Sign and Time-Reversal Symmetries of Brownian Motion

Two Symmetries, One Gaussian Calculation

Brownian motion has two useful symmetries that can look similar but involve different transformations. Negating every value, \(B_t\mapsto -B_t\), preserves its law because its Gaussian distributions are centered. Reversing the increments over a fixed interval, \(B_T-B_{T-t}\), also produces Brownian motion on that interval, but for a different reason: its covariance is again the Brownian covariance. Simply reading the values backward, \(B_{T-t}\), does not have this property because it does not start at zero.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(\mathbb{R}\) and continuous time index \(t\ge0\). We use its defining properties: \(B_0=0\) almost surely; its paths are continuous almost surely; and it is a centered Gaussian process with covariance \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). A Gaussian process means that every finite vector of its values has a multivariate Gaussian distribution, including the possibility of a degenerate Gaussian distribution.

We will use the standard Gaussian-vector fact established in the earlier material on Gaussian vectors: a Gaussian vector’s law is determined by its mean vector and covariance matrix, even when that matrix is singular. Thus, to compare the finite-dimensional distributions below, it is enough to show that the transformed vectors are Gaussian and have the required means and covariances.

Changing the Sign

Proof. Fix any finite list \(t_1,\ldots,t_m\ge0\). The vector \((B_{t_1},\ldots,B_{t_m})\) is Gaussian, so its linear transformation \((-B_{t_1},\ldots,-B_{t_m})\) is Gaussian. Its mean in coordinate \(r\) is \(-\mathbb{E}[B_{t_r}]=0\). For any \(r,q\), its covariance is \(\operatorname{Cov}(-B_{t_r},-B_{t_q})=\operatorname{Cov}(B_{t_r},B_{t_q})=\min(t_r,t_q)\). Therefore the transformed vector has the same mean and covariance matrix as \((B_{t_1},\ldots,B_{t_m})\), and hence the same Gaussian law. This holds for every finite list of times, proving equality of finite-dimensional distributions. Both processes start at zero and have continuous paths almost surely, since negation preserves continuity. The standard measure-uniqueness fact for continuous-path processes—that their law on \(C[0,T]\) is determined by evaluations at a countable dense set of times—then gives equality of path laws on every finite interval \([0,T]\). \(\square\)

This symmetry does not assert that \(B_t\) and \(-B_t\) are equal on a given sample path. It says their distributions agree. For instance, at a fixed positive time their values have the same centered normal distribution, but a particular sample path will generally differ from its negative.

Reversing Increments over a Fixed Horizon

Fix a deterministic horizon \(T>0\), and define a process \((Y_t)_{0\le t\le T}\) on the same probability space by \(Y_t=B_T-B_{T-t}\). The time domain is the closed interval \([0,T]\), and the state space is \(\mathbb{R}\). The subtraction is essential: it makes \(Y_0=0\). The reversal theorem concerns this fixed deterministic horizon; it does not assert a result for a random horizon.

Proof. First, \(Y_0=B_T-B_T=0\), and \(Y\) has continuous paths almost surely because \(t\mapsto B_{T-t}\) is continuous whenever the Brownian path is continuous. Now fix \(m\ge1\) and times \(t_1,\ldots,t_m\in[0,T]\). The vector \((Y_{t_1},\ldots,Y_{t_m})\) is a linear transformation of the Gaussian vector \((B_T,B_{T-t_1},\ldots,B_{T-t_m})\), so it is Gaussian. Its mean is zero. For \(s,t\in[0,T]\), bilinearity of covariance gives

Here \(\min(T-s,T-t)=T-\max(s,t)\), since \(s,t\le T\). Thus every finite vector of \(Y\) has the same zero mean vector and Brownian covariance matrix as the corresponding vector of \(B\). The Gaussian-vector fact implies equality of all finite-dimensional distributions. The paths of \(Y\) and \(B\) on \([0,T]\) are continuous almost surely, so the standard measure-uniqueness fact for continuous-path processes gives equality of their laws on \(C[0,T]\). \(\square\)

The covariance calculation also explains the time reversal directly. If \(s\le t\), then \(\min(s,t)=s\): the covariance of the earlier reversed value and the later one is the earlier time, exactly as for Brownian motion. The theorem is about the whole process, not only the one-dimensional marginal distributions.

Why the Centering Matters

A tempting but incorrect reversal is \(Z_t=B_{T-t}\). It reads the values backward, but its initial value is \(Z_0=B_T\), which is a nondegenerate \(N(0,T)\) variable for \(T>0\). Standard Brownian motion starts at zero almost surely, so \(Z\) cannot have its law. Its covariance also has the wrong time dependence: \(\operatorname{Cov}(Z_s,Z_t)=\min(T-s,T-t)=T-\max(s,t)\), not \(\min(s,t)\) in general.

Common mistake. For \(T=4,s=1,t=2\), one might incorrectly treat the reversed values \(B_{3}\) and \(B_2\) as if their covariance were \(\min(1,2)=1\). In fact \(\operatorname{Cov}(B_3,B_2)=\min(3,2)=2\). For the correctly centered reversed process, the covariance is \(\operatorname{Cov}(B_4-B_3,B_4-B_2)=4-2-1+2=3\)? This expression must be evaluated using the correct pair: \(\operatorname{Cov}(B_4,B_4)=4\), \(\operatorname{Cov}(B_4,B_2)=2\), \(\operatorname{Cov}(B_3,B_4)=3\), and \(\operatorname{Cov}(B_3,B_2)=2\), giving \(4-2-3+2=1\). Thus the corrected covariance is \(1=\min(1,2)\), as required. Keeping the four terms in their proper order prevents a sign or indexing error.

These symmetries are useful whenever a distributional calculation depends on the direction in which a Brownian path is viewed. Sign symmetry permits replacing \(B\) by \(-B\) in a law calculation. Fixed-horizon reversal permits converting a segment of increments read backward into an ordinary Brownian segment. Neither statement identifies transformed paths pointwise with the original paths; both are statements about distributions.