Ito Integral

Let $(\Omega,\mathcal{F},P)$ be a probability space. We first define a stochastic process.

A stochastic process is a measurable map from $[0,\infty) \times \Omega \to \mathbb{R}$.

Therefore, if $X(t,\omega)$ is a stochastic process, $X^{-1}(B) \in \mathcal{B}[0,\infty)\times \mathcal{F}$ for every Borel set $B \in \mathcal{B}(\mathbb{R})$. Hence, whenever we say "stochastic process" we automatically assume measurability.

We also have a notion of adaptedness w.r.t. increasing $\sigma$-fields (also known as a filtration). Filtrations are useful in giving mathematical meaning to phrases such as "less information"/"more information" etc.

A stochastic process $X$ is said to be adapted to a increasing family of $\sigma$-fields $\mathcal{F}_t$, if $X(t,\cdot)$ is $\mathcal{F}_t$-measurable

A more stronger notion than the joint measurability required in the definition of a stochastic process is that of progressive measurability.

A stochastic process $X(t,\omega) : [0,\infty) \times \Omega \to \mathbb{R}$ is said to be progressively measurable if the restricted map $X|_t : [0,t] \times \Omega \to \mathbb{R}$ of $X$ is measurable w.r.t $\mathcal{B}[0,t] \times \mathcal{F}_t$ for every $t \in \mathbb{R}$

A process is continuous if the sample paths, i.e. the (random) functions $X(\cdot,\omega)$ are continuous almost surely (in $\omega$). Left/Right-continuous processes are defined similarly.

Progressive measurability implies i) measurability and ii) adaptedness.

As $X$ is progressively measurable, we have, for every Borel set $B\in \mathcal{B}(\mathbb{R})$, $(X|_t)^{-1} (B) \in \mathcal{B}[0,t] \times \mathcal{F}_t$. We also have $X^{-1}(B) = \cup_{t=1}^\infty (X|_t)^{-1} (B)$. To prove measurability, it suffices to note that $\mathcal{B}[0,t] \times \mathcal{F}_t \subset \mathcal{B}[0,\infty) \times \mathcal{F}$ for every $t$. As $((X)^{-1}(B))_t = ((X|_t)^{-1}(B))_t \in \mathcal{F}_t $ (where the second subscript denotes the cross-section of the set) we get adaptedness.

Adaptedness and right-continuity imply progressive measurability. The proof can be found in Kallianpur and is not too difficult. This is stated as a lemma next.

Let $X$ be a stochastic process, which is $\mathcal{F}_t$ adapted and right-continuous. Then $X$ is progressively measurable.

More important is the following theorem which generalizes the above proposition (when the assumption of right continuity is dropped). Before, this theorem is given, the notion of modification of a process is defined.

Modification: $Y$ is a modification of $X$ if $P(Y(t,\cdot) = X(t,\cdot))$ for each $t$.

Let $X$ be an $\mathcal{F}_t$-adapted stochastic process. Then there exists a modification of $X$ which is progressively measurable.

The proof of this can be found in P.A. Meyer and is given here.

Now, basic ideas about sections of sets are given. Let $(\Omega_1,\Sigma_1,\mu_1), (\Omega_2,\Sigma_2,\mu_2)$ be two finite measure spaces. Let the product space be denoted by $(\Omega,\Sigma)$ where $\Omega =\Omega_1 \times \Omega_2$ and $\Sigma = \Sigma_1 \times \Sigma_2$, the product $\sigma$-albegra.

Section of a set : Let $E\in \Sigma$. Then the section of $E$ at $\omega_1$ (denoted by $(E)_{\omega_1})$ is defined as $(E)_{\omega_1} = \{\omega_2 | (\omega_1,\omega_2) \in E\}$

Suppose we define a new function $\Omega_1 \to \mathbb{R}$ by $\omega_1 \to \mu_2((E)_{\omega_1})$ for some fixed $E \in \Sigma$. Is this function measurable? It is not directly obvious but the proof given next shows that this is the case.

