Let $(W_t)_{t\in[0,1]}$ be a Brownian motion and let $$\mathscr{G}_t = \sigma(W_1) \vee \sigma\left(W_s : s\in [0,t]\right), t \in [0,1]$$ Calculate $E[W_t|\mathscr{G_s}]$ for $0 \le s \le t \le 1$, and prove that the Brownian motion $(W_t)_{t\in[0,1]}$ is not a $\mathscr{G}$-martingale.
Intuitively, it makes perfect sense for me, since $\mathscr{G}_t$ contains not only the information prior to time $t$ but also the information about $W_1$. But I am struggling on how to write the proof explicitly. Can anyone help me with this? Thanks a lot!
It is easy to show that $W$ is not a $\mathcal{G}$ martingale if you take $t=1$: $$ E[W_1\mid\mathcal G_{0.5}] =W_1 \neq W_{0.5}, $$ so $W$ is not a $\mathcal{G}$ martingale.
Edit, changed the link:
To compute $E[W_t\mid\mathcal G_{s}]$ for general $0\leq s \leq t \leq 1$, this might help