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2. Itô Formula and Martingale Representation Theorem

2.1 The 1-dimensional Itô Formula

**Definition 2.1.1: **

Let $B_t$ be $1$-dimensional Brownian motion on $(\Omega,\msc{F},\prob)$. A $1$-dimensional Itô process is a stochastic process $X_t$ on $(\Omega,\msc{F},\prob)$ of the form

\[ X_t = X_0 +\int_0^t u(s,\omega)\ \rmd s + \int_0^t v(s,\omega)\ \rmd B_s\ , \]

where $v\in\msc{W_H}$ is square-integrable (weak condition the third):

\[ \prob\lr[{\int_0^t v(s,\omega)^2\ \rmd s<\infty}]=1\ ,\quad \forall\ t\geq0\ . \]

We also assume that $u$ is $\msc{H}_t$-adapted, where $\msc{H}_t$ is as in weak condition the second, and

\[ \prob\lr[{\int_0^t |u(s,\omega)|\ \rmd s<\infty}]=1\ ,\quad \forall\ t\geq0\ . \]

The equation for the Itô process $X_t$ can also be written in the shorter differential form:

\[ \rmd X_t = u\ \rmd t + v\ \rmd B_t \]

Theorem 2.1.2 (The $1$-dimensional Itô Formula)

Let $X_t$ be an Itô process given by

\[ \rmd X_t = u\ \rmd t + v\ \rmd B_t\ . \]

Let $g\in(t,x)\in C^2([0,\infty)\times\R)$, then

\[ Y_t = g(t,X_t) \]

is again an Itô process, and

\[ \rmd Y_t = \frac{\partial}{\partial t}g(t,X_t)\ \rmd t + \frac{\partial}{\partial x}g(t,X_t)\ \rmd X_t + \frac{1}{2}\frac{\partial^2}{\partial x^2}g(t,X_t)\cdot(\rmd X_t)^2\ , \]

where $(\rmd X_t)^2 = (\rmd X_t)\cdot (\rmd X_t)$ is computed according to the rules

\[ \begin{aligned} \rmd t\cdot\rmd t &= \rmd t\cdot \rmd B_t =\rmd B_t\cdot\rmd t=0\ ,\\ \rmd B_t\cdot \rmd B_t &= \rmd t\ . \end{aligned} \]

Actually it is enough that $g(t,x)$ is $C^2$ on $[0,\infty)\times U$ if $U\sube \R$ is an open set such that $X_t(\omega)\in U$ for all $t \geq 0,\omega \in \Omega$. Moreover, it is sufficient that $g(t,x)$ is $C^1$ with respect to $t$ and $C^2$ with respect to $x$.

Proposition 2.1.3:

Assume that the differential equation for $X_t$ is of the form

\[ \rmd X_t = u\ \rmd t + v\ \rmd B_t\ . \]

And $Y_t = g(t,X_t)$. Then we have

\[ \begin{aligned} g(t,X_t)= & Y_0 + \int_0^t\rmd Y_s \\ = & g(0,X_0)+\int_0^t \lr({ \frac{\partial}{\partial s}g(s,X_s)\ \rmd s + \frac{\partial}{\partial x}g(s,X_s)\ \rmd X_s + \frac{1}{2}\frac{\partial^2}{\partial x^2}g(s,X_s)\cdot(\rmd X_s)^2 })\\ = & g(0,X_0)+\int_0^t \frac{\partial}{\partial s}g(s,X_s)\ \rmd s + \int_0^t \frac{\partial}{\partial x}g(s,X_s)\ (u_s\ \rmd s + v_s\ \rmd B_s)+ \int_0^t \frac{1}{2}\frac{\partial^2}{\partial x^2} g(s,X_s)(u_s\ \rmd s + v_s\ \rmd B_s)^2\\ = & g(0,X_0)+\int_0^t \lr({ \frac{\partial}{\partial s}g(s,X_s) + u_s \cdot \frac{\partial}{\partial x} g(s,X_s) + \frac{1}{2}{v_s}^2\cdot\frac{\partial^2}{\partial x^2} g(s,X_s) })\ \rmd s + \int_0^t v_s \cdot \frac{\partial}{\partial x} g(s,X_s)\ \rmd B_s\ . \end{aligned} \]

