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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 $$f(x)=x^3/3$$. –  Flounderer Nov 28 '11 at 1:12
    
@Flounderer: Thanks! I hope to prove it by definition. –  steveO Nov 28 '11 at 1:26
    
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. –  Ilya Nov 28 '11 at 9:21
    
@Ilya: Yes, I should. Thanks for any idea. –  steveO Nov 28 '11 at 14:23
    
Could you calculate the expectation and variance of the expression you've obtained? –  Ilya Nov 28 '11 at 14:43
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1 Answer

use the following identity: $3\cdot B^{2}_{t_{i}}(B_{t_{i+1}}-B_{t_{i}})=B^{3}_{t_{i+1}}-B^{3}_{t_{i}}-(B_{t_{i+1}}-B_{t_{i}})^{3}-3\cdot B_{t_{i}}(B_{t_{i+1}}-B_{t_{i}})^{2}$.

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