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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathbb F=(\mathcal F)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
  • $B=(B_t)_{t\ge 0}$ be an $\mathbb F$-adapted Brownian motion with respect to $\mathbb F$

Let $H=(H_t)_{t\ge 0}$ be $\mathbb F$-adapted, locally bounded and of the form $$H_t(\omega)=\sum_{i=1}^nH_{t_{i-1}}(\omega)1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)\;$$ for some $0=t_0<\ldots<t_n$. Let $\mathcal E$ be the space of all such $H$.

The Itô integral of $H$ with respect to $B$ is defined as $$I_\infty^B(H):=\sum_{i=1}^nH_{t_{i-1}}\left(B_{t_i}-B_{t_{i-1}}\right)\;.$$ Now, if $H$ is more general, but $\mathbb F$-progressively measurable and $$\left\|H\right\|^2:=\operatorname E\left[\int_0^\infty H_t^2\;dt\right]<\infty\;,$$ then there is a sequence of $H^n\in\mathcal E$ with $\left\|H^n-H\right\|\stackrel{n\to\infty}{\to}0$ and one defines $$\int_0^\infty H_t\;dB_t:=I_\infty^B(H):=\lim_{n\to\infty}I_\infty^B(H^n)\;\;\;\text{in }L^2(\operatorname P)\;.$$ Now, let $$H_t^{\tau}:=H_t1_{\left\{t\le \tau\right\}}\;\;\;\text{for }t\ge 0$$ for a $\mathbb F$-stopping time $\tau$ and let $I^B(H)$ be the continuous modification of the $\mathbb F$-martingale $\left(I_\infty^B(H^{(t)})\right)_{t\ge 0}$. One defines the Itô integral from $s$ to $t$ by $$\int_s^t H_r\;dB_r:=I_t^B(H)-I_s^B(H)\;,$$ if $0\le s<t$.


Question: $\;$ Let $H$ be continuous and $\mathbb F$-progressively measure with $$\int_0^TH_s^2\;ds<\infty\;\;\;\text{for all }T\ge 0\;$$ How can we show, that $$\int_0^TH_s\;dB_s=\lim_{n\to\infty}\sum_{t\in\mathcal P_T^n}H_t(B_{t'}-B_t)\;\;\;\text{for all }T\ge 0\;?\tag{1}$$ Here, $\mathcal P_T^n$ is constructed as follows:

  • $\mathcal P$ is a sequence of countable subsets $\mathcal P^n$ of $[0,\infty)$
  • $0\in\mathcal P^n\subset\mathcal P^{n+1}$
  • $\sup\mathcal P^n=\infty$
  • $\displaystyle |\mathcal P^n|:=\sup_{t\in\mathcal P^n}\min_{s\in\mathcal P^n:s\ne t}|s-t|\stackrel{n\to\infty}{\to}0$
  • $\mathcal P_T^n:=\mathcal P^n\cap [0,T)$ for $T>0$
  • $t':=t_{k+1}\wedge T$ if $t=t_k\in\mathcal P_T^n=\left\{t_0,t_1,\ldots\right\}$ with $t_0<t_1<\ldots$

For $G\in C^0(\mathbb R_{\ge o})$ one calls $$\langle G\rangle _T:=\lim_{n\to\infty}\sum_{t\in\mathcal P_T^n}|G_{t'}-G_t|^2\;\;\;\text{for }T\ge 0$$ quadratic variation of $G$ with respect to $\mathcal P$.

One can show, that $\langle B\rangle_T=T$ almost surely (independent of $\mathcal P$).

I want to show, that we can choose a sub-sequence of $\mathcal P$ such that $(1)$ holds almost surely.

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  • $\begingroup$ what do you mean by $G \in C^0(R_{\ge 0}$ $\endgroup$ – Conrado Costa Jul 8 '15 at 1:13
  • $\begingroup$ you meant in the introduction $X$ or $H$ is progressively measurable? $\endgroup$ – Conrado Costa Jul 8 '15 at 1:19
  • $\begingroup$ your question is strange, why are you defining the quadratic variation at this point? $\endgroup$ – Conrado Costa Jul 8 '15 at 1:26
  • $\begingroup$ @ConradoCosta $G\in C^0(\mathbb R_{\ge 0})$ means that $G:[0,\infty)\to\mathbb{R}$ is continuous. I've defined the quadratic variation cause it might be crucial for the question. $\endgroup$ – 0xbadf00d Jul 8 '15 at 13:59
  • $\begingroup$ @ConradoCosta I thought there might be more than once plausible way to the define the Itô integral and it would be important to give you my construction. $\endgroup$ – 0xbadf00d Jul 8 '15 at 14:04
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Well the convergence in $(1)$ occurs in probability, uniformly on compact time intervals (The proof of this is rather involved). You have the elementary result that convergence in probability implies almost surely convergence along a subsequence. So there is such a subsequence.

But here is an easier way to show the existence of such a subsequence. Be warned: There is some handwaving. First fix $T$. The triangle inequality gives you the following. Proving that the triangle inequality holds for the norm you defined might be a good exercise by the way.

$$\|H_n - H_m\| \leq \|H_n - H\| + \|H_m - H\|$$

Next we make use of the Ito isometry. $H_m\cdot B$ is short hand for $\int_0^TH_m \, dB$.

$$\|H_n - H_m\| = E\left[\left((H_n - H_m)\cdot B\right)^2\right]$$

Combined with the triangle inequality this shows that the sequence of $L^2$ martingales $(H_n\cdot B)$ is a Cauchy sequence if you use the expression on the RHS of the isometry above as some kind of distance metric. Since the space of continuous $L^2$ martingales is complete under this metric, $H_n\cdot B \rightarrow X$ for some $L^2$ martingale $X$. It is this $X$ that we refer to as $H\cdot B$. All this boils down to

$$\lim_{n \rightarrow \infty}E\left[(H_n\cdot B - X)^2\right] = 0$$

I hope this looks familiar to you. This is just ordinary $L^2$ convergence. $L^2$ convergence implies convergence in probability, which in turn implies convergence almost surely along a subsequence.

Now look at the sum you defined for the Ito integral. Pick your $\mathcal{P}^n$ such that the resulting $H_n$ approximates $H$ in the appropriate sense. This is a bit restrictive but for the general case you would have to consider the result I mentioned in the first paragraph.

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  • $\begingroup$ Why does $(1)$ hold at all? $\endgroup$ – 0xbadf00d Jul 8 '15 at 15:34
  • $\begingroup$ @0xbadf00d i will get back to you in a few hours $\endgroup$ – Calculon Jul 8 '15 at 16:09

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