# From brownian bridge to brownian motion proof

Let $B_t$ be a brownian motion. and let $\{W_t=B_t-tB_1:0\le t\le 1\}$ be a brownian bridge. Now let $Y_t=(1+t)W_{t\over 1+t}$. Proof that $Y_t$ is a brownian motion in $[0, \infty)$

My attempt:

1) $Y_0=W_0=B_0=0$

2) $Y_t$ has continuous trajectories because $B_t$ has continous trajectories

3) I need to prove that $Y_t$ has independent and stationary increments and this is the part where I´m having trouble: for the independent increments I need to prove that $$P[Y_{t_1}=y_1,...,Y_{t_n}-Y_{t_{n-1}}=y_n]=P[Y_{t_1}=y_1]...P[Y_{t_n}-Y_{t_{n-1}}=y_n]$$

But it really gets hard. Maybe I can use the Paul-Levy characterization for brownian motion; in this case I´ll need to prove that $\{Y_t: t\ge 0\}$ and $\{Y_t^2-t: t\ge 0\}$ are martingales, but I´ll also find it really difficult.

So I would really appreciate if you can help me with this problem. Any comments or suggestions would be highly appreciated

Easy way: we know that $B_t$ is a Gaussian process, i.e. for any $t_1, \dots, t_n$ the random vector $(B_{t_1}, \dots, B_{t_n})$ has a jointly Gaussian distribution. Since any linear transformation of a jointly Gaussian vector is again jointly Gaussian, it follows that $W_t$ is also a Gaussian process, and likewise so is $Y_t$. Then it's easy to compute that $Y_t$ has mean 0, and that the covariance of $Y_s$ and $Y_t$ is $s \wedge t$, so $Y_t$ is Brownian motion.