Let $B$ be a standard Brownian motion. Define a Brownian bridge $b$ by $b_t=B_t-tB_1$. Let $\mathbb{W'}$ be the law of this process.
According to Wikipedia,
A Brownian bridge is a continuous-time stochastic process B(t) whose
probability distribution is the conditional probability distribution
of a Wiener process W(t) (a mathematical model of Brownian motion)
given the condition that B(0) = B(1) = 0.
Surely it makes no sense to condition on a probability 0 event? So I'm trying to show that $\mathbb{W'}$ is the weak limit as $\epsilon\to 0$ of Brownian motion conditioned upon the event $\{|B_1|\leq \epsilon\}$. How do we prove this?
Thank you.
Best Answer
Brownian motion $B_t$ over the interval $[0,1]$ can be decomposed into two independent terms. That is, the process $X_t=B_t-tB_1$ and the random variable $Y=B_1$. As these are joint normal, to prove that they are independent, it is enough to show that the covariance ${\rm Cov}(X_t,Y)={\rm Cov}(B_t,B_1)-t{\rm Var}(B_1)=t-t$ vanishes.
The distribution of $B$ conditional on $\vert B_1\vert < \epsilon$ is just the same as that of $X$ plus the independent process $tY$ (conditioned on $\vert Y\vert < \epsilon$). As $\epsilon$ goes to zero, this converges to the distribution of $X$, which is a Brownian bridge.