Three Models for Constrained or Observed Paths
A prediction between fixed endpoints is not the same problem as a prediction from noisy measurements. In the first case, the path is required to match its endpoints exactly; in the second, the observed values include measurement error, so the latent process need not pass through them. The covariance model matters too: Brownian motion and an Ornstein–Uhlenbeck process respond differently to the same endpoint constraints. This tutorial works through each case by conditioning a Gaussian vector of endpoint or observation values together with the quantity to be predicted.
All processes below are defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), have continuous time \(t\ge0\), and take values in \(S=\mathbb{R}\). We use the finite-dimensional Gaussian conditioning formula: if \(U\) and \(V\) are jointly Gaussian, \(\operatorname{Cov}(V)\) is invertible, and their means and covariance blocks are known, then
and the conditional covariance is \(\operatorname{Cov}(U)-\operatorname{Cov}(U,V)\operatorname{Cov}(V)^{-1}\operatorname{Cov}(V,U)\). For conditioning at a finite set of process times, apply this formula to the corresponding Gaussian vectors.
When an endpoint value is specified, the event that a continuous Gaussian variable equals that particular value usually has probability zero. We use the Gaussian conditional distribution as a regular conditional law, with the continuous version of its mean and covariance formulas. This gives a well-defined bridge model for each specified endpoint value; it is not ordinary conditioning on a positive-probability event.
Brownian Motion Conditioned at Its Endpoint
Let \((B_t)_{t\ge0}\) be standard Brownian motion, so \(\mathbb{E}[B_t]=0\) and \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). Fix \(T>0\) and an endpoint \(b\in\mathbb{R}\). For \(0\le t\le T\), conditioning on \(B_T=b\) gives
Derivation. The endpoint has mean zero and variance \(T\), while \(\operatorname{Cov}(B_t,B_T)=t\) for \(t\le T\). The conditional mean formula therefore gives \(tT^{-1}b\). For \(s,t\le T\), the conditional covariance is \(\min(s,t)-sT^{-1}t\). In particular, the mean reaches \(b\) at \(T\), and the conditional variance there is \(T-T^2/T=0\). The covariance describes uncertainty remaining between the fixed endpoints.
Worked Example: A Brownian Bridge to a Fixed Value
Suppose standard Brownian motion is conditioned to satisfy \(B_2=3\). Find its conditional mean at times \(1/2\) and \(1\), and its conditional variance at time \(1/2\) and covariance between times \(1/2\) and \(1\).
Here \(T=2\) and \(b=3\). The mean at time \(1/2\) is \((\tfrac12/2)3=\tfrac34\). The mean at time \(1\) is \((1/2)3=\tfrac32\). The variance at \(1/2\) is \(\min(\tfrac12,\tfrac12)-(\tfrac12)^2/2=\tfrac12-\tfrac18=\tfrac38\). The covariance between times \(1/2\) and \(1\) is \(\tfrac12-(\tfrac12)(1)/2=\tfrac14\).
Check: At the endpoint, the conditional mean is \((2/2)3=3\), and the conditional variance is \(2-2\cdot2/2=0\), as required. The positive covariance \(\tfrac14\) is consistent with the two interior values sharing uncertainty about the path.
Conditioning an Ornstein–Uhlenbeck Process at Both Ends
Now consider a stationary, mean-zero Ornstein–Uhlenbeck process with covariance \(C(s,t)=c e^{-\theta|t-s|}\), where \(c=\sigma^2/(2\theta)>0\) and \(\theta>0\). Fix \(T>0\) and condition on \(X_0=a\) and \(X_T=b\). These endpoint values have a nonsingular joint Gaussian distribution because the endpoint covariance matrix has determinant \(c^2(1-e^{-2\theta T})>0\).
