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I am working out some of the properties for the Ito integral with Brownian motion and I am trying to use the definition to verify that $$ \int _0 ^t s \, dB_s = tB_t - \int _0 ^t B_s\, ds $$ and

$$ \int _0 ^t B_s^2 \, dB_s = \frac{1}{3}(B_t^3) - \int _0 ^t B_s\, ds $$

I am having trouble thinking about how to attack this so any help or suggestions on how to get started would be great!

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  • $\begingroup$ I'm confused: if you subtract your two equations you get: $B_t^2/3 = tB_t$. $\endgroup$ – muaddib Jun 10 '15 at 23:12
  • $\begingroup$ sorry typo, should be $$ B_t^3 $$. ill fix now $\endgroup$ – EE_13 Jun 10 '15 at 23:31
  • $\begingroup$ it still doesn't look right. $\endgroup$ – ZanCoul Jun 11 '15 at 0:00
  • $\begingroup$ Really sorry for the typos. I am new to uses latex type text. I have verified that it is now fully correct! $\endgroup$ – EE_13 Jun 11 '15 at 0:07
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EDITED to meet edit of question

The first equation is (after the edit) true. Consider the twodimensional continuous semimartingale $\left( t,B_t\right)$, and function $f(x,y)=xy$ we get $$D_xf(x,y)=y\quad D_yf(x,y)=x\quad D_1D_1f=D_2D_2f=0\quad D_1D_2f=D_2D_1f=1$$ And therefore ITO's formula gives $$tB_t=0+\int_0^t s\; dB_s+\int_0^t B_s\; ds.$$

The 2nd equation is (after the edit) true as well. You can convince yourself with an argument similar to the one i gave. Apply the 1-dimensional ITO's formula on $B_t$ with $f(x)=x^3$

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  • $\begingroup$ Sorry! thank you for your answer. I have fixed the typos in the original question. $\endgroup$ – EE_13 Jun 11 '15 at 0:08
  • $\begingroup$ @user3692250 No problem. If you have further questions on this topic, write them in the comments. If you do not, mark the answer as an answer. $\endgroup$ – Conformal Jun 11 '15 at 0:21
  • $\begingroup$ I understand your solution but do you know how to prove those properties from a general standpoint? Maybe just using the definition of the Ito integral. Thanks! $\endgroup$ – EE_13 Jun 11 '15 at 2:02
  • $\begingroup$ I don't think you can prove these equalities only using the definition of the Ito integral. Well in some sense you can: derive Ito's formula from the definition of the Ito integral, then use it. $\endgroup$ – Martin Jun 11 '15 at 15:45
  • $\begingroup$ Yes, I have realized that you have to use Ito's formula in some way so the above answer is correct. thanks! $\endgroup$ – EE_13 Jun 11 '15 at 15:57

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