I understand the concept of the 1 dimensional Brownian bridge with the form of:


$$dx\_t=\frac{-1}{1-t}x\_t \, dt + dw\_t$$


s.t. $x\_0=0$ and $x\_1=0$


where $dw\_t$ is a Wiener process.


I am thinking about the Brownian bridge in the phase space with arbitrary boundary conditions:


$$dx\_t=v\_t \, dt$$


$$dv\_t=a\_t\,dt+dw\_t$$


s.t. $x\_0=\phi\_0, v\_0=\psi\_0$ and $x\_1=\phi\_1, v\_1=\psi\_1$


Is there any clean close form of $a\_t$ that yields similar results to the Brownian bridge?


Any comments, or suggestions are welcome!



Phase space Brownian bridge
I use capital letters for random variables and small letters for possible values.


Let $W$ be a brownian motion, defined on the canonical space $\mathcal{C}(\mathbb{R}\_+)$ endowed with the Wiener measure $\mathbb{P}$ and with the canonical filtration $\mathcal{F}$. For $t>0$, call $p\_t$ the density of the random variable $(W\_t,\int\_0^t W\_s~ds)$, namely, the density of $\mathcal{N}(0,C\_t)$, where
$$C\_t = \left(\begin{array}{cc}
t & t^2/2 \\
t^2/2 & t^3/3 
\end{array} \right), 
\text{ so } C\_t^{-
1} = \frac{12}{t^4}\left(\begin{array}{cc}
t^3/3 & -t^2/2 \\
-t^2/2 & t 
\end{array} \right)
= \left(\begin{array}{cc}
4/t & -6/t^2 \\
-6/t^2 & 12/t^3 
\end{array} \right).$$
Thus $p\_t(x,y)$ is proportional to $\exp(2x^2/t - 6xy/t^2 + 6y^2/t^3)$.


Let $V,U$ the processes defined by
$$V\_t := v\_0+W\_t \text{ and } U\_t := u\_0+\int\_0^t W\_s~ds := u\_0+tv\_0+\int\_0^t W\_s~ds.$$
We want to condition $(V,U)$ by $(V\_1,U\_1) = (v,u)$.
The idea is to compute for each $t \in [0,1[$ the probability
$\mathbb{P}\big|\_{\mathcal{F}\_t} \big[\cdot \big|(V\_1,U\_1) = (v,u)\big]$ and to apply Girsanov Theorem.


Let $A \in \mathcal{F}\_t$ and $f : \mathbb{R^2} \to \mathbb{R}$ measurable and bounded. Observe that $V\_1 = V\_t + W\_1 - W\_t$ and  

$$U\_1 = U\_t + (1-t)v\_0 + \int\_t^1 W\_s~ds = U\_t + (1-t)V\_t + \int\_t^1 (W\_s-W\_t)~ds.$$
Since $W\_{s+\cdot}-W\_s$ is a Brownian motion, independent of $\mathcal{F}\_s$, we get
\begin{eqnarray\*}
\mathbb{E}[\mathbb{1}\_A f(V\_1,U\_1)] 
&=& \int\_{\mathbb{R^2}}
\mathbb{E}[\mathbb{1}\_A f(V\_t + x,U\_t + (1-t)V\_t + y) p\_{1-t}(x,y)] dxdy \\
&=& \mathbb{E}\Big[\int\_{\mathbb{R^2}}
\mathbb{1}\_A f(v,u) p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) dvdu \Big]
\end{eqnarray\*}
Therefore
$$\mathbb{P}\big|\_{\mathcal{F}\_t}\big[\cdot\big|(V\_1,U\_1) = (v,u)\big] 
= D\_t\mathbb{P}\big|\_{\mathcal{F}\_t}, \text{ where }
D\_t = \frac{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_1(v-v\_0,u-u\_0-v\_0)}.$$
The process $D$ thus defined on the time interval $[0,1[$ is a martingale.


Girsanov Theorem yields that under $\mathbb{P}\big[\cdot\big|(V\_1,U\_1) = (v,u)\big]$, the process
$$\hat{W} := W - \int\_0^\cdot \frac{d\langle D,W \rangle\_s}{D\_s}$$
is a local martingale, hence a Brownian motion, since it has the same quadratic variation as $W$.


Ito calculus yields
\begin{eqnarray\*}
dD\_t 
&=& \frac{-1}{p\_1(v-v\_0,u-u\_0-v\_0)} 
\Big( \partial\_1p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) \\
& & \hspace 5 cm +(1-t)\partial\_2p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t) \Big) dW\_t \\
& & \quad + \textrm{ process with locally bounded variation.}
\end{eqnarray\*}
Thus
\begin{eqnarray\*}
\hat{W} &=& W + \int\_0^\cdot \frac{\partial\_1p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)} dt \\
& & + (1-t)\int\_0^\cdot \frac{\partial\_2p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)}{p\_{1-t}(v-V\_t,u-U\_t-(1-t)V\_t)} dt.
\end{eqnarray\*}
Since the density $p\_{1-t}$ is well-known, this computation can be continued.


Hence, under $\mathbb{P}\big[\cdot\big|(V\_1,U\_1) = (v\_1,u\_1)\big]$, the process $W$ can be written as the sum of the Brownian motion $\hat{W}$ and an explicit drift.


