Brownian Paths Are Not Monotone on Any Interval

From Gaussian Signs to Pathwise Oscillation

Brownian motion has increments that are independent and symmetrically distributed around zero. On any fixed partition, each increment therefore has probability one-half of being positive and one-half of being negative. A path that is nondecreasing must have nonnegative increments on every partition, however fine. The probability of that sign pattern on an \(n\)-piece partition is \(2^{-n}\), which tends to zero. A further countability argument upgrades this fixed-interval calculation to a statement about every time interval simultaneously.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion with state space \(\mathbb{R}\) and continuous paths almost surely. Thus \(B_0=0\), increments over disjoint time intervals are independent, and for \(0\le s<t\), \(B_t-B_s\) has distribution \(N(0,t-s)\). We may use the natural filtration \(\mathcal{F}_t=\sigma(B_u:0\le u\le t)\); the argument below uses independent increments but does not require conditioning on this filtration.

We consider intervals of positive length. A singleton contains no pair of distinct times, so no statement asserting that it contains an upward and a downward fluctuation could hold. The aim is stronger than saying that a typical fixed interval has fluctuations: with probability one, the same path has both kinds of fluctuation in every interval of positive length.

The Fixed-Interval Argument

Proof. Fix a positive integer \(n\), and divide \([a,b]\) into \(n\) equal pieces with endpoints \(t_r=a+r(b-a)/n\), \(r=0,\ldots,n\). The increments \(B_{t_r}-B_{t_{r-1}}\), \(r=1,\ldots,n\), are independent, and each has distribution \(N(0,(b-a)/n)\). Since this is a centered Gaussian distribution with positive variance, it has no atom at zero and is symmetric, so each increment is nonnegative with probability \(1/2\), and nonpositive with probability \(1/2\). Independence gives probability \(2^{-n}\) that all \(n\) increments are nonnegative, and also probability \(2^{-n}\) that all \(n\) increments are nonpositive. If the path is nondecreasing on \([a,b]\), all the increments in this partition must be nonnegative. Consequently the probability of nondecreasing paths is at most \(2^{-n}\) for every positive integer \(n\). Since \(2^{-n}\to0\), that probability is zero. The nonincreasing case follows by the same argument using nonpositive increments. \(\square\)

The partition in this proof is chosen for each \(n\) separately. There is no claim that the sign events for different values of \(n\) are independent; they need not be. The argument only uses that the monotonicity event is contained in each one of these events, so its probability is bounded by \(2^{-n}\) for every \(n\).

Making the Statement Simultaneous

The proposition handles each fixed interval, but there are uncountably many intervals. It would not be valid simply to take an uncountable union of probability-zero events. Instead, use intervals with rational endpoints. There are only countably many such intervals, and every interval of positive length contains one of them.

Proof. For each pair of rational numbers \(a,b\) with \(0\le a<b\), let \(E^+_{a,b}\) be the event that the path is nondecreasing on \([a,b]\), and let \(E^-_{a,b}\) be the event that it is nonincreasing there. The proposition gives \(\mathbb{P}(E^+_{a,b})=\mathbb{P}(E^-_{a,b})=0\). There are countably many rational pairs, so the union of all these events also has probability zero: by countable subadditivity, its probability is at most the sum of countably many zeros. Outside this null event, the path is monotone in neither direction on any rational-endpoint interval of positive length.

Now take any interval \(J\subseteq[0,\infty)\) of positive length. It contains rational numbers \(a<b\) such that the closed interval \([a,b]\) lies inside \(J\). On this rational interval, failure to be nondecreasing means, by the definition of nondecreasing, that there exist \(s<t\) in \([a,b]\) with \(B_s>B_t\); this is a downward fluctuation in \(J\). Failure to be nonincreasing means there exist \(u<v\) in \([a,b]\) with \(B_u<B_v\); this is an upward fluctuation in \(J\). Both conclusions hold for every \(J\) on the same probability-one event. \(\square\)

For measurability, on the probability-one event of continuous paths, monotonicity on a fixed interval can be tested by comparing values at rational times in that interval: continuity extends those inequalities to all times. Thus the events used in the countable argument are measurable up to the irrelevant null set on which continuity might fail.

Why a Single Sign Check Is Not Enough

A common mistake is to use only the increment across the endpoints of an interval. For example, \(B_1-B_0\) is \(N(0,1)\), so the probability that it is nonnegative is \(1/2\). It is incorrect to conclude from this that the probability of a nondecreasing path on \([0,1]\) is \(1/2\). The endpoint comparison is necessary for monotonicity, but it does not check what happens between the endpoints.

Correction: for every positive integer \(n\), nondecreasing behavior on \([0,1]\) requires all \(n\) increments on the equal partition to be nonnegative. Those increments are independent, each with probability \(1/2\) of being nonnegative, so the probability of that necessary condition is \(2^{-n}\). In particular, for \(n=1\) the bound is \(1/2\); for \(n=2\) it is \(1/4\); and for \(n=5\) it is \(1/32\). Since the same pathwise monotonicity event must satisfy every one of these bounds, its probability is zero. This argument also explains why one fixed finite mesh, however useful as a check, does not establish a path property.

The result is a pathwise statement, not merely a claim about increments at one selected collection of times. Its force comes from two steps: independent increments make the probability of a one-sided sign pattern small on increasingly fine partitions, and rational intervals reduce the simultaneous assertion over all intervals to a countable union. The conclusion supplies both signs of fluctuation in every interval, while making no claim about the size of those fluctuations.