And the differential form:

\[ \rmd Y_t = \lr({ \frac{\partial}{\partial t}g(t,X_t) + u_t \cdot \frac{\partial}{\partial x} g(t,X_t) + \frac{1}{2}{v_t}^2\cdot\frac{\partial^2}{\partial x^2} g(t,X_t) })\ \rmd t + v_t \cdot \frac{\partial}{\partial x} g(t,X_t)\ \rmd B_t \]

Example 2.1.4:

Let us return to the integral

\[ I = \int_0^t B_s \ \rmd B_s\ . \]

Choose $X_t = B_t$ and $Y_t = g(t,X_t) = \frac{1}{2}{B_t}^2$. Then by Itô formula,

\[ \rmd Y_t = 0 + B_t\ \rmd B_t + \frac{1}{2} (\rmd B_t)^2 = B_t\ \rmd B_t + \frac{1}{2}\ \rmd t\ . \]

It is corresponding to the result in the integral form:

\[ \int_0^t B_s\ \rmd B_s = \int_0^t \rmd Y_s - \int_0^t \frac{1}{2} \rmd s = Y_t-\frac{1}{2}t=\frac{1}{2}{B_t}^2-\frac{1}{2}t\ . \]

**Theorem (Integration by Parts) 2.1.5: **

Suppose $f(s,\omega) = f(s)$ only depends on $s$ and that $f$ is continuous and of bounded variation in $[0,t]$. Then

\[ \int_0^t f(s) \ \rmd B_s = f(t) B_t - \int_0^t B_s\ \rmd f_s\ . \]

Note that it is crucial for the result to hold that $f$ does not depend on $\omega$.

2.2 The Multi-dimensional Itô Formula

Definition 2.2.1:

Let $B(t,\omega) = (B_1(t,\omega),\cdots,B_m(t,\omega))$ denote $m$-dimensional Brownian motion. If each of the processes $u_i(t,\omega)$ and $v_{ij}(t,\omega)$ satisfies the conditions of Itô formula, then we can form the following $n$ Itô processes:

\[ \begin{cases} \rmd X_1 = & u_1\ \rmd t + v_{11}\ \rmd B_1 + \cdots + & v_{1m}\ \rmd B_m \\ \quad \vdots \quad & \vdots & \vdots \\ \rmd X_n = & u_n\ \rmd t + v_{n1}\ \rmd B_1 + \cdots + & v_{nm}\ \rmd B_m \end{cases} \]

Or, in matrix notation simply

\[ \rmd \bs{X}(t) = \bs{u}\ \rmd t + \bs{v}\ \rmd \bs{B}(t)\ , \]

where

$$ \bs{X}(t) = \begin{pmatrix} X_1(t) \ \vdots \ X_n(t) \end{pmatrix}\ ,\quad

\bs{u} = \begin{pmatrix} u_1 \ \vdots \ u_n \end{pmatrix}\ ,\quad

\bs{v} = \begin{pmatrix} v_{11} & \cdots & v_{1m} \ \vdots & & \vdots \ v_{n1} & \cdots & v_{nm} \end{pmatrix}\ ,\quad

\rmd \bs{B}(t) = \begin{pmatrix} \rmd B_1(t) \ \vdots \\rmd B_m(t) \end{pmatrix}\ . $$

Such a process $X(t)$ is called an $n$-dimensional Itô process.

Theorem 2.2.2 (The General Itô Formula):

Let

\[ \rmd \bs{X}(t) = \bs{u}\ \rmd t + \bs{v}\ \rmd \bs{B}(t) \]

be an $n$-dimensional Itô process as above. Let $\bs{g}(t,\bs{x}) = (g_1(t,\bs{x}),\cdots,g_p(t,\bs{x}))$ be an $C^2$ map from $[0,\infty) \times \R^n$ into $\R^p$. Then the process

\[ \bs{Y}(t,\omega) = g(t,\bs{X}(t)) \]

is again an Itô process, whose component number $k, Y_k$ is given by

\[ \rmd Y_k = \frac{\partial}{\partial t} g_k(t,\bs{X})\ \rmd t + \sum_i \frac{\partial}{\partial x_i} g_k(t,\bs{X})\ \rmd X_i + \frac{1}{2} \sum_{i,j} \frac{\partial^2}{\partial x_i\partial x_j} g_k(t,\bs{X})\ \rmd X_i \rmd X_j \quad , \]

where $\rmd B_i \rmd B_j = \delta_{ij}\ \rmd t$, and $\rmd B_i\rmd t = \rmd t \rmd B_i =\rmd t \rmd t = 0$.

