# Ito Integral Properties with Brownian Motion

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!

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

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 $f(x)=x^3$