Brownian Scaling and Self-Similarity

Brownian Motion at Different Scales

Brownian motion has fluctuations whose typical size grows like the square root of elapsed time. Scaling time by a factor \(c\) therefore requires scaling space by \(c^{-1/2}\) to recover the same distribution. This self-similarity is a statement about probability laws: it does not say that a Brownian path is unchanged when its time and space coordinates are rescaled.

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 paths almost surely. In particular, its finite-dimensional vectors are centered Gaussian, and for \(s,t\ge0\), \(\operatorname{Cov}(B_s,B_t)=\min(s,t)\). The distribution of a Gaussian vector is determined by its mean vector and covariance matrix, including when that matrix is singular.

The Scaling Theorem

Proof. Fix \(c>0\), an integer \(m\ge1\), and times \(t_1,\ldots,t_m\ge0\). The vector \((B_{ct_1},\ldots,B_{ct_m})\) is Gaussian, so its deterministic scalar multiple \((Y_{t_1},\ldots,Y_{t_m})\) is Gaussian as well. Each coordinate has mean zero. For any indices \(r,q\in\{1,\ldots,m\}\), its covariance is

Thus the vector of \(Y\)-values has the same mean vector and covariance matrix as \((B_{t_1},\ldots,B_{t_m})\). Uniqueness of the Gaussian law from its mean and covariance gives equality of these joint distributions. This reasoning allows repeated times and times equal to zero; a resulting covariance matrix may be singular, but Gaussian-law uniqueness still applies. Since \(t\mapsto ct\) and multiplication by \(c^{-1/2}\) preserve continuity, every continuous sample path of \(B\) gives a continuous sample path of \(Y\). Hence \(Y\) has the stated Brownian properties. \(\square\)

The covariance calculation is the essential point: scaling the time argument multiplies Brownian covariance by \(c\), while scaling both spatial coordinates by \(c^{-1/2}\) multiplies covariance by \(c^{-1}\). The two factors cancel. The conclusion holds for every finite collection of times, not just for one-time distributions.

Rescaling Terminal Interval Probabilities

At a single time \(T>0\), \(B_T\) is normal with mean zero and variance \(T\). Scaling gives \(B_T/\sqrt{T}\) the same distribution as \(B_1\). If \(\Phi(x)\) denotes the standard normal distribution function, then for real numbers \(a\le b\),

The endpoints are included in the event as written. A nondegenerate normal variable assigns probability zero to either endpoint, so replacing both weak inequalities by strict ones gives the same probability when \(T>0\). At \(T=0\), however, \(B_0=0\) with probability one, so the interval probability must instead be checked directly against whether \(0\) belongs to the interval.

The same idea applies to a finite vector of observations and hence to events that impose interval restrictions at several times. All observation times and all spatial endpoints must be rescaled consistently. Scaling only the times, or only the spatial interval, changes the event.

Confinement to an Interval Over a Time Horizon

Scaling also transfers probabilities for an entire continuous path to remain in a fixed interval. This requires more than the one-time normal formula: the event refers to infinitely many times. Continuity and finite-dimensional distributions suffice for the specific transfer below.

Proof. Set \(Y_u=T^{-1/2}B_{Tu}\) for \(u\in[0,1]\). The scaling theorem gives equality of the finite-dimensional distributions of \(Y\) and \(B\), and both processes have continuous paths almost surely. Enumerate the rational numbers in \([0,1]\) as \(r_1,r_2,\ldots\). For each positive integer \(m\), let \(A_m\) be the event that \(-a/\sqrt T\le Y_{r_k}\le b/\sqrt T\) for \(1\le k\le m\), and let \(C_m\) be the event that \(-a/\sqrt T\le B_{r_k}\le b/\sqrt T\) for \(1\le k\le m\). Finite-dimensional distribution equality gives \(\mathbb{P}(A_m)=\mathbb{P}(C_m)\) for every \(m\). These events decrease as \(m\) increases, so continuity of probability for decreasing events gives \(\mathbb{P}(\bigcap_m A_m)=\mathbb{P}(\bigcap_m C_m)\). On continuous paths, satisfying the interval bounds at every rational time is equivalent to satisfying them at every time in \([0,1]\): for any time, take a sequence of rational times converging to it and use continuity and the closedness of the interval. Thus the two probabilities are equal for the full path events. Finally, \(Y_u=T^{-1/2}B_{Tu}\), so its confinement to \([-a/\sqrt T,b/\sqrt T]\) over \([0,1]\) is exactly the confinement of \(B_s\) to \([-a,b]\) over \([0,T]\), under the substitution \(s=Tu\). This proves the identity. \(\square\)

What Scaling Does—and Does Not—Say

Brownian scaling is an equality in distribution, not a pathwise identity. The theorem says that for each fixed \(c>0\), the random vectors \((c^{-1/2}B_{ct_1},\ldots,c^{-1/2}B_{ct_m})\) and \((B_{t_1},\ldots,B_{t_m})\) have the same law. It does not say their coordinates are equal for the same outcome \(\omega\in\Omega\). For instance, at any fixed \(t>0\), the two random variables \(c^{-1/2}B_{ct}\) and \(B_t\) have the same normal distribution, but the theorem gives no reason for them to coincide pointwise.

Common mistake. A tempting but incorrect calculation is \(\operatorname{Var}(c^{-1/2}B_{ct})=c\operatorname{Var}(B_t)=ct\). This omits the square of the spatial multiplier. The correct computation is

Another error is to rescale time while leaving a spatial interval unchanged. For example, changing the horizon from \(T\) to \(1\) requires dividing the interval endpoints by \(\sqrt T\); otherwise the event is different. Scaling also does not, by itself, establish that increments at different locations have the same distribution. The result proved here is self-similarity of finite-dimensional laws, and its interval-probability consequences follow by applying the same time and space transformation consistently.