Justification. Write \(r=e^{-\theta T}\). The endpoint covariance matrix is \(c\begin{bmatrix}1&r\\r&1\end{bmatrix}\), whose inverse is \(\{c(1-r^2)\}^{-1}\begin{bmatrix}1&-r\\-r&1\end{bmatrix}\). The covariance of \(X_u\) with the endpoints is \(c(e^{-\theta u},e^{-\theta(T-u)})\). Multiplying this row by the inverse and by \((a,b)^\mathsf{T}\) gives the conditional mean. Replacing the exponentials by their definitions in terms of \(\sinh\), and using \(\sinh(\theta T)=\sinh(\theta u)e^{\theta(T-u)}+\sinh(\theta(T-u))e^{-\theta u}\), gives the displayed expression. For the covariance, the Gaussian conditioning formula subtracts the endpoint-mediated covariance from \(c e^{-\theta(t-s)}\). Substitution of the same inverse and the identities \(2\sinh x\sinh y=\cosh(x+y)-\cosh(x-y)\) gives \(c\sinh(\theta s)\sinh(\theta(T-t))/\sinh(\theta T)\). The formula is symmetric in \(s,t\) and is zero when either time is an endpoint.
The hyperbolic-sine weights are determined by the OU covariance, not by a linear interpolation rule. In particular, the mean is pulled toward zero in the middle according to the mean-reverting covariance structure. The bridge covariance also depends on \(\theta\): even with the endpoints fixed, the amount and timing of uncertainty differ from the Brownian bridge.
Worked Example: An OU Bridge with Unequal Endpoints
Take \(\theta=1\), \(\sigma^2=2\), \(T=2\), \(a=1\), and \(b=0\). Then \(c=\sigma^2/(2\theta)=1\). Find the conditional mean and variance at time \(u=1\).
The mean formula gives \(\{\sinh(1)+0\cdot\sinh(1)\}/\sinh(2)=\sinh(1)/\sinh(2)=1/(2\cosh(1))\), using \(\sinh(2)=2\sinh(1)\cosh(1)\). The variance is the covariance formula with \(s=t=1\): \(1\cdot\sinh(1)\sinh(1)/\sinh(2)=\sinh(1)/(2\cosh(1))\).
Check: The mean formula gives \(1\) at \(u=0\), since \(\sinh(2)/\sinh(2)=1\), and \(0\) at \(u=2\). The covariance formula gives zero variance at both endpoints because one of its numerator factors is \(\sinh(0)=0\). The interior variance is positive because \(\sinh(1)>0\) and \(\cosh(1)>0\).
Noisy Gaussian Process Observations
Let \(f(t)\) be a Gaussian process with mean \(m(t)\) and covariance \(C(s,t)\). At observation times \(t_1,\ldots,t_n\), suppose we observe \(Y_i=f(t_i)+\varepsilon_i\), where the errors are independent Gaussian variables, independent of the process, with mean zero and variances \(r_i>0\). Let \(K\) be the \(n\) by \(n\) matrix with entries \(K_{ij}=C(t_i,t_j)\), let \(R\) be diagonal with entries \(r_i\), and let \(k_*\) be the column vector with entries \(C(t_i,t_*)\). The observation covariance is \(K+R\), which is invertible when \(R\) is positive definite. The conditional distribution of the latent value \(f(t_*)\) has mean and variance
where \(m_X\) is the column vector with entries \(m(t_i)\). These expressions follow by applying finite-dimensional Gaussian conditioning to the joint vector of \(f(t_*)\) and \(Y\). Independence of the errors leaves the cross-covariance \(\operatorname{Cov}(f(t_*),Y)=k_*^\mathsf{T}\), while adding the errors changes the observation covariance from \(K\) to \(K+R\). For noiseless observations, take \(R=0\) when \(K\) is invertible; the latent process is then constrained to match those observed values.
Worked Example: Predicting from Two Noisy Measurements
Let the two latent values \(f(t_1),f(t_2)\) have mean zero and covariance matrix \(K=\begin{bmatrix}1&\tfrac12\\\tfrac12&1\end{bmatrix}\). Observe \(y=(1,0)^\mathsf{T}\) with independent error variances \(1\), and predict \(f(t_1)\). Compute its conditional mean and variance.
