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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!

SBF
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Tim
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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 01:12
  • @Flounderer: Thanks! I hope to prove it by definition. – Tim Nov 28 '11 at 01:26
  • 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. – SBF Nov 28 '11 at 09:21
  • @Ilya: Yes, I should. Thanks for any idea. – Tim Nov 28 '11 at 14:23
  • Could you calculate the expectation and variance of the expression you've obtained? – SBF Nov 28 '11 at 14:43
  • @Ilya: Thank you! For the LHS, its expectation is 0 and its variation is complicated. So what is the idea here? – Tim Nov 28 '11 at 14:47
  • The only method we used to prove result of Ito integration be definition - verifying that expectation is zero and variance tends to zero with $\max\Delta t_i\to 0$. – SBF Nov 28 '11 at 14:59
  • @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? – Tim Nov 28 '11 at 15:12
  • let us continue this discussion in chat – SBF Nov 28 '11 at 15:32

1 Answers1

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

Peter Moor
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