I have no ideas on this problem:
Let $(B_t, 0 \leq t \leq 1)$ be a standard Brownian motion in $1$ dimension. Let $Z^y_t = yt+ (B_t -tB_1)$. We call $\{Z^y_t\}_{0 \leq t \leq 1}$ a Brownian Bridge from $0$ to $y$. Let $W^y_0$ be the law of $(Z^y_t, 0 \leq t \leq 1)$ on $\mathcal{C} [0,1]$. We want to show that for any non-negative measurable function $F: \mathcal{C} [0,1] \rightarrow \mathbb{R}_{+}$, if $f(y)= W^y_0 (F)$, then we have $$ \mathbb{E} [F(B) | B_1] =f(B_1) \quad \text{ a.s.} $$ Any ideas of how to approach this problem?