Given $B_t$, $t \ge 0$ a Brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$, computer the covariance of the process $$X_t=\int_0^t \text{sgn}(Bs)dBs$$
It was recommended in the question that I use Itô's Isometry and the hint that $xy=\frac{1}{2}(||x+y||_2^2 -||x||_2^2 -||y||_2^2)$. I didn't use the isometry or the hint. Is my solution correct?
First I should convince myself that the mean of this thing is zero. $$E[X_t] = \int_0^t E[\text{sgn}(Bs)]dBs = 0$$ I am still a little loose on the best way to show this. My argument is that we look at the intervals of both cases, when $B_s \ge 0$ and when $B_s < 0$. In the first case we would have for every interval $[a,b]$ where $B_s \ge 0$ for every $s \in [a,b]$, then $\int_a^b E[B_s] dB_s = 0$, and in the later case a negative 1 multiplier doesn't change anything, $-\int_c^dE[B_s]dBs=0$. So in either case the mean is zero, so the mean is zero. Any way to better formalize this would be appreciated.
Assume without loss of generality that $s<t$, the covariance calculation reduces to $$E[X_sX_t] = E\left[\left( \int_0^s \text{sgn}(Bv)dBv \right) \left( \int_0^t \text{sgn}(Bw)dBw \right) \right]$$ $$= E\left[\left( \int_0^t \mathbb{1}_{v \le s}\text{sgn}(Bv)dBv \right) \left( \int_0^t \text{sgn}(Bw)dBw \right) \right]$$ $$= E\left[\left( \int_0^t (\mathbb{1}_{v \le s}\text{sgn}(Bw)) \text{sgn}(Bw)dBw \right) \right]$$ Since the indicator function returns zero for all $w > s$, we can reduce to the following. $$= E\left[\left( \int_0^s \text{sgn}(Bw)^2 dBw \right) \right]$$ the sign function squared is always one. $$= E\left[\left( \int_0^s 1 dBw \right) \right]=s$$
Since I didn't use the Isometry or the hint I think this must be wrong.
Update I went through the notes about this question, and the part that didn't occur to me is that in our text the Itô's isometry is given like this: $$E\left[ \left( \int_a^b f(\omega,u) dBu \right)^2 \right]=E\left( \int_a^b f^2(\omega,u)du \right)$$ And apparently, what the hint was hinting at is that you can also apply this to scalar products of stochastic integrals not just the square of one stochastic integral. If we assume this works, then the following step becomes simpler.
$$E\left[\left( \int_0^t \mathbb{1}_{v \le s}\text{sgn}(Bv)dBv \right) \left( \int_0^t \text{sgn}(Bw)dBw \right) \right]$$ $$=E\left[\left( \int_0^t \mathbb{1}_{v \le s}\text{sgn}(Bv) \text{sgn}(Bv)dv \right) \right]$$
Now, since it is a "normal" integral in $L^2$ then we can treat it as a sum, and put the expected value inside. Which simplifies to $$=\left( \int_0^s E[ \text{sgn}(Bv)^2 ] dv \right)=\left( \int_0^s 1 dv \right)=s$$