Ornstein-Uhlenbeck Bridges Between Fixed Endpoints

Conditioning on Both Endpoints

An Ornstein–Uhlenbeck (OU) forecast uses the current value to describe the future. A bridge answers a different question: what is the law of the process at intermediate times when both its starting and ending values are specified? Conditioning on the terminal value changes both the forecast mean and its uncertainty. The result is a Gaussian process pinned at the two endpoint values.

Fix $\theta>0$, $\sigma>0$, $m\in\mathbb{R}$, and $T>0$. On a probability space $(\Omega,\mathcal{F},\mathbb{P})$, take a two-sided standard Brownian motion $(W_t)_{t\in\mathbb{R}}$, and define, for $t\geq0$, the process with state space $S=\mathbb{R}$ and continuous time index $t$ by

X_t=m+\sigma\int_{-\infty}^{t}e^{-\theta(t-u)}\,dW_u .

The integral exists in mean square. This construction makes $X$ a stationary Gaussian OU process; in particular, it does not require a nondegenerate $X_0$ to be measurable with respect to the usual natural filtration of a one-sided Brownian motion. Its mean and covariance are $\mathbb{E}[X_t]=m$ and $\operatorname{Cov}(X_s,X_t)=v e^{-\theta|t-s|}$, where $v=\sigma^2/(2\theta)$. Indeed, the two stochastic integrals share noise only up to $\min(s,t)$, and integrating the product of their kernels gives this covariance.

For conditioning, write $\mathcal{G}_t=\sigma(X_u:0\leq u\leq t)\vee\sigma(X_T)$, completed by null sets. This is a filtration that includes knowledge of the terminal value. Given endpoint values $X_0=x_0$ and $X_T=x_T$, the bridge is the regular conditional Gaussian law of $(X_t)_{0\leq t\leq T}$ given the endpoint vector. The endpoint vector has a nonsingular density because $T,\theta,\sigma>0$; thus conditioning at particular values is understood through the Gaussian conditional-distribution version, not as conditioning on an event of positive probability.

Definition: The OU bridge from $x_0$ to $x_T$ over $[0,T]$ is the conditional law of the stationary OU process given $X_0=x_0$ and $X_T=x_T$. Its mean function is the conditional expectation at each time, and its covariance function is the conditional covariance of pairs of process values.

The Gaussian Conditioning Calculation

The bridge formulas follow from conditioning a finite Gaussian vector. For clarity, if $Y$ is a Gaussian column vector with mean $\eta$ and invertible covariance matrix $\Sigma$, and $Z$ is jointly Gaussian with $Y$, then

\mathbb{E}[Z\mid Y] =\mathbb{E}[Z]+\operatorname{Cov}(Z,Y)\Sigma^{-1}(Y-\eta),

and the conditional covariance of $Z$ is $\operatorname{Var}(Z)-\operatorname{Cov}(Z,Y)\Sigma^{-1}\operatorname{Cov}(Y,Z)$.

Proof. Set $R=Z-\mathbb{E}[Z]-\operatorname{Cov}(Z,Y)\Sigma^{-1}(Y-\eta)$. Direct covariance multiplication gives $\operatorname{Cov}(R,Y)=0$. Since $(R,Y)$ is jointly Gaussian, its joint characteristic function factors when its cross-covariance is zero, so $R$ and $Y$ are independent. Therefore $\mathbb{E}[R\mid Y]=\mathbb{E}[R]=0$, yielding the conditional mean formula. Expanding $\operatorname{Var}(R)$ and using $\operatorname{Cov}(Y)=\Sigma$ gives $\operatorname{Var}(R)=\operatorname{Var}(Z)-\operatorname{Cov}(Z,Y)\Sigma^{-1}\operatorname{Cov}(Y,Z)$. Independence also shows this is the conditional covariance. $\square$

Theorem: For $0\leq t\leq T$, the mean of the OU bridge is $$ m+\frac{\sinh(\theta(T-t))}{\sinh(\theta T)}(x_0-m) +\frac{\sinh(\theta t)}{\sinh(\theta T)}(x_T-m). $$ For $0\leq s\leq t\leq T$, its covariance is $$ \operatorname{Cov}(X_s,X_t\mid X_0=x_0,X_T=x_T) =\frac{\sigma^2}{\theta} \frac{\sinh(\theta s)\sinh(\theta(T-t))}{\sinh(\theta T)}. $$ For $t<s$, use symmetry: the covariance at $(s,t)$ equals that at $(t,s)$.

