Why Orthogonal Invariance Matters
Standard Brownian motion has no preferred direction in Euclidean space. In two dimensions, rotating the coordinate axes changes the coordinates used to describe a path, but it does not change the path's probability law. The same principle holds in every finite dimension, and it includes reflections as well as rotations. The precise statement concerns all finite collections of times: the joint distribution of the process values is unchanged after applying any fixed orthogonal matrix.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) have state space \(\mathbb{R}^d\), where \(d\) is a positive integer. Standard \(d\)-dimensional Brownian motion means that \(B_0=0\), its paths are continuous almost surely, and its increments over disjoint time intervals are independent, with \(B_t-B_s\) distributed as \(N_d(0,(t-s)I_d)\) for \(0\le s<t\). Here \(I_d\) is the \(d\) by \(d\) identity matrix. Equivalently, its \(d\) coordinate processes are independent standard one-dimensional Brownian motions.
For a square orthogonal matrix, \(U^T U=I_d\) also implies \(UU^T=I_d\): the first identity makes \(U\) invertible, with inverse \(U^T\), so multiplying in the other order gives the second identity. In particular, \(U\) preserves Euclidean lengths, since \(\|Ux\|^2=x^T U^T Ux=\|x\|^2\) for every \(x\in\mathbb{R}^d\).
The Invariance Theorem
Proof. Fix an integer \(m\ge1\) and times \(0\le t_1\le\cdots\le t_m\). Reordering any finite list of times only reorders the corresponding vector entries, so considering ordered times is sufficient. Set \(t_0=0\) and \(Z_r=B_{t_r}-B_{t_{r-1}}\) for \(1\le r\le m\). If an interval has length zero, its increment is zero with probability one. Otherwise, \(Z_r\) has distribution \(N_d(0,(t_r-t_{r-1})I_d)\). The vectors \(Z_1,\ldots,Z_m\) are independent by the independent-increments property.
For any \(\theta\in\mathbb{R}^d\), the characteristic function of \(UZ_r\) is
The first equality is the characteristic function of the centered Gaussian increment, applied to \(U^T\theta\); the second uses \(\|U^T\theta\|=\|\theta\|\). When \(t_r=t_{r-1}\), both sides equal \(1\), as required for the zero increment. Thus \(UZ_r\) has the same distribution as \(Z_r\). Also, the vectors \(UZ_1,\ldots,UZ_m\) remain independent: applying a measurable function separately to each member of an independent collection preserves independence. Consequently, the whole list of transformed increments has the same joint distribution as the original list.
For each \(r\), \(B_{t_r}=\sum_{q=1}^r Z_q\), and \(Y_{t_r}=\sum_{q=1}^r UZ_q\). Both lists of process values are obtained from their corresponding increment lists by the same deterministic cumulative-sum map. Applying that map to vectors with the same distribution gives vectors with the same distribution. This proves equality of finite-dimensional distributions for all choices of times.
To verify the Brownian-motion assertion directly, \(Y_0=UB_0=0\), and \(Y\) has continuous paths because it is a fixed linear transformation of a continuous path. Its increments over disjoint intervals are independent, as they are transformations of independent increments of \(B\). For \(s<t\), the characteristic-function calculation above shows that \(Y_t-Y_s=U(B_t-B_s)\) has distribution \(N_d(0,(t-s)I_d)\). These are precisely the stated defining properties of standard \(d\)-dimensional Brownian motion. \(\square\)
The covariance calculation gives a compact way to remember the result. For \(s,t\ge0\), the covariance matrix of \(B_s\) and \(B_t\), with entries \(\operatorname{Cov}(B_s^a,B_t^b)\), is \(\min(s,t)I_d\). Therefore
This calculation is a useful check, but covariance alone would not in general determine a distribution. Here the full conclusion follows because Brownian motion is Gaussian, or, as in the proof, because the transformed increments have the same Gaussian distributions and independence.
Worked Examples
Worked Example: Rotating Planar Brownian Motion
Take \(d=2\) and rotate by angle \(\pi/3\). The rotation matrix is
Its first row has squared length \((\tfrac12)^2+(-\tfrac{\sqrt3}{2})^2=\tfrac14+\tfrac34=1\); its second row has squared length \((\tfrac{\sqrt3}{2})^2+(\tfrac12)^2=\tfrac34+\tfrac14=1\). Their inner product is \(\tfrac12\cdot\tfrac{\sqrt3}{2}+(-\tfrac{\sqrt3}{2})\cdot\tfrac12=0\), so the rows are orthonormal and \(U\) is orthogonal.
