Brownian Bridge as Conditioned Brownian Motion

Fixing Brownian Motion at an Endpoint

Brownian motion is random both at intermediate times and at the end of an interval. A Brownian bridge describes its path when the endpoint is fixed: over \([0,T]\), the path starts at \(0\) and is conditioned to end at a prescribed value \(b\). This construction is useful when a model needs continuous random paths subject to endpoint constraints, or when the effect of observing an endpoint must be separated from the remaining path variation.

Let \((\Omega,\mathcal F,\mathbb P)\) support a standard Brownian motion \((B_t)_{t\ge0}\), with its natural filtration \((\mathcal F_t)_{t\ge0}\), where \(\mathcal F_t=\sigma(B_u:0\le u\le t)\) (with the usual completion if desired). Thus \(B_0=0\), paths are continuous, and increments over disjoint time intervals are independent centered normal variables, with \(\operatorname{Var}(B_t-B_s)=t-s\) for \(0\le s\le t\). In particular, \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). Fix \(T>0\) and an endpoint \(b\in\mathbb R\).

We will obtain that kernel directly from Brownian motion. The central device is to remove from each \(B_t\) the linear component associated with \(B_T\). The resulting process is independent of the endpoint, so fixing the endpoint changes only the removed linear component.

Separating the Endpoint from the Remaining Path

Proof. For any finite selection \(t_1,\ldots,t_m\in[0,T]\), the vector \((R_{t_1},\ldots,R_{t_m},B_T)\) is jointly Gaussian: each coordinate is a linear combination of a finite Gaussian vector of Brownian observations. For every \(t\in[0,T]\),

A jointly Gaussian vector with zero cross-covariances between two subvectors has independent subvectors: its joint characteristic function factors into the product of the two marginal characteristic functions, since the cross terms in the Gaussian exponent vanish. Consequently, each finite vector of \(R\)-values is independent of \(B_T\). The cylinder events determined by finite collections of \(R_t\)'s generate \(\sigma(R_t:0\le t\le T)\); the standard extension of independence from these generating events gives independence of this sigma-field and \(\sigma(B_T)\).

The process \(R\) is centered because both terms in its definition have mean zero. For \(s,t\in[0,T]\), expansion of the covariance gives

Each finite vector of \(R\)-values is Gaussian by linearity, so \(R\) is a Gaussian process. These computations prove all the claims. \(\square\)

Proof. For every path and every \(t\in[0,T]\), the identity \(B_t=R_t+\frac{t}{T}B_T\) holds. The lemma shows that \(R\) is independent of \(B_T\). Thus, for a fixed endpoint value \(y\), the law of the path \(R_t+\frac{t}{T}y\) supplies a conditional distribution kernel: averaging this law over the distribution of \(B_T\) recovers the original Brownian path law by the identity. This is precisely the defining property of a regular conditional law, and the kernel is defined for every \(y\), including values whose individual endpoint events have probability zero.

Under the kernel at \(y=b\), the centered process \(R\) is unchanged, and the added function \(tb/T\) is deterministic. Hence the mean at time \(t\) is \(0+tb/T\), while adding a deterministic function does not change covariance. The lemma therefore gives the stated covariance. The process is Gaussian because its finite vectors are translations of Gaussian vectors, and its paths are continuous because \(R\) and \(t\mapsto tb/T\) are continuous. \(\square\)

The formula separates two effects. The endpoint value determines the straight-line conditional mean from \(0\) to \(b\); the covariance describes random departures from that line and does not depend on \(b\). At \(t=0\) and \(t=T\), the covariance with every time is zero, so the process is fixed at \(0\) and \(b\), respectively.

Calculating Bridge Moments

Why Conditioning Changes the Covariance

Conditioning on \(B_T=b\) removes the part of each intermediate observation that is predictable from the endpoint. In the representation \(B_t=R_t+\frac{t}{T}B_T\), the second term is fixed once the endpoint is fixed, while the independent residual \(R_t\) retains its randomness. That is why the conditional covariance is the covariance of \(R\), rather than the unconditioned covariance \(\min(s,t)\).

Common mistake. A tempting but incorrect calculation is to use \(\operatorname{Var}(B_{T/2}\mid B_T=b)=\operatorname{Var}(B_{T/2})=T/2\), on the grounds that Brownian motion has variance \(T/2\) at the midpoint. This ignores the information supplied by the endpoint. The bridge formula instead gives

For example, when \(T=2\), the incorrect value is \(1\), whereas substitution into the conditional covariance formula gives \(1-\frac{1}{2}=\frac12\). The endpoint information reduces the midpoint variance by half in this case.

A second point of care is that \(\mathbb P(B_T=b)=0\) does not make the bridge undefined. It means only that conditioning cannot be performed by dividing probabilities by \(\mathbb P(B_T=b)\). The independent-residual construction supplies a regular conditional law at every \(b\), and its formulas specify the version used here.