Let $\mathcal{D} = \{E \subset \Omega| \omega_1 \to \mu_2((E)_{\omega_1}) \text{ is measurable }\}$. The following can now be observed

• If $E = E_1\times E_2$ (called a rectangle) where $E_1 \in \Sigma_1, E_2 \in \Sigma_2$ then $\mu_2((E)_{\omega_1}) = \mu_2(E_2)I_{E_1}(\omega_1)$, which is measurable.
• Let $E^1, E^2$ be two rectangles then it can be seen that $E^1 \cap E^2$ belongs to $\mathcal{D}$ and that $(E^1)^c$ is also in $\mathcal{D}$
• If $E^i \in \mathcal{D}$ and $E^i \uparrow$ then using monotonicity of the measure it is clear that $\cup E^i \in \mathcal{D}$. Similalry, $\cap E^i \in \mathcal{D}$ if $E^i \downarrow$

Above, we have shown that $\mathcal{D}$ is a monotone class and that it contains a field which generates the product $\sigma$-field $\Sigma$. Now, we use the fact that the smallest monotone class and the smallest $\sigma$-field containing a given field are the same. Therefore, $\mathcal{D}$ contains all measurable sets.

We now proceed with the construction of the Ito integral. This integral is developed in much the same way as the Lebesgue integral, for example, i.e. defining the integral for "simple" functions first and then extending the definition for the general case.

Before, we start it is important that we already have constructed a Brownian Motion, B, on some complete probability space $(\Omega,\mathcal{F},P)$. Some of the simplest constructions are given in 1) Brownian Motion by Peres and Morters and 2) Stochastic Filtering Theory by G. Kallianpur. Let $\mathcal{F}_t$ be the smallest $\sigma$-field containing $\sigma(B_s, 0\leq s\leq t)$ and all the $P$-null sets of $\mathcal{F}$. (Note: $\mathcal{F}_t = \{A \cup B | A \in \sigma(B_s, 0\leq s\leq t), B \text{ a P-null set of } \mathcal{F}\}$). Adding the $P$-null sets of $\mathcal{F}$ does not disturb the properties of the Brownian Motion, i.e., $B_s - B_t$ is still independent of $\mathcal{F}_t$.

This following is based on the texts by Bernt Oksendal, Kartazas Shreve and attempts to give any missing details

Let $\mathcal{V} = \mathcal{V}(S,T)$ be the class of functions $f(t,\omega) : [0,\infty) \times \Omega \to \mathbb{R}$ which satisfy

• $(t,\omega) \to f(t,\omega)$ is $\mathcal{B}[0,\infty) \times \mathcal{F}$ measurable.
• $f(t,\omega)$ is $\mathcal{F}_t$-adapted
• $E(\int_S^T f(t,\omega)^2 dt) < \infty$

The integral is defined in steps:

Step 1 : Bounded Step Processes :- First, the integral is defined for step processes. Let $S = t_0 < t_2 < \cdots < t_n = T$ and $e_j : \Omega \to \mathbb{R}$ be $\mathcal{F}_{t_j}$-measurable and bounded. Now, define $\phi(t,\omega) = \Sigma_{j=0}^{n-1} e_j(\omega)\chi_{[t_j,t_{j+1})}(t)$. It needs to be checked if the process $\phi(t,\omega)$ satisfies the requirements of the definition above

• $\phi^{-1}(B)= \cup_{j=0}^{n-1} \left ([t_j, t_{j+1})\times e_j^{-1}(B) \right) \in \mathcal{B}[0,\infty) \times \mathcal{F}$ for every Borel set $B \in \mathcal{B}(\mathbb{R})$.
• If $t\in [t_j,t_{j+1})$ then $\phi(t,\omega) = e_j(\omega)$. $e_j$ is $\mathcal{F}_{t_j}$-measurable and hence $\mathcal{F}_t$-measurable.
• As $\phi(t,\omega)$ is bounded, the expectation $E(\int_S^T \phi(t,\omega)^2 dt)$ is finite.

The stochastic integral for these "step processes" $\phi(t,\omega)$ is defined as \[\int_S^T\phi(t,\omega)dB_t(\omega) = \sum_{j=0}^ne_j(\omega) (B_{t_{j+1}}(\omega) - B_{t_j}(\omega))\] The first thing to note here is that the stochastic integral is a random variable.

Ito Isometry for bounded step processes : \[E\left[\left(\int_S^T\phi(t,\omega)dB_t(\omega)\right)^2\right] = E\left(\int_S^T \phi(t,\omega)^2dt\right)\]

The proof is straight-forward and uses independence of increments of Brownian motion and the fact that $e_j$ is independent of $B_{t_{j+1}} - B_{t_j}$. This Lemma is used in later steps to show uniqueness of the integral.

