# I want to calculate $\int B(t)^2 dB(t)$ where $B(t)$ is Brownian motion

Let $B(t)$ be Brownian motion. I want to calculate $\int B(t)^2 dB(t)$.

definition.A process $\{X(t),0\le t \le T \}$ is called a simple adapted process if there exist times $0=t_{0}<t_{1}<t_{2}<\cdots<t_{n}=T$ and random variables $\eta_{0},\eta_{1},\cdots,\eta_{n}$ such that $\eta_{0}$ is a constant,$\eta_{i}$ is $\mathcal F_{i}$-measurable,For simple adapted processes Ito integral $\int X dB$ is defined as a sum $$\int_{0}^{T}X(t)dB(t)=\sum_{i=0}^{n-1}\eta_{i}(B(t_{i+1}-B(t_{i}))$$

• What are the integration limits? Also if I recall correctly you can find the answer to this question in any stochastic calculus textbook as an example of applying Ito's lemma. – mpiktas Nov 29 '13 at 7:03
• @mpiktas.if you give me a link or Introduce a book you help me so much – pual ambagher Nov 29 '13 at 7:11
• And what is $T$ ? – Stéphane Laurent Nov 29 '13 at 7:42
• Same question also posted on math.stackexchange.com, see math.stackexchange.com/questions/586534/…. – UwF Nov 30 '13 at 10:13
• It seems that all of @pualambagher 's questions are double-posted on math and stat-exchange. – Drew75 Nov 30 '13 at 13:04

Not an answer, at least not to your question, but an example of how to use Ito's formula (http://en.wikipedia.org/wiki/It%C5%8D_calculus).

Since $[B]_t=t$, we have

$$B_t^n = \int_0^t nB_s^{n-1}dB_s + \frac{n(n-1)}{2} \int_0^t B_s^{n-2} ds$$

for $n\ge 2$.

In particular,

$$B_t^3 = 3 \int_0^t B_s^2 dB_s + 3 \int_0^t B_sds$$

so that

$$\int_0^t B_s^2 dB_s = \frac{B_t^3}{3}-\int_0^t B_sds$$

Using

$$tB_t = \int_0^tsdB_s+\int_0^t B_sds$$

you can transform the term $\int_0^t B_sds$ into $tB_t - \int_0^tsdB_s$.