Quadratic variation of Ito integral : Relationship between Ito Integral and Brownian Motion

Question

This is theorem 4.2.3 from Shreve's Stochastic Calculus for finance II.

Statement of the theorem:

The quadratic variation accumulated up to time $t$ by the Ito integral is: $$ [ I,I](t) = \int_{0}^{t} \Delta^{2}(u) du.$$

The proof revolves around computing quadratic variation accumulated by the Ito integral on one of the subintervals $[t_j, t_{j+1}]$ on which $\Delta(u)$ is constant. They choose the partition points:

$$t_{j} = s_{0} < s_{1} < \ldots < s_{m} = t_{j+1} $$

and compute

$$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2 $$

I can't figure out how we got the aforementioned equality involving Ito integral and Brownian Motion. Any help appreciated.

Here we have the following definitions: if $t_{k} \leq t \leq t_{k+1}$ then $$I(t) = \sum_{j=0}^{k-1}\Delta(t_{j})[W(t_{j+1} - W(t_j)] + \Delta(t_{k})[W(t) - W(t_{k})]$$ The process $I(t)$ is the Ito integral of the simple process $\Delta(t)$ that is written as $I(t) = \int_{0}^{t}\Delta(u)dW(u).$

Suppose $\Delta(t)$ is an adapted stochastic process i.e. it is $\mathcal{F}(t)$ measurable for each $t \geq 0.$ Assume further that $\Delta(t)$ is constant in $t$ on each subinterval $[t_{j}, t_{j+1}).$ Then $\Delta(t)$ is a simple process.

Here u is just the dummy variable and W(t) is the Brownian Motion.

Answer 1

It's not clear if $\Delta (u)$ is deterministic or not. I'll suppose it is, and if it's not, I let you adapt the proof. I denote $\sum_{j}$ for $\sum_{j=0}^{m-1}$. Then,

\begin{align*} \mathbb E\left[\left(\sum_{j}(I(t_{j+1})-I(t_j))^2-\int_{0}^t\Delta (u)^2\,\mathrm d u\right)^2\right]&=\mathbb E\left[\left(\sum_{j}\big(\Delta (t_j)(W_{t_{j+1}}-W_{t_j})\big)^2-\Delta (t_j)^2(t_{j+1}-t_j)\right)^2\right]\\ &=\mathbb E\left[\left(\sum_{j}\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)\right)^2\right]\\ &=\sum_{i,j}\mathbb E[\Delta (t_i)^2\big((W_{t_{i+1}}-W_{t_i})^2-(t_{i+1}-t_i)\big)\Delta (t_j)^2\big((W_{t_{j+1}}-W_{t_j})^2-(t_{j+1}-t_j)\big)]\\ &\underset{(*)}{=}2\sum_{i}\Delta (t_i)^4(t_{i+1}-t_i)^2\\ &\leq 2\max_{i}|t_{i+1}-t_i|\int_0^t\Delta (u)^4\,\mathrm d u\underset{m\to \infty }{\longrightarrow }0, \end{align*} where in $(*)$ we used the fact that Brownian motion has independent increments and that if $X\sim \mathcal N(0,1)$, then $\mathbb E[X^4]=3$.

Answer 2

Following the definitions, we have $$I(s_{i+1})=I(t_j)+ \int_{t_j}^{s_{i+1}} \Delta(t_j)dW(u) = I(t_j)+\Delta(t_j)(W(s_{i+1})-W(t_j)).$$ Similarly $$I(s_i)=I(t_j)+\Delta(t_j)(W(s_i)-W(t_j)).$$ Thus $$I(s_{i+1})-I(s_i)=\Delta(t_j)(W(s_{i+1})-W(s_i))$$ and therefore $$\sum_{i=0}^{m-1}[I(s_{i+1}) - I(s_{i})]^2 = \sum_{i=0}^{m-1}[\Delta(t_j)(W(s_{i+1}) - W(s_{i}))]^2. $$

EDIT: In the first expression I am using the linearity property of the Ito integral. Remember that $$t_j = s_0 < s_1 < ... < s_m = t_{j+1},$$ so \begin{align} I(s_{i+1})=\int_0^{s_{i+1}} \Delta(u)dW(u)&=\int_0^{t_j} \Delta(u)dW(u)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u) \\ &=I(t_j)+\int_{t_j}^{s_{i+1}} \Delta(u)dW(u) \end{align}