Given a filtered probability space $(\Omega, \mathcal{F}, \mathcal{F}_{t\geq 0}, P)$:

If $B_1, B_2, \dots, B_m $ are all real $\mathcal{F}_t$ Brownian motions, jointly independent. Is the resulting vector $B = (B_1, B_2, \dots, B_m )$ an $\mathcal{F}_t$ $m$-dimensional Brownian motion or just a natural one?

To me it seems that I can't conclude that $B_t - B_s$ is independent of $\mathcal{F}_s$ even though all components are.

In fact $P(\{B_{1,t}-B_{1,s} \in \Lambda_1\}, \dots,\{B_{m,t}-B_{m,s} \in \Lambda_m\}, A)$ for $\Lambda_i$ a borel set, and $A \in \mathcal{F}_s$ can't be factorized, because all events are only pairwise independent with $A$.

Am I missing something or am I correct?

To summarise, if $X$ and $Y$ are independent, and also pairwise independent of a sigma-algebra $\mathcal{D}$, I need that $\sigma(X,Y)$ independent of $\mathcal{D}$ to factorize the probabilities. Or in other words a random vector whose components are independent of a sigma algebra is not necessarily independent itself.

  • $\begingroup$ Related: math.stackexchange.com/questions/2958680/… $\endgroup$ – Angela Pretorius May 26 '19 at 7:56
  • 1
    $\begingroup$ Thanks, I already know that it is a natural Brownian motion, but I am wondering if it still is when assigned a given filtration. The definition I know of a multivariate Brownian motion assumes the vector increment $B_t - B_s$ independent of the filtration at time $s$. $\endgroup$ – lucmobz May 26 '19 at 21:03

Take $m=2$. Let $\mathcal F_t$ be the natural filtration and consider a variable $X$ such that $$F_{X,B_{1,1},B_{2,1}}(x,y,z) = 2F(x)F(y)F(z) \text{ if } xyz>0, 0 \text{ otherwise}$$ where $F$ stands for the standard normal cdf, and this definition is extended by "independence" (i.e. you build the rest of your Brownian motions by drawing mutually independent normal variables).

Consider the filtration $\mathcal G_t = \sigma(\mathcal F_t \cup \sigma_X)$.

Then you can show that $\mathcal G_t$ is a filtration for $B_1$ and $B_2$, essentially because $X$ is independent with $B_{1,1}$ and with $B_{2,1}$, however it is not a filtration for $(B_1, B_2)$, because the sign of $X$ determines the sign of the product $B_{1,1}.B_{2,1}$, so $(B_{1,1}, B_{2,1})$ is not independent of $\mathcal G_0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.