Lecture 16. Itō integral

Since a Brownian motion $(B_t)_{t \ge 0}$ does not have absolutely continuous paths, we can not directly use the theory of Riemann-Stieltjes integrals to give a sense to integrals like $\int_0^t \Theta_s dB_s$ for every continuous stochastic process $(\Theta_s)_{s \ge 0}$ . However, if $(\Theta_s)_{s \ge 0}$ is regular enough in the Hölder sense, then $\int_0^t \Theta_s dB_s$ can still be constructed as a limit of Riemann sums by using a celebrated result of L.C. Young. In the sequel, we shall denote by $\mathcal{C}^\alpha(I)$ the space of $\alpha$ – Hölder continuous functions that are defined on an interval $I$ .

Theorem.(Young’s integral) Let $f \in \mathcal{C}^\beta([0,T])$ and $g \in \mathcal{C}^\gamma([0,T])$ . If $\beta+\gamma>1$ , then for every subdivision $\Delta_n [0,T]$ , whose mesh tends to 0, the Riemann sums

$\sum_{i=0}^{n-1} f(t^n_i)( g(t^n_{i+1}) - g(t^n_i))$

converge, when $n \to \infty$ to a limit which is independent of the subdivision $t^n_i$ . This limit is denoted $\int_0^T f dg$ and called the Young’s integral of $f$ with respect to $g$ .

As a consequence of the previous result, we can therefore use Young’s integral to give a sense to the integral $\int_0^t \Theta_s dB_s$ as soon as the stochastic process $(\Theta_s)_{s \ge 0}$ has $\gamma$ -Hölder paths with $\gamma >1/2$ . This is not satisfying enough, since for instance the integral $\int_0^t B_s dB_s$ is not even well defined. The alternative to using Riemann sums to and studying its almost sure convergence is to take advantage of the quadratic variation of the Brownian motion paths and use the full power of probabilistic methods.

In what follows, we consider a Brownian motion $(B_t)_{t\ge 0}$ which is defined on a filtered probability space $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ . $(B_t)_{t \ge 0}$ is assumed to be adapted to the filtration $(\mathcal{F}_t)_{t \ge 0}$ . We also assume that $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ satisfies the usual conditions as they were defined at the begining of Lecture 10. These assumptions imply in particular the following facts that we record here for later use :

A limit (in $L^2$ or in probability) of adapted processes is still adapted;
A modification of a progressively measurable process is still a progressively measurable process.

Proposition. Let $(B_t)_{t \ge 0}$ be a standard Brownian motion. We denote by $(\mathcal{F}^B_t)_{t \ge 0}$ its natural filtration: $\mathcal{F}^B_\infty=\sigma ( B_u,u \ge 0)$ and by $\mathcal{N}$ the null sets of $\mathcal{F}^B_\infty$ .
Show that the filtration $(\sigma (\mathcal{F}^B_t, \mathcal{N}))_{t \ge 0}$ satisfies the usual conditions.

We denote by $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ the set of processes $(u_t)_{t \ge 0}$ that are progressively measurable with respect to the filtration $(\mathcal{F}_t)_{t \ge 0}$ and such that

$\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right)<+\infty.$

Exercise. Show that the space $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ endowed with the norm

$\| u \|^2=\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right)$
is a Hilbert space.

We now denote by $\mathcal{E}$ the set of processes $(u_t)_{t \ge 0}$ that may be written as:

$u_t=\sum_{i=0}^{n-1} F_i 1_{(t_i,t_{i+1}]} (t),$

where $0\le t_0 \le ... \le t_n$ and where $F_i$ is a random variable that is measurable with respect to $\mathcal{F}_{t_i}$ and such that $\mathbb{E}( F_i^2)<+\infty$ . The set $\mathcal{E}$ is often called the set of simple previsible processes. We first observe that it is straightforward that

$\mathcal{E} \subset L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P}).$

The following theorem provides the basic definition of the so-called Itō integral.

