Distribution of Brownian bridge (pinned Brownian motion) Let $X=(X(t))_{t \geq 0}$ be a one-dimensional Brownian motion such that $X(0)=0$.
For fixed $t_{0}>0$ and $x,y \in {\mathbb R}^{1}$, define the process $X_{x}^{t_{0},y}=(X_{x}^{t_{0},y}(t))_{0 \leq t \leq t_{0}}$ by
\begin{align*}
X_{x}^{t_{0},y}(t)&=x+X(t)+\frac{t}{t_{0}}\left(-X(t_{0})+(y-x)\right) \\
&=x+\frac{t}{t_{0}}(y-x)+X_{0}^{t_{0},0}(t).
\end{align*}
My question is why the probability law of $X_{x}^{t_{0},y}$ coincides with ${\mathbb P}_{x}(\bullet \mid w(t_{0})=y)$, where ${\mathbb P}_{x}$ is the Wiener measure starting at $x$. 
 A: Let $t \in [0,t_{0}]$.
Since $X(t) \sim {\rm N}(0,t)$ and $X(t_{0})-X(t) \sim {\rm N}(0,t_{0}-t)$, we obtain
$$
\left(1-\frac{t}{t_{0}}\right)X(t) \sim {\rm N}\left(0,\left(1-\frac{t}{t_{0}}\right)^{2}t\right) \quad -\frac{t}{t_{0}}\left(X(t_{0})-X(t)\right) \sim {\rm N}\left(0,\left(\frac{t}{t_{0}}\right)^{2}t_{0}-t\right).
$$
Hence since $X(t)$ and $X(t_{0})-X(t)$ are independent, we have
\begin{align*}
(1) \quad 
X_{x}^{t_{0},y}(t)&=x+\frac{t}{t_{0}}(y-x)+\left(1-\frac{t}{t_{0}}\right)X(t)-\frac{t}{t_{0}}\left(X(t_{0})-X(t)\right) \\
&\sim {\rm N}\left(x+\frac{t}{t_{0}}(y-x),t\left(1-\frac{t}{t_{0}}\right)\right).
\end{align*}
On the other hand, we obtain
\begin{align*}
(2) \quad 
&{\mathbb P}(x+X(t) \in {\rm d}z \mid x+X(t_{0})=y)
=\frac{\frac{1}{\sqrt[]{2\pi t}}\exp\{-\frac{(z-x)^{2}}{2t}\}\frac{1}{\sqrt[]{2\pi (t_{0}-t)}}\exp\{-\frac{(y-z)^{2}}{2(t_{0}-t)}\}}{\frac{1}{\sqrt[]{2\pi t_{0}}}\exp\{-\frac{(y-x)^{2}}{2t_{0}}\}}{\rm d}z \\
&=\frac{1}{\sqrt[]{2\pi t\left(1-\frac{t}{t_{0}}\right)}}\exp\left\{-\frac{\left(1-\frac{t}{t_{0}}\right)(z-x)^{2}+\frac{t}{t_{0}}(y-z)^{2}-\frac{t}{t_{0}}\left(1-\frac{t}{t_{0}}\right)(y-x)^{2}}{2t\left(1-\frac{t}{t_{0}}\right)}\right\}{\rm d}z
\end{align*}
and
\begin{align*}
(3) \quad 
&\left(1-\frac{t}{t_{0}}\right)(z-x)^{2}+\frac{t}{t_{0}}(y-z)^{2}-\frac{t}{t_{0}}\left(1-\frac{t}{t_{0}}\right)(y-x)^{2} \\
&=(z-x)^{2}-2\frac{t}{t_{0}}(z-x)(y-x)+\left(\frac{t}{t_{0}}\right)^{2}(y-x)^{2} \\
&=\left(z-x-\frac{t}{t_{0}}(y-x)\right)^{2}.
\end{align*}
Consequently, by (1), (2) and (3), we obtain
$$
{\mathbb P}(X_{x}^{t_{0},y}(t) \in \bullet)={\mathbb P}(x+X(t) \in \bullet \mid x+X(t_{0})=y).
$$
Thus this implies that
$$
{\mathbb P}(X_{x}^{t_{0},y} \in \bullet)={\mathbb P}(x+X \in \bullet \mid x+X(t_{0})=y)={\mathbb P}_{x}(\bullet \mid w(t_{0})=y)
$$
(cf. page 204 or 241 in Ikeda-Watanabe).
