Brownian Bridges as Conditioned Gaussian Paths

Conditioning Brownian Motion on Its Endpoint

Brownian motion is usually described by its distribution without fixing its terminal value. A Brownian bridge describes the corresponding path law when the terminal value is fixed. The conditioning event \(B_T=y\) has probability zero, so this is not ordinary conditioning on an event of positive probability. Instead, we use the regular conditional distribution of the Brownian path given \(B_T\); its Gaussian formulas define a conditional law for each \(y\in\mathbb{R}\).

Throughout, let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space carrying a standard Brownian motion \((B_t)_{t\ge0}\), and fix \(T>0\). We use the standard facts that finite collections of Brownian values are jointly Gaussian, \(\mathbb{E}[B_t]=0\), and \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). No stochastic integrals are needed here.

Conditional Mean and Covariance

Proof. Put \(R_i=B_{t_i}-(t_i/T)B_T\). The vector \((R_1,\ldots,R_n,B_T)\) is jointly Gaussian because it is a linear transformation of a jointly Gaussian vector. For each \(i\), $$ \operatorname{Cov}(R_i,B_T) =\operatorname{Cov}(B_{t_i},B_T)-\frac{t_i}{T}\operatorname{Var}(B_T) =t_i-\frac{t_i}{T}T=0. $$ Thus the vector of residuals is independent of \(B_T\): for jointly Gaussian vectors, zero cross-covariance makes the joint characteristic function factor into the two marginal characteristic functions, which is independence. Consequently, conditioning on \(B_T=y\) leaves the distribution of the residual vector unchanged and gives \(B_{t_i}=(t_i/T)y+R_i\). Its conditional mean is therefore \((t_i/T)y\). For the covariance, direct expansion gives $$ \begin{aligned} \operatorname{Cov}(R_i,R_j) &=\min(t_i,t_j)-\frac{t_j}{T}\operatorname{Cov}(B_{t_i},B_T) -\frac{t_i}{T}\operatorname{Cov}(B_T,B_{t_j}) +\frac{t_it_j}{T^2}\operatorname{Var}(B_T)\\ &=\min(t_i,t_j)-\frac{t_it_j}{T}-\frac{t_it_j}{T}+\frac{t_it_j}{T}\\ &=\min(t_i,t_j)-\frac{t_it_j}{T}. \end{aligned} $$ This proves both formulas. If a time is \(0\), its value is deterministically \(0\); if a time is \(T\), its value under the conditional law is deterministically \(y\), also as given by the formulas. \(\square\)

The covariance subtraction is the effect of fixing the endpoint. In particular, at a single time \(t\in[0,T]\), the bridge value is normal with mean \(ty/T\) and variance \(t-t^2/T=t(T-t)/T\). At either endpoint the variance is zero, as it must be.

A Representation Using an Independent Brownian Motion

Proof. The process \(X\) is Gaussian because each finite vector of its values is a linear transformation of a finite vector of values of \(\widetilde B\). Its paths are continuous, and \(X_0=0\), \(X_T=y+\widetilde B_T-\widetilde B_T=y\). The centered part \(X_t-(t/T)y\) has mean zero. For \(s,t\in[0,T]\), its covariance is $$ \begin{aligned} \operatorname{Cov}\!\left(\widetilde B_s-\frac{s}{T}\widetilde B_T,\, \widetilde B_t-\frac{t}{T}\widetilde B_T\right) &=\min(s,t)-\frac{t}{T}s-\frac{s}{T}t+\frac{st}{T^2}T\\ &=\min(s,t)-\frac{st}{T}. \end{aligned} $$ These are exactly the conditional means and covariances established in the preceding theorem. Gaussian vectors with the same means and covariances have the same distributions, so all finite-dimensional distributions agree. Since the processes are continuous, their laws on continuous paths are determined by their values at rational times in \([0,T]\). Hence \(X\) has the stated bridge law. \(\square\)

This representation is a construction of the conditional law, not an assertion that the original Brownian path and the constructed bridge are identical. It also makes the pinning at \(T\) explicit: the random endpoint \(\widetilde B_T\) is subtracted in exactly the proportion needed to make the terminal value deterministic.

Joint and Transition Distributions

The covariance formula gives the joint distribution at any finite collection of times. A useful related description is the bridge transition law: once the bridge has value \(x\) at time \(s\), its distribution at a later time \(t\) is still Gaussian, but its mean is pulled toward the fixed endpoint \(y\).

Proof. First take \(0<s<t<T\). From the bridge mean and covariance formulas, $$ \operatorname{Var}(X_s)=\frac{s(T-s)}{T},\quad \operatorname{Var}(X_t)=\frac{t(T-t)}{T},\quad \operatorname{Cov}(X_s,X_t)=\frac{s(T-t)}{T}. $$ For jointly Gaussian variables with \(\operatorname{Var}(X_s)>0\), conditioning on \(X_s=x\) gives conditional mean $$ \frac{s}{T}y+ \frac{\operatorname{Cov}(X_s,X_t)}{\operatorname{Var}(X_s)} \left(x-\frac{s}{T}y\right) =\frac{s}{T}y+\frac{T-t}{T-s}\left(x-\frac{s}{T}y\right) =x+\frac{t-s}{T-s}(y-x). $$ The conditional variance is $$ \begin{aligned} \operatorname{Var}(X_t) -\frac{\operatorname{Cov}(X_s,X_t)^2}{\operatorname{Var}(X_s)} &=\frac{t(T-t)}{T} -\frac{s(T-t)^2}{T(T-s)}\\ &=\frac{(T-t)\big(t(T-s)-s(T-t)\big)}{T(T-s)}\\ &=\frac{(T-t)T(t-s)}{T(T-s)} =\frac{(t-s)(T-t)}{T-s}. \end{aligned} $$ At \(s=0\), the starting value is fixed at \(0\), and the formulas reduce to the one-time bridge mean \(ty/T\) and variance \(t(T-t)/T\). At \(t=T\), the endpoint is fixed at \(y\), so its conditional variance is zero. This verifies the stated boundary cases. \(\square\)

Common Pitfall and Interpretation

A frequent error is to keep the unconditioned Brownian variance \(t\) after fixing the endpoint. For example, at \(t=2\) on \([0,4]\), that incorrect calculation would assign variance \(2\). The correct conditional variance is \(t-t^2/T=2-4/4=1\). The reduction is not optional: observing the endpoint supplies information about the value at the intermediate time, and the covariance subtraction accounts for that information.

Another mistake is to interpret \(B_T=y\) as an event of positive probability and use a ratio of probabilities. For \(T>0\), \(B_T\) has a density, so \(\mathbb{P}(B_T=y)=0\). The bridge formulas instead specify the regular conditional distribution at the value \(y\). They apply for every real \(y\), and they also agree with the endpoint constraints: the mean at \(T\) is \(y\), while the variance at \(T\) is zero.

Brownian bridges are useful whenever a path is studied between known starting and ending values, including conditional simulation and Gaussian interpolation. The representation with \(\widetilde B\) gives a direct way to generate bridge paths, while the transition formula gives the distribution of a later value once an intermediate value is known.