The following proposition is easy to see.

The integral for bounded step-processes is Linear, i.e., if $\phi(t,\omega), \eta(t,\omega)$ are two bounded step processes then $\int_S^T(\phi(t,\omega)+\eta(t,\omega))dB_t(\omega) = \int_S^T\phi(t,\omega)dB_t(\omega) +\int_S^T\eta(t,\omega)dB_t(\omega)$

Step 2: Bounded Continuous Processes : Now, the integral is defined for bounded, continuous processes.

Let $g\in \mathcal{V}$ be bounded and $g(\cdot,\omega)$ continuous for each $\omega$. Then, there exist bounded step processes $\phi_n \in \mathcal{V}$ such that \[E\left(\int_S^T (g-\phi_n)^2dt\right) \to 0~\text{as } n \to \infty\]

Let $\phi_n(t,\omega) = \sum_{j=0}^{\lfloor (T-S)(2^n) \rfloor -1}g(S+\frac{j}{2^n},\omega)\chi_{[S+\frac{j}{2^n}, S+\frac{j+1}{2^n})}(t)$. As the dyadic rationals are dense in $\mathbb{R}$ and $g$ is continuous we get that $\phi_n(t,\omega) \to g(t,\omega)$ as $n \to \infty$ for every $t \in [S,T)$ and $\omega \in \Omega$. Therefore, $(\phi_n(t,\omega) - g(t,\omega))^2 \to 0$ as $n \to \infty$. As $g$ and $\phi_n$ are bounded processes, using dominated convergence theorem, we get \[\int_S^T (g(t,\omega)-\phi_n(t,\omega))^2 dt \to 0 \text{ as } n \to \infty\] Now,$\int_S^T (g(t,\omega)-\phi_n(t,\omega))^2 dt$ is uniformly bounded for each $n$ and therefore using dominated convergence once again, we get that \[E\left(\int_S^T (g(t,\omega)-\phi_n(t,\omega))^2 dt\right) \to 0 \text{ as } n \to \infty\]

$\{\int_S^T \phi_n(t,\omega)dB_t(\omega)\}$ form a Cauchy sequence in $\mathcal{L}^2(\Omega,\mathcal{F},P)$

\begin{array}\\ E\left(\int_S^T \phi_n(t,\omega)dB_t(\omega) - \int_S^T \phi_m(t,\omega)dB_t(\omega)\right)^2 &= E\left(\int_S^T (\phi_n(t,\omega)- \phi_m(t,\omega))dB_t(\omega)\right)^2 \\ &= E\left(\int_S^T (\phi_n(t,\omega)- \phi_m(t,\omega))^2dt\right) \end{array} The first equality is due to linearity of the integral for bounded step processes. The second equality is because of Ito isometry proved above. By adding and subtracting $g(t,\omega)$ in the integral, it can be seen that the left hand quantity above can be made arbitrarily small by choosing large enough $n,m$.

As the sequence $\{\int_S^T \phi_n(t,\omega)dB_t(\omega)\}$ is cauchy, there exists a subsequence which converges almost surely (Note: this is the main part in the proof that $\mathcal{L}^2$ is complete. See, Probability with Martingales by David Williams, Page 65). This subsequence can infact be constructed. Let the limit of this subsequence be $I_g(\omega)$. Then we have $E(I_g(\omega) - \int_S^T \phi_n(t,\omega)dB_t(\omega))^2 \to 0$ as $n \to \infty$.

The integral for bounded, continuous processes is then defined as \[\int_S^Tg(t,\omega)dB_t(\omega) = I_g(\omega) = \lim_{n \to \infty}\int_S^T \phi_n(t,\omega)dB_t(\omega)\text{ (limit in } \mathcal{L}^2(\Omega,\mathcal{F},P))\]

Step 3: Bounded Progressively Measurable Processes

Suppose we have a bounded progressively measurable process, $f : [0,\infty) \times \Omega \to \mathbb{R}$ and define $F(t,\omega) = \int_0^t f(s,\omega)ds, 0\leq t$. Then, $F(t,\omega)$ is adapted and continuous (and hence progressively measurable)

