Quadratic Variation for $X_t= \int \sigma_s dW_s$ where $\sigma_s \in S$

Question

Let $\sigma_s \in S$. Setting $X_t=\int^t_0 \sigma_s dW_s$ and partitioning the interval $[0,t]$ i.e. $0=t^n_0<t^n_1... $ such that $d_n=\max_i |t^n_{i+1}-t^n_i| \rightarrow 0$ as $n \rightarrow \infty$, prove that

$\sum_i |X_{t^n_{i+1}} - X_{t^n_{i}} |^2 \rightarrow \int^t_0 \sigma^2_s ds$ (a.s)

Heres what I've got so far, I was wondering if you guys agreed with it and could help me finish solving this:

• $|X_{t_{i+1}} - X_{t_{i}} |^2 = |\int^{t_{i+1}}_{t_{i}} \sigma_s dW_s|^2$
• $E|X_{t^n_{i+1}} - X_{t^n_{i}} |^2$ via Isometry is $ \int^{t^n_{i+1}}_{t^n_{i}} E\sigma_s^2 ds$

I'm not so sure how the second bullet point can help me though....

Answer 1

You're on the right track.

You need to compute the first and second moments.

Use your ito isometry expression to get an expression for the first moment of the sum.

Then you need to compute the second central moment. This will be an expression involving $|\Delta X|^4$. If you show that the second central moment vanishes in the limit, you'll be done.