Walker whose Velocity is a Brownian Bridge

Question

Consider a continuous random walk $x (t) $, in which the velocity $v (t) = \mathrm dx/\mathrm dt $ rather than the position is described by Brownian motion, so that $v (t) = B_t $ where $B_{t+\epsilon} - B_t \sim \mathcal N (0,\epsilon^2) $. Given the initial position $x (0) = x_0$ and velocity $v (0) = v_0$, the distribution of position with time is nevertheless normally distributed, namely $x (T) \sim \mathcal N (x_0 + v_0T, T^3/3) $.

Suppose that at time $t=T$, the walker is known to have a particular velocity $v_T $, fixing a final boundary condition as well as an initial one. What then is the distribution of $x (T) $, and how does one approach the analysis?

(This question is a follow-up to a question I asked earlier about a random walk with speed boosts. As this sort of question appears easier to approach in a continuous time setting, that is the context I take in this case.)

Answer 1

As Kwaśnicki remarked, the velocity process $v_t$ is a Brownian bridge, which can be represented as: $$ v_t = v_0 (1 - \frac{t}{T}) + v_T \frac{t}{T} + (T - t) \int_0^t \frac{1}{T-s} d B_s \;. $$ (For an intro to this representation, see the first exercise of the following exercise sheet on Brownian bridges).

As before, the position process $x_T$ is obtained by integrating the velocity process: \begin{align*} x_T &= x_0 + \int_0^T v_t dt \\ &=x_0 + \frac{T}{2} (v_0 + v_T) + \int_0^T \int_0^t \frac{T-t}{T-s} d B_s dt \\ &=x_0 + \frac{T}{2} (v_0 + v_T) + \int_0^T \left( \int_s^T \frac{T-t}{T-s} dt \right) d B_s \end{align*} This double integral is Gaussian with mean zero and variance $T^3 / 12$. Hence, $$ x_T \sim \mathcal{N}( x_0 + \frac{T}{2} (v_0 + v_T), \frac{T^3}{12} ) \;. $$ Note that the variance of this process with pinned initial and final velocities is a quarter of the variance of the position process with pinned initial and unpinned final velocity -- which makes sense intuitively.