Are Brownian motions $B^i_t, B^j_t$ independent if $\langle B^i, B^j \rangle_t = 0 ~~ \forall t \geq 0$?

Question

Are two Brownian motions $B^i_t, B^j_t$ independent if $\langle B^i, B^j \rangle_t = 0 ~~ \forall t \geq 0$?

I think I can show that $B^i_t$ and $B^j_t$ are in fact independent by applying Itô's lemma to $e^{\lambda B^i_t + \mu B^j_t}$, taking expectations and solving the ODE for the moment generating function. But how do I show that the processes are independent. Are they even independent?

I believe there might be a proof by quoting a multivariate version of Levy's Characterisation of Brownian motion. Is there maybe an elementary argument?

Answer 1

The multivariate version of Levy's Characterisation of Brownian motion gives us a proof. It states:

Let $X=(X^1,\ldots,X^d)$ be a d-dimensional local martingale with $X_0=0$. Then, the following are equivalent.

1. X is a Brownian motion on the underlying filtered probability space.

2. X is continuous and $X^i_tX^j_t-\delta_{ij}t$ is a local martingale for $1\le i,j\le d$.

3. X has quadratic covariations $\langle X^i,X^j \rangle_t=\delta_{ij}t$ for $1\le i,j\le d$.

However, I am still wondering if there is a slick proof to the question in the title without this heavy machinery (and without redoing $80$% of the proof used to show the implication $3 \implies 1$ in the above theorem)