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

Question

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 $$\int_{0}^{T}X(t)dB(t)=\sum_{i=0}^{n-1}\eta_{i}(B(t_{i+1}-B(t_{i}))$$

thanks for help

Answer 1

Itô's formula indicates that for every nonnegative deterministic $t$, $$ \int_0^tB_s^2\mathrm dB_s=\tfrac13B_t^3-\int_0^tB_s\mathrm ds=\tfrac13B_t^3-tB_t+\int_0^ts\mathrm dB_s. $$ To determine whether this "computes" the random variable on the LHS or not, one would need to know what it means to "compute a random variable" (the notion escapes me).