Why Brownian Motion Is Nowhere Differentiable

Brownian motion is almost surely nowhere differentiable because its increments typically scale like

\sqrt h

over an interval of length

h

. A finite derivative would require changes near a point to be bounded by a constant times

h

. Brownian paths almost surely have fluctuations that violate that bound at every time. Financial models driven by Brownian motion use stochastic integrals and Itô's formula to handle this roughness.

Why the proof needs three neighbouring increments

The argument has to rule out a derivative at any time, including a time picked after seeing the path. Showing that the difference quotient behaves badly at each fixed deterministic time is insufficient for that conclusion. The grid proof below works simultaneously across the interval. Three small increments near a candidate point make the event rare enough even after you count every possible location on the grid.

[0,1]

use grids with spacing

h=2^{-n}

. If a path has a finite derivative at some

t

, there is an integer

M

for which

|W_u-W_t|\leq M|u-t|

for all sufficiently close

u

. On every sufficiently fine grid, choose three consecutive grid intervals whose endpoints are within

4h

t

. Each of their increments must then have absolute value at most

8Mh

by the triangle inequality.

For any given triple, the increments are independent $N(0,h)$ variables. For $Z\sim N(0,1)$ ,

\Pr(|Z|\leq 8M\sqrt h)\leq C_M\sqrt h

for a finite constant

C_M

, since the standard normal density is bounded. There are at most

1/h

candidate triples. The union bound gives a probability at most

C_M^3\sqrt h

that any triple satisfies the required inequalities. Summing this over dyadic grids is finite, so Borel–Cantelli says the event occurs only finitely often almost surely. Taking a countable union over

M

excludes every possible point of differentiability at once. The same reasoning on each unit interval covers all finite times.

The normal density of an increment with variance $h$ is bounded by $1/\sqrt{2\pi h}$ . The chance of landing in an interval of width proportional to $h$ is therefore at most a constant times $\sqrt h$ . Three disjoint increments produce an upper bound of order $h^{3/2}$ for a specified triple. There are of order $1/h$ candidate triples on the grid, so even after allowing all possible locations the probability is of order $\sqrt h$ . On dyadic grids these bounds are summable. This is the step that lets the argument handle a point chosen from the path itself.

What the roughness means at market timescales

In geometric Brownian motion,

d\log S_t=(\mu-\sigma^2/2)dt+\sigma\,dW_t

. Over one day modelled as

1/252

years and with

\sigma=20\%

, the random log return has standard deviation about

1.26\%

. Over one quarter of that interval, the model's standard deviation falls to about

0.63\%

. The size of the move falls, but the standard deviation of the move per unit time doubles.

Shorter sampling intervals make the slope a noisier estimate of instantaneous drift in this model. For pricing and hedging equations, calculus works through stochastic differentials and their quadratic variation. At very high observation frequencies, empirical prices also exhibit features such as ticks and bid–ask bounce that the continuous Brownian model abstracts away.

References and next steps

MIT OpenCourseWare, Theory of Probability, Brownian motion lecture

Why the proof needs three neighbouring increments

What the roughness means at market timescales

Explore a Brownian path at different resolutions

References and next steps