## 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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