Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 26 / 4

Use the Kolmogorov continuity theorem in its one-parameter form: if on a compact interval a process satisfies $\mathbb{E}|X_t-X_s{|}^p\le C|t-s{|}^{1+\beta }$ for some $p,\beta >0$, it has a modification whose paths are Hölder continuous of every order $\gamma <\beta /p$ on that interval.

For Brownian motion, normal increments give, for every $p>0$,Every such normal moment is finite. Taking $p>2$ yields $\beta =p/2-1$, hence any order below $1/2-1/p$. For a prescribed $0<\alpha <1/2$, choose $p$ with $1/2-1/p>\alpha$.

The continuous modification and the given continuous Brownian motion agree at all rational times on one almost sure event; continuity makes them agree everywhere on the interval. To obtain all exponents and all compact intervals simultaneously, apply the theorem to integer intervals $[0,N]$ and a countable sequence of positive exponents increasing to $1/2$, then intersect these almost sure events. A bound at exponent $\gamma >\alpha$ implies a bound at $\alpha$ on a compact interval. Consequently the Brownian Hölder regularity conclusion iswith finite random constants on one common event of probability one. This uses the usual positive-exponent meaning of Hölder continuity.

A finite right derivative at zero would make the difference quotients $B_{1/n}/(1/n)$ eventually bounded. Therefore that event is contained inFor fixed $K,N$, its probability is at most $\mathbb{P}(|B_{1/n}|\le K/n)$ for each $n\ge N$. By the normal distribution of the Brownian increment,Thus every event in this countable union has probability zero. With probability one, Brownian motion has no finite right derivative at zero, which is the appropriate derivative for its time domain. No independence of the quotients is assumed or needed. This proves the required endpoint case of nowhere differentiability of Brownian motion.

Fix rational $0<p<q$. Divide $[p,q]$ into $N$ equal intervals. Its Brownian increments are independent centered normal variables, so the probability that all $N$ are nonnegative is $2^{-N}$. A nondecreasing path would force this event for every $N$, hence its probability is zero. The same argument with nonpositive increments excludes a nonincreasing path.

There are countably many rational pairs $p<q$, so with probability one none of these intervals supports a monotone path. Every real interval $0<a<b<\infty$ contains such a rational subinterval. Monotonicity on the larger interval would imply monotonicity on that subinterval, a contradiction. Thus the nowhere monotonicity of Brownian motion assertion holds simultaneously:Both nondecreasing and nonincreasing behavior, including a constant path segment, are excluded.