Brownian bridges as conditioning

Question

Brownian bridges are interpreted as Brownian motions conditioned to start and end at given points. However, I have not seen a source that makes this precise, though this may be due to my own lack of exposure.

Let $W$ be a standard one dimensional Brownian motion on $[0, T]$, started at $0$. Denote by $\nu_y$ the conditional law of $W$ on $C[0, T]$ given $W_T$. Thus for any bounded measurable function $F$ on $C[0, T]$, the following disintegration formula is satisfied:

$$\int_{C[0, T]} F(W) \, d\mu(W) = \int_{\mathbb R} \int_{C[0, T]} F(W) \, d\nu_y (W) \, d\mu_{W_T} (y).$$

Here $\mu$ denotes the Wiener measure, and $\mu_{W_T}$ denotes the law of $W_T$.

Question: Is it true that for Lebesgue a.e. $y$, we have that $\nu_y$ is the law of a Brownian bridge started at $0$ and ending at $y$?

Answer 1

If $X$ and $Y$ are independent random variables taking values in arbitrary spaces $E$ and $F$, if $Z = f(X,Y)$ for any measurable map $f : E \times F \to G$, then the family of distributions of the random variables $(f(X,y))_{y \in F}$ provides a version of conditional law $\mathcal{L}(Z|Y)$. Indeed, an application of Fubini theorem yields for every bounded measurable function $F : G \to \mathbb{R}$ $$\int_{G} F(z) \, dP_Z(z) = \int_{E \times F} F(f(x,y)) \, dP_X(x)dP_Y(y) = \int_E \Big(\int_{E} F(f(x,y)) \, d\nu_X (x) \Big) \, dP_{Y} (y).$$

Applying this general fact to $X = (W_t-(t/T)W_T)_{t \in [0,T]}$, $Y = W_T$ and $f : \mathcal{C}([0,T]) \times \mathbb{R} \to \mathcal{C}([0,T])$ defined by $f(x,y)(t) := x(t)+(t/T)y$ yields the desired conclusion. Since the couple $(X,Y)$ is Gaussian as a linear function of the Gaussian process $W$, one proves the independence of $X$ and $Y$ by checking that $\mathrm{Cov}(X_t,Y)=0$ for all $t \in [0,T]$. See also here

https://math.stackexchange.com/questions/974724/brownian-bridge-equivalence-of-definitions

Answer 2

The standard definition of the Brownian bridge $B_t=W_t-(t/T)W_t$ can only informally be considered as conditioning the original Brownian motion $W_t$ to be at 0 at time $T$, because this condition obviously has probability 0. Moreover, this definition critically uses the fact that the state space $\mathbb R$ is a group, and that the Brownian motion respects this group structure.

An alternative general approach is based on the fact that for Markov processes conditioning amounts to taking the $h$-transform for an appropriate space-time harmonic function $h$ (non-trivial dependence on time is essential here: if $h$ is not space dependent, then the resulting process is time homogeneous). In order to obtain the classical Brownian bridge one has to take for $h$ the fundamental solution of the heat equation. This definition is then applicable to arbitrary diffusion processes without imposing any further conditions.