Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

While by definition the increments of a Brownian motion are independent, it is unclear to me whether (that implies that) the random variables $W_t$ and $W_s$ are independent for $t \neq s$. While these random variables have different density functions, they are defined on the same state space, and I am not sure whether they are independent.

share|improve this question
1  
"..whether they are independent for a given $\omega$" does not make sense. It is $W_t$ and $W_s$ that are random variables (as they are functions of $\omega$), and not $W_t(\omega)$ and $W_s(\omega)$ which are just real numbers. –  Stefan Hansen Nov 17 '12 at 11:56
    
Right, poorly formulated. –  lodhb Nov 17 '12 at 12:00
add comment

2 Answers

up vote 1 down vote accepted

Suppose $W_t$ and $W_s$ are independent for $t\neq s$. Then for $0\leq s<t$ we would have that $W_t$ is independent of $\mathcal{F}^W_s=\sigma(W_u\mid u\leq s)$. Now the martingale property of the Brownian motion yields $$ W_s=E[W_t\mid\mathcal{F}^W_s]=E[W_t]=0\quad\text{a.s.}, $$ which certainly isn't true.

share|improve this answer
add comment

Take $t<s$, then you can write $W(s)=W(t)+(W(s)-W(t))$, from which you immediately see that they are not independent.

It is the increments which are independent. Take e.g. $t<s$, then $W(t)-W(0)$ and $W(s)-W(t)$ are independent.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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