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The covariance kernel of a standard Brownian bridge on $[0,1]$ is $c(s,t)=\min (s,t)-st$. Its integral operator has eigenfunctions $\sqrt{2}\sin (k\pi t)$ and eigenvalues $(k\pi {)}^{-2}$ for $k\ge 1$.

For standard Brownian motion, the Brownian bridge $U_t=B_t-t{B}_1$, $0\le t\le 1$, is independent of $B_1$. Every finite vector of bridge coordinates is jointly multivariate normal with $B_1$, and its cross-covariances with $B_1$ vanish. The independence of uncorrelated jointly normal variables and independence extended from generating pi-systems prove independence of the whole stochastic process.

For an empirical distribution function based on an independent sample from a continuous distribution $F$,where $B$ is a Brownian bridge.

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.

The Kolmogorov-Smirnov theorem states that, for continuous $F$,where $B$ is a Brownian bridge; the finite-sample law is also independent of continuous $F$. To test $H_0:F=F_0$, computeand reject when $\sqrt{n}{D}_n$ exceeds the appropriate null quantile.

If $t\le s$, then $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_t$. If $s<t\le 1$, Gaussian conditional expectation for the Brownian bridge between $(s,X_s)$ and $(1,X_1)$ gives

The Fokker-Planck equation corresponds to the Itô diffusionabsorbed at $0$ and reflected at $L$.

For $v(x)=x$, an Euler--Maruyama proposal isIf $Y\le 0$, kill the path. If $X_n>0$ and $Y>0$, an endpoint-only test can miss a crossing. Conditional on the endpoints, the local Brownian bridge crossing probability isKill the path with this probability; otherwise impose reflection at $L$ by replacing an overshoot $Y>L$ with $2L-Y$ and set $X_{n+1}=Y$. Repeated reflection handles very rare multiple overshoots, and the approximation converges as $\Delta t\to 0$.