Independence of stochastic process $(dB_1t)(dB_2t)$=0? What does it mean (definition) for two stochastic processes to be independent? like two independent Brownian motion $B_1(t), B_2(t)$. I come across this when I saw a solution of a problem says if $B_1(t), B_2(t)$ are independent, then $(dB_1t)(dB_2t)=0$. How do we prove this?
This is something I dont understand in part of solution from my homework problem, and here is the original problem:
Let $X(t)=B_1(t)B_2(t)$, where $B_1(t)$ and $B_2(t)$ are two independent Brownian motions, check if $X(t)$ is a martingale and find its martingale representation.
Solution: Since $\Bbb E[X(t)]=\Bbb E[B_1(t)]\Bbb E[B_2(t)]=0=X(0)$ so it is a martingale.
By Ito's product rule, $$dB_1(t)dB_2(t)=B_1(t)dB_2(t)+B_2(t)dB_1(t)+dB_1(t)dB_2(t)$$
since $B_1(t),B_2(t)$ are independent so we take $dB_1(t)dB_2(t)=0$
...
 A: First of all, note that $\mathbb{E}(X_t)=0$ does in general not imply that $(X_t)_{t \geq 0}$ is a martingale. In order to prove that $(X_t)_{t \geq 0}$ is a martingale, i.e. satisfies
$$\mathbb{E}(X_t \mid \mathcal{F}_s ) = X_s, \qquad s  \leq t, \tag{1}$$
you have to use the independence of the processes as well as the independence of the increments of a Brownian motion.
Turnining to your original question: Itô's formula states that
$$d(B_1(t) B_2(t)) = B_1(t) \, dB_2(t) + B_2(t) \, dB_1(t) + d\langle B_1,B_2 \rangle_t$$
where $\langle B_1,B_2 \rangle_t$ is the quadratic covariation of the processes $(B_1(t))_{t \geq 0}$ and $(B_2(t))_{t \geq 0}$. By definition, the quadratic covariation is the unique process (with some nice properties) such that
$$B_1(t) \cdot B_2(t)- \langle B_1,B_2 \rangle_t=X_t-\langle B_1,B_2 \rangle_t$$
is a martingale. Since we have already seen in $(1)$ that $(X_t)_{t \geq 0}$ is a martinale, we get $\langle B_1,B_2\rangle_t=0$. Consequently,
$$d(B_1(t) B_2(t)) = B_1(t) \, dB_2(t) + B_2(t) \, dB_1(t).$$
Remark As already pointed out by @Did you have to be careful with the notations. You cannot simply multiply differentials of stochastic processes. See e.g. this question for some (heuristical) explanations in a similar context.
