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 $X_1 = 0$?
Best Answer
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 $X_1=0$