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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.

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What is $S_n$? $ $ – Did Jan 14 '13 at 21:31
I think it is some simple process, that approxiamtes Brownian Motion (A kind of step function) – Alexander Jan 14 '13 at 21:34
Yes, sorry I forgot to say that, S_n is some simple process – Voyage Jan 14 '13 at 21:35
Which simple process? Since choosing any simple process will not do, the statement of the question is incomplete. – Did Jan 16 '13 at 20:19

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 $\int B_s\mathrm dB_s$.

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Thanks, with E[X]^2 you mean E[X]*E[X], correct? Can i simply put the E inside the integral and say E[B_t-S_n]=0 ? – Voyage Jan 15 '13 at 12:39
@Voyage: no, I rather meant that the square is under the expectation - made it more clear now, take a look on the edited version. – Ilya Jan 15 '13 at 12:53
Ok, I have one more problem: Why should $\lim\mathbb E\int_0^TS_ndB_t\int_0^TS_ndB_t=B_T^2-\frac{1}{2}T$ – Voyage Jan 15 '13 at 13:10
@Voyage: only in limit. write it as an expectation of the product of two sums – Ilya Jan 15 '13 at 16:08

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