**Example 2.2.3: **

Let $\bs{B} = (B_1, \cdots ,B_n)$ be Brownian motion in $\R^n$, $n \geq 2$, and consider

\[ \bs{R}(t,\omega) = |\bs{B}(t,\omega)| = ({B_1}^2(t,\omega) + {B_n}^2(t,\omega))^\frac{1}{2}\ . \]

Then we have

\[ \rmd \bs{R} = \sum_{i=1}^n \frac{B_i\ \rmd B_i}{\bs{R}} + \frac{n-1}{2\bs{R}}\ \rmd t\ . \]

The process $\bs{R}$ is called the $n$-dimensional Bessel process because its generator is the Bessel operator

\[ A f(x) = \frac{1}{2} f^{\prime\prime} (x) + \frac{n-1}{2x} f^\prime(x)\ . \]

See the following chapters.

2.3 The Martingale Representation Theorem

Let $\bs{B}(t) = (B_1(t),\cdots,B_n(t))$ be $n$-dimensional Brownian motion. In corollary 1.2.6 we have proved that if $\bs{v}\in\msc{V}^n$ then

\[ \bs{X}_t = \bs{X}_0 + \int_0^t \bs{v}(s,\omega)\ \rmd \bs{B}(s) \]

is always a martingale with respect to filtration $\msc{F}_t^{(n)}$. In this section we will prove that the converse is also true: Any $\msc{F}_t^{(n)}$-martingale can be presented as an Itô integral (The martingale representation theorem). For simplicity we prove the result only when $n=1$.

Lemma 2.3.1

Fix $T>0$. The set of random variables

\[ \lr\{{\phi(B_{t_1},\cdots,B_{t_n}) \ \big| \ t_i\in[0,T],\ \phi\in C_0^\infty(\R^n)}\} \]

is dense in $L^2(\msc{F}_t,\prob)$. By the term "$A$ is dense in $B$" we say that any element in $B$ can be approximated by a sequence in $A$.

Proof: Let $\set{t_i}_{i=1}^\infty$ be a dense subset of $[0,T]$ and for each $n$, let $\msc{H}_n$ be the $\sigma$-algebra generated by $B_{t_1}(\cdot),\cdots,B_{t_n}(\cdot)$. Then $\set{H_n}$ is a filtration. And $\msc{F}_T$ is the smallest $\sigma$-algebra containing all the $\msc{H}_n$'s. Choose an arbitrary function $g\in L^2(\msc{F}_T,\prob)$. Then by the martingale convergence theorem we have that

\[ g = \Exp[g|\msc{F}_T] = \lim_{n\to\infty} \Exp[g|\msc{H}_n]\ . \]

The limit is pointwise a.e.($\prob$) and in $L^2(\msc{F}_T,\prob)$. By the Doob-Dynkin Lemma we can write, for each $n$,

\[ E[g|\msc{H}_n] = g_n(B_{t_1},\cdots,B_{t_n}) \]

for some Borel measurable function $g_n:\R^n\to\R$. Each such $g_n(B_{t_1},\cdots,B_{t_n})$ can be approximated in $L^2(\msc{F}_T,\prob)$ by functions $\phi_n(B_{t_1},\cdots,B_{t_n})$ where $\phi_n\in C_0^\infty (\R^n)$.

Lemma 2.3.2:

The linear span of random variables of the type

\[ \exp\lr\{ { \int_0^T h(t)\ \rmd B_t(\omega) -\frac{1}{2}\int_0^T h^2(t)\ \rmd t }\}\ , \quad h \in L^2[0,T] \]

is dense in $L^2(\msc{F}_T,\prob)$.

