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I want to prove $$ \int_0^T B_t^2 dB_t = \frac{B_T^3}{3} - \int_0^T B_t dt $$ by the definition of Ito integral.

I have tried this so far. Given a partition $0=t_0 < t_1 < ... < t_n=T$, I want to have $$ \sum_i B_{t_i}^2 (B_{t_{i+1}} - B_{t_i}) - \sum_i \frac{B_{t_{i+1}}^3 - B_{t_i}^3}{3} + \sum_i B_{t_i} (t_{i+1} - t_i) \to 0 $$ as the partition becomes finer and finer.

But I am stuck here. How shall I proceed? Thanks a lot!

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    It looks like it follows from a direct application of the Ito-Doeblin formula $$f(B_T)-f(B_0) = \int_0^T f'(B_t) dB_t + (1/2) \int_0^T f''(B_t) dt$$ with $$f(x)=x^3/3$$.2011-11-28
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    @Flounderer: Thanks! I hope to prove it by definition.2011-11-28
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    Do you mean that you should prove it by definition? Otherwise, it is a very simple example for the application of Ito formula and @Flounderer told you.2011-11-28
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    @Ilya: Yes, I should. Thanks for any idea.2011-11-28
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    Could you calculate the expectation and variance of the expression you've obtained?2011-11-28
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    @Ilya: Thank you! For the LHS, its expectation is 0 and its variation is complicated. So what is the idea here?2011-11-28
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    The only method we used to prove result of Ito integration be definition - verifying that expectation is zero and variance tends to zero with $\max\Delta t_i\to 0$.2011-11-28
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    @Ilya: Thank you! By Ito integral definition, the convergence is in terms of $L^2$ norm. Is it equivalent to expectation cvging to 0 and variance also cvging to 0?2011-11-28
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1871/discussion-between-ilya-and-steveo)2011-11-28

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