Stationary Increments, Symmetry, and Sign Reversal

Stationary Increments Are Not Stationary Levels

Brownian motion changes its distribution as time passes: \(B_t\) has variance \(t\), so its level at time \(t\) does not have the same law as its level at an earlier time. Its increments have a different property. The change over an interval has a distribution determined by the interval’s length, not by its starting time. This is called stationary increments. The centered Gaussian increment law is also symmetric about zero, which leads to an exact symmetry of the entire Brownian path law under sign reversal.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \((B_t)_{t\ge0}\). Thus \(B_0=0\), its paths are continuous almost surely, and for \(0\le s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). In particular, increments over disjoint time intervals are independent.

For Brownian motion the defining Gaussian increment property gives the precise statement \(B_{s+h}-B_{t+h}\sim N(0,t-s)\) when \(s\le t\). More generally, shifting a finite collection of ordered disjoint intervals by the same amount preserves the joint law of their increments. This follows because each increment remains centered Gaussian with the same interval length, and the increments remain independent.

Joint Laws on Shifted Intervals

Proof. The shifted intervals \([a_i+h,b_i+h]\) are disjoint except possibly for shared endpoints, just as the original intervals are. Brownian increments over these intervals are therefore independent. By the Gaussian increment property, \(B_{b_i+h}-B_{a_i+h}\) has law \(N(0,(b_i+h)-(a_i+h))=N(0,b_i-a_i)\). The unshifted increments are also independent and have these same respective distributions. Independent random vectors with matching coordinate distributions have the same joint distribution, proving the assertion. \(\square\)

This is a statement about increments, not about the levels \(B_t\). In particular, it does not assert that shifting time leaves the process unchanged. The distinction matters: the distribution of \(B_t\) changes with \(t\), while the distribution of an increment depends only on the elapsed time.

Sign Reversal of Brownian Motion

A centered normal distribution is unchanged when its sign is reversed: if \(Z\sim N(0,v)\), then \(-Z\sim N(0,v)\). Applying this fact to Brownian increments shows that \(-B\) has the same Brownian increment laws as \(B\). There is also a pathwise way to see the transformation: a continuous path \(t\mapsto B_t(\omega)\) is sent to the continuous path \(t\mapsto -B_t(\omega)\).

Proof. First, \(\widetilde B_0=-B_0=0\), and its paths are continuous wherever the paths of \(B\) are continuous. It is adapted because \(B_t\) is \(\mathcal{F}_t\)-measurable, so \(-B_t\) is also \(\mathcal{F}_t\)-measurable. For \(s<t\), \(\widetilde B_t-\widetilde B_s=-(B_t-B_s)\). This increment is independent of \(\mathcal{F}_s\), since \(B_t-B_s\) is independent of \(\mathcal{F}_s\). Its law is \(N(0,t-s)\), because a centered normal variable has the same law as its negative. These are the defining properties of Brownian motion. Finally, for any finite set of times \(t_1,\ldots,t_n\), the vector \((\widetilde B_{t_1},\ldots,\widetilde B_{t_n})\) is therefore a Brownian finite-dimensional vector. It has the same law as \((B_{t_1},\ldots,B_{t_n})\), as can also be seen by expressing each level as a sum of independent centered Gaussian increments and negating all of them. \(\square\)

Equality of finite-dimensional distributions also gives equality of the laws of the continuous paths on a bounded time interval. Here is the precise link. Regard each continuous path on \([0,T]\) as a point of \(C([0,T])\), equipped with the uniform norm. Evaluation at any fixed time is measurable. Moreover, evaluations at rational times in \([0,T]\), together with the endpoint \(T\), determine the uniform distance between continuous paths: the supremum of their pointwise distance over this dense set equals the supremum over all times, by continuity. Thus these evaluations generate the Borel sets of \(C([0,T])\). The standard uniqueness theorem for probability measures on a sigma-algebra generated by a collection closed under finite intersections (the pi-lambda uniqueness theorem) then implies that two probability measures agreeing on all finite-time cylinder sets agree on the Borel sets of \(C([0,T])\). Applying it to \(B\) and \(-B\) proves equality of their path laws.

Path Events and What the Symmetry Says

For a path \(x\in C([0,T])\), define its sign reversal by \(R(x)(t)=-x(t)\). If \(A\) is a Borel set of paths, then \(R^{-1}(A)=\{x:R(x)\in A\}\) is also a Borel set, since \(R\) is continuous in the uniform norm. Equality of the path laws of \(B\) and \(-B\) gives \(\mathbb{P}(B\in A)=\mathbb{P}(-B\in A)=\mathbb{P}(B\in R^{-1}(A))\). This relates path events without requiring their probabilities to be explicitly computed.

For a concrete path illustration, take the continuous function \(x(t)=t\) on \([0,1]\). Its sign reversal is \(R(x)(t)=-t\); the endpoint changes from \(1\) to \(-1\), while the uniform norm remains \(\sup_{0\le t\le1}|x(t)|=1\). This illustrates the transformation of paths, not a claim that either specific function is a Brownian sample path or has positive probability of occurring exactly.

Common Mistakes

Mistaking stationary increments for stationary levels. A common incorrect calculation is to infer that \(B_{s+t}\) has the same distribution as \(B_t\) because the increment over a length-\(t\) interval has the same law wherever it starts. For \(s=2\) and \(t=1\), \(B_3\sim N(0,3)\), whereas \(B_1\sim N(0,1)\); these distributions differ. The correct statement is \(B_3-B_2\sim N(0,1)\), matching the law of \(B_1-B_0\).

Reversing only some signs. The symmetry theorem negates every coordinate of a Brownian finite-dimensional vector, or every value of a path. It does not say that a selected coordinate may be negated while the others are held fixed without changing the joint law. For example, \(B_1\) and \(B_2-B_1\) are independent, but the level \(B_2\) includes both increments; changing one increment affects the joint vector of levels.

Using symmetry to evaluate a path probability without a further argument. Sign reversal equates probabilities of corresponding events, such as an upper crossing and a lower crossing. It does not by itself give their numerical values. Computing a maximum or a hitting probability requires additional results beyond this symmetry.