Theorem. (Itō integral) There is a unique linear map

$\mathcal{I}:L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P}) \rightarrow L^2 (\Omega, \mathcal{F},\mathbb{P})$

such that:

For $u=\sum_{i=0}^{n-1} F_i 1_{(t_i,t_{i+1}]} \in \mathcal{E}$ , $\mathcal{I} (u)=\sum_{i=0}^{n-1} F_i (B_{t_{i+1}} -B_{t_i});$

For $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ , $\mathbb{E} \left( \mathcal{I} (u)^2\right)=\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right).$

The map $\mathcal{I}$ is called the Itō integral and for $u \in L^2 (\Omega,(\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ , we will use the notation

$\mathcal{I} (u)=\int_0^{+\infty} u_s dB_s.$

Proof: Since $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ endowed with the norm $\| u \|^2=\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right)$
is a Hilbert space, from the isometries extension theorem we just have to prove that

For $u=\sum_{i=0}^{n-1} F_i 1_{(t_i,t_{i+1}]} \in \mathcal{E}$ ,
$\mathbb{E} \left( \left(\sum_{i=0}^{n-1} F_i (B_{t_{i+1}} -B_{t_i})\right)^2\right)=\mathbb{E} \left(\int_0^{+\infty} u_s^2 ds \right);$
The set $\mathcal{E}$ is dense in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ .

Let $u=\sum_{i=0}^{n-1} F_i 1_{(t_i,t_{i+1}]} \in \mathcal{E}$ . Due to the independence of the Brownian motion increments, we have:

$\mathbb{E} \left( \left(\sum_{i=0}^{n-1} F_i (B_{t_{i+1}} -B_{t_i})\right)^2\right)$
$= \mathbb{E} \left( \sum_{i,j=0}^{n-1} F_i F_j (B_{t_{i+1}} -B_{t_i})(B_{t_{j+1}} -B_{t_j})\right)$
$= \mathbb{E} \left( \sum_{i=0}^{n-1} F_i^2 (B_{t_{i+1}} -B_{t_i})^2\right)+2 \mathbb{E} \left(\sum_{0\le i<j \le n-1}F_i F_j (B_{t_{i+1}} -B_{t_i})(B_{t_{j+1}} -B_{t_j})\right)$
$= \mathbb{E} \left( \sum_{i=0}^{n-1} F_i^2 (t_{i+1} -t_i)\right)$
$=\mathbb{E} \left( \int_0^{+\infty} u_s^2 ds \right).$

Let us now prove that $\mathcal{E}$ is dense in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ . We proceed in several steps. As a first step, let us observe that the set of progressively measurable bounded processes is dense in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ . Indeed, for $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ , the sequence $(u_t \mathbf{1}_{[0,n]} (\mid u_t \mid))_{t \ge 0}$ converges to $u$ .

As a second step, we remark that if $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ is a bounded process, then $u$ is a limit of bounded processes that are in $L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ and such that almost every paths are supported in a fixed compact set (consider the sequence $(u_t \mathbf{1}_{[0,n]} (t)_{t \ge 0}$ ).

As a third step, if $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ is a bounded process such that almost every paths are supported in a fixed compact set, then the sequence $\left( \frac{1}{n} \int_{t-\frac{1}{n}} ^t u_s ds \mathbf{1}_{(\frac{1}{n},+\infty)} (t) \right)_{t \ge 0}$ is seen to converge toward $u$ . Therefore, $u$ is a limit of left continuous and bounded processes that are in $L^2(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ and such that almost every paths are supported in a fixed compact set.

Finally, it suffices to prove that if $u \in L^2 (\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ is a left continuous and bounded process such that almost every paths are supported in a fixed compact set, then $u$ is a limit of processes that belong to $\mathcal{E}$ . This may be proved by considering the sequence:
$u^n_t=\sum_{i=0}^{+\infty} u_{\frac{i}{n}} 1_{(\frac{i}{n},\frac{i+1}{n}]} (t) \square$

4 Responses to Lecture 16. Itō integral

alabair says:

August 29, 2012 at 7:36 pm

With regard to the fractional Brownian motion, what are the conditions on the coefficient of Husrt, in order to define a stochastic integral in the sense of Young?

- Fabrice Baudoin says:
  
  August 29, 2012 at 10:36 pm
  
  To define the integral $\int_0^t f(B_s)dB_s$ when $f$ is Lipschitz and B is a fractional Brownian motion with parameter $H$ , you would need that $H+H >1$ , that is $H> 1/2$ .
  
Alsichkann says:

December 7, 2013 at 7:17 am

In the third step of the proof of Itō integral theorem, the constructed sequence seems to be not continuous at the time $t=\frac{1}{n}$. I think it should be right continuous. Am I wrong for this?

- Fabrice Baudoin says:
  
  December 8, 2013 at 12:05 am
  
  Thanks for the remark. I corrected to make it left continuous which is enough to conclude.