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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 $$=\left( \int_0^s E[ \text{sgn}(Bv)^2 ] dv \right)=\left( \int_0^s 1 dv \right)=s$$

jeffery_the_wind
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    This is perhaps the smallest issue, but if $Y_t = \int_0^t 1, dB_s$, can we conclude $\mathbb{E}[Y_t] = \int_0^t \mathbb{E}[1],dB_s$? – Brian Moehring Feb 20 '21 at 23:02
  • @BrianMoehring Thank you for the help. The last line there, I (in my head) evaluated the integral before taking the expected value. I figured that 1 doesn't depend of $B_w$ so we can just evaluate it like $\int_0^t1 ds$? I assume this is the step where the isometry is necessary. However the isometry only works for the 2 norm, again probably why we need to use the hint. – jeffery_the_wind Feb 20 '21 at 23:09
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    $\mathbb E(\int_0^t Z_sdB_s)$ is a number whereas $\int_0^s \mathbb E(Z_s)dB_s$ is a random variable. So, something is wrong in what you did. Nevertheless $\mathbb E((\int_0^tZ_zdB_s)^2)=\int_0^t \mathbb E(Z_s^2)ds$ (by ito isometry) – user841366 Feb 20 '21 at 23:16
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    I didn't even notice my comment's similarity to your last line. I was trying to point out that when you find the expectation of a stochastic integral $$\mathbb{E}\left[\int_0^t h(B_s,s), dB_s\right]$$ (like when you're showing $\mathbb{E}[X_t]=0$) you cannot end up with $\int_0^t \mathbb{E}[h(B_s,s)], dB_s$ – Brian Moehring Feb 20 '21 at 23:19
  • @user841366 Thank you for that. I see now that the calculation of the mean is wrong. I was just thinking we could always put the expectation inside the integral, but I see this is a stochastic integral so it's different. Is there something wrong with the covariance calculation? – jeffery_the_wind Feb 20 '21 at 23:24
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    Yes, covariance calculation is wrong (but surprisingly the answer is correct). You should stop thinking stochastic integral as a normal integral $\nu(A):=\int_AdB_s$ is not a measure). One can prove that $\mathbb E(\int_0^tX_sdB_s\int_0^u Y_sdB_s)=\int_0^{\min{t,u}}\mathbb E[X_sY_s]ds$ – user841366 Feb 20 '21 at 23:36
  • @BrianMoehring I see. I checked the solution for this, and the professor simply said its just an accepted fact that a stochastic integral of a function (in $H^2$) always has zero mean :-|. It makes sense since any function there augments the variance of the distribution of the paths but not the "drift". – jeffery_the_wind Feb 21 '21 at 14:12
  • @user841366 I added an update there which uses this identity that you put there. I realized that the hint was namely to help use show this exact thing that you put there. – jeffery_the_wind Feb 21 '21 at 14:13

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