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 $y$?



Brownian bridges as conditioning
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


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


