3
$\begingroup$

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!

  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1871/discussion-between-ilya-and-steveo)2011-11-28

1 Answers 1

1

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