Let \[f_m(s,\omega) = \sum_{j=-m2^m+1}^{m2^m} \left(\frac{j-1}{2^m}\right) \chi_{[\frac{j-1}{2^m},\frac{j}{2^m})}(f(s,\omega))\] We have $f_m(s,\omega) \to f(s,\omega)$ as $m \to \infty$. Hence, it suffices to show that $\int_0^t f_m(s,\omega)ds$ is progressively measurable for each $m$ (as $\int_0^t f_m(s,\omega)ds \to \int_0^t f(s,\omega)ds$ due to boundedness). Then, \[\int_0^t f_m(s,\omega)ds = \sum_{j=-m2^m+1}^{m2^m} \left(\frac{j-1}{2^m}\right) \lambda\left(\left(f^{-1}\left[\frac{j-1}{2^m}, \frac{j}{2^m}\right)\right)_\omega\cap [0,t]\right)\] In the above, $\left(\left(f^{-1}\left[\frac{j-1}{2^m}, \frac{j}{2^m}\right)\right)_\omega\cap [0,t]\right)$ equals $\left((f|_t)^{-1}\left[\frac{j-1}{2^m}, \frac{j}{2^m}\right)\right)_\omega \in \mathcal{B}[0,t] \times \mathcal{F}_t$ due to the progressive measurability of $f$. The map $\Omega \to \mathbb{R}$, given by $\omega \to \lambda(E_\omega)$ is $\mathcal{F}_t$ measurable, if $ E \in \mathcal{B}[0,t] \times \mathcal{F}_t$. This can be shown similarly to the method described before. Hence, $\int_0^t f_m(s,\omega)ds$ and therefore $\int_0^t f(s,\omega)ds$ is adapted. Continuity is stratight forward due to boundedness.

Now, let $f(t,\omega)$ be a bounded progressively measurable process and define $F(t,\omega) = \int_0^t f(s,\omega)ds$ as before. Also let $f_m(t,\omega) =(F(t,\omega) - F(\max((t-\frac{1}{m}),0),\omega))/(1/m)$. Then, $f_m \in \mathcal{V}$ and continuous. Therefore, by the previous step we have a sequence of bounded step processes $f_{m,n}$ such that \[E\int_S^T |f_m(t,\omega)-f_{m,n}(t,\omega)|^2dt \to 0\text{ as } n \to \infty\]

By the fundamental theorem of calculus $F^\prime(t,\omega) = f(t,\omega)$ a.e. w.r.t. the Lebesgue measure on $[0,\infty)$. However, by definition $F^\prime(t,\omega) = \lim_{m \to \infty} f_m(t,\omega)$. This gives \[E\int_S^T |f(t,\omega)-f_{m,n}(t,\omega)|^2dt \to 0\text{ as } m,n \to \infty\]

Step 4 : Bounded, Measurable and Adapted process

In this case, the process $F(t,\omega)$ may not be adapted (This is because $\left(\left(f^{-1}\left[\frac{j-1}{2^m}, \frac{j}{2^m}\right)\right)_\omega\cap [0,t]\right)$ might not have a nice representation as we had in the progressively measurable case. However, we know that there exists a bounded progressively measurable modification $g(t,\omega)$ of $f(t,\omega)$. (The proof of this statement can be found in Probabilty and Potentials by P.A. Meyer. A post on this can be found here. This is also proposition 1.1.4. in Gopinath Kallianpur's text). From the previous step, we then have $g_n(t,\omega)$ such that $E\int_S^T |g(t,\omega)-g_{n}(t,\omega)|^2dt \to 0\text{ as } n \to \infty$. As $g,f$ are modifications of each other, we get $E\int_S^T |f(t,\omega)-g_{n}(t,\omega)|^2dt \to 0\text{ as } n \to \infty$.

Step 5 : $f\in \mathcal{V}$ . Let $h_n(t,\omega) = (-n)1_{f(t,\omega) < -n} + f(t,\omega)1_{-n < f(t,\omega) \leq n} + (n)1_{f(t,\omega) > n}$. We have $h_n(t,\omega) \to f(t,\omega)$ and therefore $E\int_S^T (f-h_n)^2 \to 0$ as $n \to \infty$ from dominated convergence theorem. As each $h_n$ is a bounded, measurable and adapted process, from the previous step it can be approximated by a sequence of bounded step processes. Therefore, the same holds true for $f$.