Lecture 23. Itō’s representation theorem

Posted on September 23, 2012 by Fabrice Baudoin

In this Lecture we show that, remarkably, any square integrable integrable random variable which is measurable with respect to a Brownian motion, can be expressed as a stochastic integral with respect to this Brownian motion. A striking consequence of this result, which is known as Itō’s representation theorem, is that any square integrable martingale of a Brownian filtration has a continuous version.

Let (B_t)_{t \ge 0} be a Brownian motion. In the sequel, we consider the filtration (\mathcal{F}_t)_{t \ge 0} which is the usual completion of the natural filtration of (B_t)_{t \ge 0} (such a filtration is called a Brownian filtration).

The following lemma is a straightforward consequence of Itō’s formula.

Lemma. Let f : \mathbb{R}_{\ge 0} \to \mathbb{R} be a locally square integrable function. The process \left( \exp\left( \int_0^t f(s) dB_s -\frac{1}{2} \int_0^t f(s)^2 ds \right)\right)_{t \ge 0} is a square integrable martingale.

Proof. From Itō’s formula we have
\exp\left( \int_0^t f(s) dB_s -\frac{1}{2} \int_0^t f(s)^2 ds \right)=1+\int_0^t f(s) \exp\left( \int_0^s f(u) dB_u -\frac{1}{2} \int_0^s f(u)^2 du \right)dB_s.
The random variable \int_0^s f(u) dB_u is a Gaussian random variable with mean 0 and variance \int_0^s f(u)^2 du. As a consequence
\mathbb{E}\left( \int_0^t f(s)^2 \exp\left( 2\int_0^s f(u) dB_u \right) ds \right)<+\infty
and the process
\int_0^t f(s) \exp\left( \int_0^s f(u) dB_u -\frac{1}{2} \int_0^s f(u)^2 du \right)dB_s
is a martingale. \square

Lemma. Let \mathcal{D} be the set of compactly supported and piecewise constant functions \mathbb{R}_{\ge 0} \rightarrow \mathbb{R}, i.e. the set of functions f that can be written as f=\sum_{i=1}^n a_i \mathbf{1}_{(t_{i-1}, t_i]}, for some 0\le t_1 \le \cdots \le t_n and a_1, \cdots , a_n \in \mathbb{R}. The family \left\{ \exp\left( \int_0^{+\infty} f(s) dB_s -\frac{1}{2} \int_0^{+\infty} f(s)^2 ds \right), f \in \mathcal{D} \right\} is total in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ).

Proof.
Let F \in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ) such that for every f \in \mathcal{D},
\mathbb{E} \left(F \exp\left( \int_0^{+\infty} f(s) dB_s -\frac{1}{2} \int_0^{+\infty} f(s)^2 ds \right) \right)=0.
Let t_1,\cdots, t_n \ge 0. We have for every \lambda_1,\cdots,\lambda_n \in \mathbb{R},
\mathbb{E} \left(F \exp\left( \sum_{i=1}^n \lambda_i (B_{t_i}-B_{t_{i-1}}) \right) \right)=0.
By analytic continuation, we see that
\mathbb{E} \left(F \exp\left( \sum_{i=1}^n \lambda_i (B_{t_i}-B_{t_{i-1}}) \right) \right)=0.
actually also holds for every \lambda_1,\cdots,\lambda_n \in \mathbb{C}. By using the Fourier transform, it implies that
\mathbb{E} \left(F \mid B_{t_1},\cdots , B_{t_n} \right)=0.
Since t_1,\cdots, t_n were arbitrary, we conclude that \mathbb{E}(F \mid \mathcal{F}_\infty)=0. As a conclusion F=0 \square

We are now in position to state the representation theorem.

Theorem. For every F \in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ), there is a unique progressively measurable process (u_t)_{t \ge 0} such that \mathbb{E} \left(\int_0^\infty u_s^2 ds \right)<+\infty and F=\mathbb{E} (F)+ \int_0^{+\infty} u_s dB_s.

Proof. The uniqueness is immediate as a consequence of the Itō’s isometry for stochastic integrals. Let \mathcal{A} be the set of random variables F \in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ) such that there exists a progressively measurable process (u_t)_{t \ge 0} such that \mathbb{E} \left(\int_0^\infty u_s^2 ds \right)< +\infty and F=\mathbb{E} (F)+ \int_0^{+\infty} u_s dB_s. From the above lemma, it is clear that \mathcal{A} contains the set of set of random variables
\left\{ \exp\left( \int_0^{+\infty} f(s) dB_s -\frac{1}{2} \int_0^{+\infty} f(s)^2 ds \right), f \in \mathcal{D} \right\}.
Since this set is total in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ), we just need to prove that \mathcal{A} is closed in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ). So, let (F_n)_{n \in \mathbb{N}} be a sequence of random variables such that F_n \in \mathcal{A} and F_n \to_{n \to \infty} F in \mathbf{L}^2 (\mathcal{F}_\infty , \mathbb{P} ). There is a progressively measurable process (u^n_t)_{t \ge 0} such that \mathbb{E} \left(\int_0^\infty (u^n_s)^2 ds \right)<+\infty and F_n=\mathbb{E} (F_n)+ \int_0^{+\infty} u^n_s dB_s. By using Itō’s isometry, it is seen that the sequence u^n is a Cauchy sequence and therefore converges to a process u which is seen to satisfy
F=\mathbb{E} (F_n)+ \int_0^{+\infty} u^n_s dB_s \square

As a consequence of the representation theorem, we obtain the following description of the square integrable martingales of the filtration (\mathcal{F}_t)_{t \ge 0}.

Corollary. Let (M_t)_{t \ge 0} be a square integrable martingale of the filtration (\mathcal{F}_t)_{t \ge 0}. There is a unique progressively measurable process (u_t)_{t \ge 0} such that for every t \ge 0, \mathbb{E} \left(\int_0^t u_s^2 ds \right)<+\infty and M_t=\mathbb{E} (M_0)+ \int_0^{t} u_s dB_s. In particular, (M_t)_{t \ge 0} admits a continuous version.

Exercise. Show that if (M_t)_{t \ge 0} is a local martingale of the filtration (\mathcal{F}_t)_{t \ge 0}, then there is a unique progressively measurable process (u_t)_{t \ge 0} such that for every t \ge 0, \mathbb{P} \left(\int_0^t u_s^2 ds < +\infty \right)=1 and M_t=\mathbb{E} (M_0)+ \int_0^{t} u_s dB_s.

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