Proof. Center the endpoint vector by setting $Y=(X_0-m,X_T-m)^{\mathsf T}$. Its covariance is $v\begin{bmatrix}1&r\\r&1\end{bmatrix}$, where $r=e^{-\theta T}$. Its determinant is $v^2(1-r^2)>0$, so the conditioning formula applies. For $Z=X_t-m$, its covariance with the endpoints is $v(a,b)$, where $a=e^{-\theta t}$ and $b=e^{-\theta(T-t)}$. Inverting the endpoint covariance matrix gives the coefficient of $x_0-m$ as $(a-rb)/(1-r^2)$, and the coefficient of $x_T-m$ as $(b-ra)/(1-r^2)$. Substitution of the exponentials gives, respectively, $\sinh(\theta(T-t))/\sinh(\theta T)$ and $\sinh(\theta t)/\sinh(\theta T)$. This proves the mean formula.

For the covariance, take $s\leq t$, and put $q=e^{-\theta s}$, $u=e^{-\theta(t-s)}$, and $w=e^{-\theta(T-t)}$. The unconditioned covariance divided by $v$ is $u$. The covariance correction from conditioning on the endpoints, divided by $v$, is $$ \frac{q^2u+uw^2-q^2uw^2-q^2u^3w^2}{1-q^2u^2w^2}. $$ Subtracting this expression from $u$ leaves $u(1-q^2)(1-w^2)/(1-q^2u^2w^2)$. Replacing $q,u,w$ by their exponentials and using $1-e^{-2z}=2e^{-z}\sinh(z)$ reduces the conditional covariance to $2v\,\sinh(\theta s)\sinh(\theta(T-t))/\sinh(\theta T)$. Since $2v=\sigma^2/\theta$, this is the claimed formula. Symmetry of covariance covers $t<s$. $\square$

The theorem also checks the pinning directly. At $t=0$, the mean is $x_0$, and at $t=T$, it is $x_T$. The covariance is zero if either time is an endpoint, since the corresponding hyperbolic-sine factor is zero. At interior times, the mean generally depends on both endpoints, while the covariance depends on the times and process parameters, not on the observed endpoint values.

Worked Bridge Calculations

Worked Example: The Midpoint of an OU Bridge

Take $\theta=1$, $\sigma^2=2$, $m=0$, $T=2$, and endpoints $x_0=1$, $x_T=3$. Find the conditional mean and variance at $t=1$.

Both mean weights are $\sinh(1)/\sinh(2)$. Using $\sinh(2)=2\sinh(1)\cosh(1)$, each weight is $1/(2\cosh(1))$. Therefore the conditional mean is $$ \frac{1}{2\cosh(1)}(1)+\frac{1}{2\cosh(1)}(3) =\frac{2}{\cosh(1)}. $$ The variance is $$ \frac{2}{1}\frac{\sinh(1)\sinh(1)}{\sinh(2)} =\frac{\sinh(1)}{\cosh(1)} =\tanh(1). $$ Check: The mean is the sum of the two endpoint contributions, and the variance is positive because $\sinh(1)$ and $\cosh(1)$ are positive. The endpoint covariance formula at $s=0$ or $t=2$ gives zero, as required.

Worked Example: Covariance at Two Interior Times

Let $\theta=1$, $\sigma^2=2$, $m=0$, $T=2$, and take endpoints $x_0=0$, $x_T=2$. Compute the bridge means at $s=\tfrac12$ and $t=\tfrac32$, along with their variances and covariance.

At $s=\tfrac12$, the mean is $2\sinh(\tfrac12)/\sinh(2)$. At $t=\tfrac32$, it is $2\sinh(\tfrac32)/\sinh(2)$. The variance at $s$ is $$ \frac{2\sinh(\tfrac12)\sinh(\tfrac32)}{\sinh(2)}. $$ At $t$, the variance is the same, because the formula gives $\sinh(\tfrac32)\sinh(\tfrac12)$ again. Their covariance is $$ \frac{2\sinh(\tfrac12)\sinh(\tfrac12)}{\sinh(2)} =\frac{2\sinh^2(\tfrac12)}{\sinh(2)}. $$ Check: Substituting $s=\tfrac12$, $t=\tfrac32$ into the covariance formula gives the two factors $\sinh(\tfrac12)$ and $\sinh(2-\tfrac32)=\sinh(\tfrac12)$. Interchanging $s$ and $t$ leaves the covariance unchanged, as symmetry requires. The endpoint values affect the two means but do not enter any of these covariance expressions.

