\infty}{\to}\int_0^Tf(B_s)\;dB_s\tag{1}$$ in probability and why does that imply $$I=\int_0^Tf(B_s)\;dB_s\;\;\;\text{almost surely}\;?\tag{2}$$

I've tried the following: Let $T\ge 0$, $H:=f\circ B$ and $$H^n_s(\omega):=\sum_{t\in\mathcal P_T^n}H_t(\omega)1_{(t,t']}(s)\;.$$ Since $B$ is almost surely continuous, $$\left\{B_s:s\in [0,T]\right\}$$ is almost surely compact and thereby $H$ almost surely bounded on $[0,T]$. Hence, each $H^n$ is a elementary predictable process. Now, let $n\in\mathbb{N}$ and $s\in (0,T]$. Then, there is a unique $t_0\in\mathcal P_T^n$ with $s=t_0'$ and $$H_s^n-H_s=H_{t_0}-H_{t_0'}\;.$$ Since $H$ is almost surely continuous and $|\mathcal P^n|\stackrel{n\to\infty}{\to}0$, we've got $$H^n_s-H_s\stackrel{n\to\infty}{\to}0\;\;\;\text{almost surely}\tag{3}$$ for all $s\in (0,T]$. Since $(3)$ holds for all $T>0$, we can conclude $$\left\|H^n-H\right\|\stackrel{n\to\infty}{\to}0$$ by the dominated convergence theorem, where $$\left\|X\right\|^2:=\operatorname E\left[\int_0^\infty X_s^2\;ds\right]\;.$$ Since by the definition of the Itō integral for elementary predictable processes $$\int_0^\infty H^n\;dB_s=\sum_{t\in\mathcal P_T^n}H_t(B_{t'}-B_t)$$ and by definition of the Itō integral for progressively measurable processes $$\int_0^\infty H_s\;dB_s=\lim_{n\to\infty}\int_0^\infty H^n\;dB_s\;\;\;\text{in }L^2(\operatorname P)\;,\tag{4}$$ we should be able to conclude $(1)$. Why? Well, $$\int_0^T H\;dB_s\stackrel{\text{def}}{=}\int_0^\infty 1_{\left\{s\le T\right\}}H_s\;dB_s\;.$$ Now, I got still problems to conclude $(2)$. Sure, we can choose a subsequence of $(H^n)_{n\in\mathbb{N}}$ such that the convergence in $(4)$ holds almost surely. But does that mean, that $(2)$ holds? For some reasons, I'm unsure.

  • $\begingroup$ what about $f$; is it continuous or differentiable or...? $\endgroup$ – saz Jul 12 '15 at 12:00
  • $\begingroup$ Well, continuity doesn't suffice to ensure the existence of the integral. Some additional assumption is needed (boundedness, mean sqaure continuity,...) $\endgroup$ – saz Jul 12 '15 at 15:15
  • $\begingroup$ How is $H^n$ related to the sum $\sum f(B_t) (B_{t'}-B_t)$....? And how exactly do you want to apply the dominated convergence theorem? What is your dominating function? $\endgroup$ – saz Jul 12 '15 at 18:09
  • $\begingroup$ the constant depends on $\omega $... $\endgroup$ – saz Jul 12 '15 at 19:34
  • $\begingroup$ How do you get a uniform bound for all $\omega $? Continuity gives $|H_t (\omega)| \leq c (\omega) $. $\endgroup$ – saz Jul 12 '15 at 19:56

No, we cannot apply the dominated convergence theorem in this way (see all the comments above).

For fixed $R>0$ denote by

$$\tau := \inf\{t>0; |B_t| \geq R\}$$

the exit time from $(-R,R)$. Moreover, we denote by

$$H \bullet B(T) := (H \bullet B)(T) := \int_0^T H(s) \, dB_s$$

the stochastic integral of $H$. By Markov's inequality and Itô's isometry,

$$\begin{align*} \mathbb{P}(|H^n \bullet B(T)-H \bullet B(T)| \geq \epsilon) &\leq \mathbb{P}(|H^n\bullet B(T)-H\bullet B(T)| \geq \epsilon, \tau > T) + \mathbb{P}(\tau \leq T) \\ &\leq \mathbb{P}(|H^n\bullet B(T \wedge \tau)-H\bullet B(T \wedge \tau)| \geq \epsilon) + \mathbb{P}(\tau \leq T) \\ &\leq \frac{1}{\epsilon^2} \mathbb{E} \left( \int_0^{T \wedge \tau} |H^n(s)-f(B_s)|^2 \, ds \right) + \mathbb{P}(\tau \leq T). \end{align*}$$

Since $f|_{[-R,R]}$ is bounded, we can let $n \to \infty$ using the dominated convergence theorem and the continuity of $f$ and obtain

$$\limsup_{n \to \infty} \mathbb{P}(|H^n \bullet B(T)-H \bullet B(T)| \geq \epsilon) \leq \mathbb{P}(\tau \leq T).$$

Finally, we can let $R \to \infty$ and conclude

$$\limsup_{n \to \infty} \mathbb{P}(|H^n \bullet B(T)-H \bullet B(T)| \geq \epsilon)=0.$$

This shows that

$$\begin{align*} I &:=\mathbb{P}-\lim_{n \to \infty} \sum_{t \in P_T^n} f(B_t) (B_{t'}-B_t) \\ &= \mathbb{P}-\lim_{n \to \infty} (H^n \bullet B)(T) \\ &= \int_0^T H(s) \, dB_s = \int_0^T f(B_s) \, dB_s \end{align*} $$

almost surely.

  • 1
    $\begingroup$ What do you mean by $ \mathbb{P}-\lim$? $\endgroup$ – 0xbadf00d Jul 13 '15 at 10:47
  • 1
    $\begingroup$ @0xbadf00d limit in probabiliy $\endgroup$ – saz Jul 13 '15 at 10:53

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