Here \(R=I\), so \(K+R=\begin{bmatrix}2&\tfrac12\\\tfrac12&2\end{bmatrix}\), with determinant \(4-\tfrac14=\tfrac{15}{4}\). Its inverse is \(\begin{bmatrix}\tfrac{8}{15}&-\tfrac{2}{15}\\-\tfrac{2}{15}&\tfrac{8}{15}\end{bmatrix}\). For prediction at \(t_1\), \(k_*=(1,\tfrac12)^\mathsf{T}\). Multiplying the inverse by \(y\) gives \((\tfrac{8}{15},-\tfrac{2}{15})^\mathsf{T}\), so the conditional mean is \(1\cdot\tfrac{8}{15}+\tfrac12(-\tfrac{2}{15})=\tfrac{7}{15}\). Multiplying the inverse by \(k_*\) gives \((\tfrac{7}{15},\tfrac{2}{15})^\mathsf{T}\). Thus the conditional variance is \(1-(1\cdot\tfrac{7}{15}+\tfrac12\cdot\tfrac{2}{15})=1-\tfrac{8}{15}=\tfrac{7}{15}\).
Check: The observation covariance times its inverse has diagonal entries \(2\cdot\tfrac{8}{15}+\tfrac12(-\tfrac{2}{15})=1\) and off-diagonal entries \(2(-\tfrac{2}{15})+\tfrac12(\tfrac{8}{15})=0\). Also, the predicted latent mean \(\tfrac{7}{15}\) is not forced to equal the noisy measurement \(1\), and its remaining variance is positive.
Choosing the Model and Avoiding a Common Error
The conditioning operation is the same in each case, but its inputs encode different assumptions. A Brownian bridge uses the Brownian covariance and one fixed endpoint; an OU bridge uses the OU covariance and two fixed endpoints; noisy-data prediction adds the measurement-error covariance before inversion.
| Situation | Conditioning values | Observation covariance | Must the latent path match the data? |
|---|---|---|---|
| Brownian bridge | One Brownian endpoint | Endpoint variance \(T\) | Yes, at the fixed endpoint |
| OU bridge | Two OU endpoints | Endpoint covariance matrix | Yes, at both fixed endpoints |
| Noisy Gaussian process | Measurements plus error | \(K+R\) | No |
Common mistake: treating noisy measurements as exact. In the two-measurement example, using \(K^{-1}\) instead of \((K+R)^{-1}\) would treat the data as noiseless. Since \(f(t_1)\) is then observed exactly as \(1\), that calculation gives conditional mean \(1\) and conditional variance \(0\). But the actual measurements have error variance \(1\), so the correct calculations use \(K+R\) and give mean \(\tfrac{7}{15}\) and variance \(\tfrac{7}{15}\). The error is not a small algebraic discrepancy: it replaces an uncertain measurement with a hard constraint.
Before conditioning, specify which values are fixed and which are noisy, and choose a covariance model appropriate to the process. Then form the cross-covariance with the target and the covariance of the conditioning vector. This keeps endpoint constraints, measurement error, and latent-process uncertainty in their proper roles.
Check Your Understanding
Use the conditioning formulas to answer each question, and check endpoint or covariance claims by substitution.
- For a standard Brownian bridge on \([0,4]\) conditioned to end at \(b=2\), find the conditional mean at time \(1\) and variance at time \(2\).
- In the OU bridge formula, what happens to the conditional variance when \(s=0\) or \(t=T\), and which factor makes it happen?
- For noisy observations, explain why the cross-covariance between \(f(t_*)\) and \(Y\) is \(k_*^\mathsf{T}\), not \(k_*^\mathsf{T}+R\).
- In the two-measurement example, calculate the conditional mean of \(f(t_2)\) using the same inverse matrix and verify it by multiplication.
- State the modeling change made when \(R=0\), assuming the observation covariance matrix \(K\) is invertible.