Stochastic Integrals – Integral of Square of Brownian Motion

Question

While trying to compute $\int_0^TB_t^2\ dB_t$, $B$ being the standard Brownian motion, I got stuck at showing the following.
$$\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{t_i})^2 \rightarrow \int_0^TB_t\ dt \quad \text{in} \quad L^2$$
as $mesh(\pi) \rightarrow 0$ where $\pi: 0 = t_0 < t_1 < \ldots < t_n = T$.

Since I have that
$$\sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i) \rightarrow \int_0^TB_t\ dt \quad \text{in} \quad L^2$$
I figured if I can show
$$\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{t_i})^2 \rightarrow \sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i) \quad \text{in} \quad L^2$$
I will have the result that I want. It is easy to see that
$$E\left[\sum_{i=0}^{n-1}B_{t_i}(B_{t_{i+1}}-B_{{t_i}})^2\right] = E\left[\sum_{i=0}^{n-1}B_{t_i}(t_{i+1}-t_i)\right]$$

But when I involve square of the difference of the sums to show convergence in $L^2$, things get out of hand (my hand at least). Could someone tell me if I am on the right track up until now? If yes, how do I proceed from this point on? I would appreciate any help/hint.

Answer 1

Steps:

1. It suffices to show that $$\left\| \sum_{i=0}^{n-1} B_{t_i} \bigg( (B_{t_{i+1}}-B_{t_i})^2 - (t_{i+1}-t_i) \bigg) \right\|_{L^2} \to 0.$$
2. Recall that $M_t := B_t^2-t$ is a martingale (with respect to the canonical filtration). For brevity of notation set $\Delta_i := B_{t_{i+1}}-B_{t_i}$ and $\Delta t_i := t_{i+1}-t_i$. Show that $$\mathbb{E}(B_{t_i} B_{t_j} \cdot (\Delta B_i^2 - \Delta t_i)(\Delta B_j^2- \Delta t_j))=0$$ whenever $i \neq j$.
3. Conclude from step 2 that $$\left\| \sum_{i=0}^{n-1} B_{t_i} \bigg( (B_{t_{i+1}}-B_{t_i})^2 - (t_{i+1}-t_i) \bigg) \right\|_{L^2}^2 = \sum_{i=0}^{n-1} \mathbb{E}( B_{t_i}^2 (\Delta B_i^2-\Delta t_i)^2.).$$
4. Use the independence of the increments and the fact that $\Delta B_i \sim B_{t_{i+1}-t_i} \sim \sqrt{t_{i+1}-t_i} B_1$ to conclude that the expression at the right-hand side (in step 3) converges to $0$.