Proof: Suppose $g\in L^2(\msc{F}_T,\prob)$ is orthogonal to all functions of that form. Then in particular

\[ G(\bs\lambda):= \int_\Omega \exp (\lambda_1 B_{t_1}+\cdots+\lambda_n B_{t_n})\cdot g(\omega)\ \rmd \prob(\omega) = 0 \]

for all $\bs{\lambda} = (\lambda_1,\cdots,\lambda_n)^\T \in \R^n$ and all $t_1,\cdots,t_n\in[0,T]$. The function $G(\bs\lambda)$ is real analytic in $\bs\lambda\in\R^n$ and hence $G$ has an analytic extension to the complex space $\C^n$ given by

\[ G(\bs{z}) = \int_\Omega \exp({z_1 B_{t_1}(\omega)+\cdots+z_n B_{t_n}(\omega)})\cdot g(\omega)\ \rmd\prob(\omega) \]

for all $\bs{z} = (z_1,\cdots,z_n)^\T \in \C^n$. Since $G = 0$ on $\R^n$ and $G$ is analytic, $G = 0$ on $\C^n$. And in particular,

\[ G(\rmi\bs{y})=G(\rmi y_1,\cdots, \rmi y_n) = 0 \]

for all $\bs{y} = (y_1,\cdots,y_n)^\T\in\R^n$. Now consider $\phi \in C_0^\infty (\R^n)$ and its Fourier transform, we have

\[ \begin{aligned} &\int_\Omega \phi(B_{t_1},\cdots,B_{t_n})g(\omega)\ \rmd \prob(\omega) \\ =& \int_\Omega \lr({\int_{\R^n}\frac{1}{(2\pi)^{n/2}} \cdot \hat{\phi}(\bs{y})\cdot\exp\lr({\rmi\cdot\sum_{k=1}^n y_k B_{t_k}})\rmd \bs{y}}) g(\omega)\ \rmd\prob(\omega)\\ =& \frac{1}{(2\pi)^{n/2}}\int_{\R^n}\hat{\phi}(\bs{y}) \lr({ \int_{\Omega} \exp\lr({\rmi\cdot\sum_{k=1}^n y_k B_{t_k}}) g(\omega) \rmd\prob(\omega) })\rmd \bs{y}\\ =& \frac{1}{(2\pi)^{n/2}}\int_{\R^n}\hat{\phi}(\bs{y})G(\rmi\bs{y})\ \rmd\bs{y} = 0\ . \end{aligned} \]

Therefore $g$ has to be $0$ if $g$ is orthogonal to a dense subset of $L^2(\msc{F}_T,\prob)$. Then the linear span of the function of that form in Lemma 2.3.2 must be dense in $L^2(\msc{F}_T,\prob)$ as claimed.

Theorem 2.3.3 (The Itô Representation Theorem):

Let $F\in L^2(\msc{F}_T,\prob)$. Then there exists a unique stochastic process $f(t,\omega)\in \msc{V}^n(0,T)$ such that

\[ F(\omega) = \Exp[F] + \int_0^T f(t,\omega)\ \rmd B(t)\ . \]

Proof: Consider the $1$-dimensional case. First assume that $F$ has the form in Lemma 2.3.2:

\[ F(\omega) = \exp\lr( { \int_0^T h(t)\ \rmd B_t(\omega) -\frac{1}{2}\int_0^T h^2(t)\ \rmd t })\ \]

for some $h(t)\in L^2[0,T]$. Define

\[ Y_t(\omega) = \exp\lr( { \int_0^t h(s)\ \rmd B_s(\omega) -\frac{1}{2}\int_0^t h^2(s)\ \rmd s })\ , \quad 0 \leq t \leq T\ . \]

Put $Y_t = g(t,X_t) = \exp(X_t)$. By Itô formula the differential form is.

$$ \begin{aligned} \rmd Y_t = &\ \lr({ \frac{\partial}{\partial t}g(t,X_t) + u_t \cdot \frac{\partial}{\partial x} g(t,X_t) + \frac{1}{2}{v_t}^2\cdot\frac{\partial^2}{\partial x^2} g(t,X_t) })\ \rmd t + v_t \cdot \frac{\partial}{\partial x} g(t,X_t)\ \rmd B_t\ =&\ \lr({0-\frac{1}{2}h^2(t)\cdot\exp(X_t)+\frac{1}{2}h^2(t)\cdot\exp(X_t) })\ \rmd t + h(t)\cdot\exp(X_t)\ \rmd B_t \ =&\ h(t)\cdot Y_t\ \rmd B_t\ .

