Phase space Brownian bridge

Question

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!

Answer 1

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.