Expectation of this stochastic process Let a stochastic process $X(t)= \int_0^t \operatorname{sign}(B(s)) \, dB(s)$, now how to show that $\Bbb E[B(t)X(t)]=0$ ?
here $\operatorname{sign}(x)=-1$ for $x<0$, and $1$ otherwise. $B(t)$ is the Wiener process. please gives more details in calculation due to my limited knowledge in this area.
I've got the idea to use Ito's product rule to derive $d[B(t)X(t)]$ and then we get $B(t)X(t)$ and then take the expectation, but I am not sure why we have to use the intergation of $dX(t)\,B(t)$ to get the expectation. 
 A: Hint: Replace all occurrences of $B(\cdot)$ by $-B(\cdot)$ in $\mathbb{E}[B(t)X(t)]$.
A: I really like @ByronSchmulands approach, but since the OP suggested to prove the equality using Itô's formula, I'll outline this alternative approach:
For any two Itô processes of the form $\DeclareMathOperator \sgn{sgn}$
$$dX_t = b_1(t) \, dt + \sigma_1(t) \, dB_t$$
and
$$dY_t = b_2(t) \, dt + \sigma_2(t) \, dB_t$$
it holds by Itô's formula that
$$d(X_t Y_t) = Y_t \, dX_t+ X_t , dY_t + d \langle X,Y \rangle_t$$
where $$\langle X,Y \rangle_t = \int_0^t \sigma_1(s) \sigma_2(s) \, ds.$$
Applying this to $dX_t = \sgn(B_t) \, dB_t$ (i.e. $b_2=0$, $\sigma_2(t)=\sgn(B_t)$) and $dY_t := dB_t$ (i.e. $b_2=0$, $\sigma_2=1$) yields
$$d(X_t Y_t) = B_t \sgn(B_t) \, dB_t + \left(\int_0^t \sgn(B_s) \, dB_s\right) \, dB_t + \sgn(B_t) \, dt.$$
Since the first two terms are stochastic integrals, hence martingales, its expectation equals $0$. Therefore,
$$\mathbb{E}(X_t Y_t) = \int_0^t \mathbb{E}(\sgn(B_s)) \, ds.$$
As $$\mathbb{E}(\sgn(B_s)) = 1 \cdot \mathbb{P}(B_s \geq 0) -1 \cdot \mathbb{P}(B_s<0) = 0$$ the claim follows.
