Proving that $X_t = W_t~ I (0<t\le T) + (2W_T - W_t) ~I(t > T)$ is a brownian motoin

Question

The steps to showing that a process is a BM are as follows:

(1)$X_0 = 0$

(2) $ \forall t ~~~X_t$ is continuous

(3)$X_t \sim N(0,t)$

(4)$X_{t+s}-X_{s} \sim N(0,t)$

(5)$X_{t+s}-X_{s} \bot \mathscr F_s=(W_u)_{u\in(0,s)} $

If $W_t$ is a Standard Brownian Motion. $X_t = W_t~ I (0<t\le T) + (2W_T - W_t) ~I(t > T)$, where I is the indicator function taking the value of 1 if t is in the specified range.

I am having trouble showing step (3)

my attempt is to just break this up into cases:

Case 1: $t<T \implies X_t=W_t$ which is clearly normal with mean zero and variance t

Case 2: $t>T \implies X_t = 2W_T - W_t$. Here, $W_T$ is a constant as we are in the period of time after T has occured. So wouldn't this process not have a zero mean? but i know that $X_t$ is in fact a brownian motion.

for this process though, any suggestions?

Answer 1

Let $(\Omega, \mathcal F,{\{\mathcal F_t\}}_{t\geqslant 0},\mathbb P)$ be the underlying filtered probability space and let $T$ be a stopping time. Consider the following processes:

1. $W_t~$ for $0<t\leqslant T$;
2. $W_{t}-W_T~$ for $T< t$;
3. $-(W_{t}-W_T)~$ for $T< t$.

By the strong Markov property of Brownian motion, processes $(2)$ and $(3)$ are Brownian motions, each independent of process $(1)$. Furthermore, adding process $(2)$ to $(1)$ gives Brownian motion $W_t$, while adding process $(3)$ to $(1)$ gives $X_t$. Since $(2)$ and $(3)$ are identically distributed processes, then $X_t$ is Brownian motion.