Lecture 23. Itō’s representation theorem

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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