I was solving questions from the Martingales chapter in "Stochastic Processes" by Richard Bass. There was a question regarding d- dimensional Brownian motions(BM):
Let $(W_t^1,…,W_t^d)$ be a d dimensional Brownian motion. Show that for $i \neq j,\quad W_t^iW_t^j$ is a Martingale.
Just to revise, a d-dimensional Brownian motion is a process of the form $(W_t^1,…,W_t^d)$
where each $W_t^i$ is a Brownian motion wrt filtration $\mathcal{F}_t$ and $W^{1}$,…,$W^{d}$ are mutually independent.
While working it out, I had the following issues:
1) Is the filtration to be chosen for this martingale $\mathcal{F_t}=\sigma(W_u^iW_u^j,\quad 0\leq u \leq t)$ ? Or is it $\sigma(W_u^i,W_u^j,\quad 0\leq u \leq t)$? I feel it is the latter but because the filtration wasn't specified, it could very well be the former.
To understand why this would be a problem, let me share my solution. (I have proved the integrability requirement for a martingale).
Consider w.l.o.g $W_t$ and $V_t$, two Brownian motions mutually independent. It suffices to show
$$\mathbb{E}[W_tV_t | \mathcal{F_s}] = W_sV_s$$
Hence consider
$$\mathbb{E}[(W_t – W_s)(V_t – V_s)|\mathcal{F}_s]$$
which after some simplification yields
$$= \mathbb{E}[W_tV_t | \mathcal{F_s}] – W_sV_s$$
Now I wanted to use the independent increments property of BM to conclude the following*:
$$\mathbb{E}[(W_t – W_s)(V_t – V_s)|\mathcal{F}_s] =\mathbb{E}[(W_t – W_s)(V_t – V_s)]$$
Due to independence of $V_t$ and $W_t$
$$=\mathbb{E}[(W_t – W_s)]\mathbb{E}[(V_t – V_s)] = 0$$
At step *, I have used that $(W_t – W_s)$, $(V_t – V_s)$ and $\mathcal{F}_s$ are MUTUALLY independent. Is this sequitur (does it follow) if I assume $\mathcal{F}_t = \sigma(W_u,V_u,\quad 0\leq u \leq t)$?
I would appreciate any help in this regard. Additionally I have googled out d dimension brownian motion martingales but this query was not there. Nor did I find anything like it here.
Best Answer
$$\mathcal F_t=\sigma(W_s\,;\,0\leqslant s\leqslant t)=\sigma(W^i_s\,;\,1\leqslant i\leqslant d,\,0\leqslant s\leqslant t)$$