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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.

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$$\mathcal F_t=\sigma(W_s\,;\,0\leqslant s\leqslant t)=\sigma(W^i_s\,;\,1\leqslant i\leqslant d,\,0\leqslant s\leqslant t)$$

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  • $\begingroup$ Thanks. Let me try it with this filtration. $\endgroup$
    – Gautam Shenoy
    Feb 15, 2013 at 11:25
  • $\begingroup$ Wait. I have tried this. @Did : Could you tell me if my arguments at the end follow? The mutual Independence one. $\endgroup$
    – Gautam Shenoy
    Feb 15, 2013 at 12:03
  • $\begingroup$ Yes, since four families are globally independent: $(W_u)_{u\leqslant t}$, $(V_u)_{u\leqslant t}$, $(W_u-W_t)_{u\geqslant t}$ and $(V_u-V_t)_{u\geqslant t}$. $\endgroup$
    – Did
    Feb 15, 2013 at 13:03
  • $\begingroup$ Great. This is what I was looking for. Thanks $\endgroup$
    – Gautam Shenoy
    Feb 17, 2013 at 16:34
  • $\begingroup$ @Did Just a curiosity I can't quite figure out on my own: what would happen if we used $\mathcal{G_t}=\sigma(W_s^iW_u^j:\ 0\leq s \leq t, 0\leq u \leq t)$ as our filtration? Would $W_u^iW_u^j$ be a $\mathcal{G_t}$-martingale? $\endgroup$
    – AdrianoMeis
    Jan 26, 2019 at 20:02

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