\end{aligned} $$

So that

\[ Y_t = Y_0 + \int_0^t Y_s\cdot h(s)\ \rmd B_s\ , \quad t\in[0,T]\ . \]

Therefore

\[ F = Y_T = Y_0 + \int_0^T Y_s\cdot h(s)\ \rmd B_s \]

and hence $\Exp[F] = Y_0$. Thus the equation holds in this case. And by linearity and lemma 2.3.2 it also holds for arbitrary $F\in L^2(\msc{F}_T,\prob)$. Rigorous proof is omitted here. The uniqueness follows from the Itô isometry. Suppose

\[ F(\omega) = \Exp[F] + \int_0^T f_1(t,\omega)\ \rmd B_t(\omega) =\Exp[F] + \int_0^T f_2(t,\omega)\ \rmd B_t(\omega) \]

with $f_1, f_2\in\msc{V}(0,T)$. Then

\[ 0 =\Exp\lr[{\lr({\int_0^T (f_1-f_2)\ \rmd B_t(\omega)})^2}] = \Exp\lr[{\int_0^T(f_1-f_2)^2\ \rmd t}] = \int_0^T\Exp\lr[{(f_1-f_2)^2}]\ \rmd t \]

and therefore $f_1 = f_2$ for a.a.$(t,\omega)\in [0,T]\times \Omega$.

Theorem 2.3.4 (The Martingale Representation Theorem):

Let $\bs{B}(t) = (B_1(t),\cdots,B_n(t))^\T$ be $n$-dimensional Brownian motion. Suppose $\bs{M}_t$ is an $\msc{F}_T^{(n)}$-martingale and that $\bs{M}_t\in L^2(\prob)$ for all $t\geq 0$. Then there exists a unique stochastic process $\bs{g}(s,\omega)$ such that $\bs{g}\in\msc{V}^{(n)}(0,t)$ for all $t\geq 0$ and

\[ \bs{M}_t(\omega) = \Exp[\bs{M}_0] + \int_0^t \bs{g}(s,\omega) \ \rmd \bs{B}(s) \quad \text{a.s. for all $t\geq 0$ .} \]

Proof (1-dimensional): By Theorem 2.3.3 applied to $T=t$ and $F=M_t$, we have that for all $t$ there exists a unique $f^{(t)}(s,\omega)\in L^2(\msc{F}_t,\prob)$ such that

\[ M_t(\omega) = \Exp[M_t] + \int_0^t f^{(t)}(s,\omega)\ \rmd B_s (\omega) = \Exp[M_0] + \int_0^t f^{(t)}(s,\omega)\ \rmd B_s(\omega)\ . \]

Assume $0 \leq t_1 <t_2$. Then

\[ \begin{aligned} M_{t_1} =&\ \Exp [M_{t_2}|\msc{F}_{t_1}] \\ =&\ \Exp\lr[{M_0+\int_0^{t_2} f^{(t_2)}(s,\omega)\ \rmd B_s(\omega) \ \Big|\ \msc{F}_{t_1}}] \\ =&\ \Exp[M_0] + \int_0^{t_1} f^{(t_2)} (s,\omega) \ \rmd B_s(\omega)\ . \end{aligned} \]

While it is also true that

\[ M_{t_1} = \Exp[M_0] + \int_0^{t_1} f^{(t_1)}(s,\omega) \ \rmd B_s(\omega)\ . \]

Therefore, we have:

\[ \int_0^{t_1} (f^{(t_2)}(s,\omega)-f^{(t_1)(s,\omega)})\ \rmd B_s(\omega) = 0\\ \]

By Itô isometry, we get

\[ 0 = \Exp\lr[{\lr({\int_0^{t_1}(f^{(t_2)}-f^{(t_1)})\ \rmd B})^2}] = \Exp\lr[{\int_0^{t_1}(f^{(t_2)}-f^{(t_1)})^2\ \rmd B}] = \int_0^{t_1}\Exp\lr[{(f^{(t_2)}-f^{(t_1)})^2}]\ \rmd B\ , \]

which implies that $f^{(t_1)} = f^{(t_2)}$ for a.a.$(s,\omega)\in[0,t_1]\times\Omega$. Thus we can define $f(s,\omega)$ for a.a.$(s,\omega)\in[0,\infty)\times\Omega$ by setting

\[ f(s,\omega) = f^{(N)}(s,\omega)\quad \text{if } s\in[0,N] \]

and then we get

\[ M_t = \Exp[M_0]+\int_0^t f(s,\omega) \ \rmd B_s(\omega)\ ,\quad \forall t\geq 0. \]