Conditional Predictions Along a Brownian Bridge

Local Prediction Along a Brownian Bridge

A Brownian bridge is constrained by its endpoint values, so predictions inside the interval must account for both what has already been observed and what the path must still reach. For a prediction at time \(u\), the key fact is local: once the bridge values immediately before and after \(u\) are known, other observations do not change the conditional distribution at \(u\). This gives a compact exact formula even when many interior values have been recorded.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) supporting a standard Brownian motion \((B_t)_{t\ge0}\). Fix \(T>0\). The standard Brownian bridge on \([0,T]\) is the continuous-time process with state space \(\mathbb{R}\) defined by \(X_t=B_t-(t/T)B_T\), for \(0\le t\le T\). It has fixed endpoints \(X_0=X_T=0\), mean zero, and covariance \(K(s,t)=\min(s,t)-st/T\). More generally, a bridge with endpoints \(a,b\in\mathbb{R}\) and scale \(\sigma\ge0\) is \(X_t=a+(b-a)t/T+\sigma(B_t-(t/T)B_T)\). Its mean is the line joining the endpoints and its covariance is \(\sigma^2K(s,t)\).

Conditioning on an exact real-valued observation is understood through the regular conditional Gaussian distribution. For distinct interior observation times, their joint Gaussian distribution is nondegenerate, so the displayed conditional law is defined for every vector of observed real values. The theorem concerns a query time different from all observed times; if \(u\) itself is observed, the conditional value is that observation with variance zero.

Proof. It is enough first to consider the standard bridge, since subtracting the deterministic endpoint line and dividing by \(\sigma\) reduces a general bridge to that case. Put \(\alpha=(r-u)/(r-l)\), \(\beta=(u-l)/(r-l)\), and define the residual \(R=X_u-\alpha X_l-\beta X_r\). Since \(\alpha+\beta=1\), this subtracts the linear interpolation between the two bracketing values.

Every other observed time \(v\) lies either at or to the left of \(l\), or at or to the right of \(r\). If \(v\le l\), then \(K(t,v)=v(1-t/T)\) for \(t\ge v\), which is linear in \(t\). Therefore \(K(u,v)=\alpha K(l,v)+\beta K(r,v)\), and \(\operatorname{Cov}(R,X_v)=0\). If \(v\ge r\), then \(K(t,v)=t(1-v/T)\) for \(t\le v\), also linear in \(t\), giving the same zero covariance. Thus \(R\) has zero covariance with every observed value, including the two bracketing values.

The residual and the observation vector are jointly Gaussian. For jointly Gaussian vectors, zero cross-covariance implies independence: their joint characteristic function, whose exponent is determined by the mean and covariance, factors into the product of the two marginal characteristic functions when the cross-covariance is zero. Hence the conditional mean of \(R\) given the observations is zero, and its conditional variance equals its unconditional variance.

For that variance, use \(K(s,t)=\min(s,t)-st/T\) in the expansion

Here \(K(u,u)=u(T-u)/T\), \(K(l,l)=l(T-l)/T\), \(K(r,r)=r(T-r)/T\), \(K(l,r)=l(T-r)/T\), \(K(u,l)=l(T-u)/T\), and \(K(u,r)=u(T-r)/T\). Substituting these six expressions and \(\alpha=(r-u)/(r-l)\), \(\beta=(u-l)/(r-l)\), and collecting terms gives \(\operatorname{Var}(R)=(u-l)(r-u)/(r-l)\). The interpolation \(\alpha X_l+\beta X_r\) is fixed when the observations are given. Adding the independent, mean-zero residual gives the claimed conditional normal law. Multiplication by \(\sigma\) multiplies the variance by \(\sigma^2\), and adding the deterministic endpoint line preserves the formula in terms of the observed endpoint values. \(\square\)

Proof. In the theorem’s proof, the residual \(R\) has zero covariance with every observed value outside \((l,r)\), as well as with \(X_l\) and \(X_r\). Joint Gaussianity makes \(R\) independent of the entire finite observation vector. Its variance and the interpolation determined by \(X_l,X_r\) are therefore unchanged by conditioning on the additional observations. \(\square\)

The formula separates two roles. The conditional mean is the straight-line interpolation of the two values that bracket the prediction time. The conditional variance depends on the spacing of those times, not on the values observed there. A query close to either bracketing observation has smaller conditional variance; a query halfway between them has the largest variance within that interval.

Why the Local Formula Matters

The local prediction formula is useful when a bridge is observed at many times but only one additional value is needed. Instead of inverting a covariance matrix whose size grows with the number of observations, identify the closest observation on each side and use two time gaps. The Gaussian conditioning calculation still explains why this shortcut is valid: the prediction residual is uncorrelated with, and hence independent of, the entire observation vector.

Do not confuse a conditional prediction with the unconditional marginal law. For the standard bridge, \(X_u\) has mean zero and variance \(u(T-u)/T\) before any interior values are observed. After conditioning, both the mean and the variance generally change. In particular, the conditional variance is determined by the bracketing times, not by the distance from \(u\) to the endpoints alone.

Common mistake. In the first example, using the unconditional law at \(u=1/2\) gives mean zero and variance \((1/2)(1-1/2)=1/4\), so one might incorrectly report \(N(0,1/4)\). That ignores the observed values. The correct bracketing calculation gives mean \(1/8\) and variance \(1/8\), hence \(N(1/8,1/8)\). The correction is to condition first: locate \(l,r\), interpolate their observed values, and calculate the variance from the two gaps.

The assumptions also matter. The query must lie strictly between its bracketing times for the positive-variance formula; if it coincides with an observation, its conditional variance is zero. The scale must be included as \(\sigma^2\), not \(\sigma\), in the variance. If \(\sigma=0\), the process is deterministic and every conditional variance is zero, so the nondegenerate Gaussian interpretation does not apply.