Writing \(B_t=(B_t^1,B_t^2)^T\), the rotated coordinates are \(Y_t^1=\tfrac12B_t^1-\tfrac{\sqrt3}{2}B_t^2\) and \(Y_t^2=\tfrac{\sqrt3}{2}B_t^1+\tfrac12B_t^2\). At \(t=2\), each original coordinate has variance \(2\), and the coordinates are independent. Hence
and
The transformed vector is centered Gaussian, so its zero coordinate covariance implies that its two coordinates are independent. The calculation agrees with \(Y_2\sim N_2(0,2I_2)\), as the theorem requires.
Worked Example: Reflection in Three Dimensions
Consider \(U=\operatorname{diag}(1,1,-1)\), which reflects the third coordinate. Its rows are \((1,0,0)\), \((0,1,0)\), and \((0,0,-1)\). Each row has squared length \(1\), and the inner product of any two distinct rows is \(0\); hence \(U\) is orthogonal. It has determinant \(-1\), so this example is a reflection, not a rotation with determinant \(1\).
If \(B_t=(B_t^1,B_t^2,B_t^3)^T\), then \(Y_t=(B_t^1,B_t^2,-B_t^3)^T\). For any \(s,t\ge0\), the covariance matrix is
The sign change in the third coordinate does not change its centered Gaussian law or its independence from the first two coordinates. Thus the reflection preserves the joint law at every finite set of times, not just the distribution at one fixed time.
Worked Example: A One-Dimensional Projection
Let \(u=(\tfrac35,\tfrac45)^T\), which has squared length \(\tfrac9{25}+\tfrac{16}{25}=1\). Complete it to the orthogonal matrix
The first row has squared length \(\tfrac9{25}+\tfrac{16}{25}=1\), the second has squared length \(\tfrac{16}{25}+\tfrac9{25}=1\), and their inner product is \(-\tfrac{12}{25}+\tfrac{12}{25}=0\). Therefore \(U\) is orthogonal. The first coordinate of \(UB_t\) is \(Z_t=\tfrac35B_t^1+\tfrac45B_t^2\). By the theorem, \(UB\) is standard two-dimensional Brownian motion, so its first coordinate \(Z\) is standard one-dimensional Brownian motion.
For a direct check at time \(t=5\), independence of \(B^1\) and \(B^2\) gives \(\mathbb{E}[Z_5]=0\) and
For \(0\le s\le t\), the same calculation gives \(\operatorname{Cov}(Z_s,Z_t)=(\tfrac9{25}+\tfrac{16}{25})s=s\). The theorem supplies the full increment and independence properties as well; this covariance check alone is not being used as a substitute for them.
What Invariance Does—and Does Not—Say
Orthogonal invariance is useful whenever a calculation becomes simpler after choosing a direction or rotating coordinates. For example, a projection onto any unit direction has the same one-dimensional Brownian law. More generally, no fixed spatial orientation is singled out by standard multidimensional Brownian motion. The statement includes every orthogonal matrix, whether its determinant is \(1\) or \(-1\).
Equality of finite-dimensional distributions is an equality of probability laws, not a claim that \(UB_t=B_t\) for each sample path. In general, the two random vectors differ. For instance, in the reflection example, equality at time \(t>0\) would require \(B_t^3=0\), an event of probability zero for a nondegenerate centered Gaussian coordinate. The theorem says instead that the reflected random vector has the same distribution.
Common mistake. A determinant of \(1\) is not enough to establish invariance. For example, \(A=\operatorname{diag}(2,\tfrac12)\) has determinant \(1\), but it is not orthogonal. At time \(1\), the covariance matrix of \(AB_1\) is
not \(I_2\). The incorrect calculation would replace \(AA^T\) by \(I_2\) based only on \(\det(A)=1\). The correction is to check \(A^T A=I_2\) (equivalently \(AA^T=I_2\)) before invoking orthogonal invariance. A general volume-preserving linear map can stretch one direction and contract another, changing the Brownian law.
Check Your Understanding
For each question, distinguish a statement about a single time from one about the full finite-dimensional law.
- What two matrix identities follow for a square orthogonal matrix, and why are they equivalent?
- For a fixed orthogonal \(U\), why do the transformed increments over disjoint intervals remain independent?
- Does the matrix \(\operatorname{diag}(1,-1)\) preserve standard planar Brownian motion? Justify your answer using its rows or its transpose.
- Let \(u\in\mathbb{R}^d\) be a unit vector. Why does the scalar projection \(u^TB_t\) have the standard one-dimensional Brownian law?
- Explain why equality in finite-dimensional distributions does not imply \(UB_t=B_t\) for every \(t\) on each sample path.