Comparison with a Brownian Bridge

A Brownian bridge is Brownian motion conditioned on its values at two fixed times. The OU bridge approaches the corresponding Brownian bridge formulas as $\theta$ tends to zero, with $T$ and $\sigma$ fixed. This is a statement about the conditional bridge formulas: the stationary OU initial variance $\sigma^2/(2\theta)$ itself diverges in this limit, so it is not a claim that the unconditioned stationary processes converge to Brownian motion.

For fixed $z$, $\sinh(\theta z)/\sinh(\theta T)$ tends to $z/T$, since $\sinh(y)/y$ tends to $1$ as $y$ tends to zero. Thus the bridge mean tends to $m+(1-t/T)(x_0-m)+(t/T)(x_T-m)$, which is linear interpolation between the endpoints. For $s\leq t$, the covariance tends to $\sigma^2s(T-t)/T$: use $\sinh(\theta s)/\theta\to s$, $\sinh(\theta(T-t))/\theta\to T-t$, and $\sinh(\theta T)/\theta\to T$ in the covariance formula. This is the Brownian bridge covariance for diffusion coefficient $\sigma$.

Worked Example: The Brownian-Bridge Limit

Use the limiting bridge formulas with $T=4$, $\sigma^2=2$, $m=0$, $x_0=1$, and $x_T=5$. Find the mean at times $1$ and $3$, and the covariance between those values.

The limiting mean at $t=1$ is $(1-\tfrac14)1+\tfrac14(5)=\tfrac34+\tfrac54=2$. At $t=3$, it is $(1-\tfrac34)1+\tfrac34(5)=\tfrac14+\tfrac{15}4=4$. Since $1\leq3$, the limiting covariance is $$ 2\frac{1(4-3)}{4}=\frac12. $$ The variance at time $1$, obtained by setting both times to $1$, is $2\cdot1\cdot(4-1)/4=\tfrac32$. Check: The mean at each time uses weights summing to $1$; the covariance uses the earlier time $1$ and later time $3$, and is positive. The formula gives zero covariance with either fixed endpoint.

A Common Conditioning Error

A frequent mistake is to use an OU forecast based only on $X_0=x_0$ and treat it as the bridge mean. For example, with $\theta=1$, $\sigma^2=2$, $m=0$, $T=2$, $x_0=0$, and $x_T=2$, the one-endpoint forecast at $t=1$ is $e^{-1}x_0=0$. But conditioning on both endpoints gives the bridge mean $\sinh(1)\,2/\sinh(2)=1/\cosh(1)$, which is nonzero. The terminal observation pulls the conditional mean upward.

The uncertainty is also different. Conditioning only on $X_0$ gives variance $v(1-e^{-2\theta t})$; at the stated parameters and time this is $1-e^{-2}$. The bridge variance, calculated above from the two-endpoint formula, is $\tanh(1)$. These expressions are not interchangeable: the former leaves the terminal value unconstrained, while the latter accounts for it. Always use the covariance matrix of all conditioned observations, including both endpoints.

Takeaway: An OU bridge is obtained by conditioning a Gaussian process on its two endpoint values. Its mean combines both endpoints with hyperbolic-sine weights, and its covariance is the unconditioned covariance after removing the information explained by the endpoints. In the zero-mean-reversion limit, these formulas become the linear mean and triangular covariance of a Brownian bridge.

Check Your Understanding

Use the bridge formulas and show the substitutions that support each answer.

For $T>0$, evaluate the bridge mean at $t=0$ and $t=T$. Why does the covariance with either endpoint vanish?
For $0<s\leq t<T$, write the OU bridge covariance and explain why it does not depend on $x_0$ or $x_T$.
With $\theta=1$, $\sigma^2=2$, and $T=2$, find the bridge variance at the midpoint.
State the limiting mean and covariance when $\theta$ tends to zero, with $T$ and $\sigma$ fixed.
In one sentence, explain why a one-endpoint OU forecast is not generally the mean of a bridge with a specified terminal value.