Suppose I have a Brownian motion $B_1(1)$ and a Brownian Bridge $BB(\tau) = B_2(\tau) – \tau B_2(1) $. How can I show that they are independent?
I know that it is enough to show that their covariance is $0$ for independence, as they are normally distributed. So, we have
$$ Cov(B_1(1),B_2(\tau)-\tau B_2(1)) = Cov(B_1(1),B_2(\tau)) – \tau Cov(B_1(1),B_2(1)). $$ How can I show that this is equal to $0$?
Best Answer
Central Limit theorem gives you convergence in law.
You may so consider an other Brownian motion $\tilde{B}^1(t)$, with the same law of $B^1(t)$ (i.e. simply a Brownian motion) and independent of $B^2(t)$.
So you can proove that $\tilde{B}^1(t)$ is independent of $BB(τ)$.