Independent increment of Brownian motion (mistakes wikipedia definition ?)

I know that Brownian motion has the property that if $$0\leq t_1\leq t_2\leq ...\leq t_n$$ then $$B_{t_1}, (B_{t_2}-B_{t_1}),...,(B_{t_n}-B_{t_{n-1}})\tag{*}$$ are independents.

In wikipedia they say that increment of Brownian motion, and they define it as for $$0\leq t_1< t_2\leq t_3, $$B_{t_2}-B_{t_1}, B_{t_4}-B_{t_3}.$$ However, I know that if $$A,B,C$$ are pairwise independent, they are not necessarily independent. So, how do we get $$(*)$$ from wikipedia definition ?

• If I am not mistaken, this is a special property of Gaussian vectors... – Daniel Sep 19 at 22:26

1 Answer

The mutual independence of the variables in (*) follows from pairwise independence because the joint distribution of a vector of normally distributed random variables is determined entirely by the means and covariances of the variables.

Assume in (*) that $$0. If the variables are known to be pairwise independent, then the covariance matrix is diagonal and invertible, which means the joint density factorizes, which implies the variables are mutually independent. (If any of the inequalities in the condition $$0\le t_1\le t_2\le \cdots\le t_n$$ are not strict, then some of the variables are zero; but constant variables are independent of everything.)

• This analysis presumes that the joint distribution of the $n$ increments is multivariate normal, which is not part of the wikipedia definition. The means and covariances may not characterize the joint distribution outside the multivariate normal case. – John Dawkins Sep 20 at 21:30