On Coupling of Brownian Bridges

Let $\{B(t),\ 0\le t\le1\}$ be a Brownian bridge. Let $Y(t)=\int^t_0 f(u)\,dB(u)$, where $f\colon[0,1]\to\{+1,-1\}$ is a nonrandom and measurable function. Problem: Is there a Brownian bridge $B^*$ such that $|Y(t)|\ge|B^*(t)|$, $0\le t\le 1$, a.s.? The answer is positive. We will prove that we can take $B^*$ to be
$$ B^*(t)=\begin{cases} Y(t),&0\le t\le\tau,\\ B(t),&\tau\le t\le 1,\ Y(\tau)=+B(\tau),\\ -B(t),&\tau\le t\le 1,\ Y(\tau)=-B(\tau), \end{cases} $$
where $\tau=\max\{t\ge 0:|Y(t)|=|B(t)|\}$. Now let $X_+(t)=\int^t_0 1_{\{f=+1\}}(u)\,dB(u)$ and $X_-(t)=\int^t_0 1_{\{f=-1\}}(u)\,dB(u)$, $0\le t\le 1$. Is there a Brownian bridge $B_*$ such that $\max_{0\le t\le 1}|B_*(t)|=\max_{0\le t\le 1}\{|X_+(t)|\vee|X_-(t)|\}$? Again, the answer is positive and will be discussed. As a corollary of these constructions, we get a sharp inequality that compares the distributions of $\max_{0\le t\le 1}|B(t)|$ and $\max_{0\le t\le 1}|Y(t)|$.