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Let $(B_t , 0 ≤ t ≤ 1)$ be a standard Brownian motion in 1 dimension. We let $(Z_t^y = yt + (B_t − tB_1 ), 0 ≤ t ≤ 1)$ for any $y \in R$ and call it the Brownian bridge from $0$ to $y$. Let $W_0^y$ be the law of $(Z_t^y, 0 ≤ t ≤ 1)$ on $C([0, 1])$. Show that for any non-negative measurable function $F : C([0, 1]) → R_+$ for $f(y)$ = $W_0^y (F)$, we have $$\mathbb{E}[F (B)|B_1 ] = f (B_1 )$$

I have a problem with $f(B_1)$ = $W_0^{B_1}(F)$, which is $\mathbb{P}(Z_t^{B_1} \in F)$ according to my lecture notes. However this is nonsense, as $F$ is not a set of functions. Any hints?

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Note that $W_0^y(F)=\int F\mathrm dW_0^y=E(F(Z^y))$ where $Z^y=(Z^y_t)_{0\leqslant t\leqslant 1}$ and that $F$ is not a subset of $C([0,1])$ but a functional defined on $C([0,1])$. –  Did Jan 15 '12 at 19:00
    
Thank you very much. Where could I find out why $\int{FdW_t^y}=\mathbb{E}[F(Z_t^y)]$? –  Tom Artiom Fiodorov Jan 15 '12 at 19:33
    
There is no $W^y_t$, only $W^y_0$, which is a measure on $C([0,1])$. And $\int F\mathrm dW^y_0=E(F(Z^y))$ (not $=E(F(Z^y_t))$) is the definition of $W^y_0$. –  Did Jan 16 '12 at 6:28
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