We want to attempt to define $\int_{S}^{T}f(t, \omega)dB_{t}(\omega)$ where > $B_{t}(\omega)$ denotes 1-dimensional Brownian motion.

Question

The following is from [1] and is attempting to define the integral of Brownian motion beginning with simpler functions.

Suppose that $0 \leq S < T$ and $f(t, \omega)$ is given. We want to attempt to define $\int_{S}^{T}f(t, \omega)dB_{t}(\omega)$ where $B_{t}(\omega)$ denotes 1-dimensional Brownian motion.

As an initial step suppose $f$ is of the form $\phi(t, \omega) = > \sum_{j \geq 0}e_{j}(\omega)\chi_{[j2^-{n}, (j+1)2^{-n}]}(t)$. Then define $\int_{S}^{T}\phi(t, \omega)dB_{t}(\omega) = \sum_{j \geq > 0}e_{j}(\omega)[B_{t_{j+1}} - B_{t_{j}}](\omega)$

\begin{equation} t_k=t_k^{(n)} = \left\{ > \begin{array}{lll} > k\cdot2^{-n} & \text{if }\quad S\leq k\cdot2^{-n}\leq T\\ > S &\text{if }\quad k\cdot2^{-n}<S\\ > T &\text{if }\quad k\cdot2^{-n}>T\\ > \end{array} \right\} \end{equation}

Example:

Choose $\phi_{2}(t, \omega) = \sum_{j \geq 0}B_{{(j+1)}2^{-n}}(\omega)\chi_{[j2^{-n},(j+1)2^{-n}]}(t)$. Then
$E[\int_{0}^{T}\phi_{2}(t,\omega)dB_{t}(\omega)] = \\ \sum_{j \geq 0}E[B_{t_{j+1}}(B_{t_{j+1}} - B_{t_{j}})] = \\ \text{How do we get the following} \sum_{j \geq 0}E[(B_{t_{j+1}} - B_{t_{j}})^{2}] = T$

How do we get the final sum and how does it equal $T$ in the example below?

[1] B. Oksendal, Stochastic Differential Equations, 5th ed.

Answer 1

In that piece of text, Oksendal demonstrates that two different choices of the step function $\phi$ which both look reasonable lead to very different expectations of the stochastic integrals.

• The choice $$ \phi_1(t)=\sum_{j\ge 1}B_{j2^{-n}}\cdot\chi_{\textstyle [j2^{-n},(j+1)2^{-n-1}\color{red}{)}}(t) $$ (your $\color{red}{]}$ is wrong here) leads to \begin{align} \mathbb E\Big[\int_0^T\phi_1(t)\,dB_t \Big]&=\sum_{j\ge 1}\mathbb E\Big[B_{t_j}(B_{t_{j+1}}-B_{t_j}) \Big]\\[2mm] &=\sum_{j\ge 1}\mathbb E\big[B_{t_j}\big]\mathbb E\Big[B_{t_{j+1}}-B_{t_j}\Big]\quad\text{(independent increments)}\\ &=\sum_{j\ge 1}\mathbb E\big[B_{t_j}\big]\cdot 0\quad\text{($N(0,t_{j+1}-t_j)$ distributed increments)}\\ &=0\,. \end{align}

• The choice $$ \phi_2(t)=\sum_{j\ge 1}B_{\color{red}{(j+1)}2^{-n}}\cdot\chi_{\textstyle [j2^{-n},(j+1)2^{-n-1})}(t) $$ leads to \begin{align} \mathbb E\Big[\int_0^T\phi_2(t)\,dB_t \Big]&=\sum_{j\ge 1}\mathbb E\Big[B_{t_\color{red}{j+1}}(B_{t_{j+1}}-B_{t_j} )\Big]\\ &=\sum_{j\ge 1}\mathbb E\Big[(B_{t_{j+1}}-B_{t_j})^2 \Big]=\sum_{j\ge 1}t_{j+1}-t_j=T \end{align} because for $s<t\,,$ $$ \mathbb E[B_t(B_t-B_s)]=\mathbb E[B_t(B_t-B_s)]-\underbrace{\mathbb E[B_s(B_t-B_s)]}_{\textstyle 0}=\mathbb E[(B_t-B_s)^2]\,. $$ (You know why that underbraced term is zero? Hint: remember the independent increments.)