Let $W=\{W(t)\}_{t \in [0,\infty)}$ be a Brownian Motion and $X=\{X(t):=W(t)-t W(1)\}_{t \in [0,1]}$ the associated Brownian Bridge. I wan't to verify that $X$ is not adapted to the natural filtration of $W$, i.e. not adapted to $\mathcal{F}^W=\{\mathcal{F}^W(t)\}_{t \in [0,1]}$ with $\mathcal{F}^W(t):= \sigma(W(r): r \leq t)$.
my attempt: Choose $t=0.5$ and show that $\sigma(W(0.5)-0.5* W(1)) \not\subset \sigma(W(r): r \leq 0.5)$. The independent increments property of the Brownian Motion gives us that $W(1)-W(0.5)$ is independent of $\sigma(W(r):r \leq 0.5)$. Additionally we know that $W(1)-W(0.5)$ is $\sim\mathcal{N}(0,0.5)$ distributed. So we can choose an $B \in \mathcal{B}(\mathbb{R})$ (e.g. $B = [0,2]$) such that $\mathbb{P}[W(1)-W(0.5) \in B] \in (0,1)$. Together with the independence of $\sigma(W(r):r \leq 0.5)$ we can conclude that $A:= \{W(1)-W(0.5) \in B\} \in \sigma(W(1)-W(0.5) )$ but $A \notin \sigma(W(r): r \leq 0.5) $, since otherwise we would have $\mathbb{P}[A]= \mathbb{P}[A \cap A]= \mathbb{P}[A]^2$ which can't be true for $\mathbb{P}[A]\in (0,1)$.
However this just holds for $W(1)-W(0.5)$ and not for $W(0.5)-0.5*W(1)$ and I see no direct argument to transfer this argumentation. It even holds that $W(0.5)-0.5*W(1)$ and $\sigma(W(r): r \leq 0.5) $ are dependent since $cov[W(0.5),W(0.5)-0.5*W(1)]=0.25\neq 0$.
I would be thankful for any hints!
