Proving that a process is Brownian Bridge

Question

This is from Durrett's Probability : Theory and Examples , excercies 8.4.2.

$B_t$ is brownian motion starting from $0$. Show that $X_t = (1-t)B(\frac{t}{1-t})$ is Brownian Bridge.

I managed to show that $X_t$ is Gaussian and has same mean and variance as Brownian Bridge.

However, I am stuck with how to prove $X_1 =0$. Since at $t=1$, $B(\frac{t}{1-t})$ is discontinuous, can we even define $X_1$?

Regarding time inversion of Brownian motion, I know $Y_t$ defined as below is Brownian motion $$Y_t = \begin{cases} 0 & t=0\\ tB(1/t) & t>0 \end{cases}$$

As in this case, can we just define $X_1 = 0$?

score 1 · Accepted Answer · answered Jun 12 '21 at 05:47

1

Use SLLN's For BM: Law of large numbers for Brownian Motion (Direct proof using L2-convergence)

Since $(1-t)B(\frac t {1-t})= t \frac {B(s)} s$ where $s=\frac t {1-t}$ it follows that we can take $X_1=0$

answered Jun 12 '21 at 05:47

Kavi Rama Murthy

359,332

Applying SLLN's for BM, I guess $X_1\rightarrow 0$ a.s. But does it mean $X_1=0$ for Brownian Bridge? – Dongri Jun 12 '21 at 05:53
Well, $X_1$ is not defined at $t=1$ but Durett is assuming that you define $X_1$ as the limiting value at $t=1$ @Dongri – Kavi Rama Murthy Jun 12 '21 at 05:56
Ah,I got it!Thanks for clear explanation! – Dongri Jun 12 '21 at 06:16

Proving that a process is Brownian Bridge

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