convergence ito integral

Question

It is easy to calculate the integral $\int_0^T B_t \, dB_t=\frac{1}{2}B_T^2-\frac{1}{2}T$

That means I showed that $\int_0^T S_n \, dB_t=\sum_{t_i\in\Pi_n}S_{t_i}(B_{t_{i+1}}-B_{t_i})=\sum_{t_i\in\Pi_n}B_{t_i}(B_{t_{i+1}}-B_{t_i})$ converges to $\frac{1}{2}B_T^2-\frac{1}{2}T$ correct?

My question is, how can it be shown that $S_n$ converges to $B$ in the $H_2$ norm, if you know what I mean? I am no sure it its $H_2$ or $H^2$, something like that.

Answer 1

Recall that if $(\Omega,\mathscr F,\mathsf P)$ is a probability space then $$ L_2\text{-}\lim_nX_n = X \quad \Leftrightarrow \quad \lim_n\mathsf E\left[\left(X - X_n\right)^2\right] = 0. $$ You shall just plug in this definition $X_n = \int_0^T S_n\mathrm dB_t$ and $X = B^2_T - \frac12T$. And by the way, this is what we were told to do on our stochastic calculus class to prove the correctness of the formula for $\int B_s\mathrm dB_s$.