Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 21 / 4

False. In fact, standard one-dimensional Brownian motion satisfies ${\mathrm{liminf}}_{t\to \infty }{B}_t=-\infty$ almost surely. For each integer $m\ge 1$, let ${\tau }_{-m}=\inf \{t:B_t=-m\}$. The Brownian reflection principle, applied to $-B$, giveswhere $\Phi$ is the standard normal distribution function. Thus each negative integer is reached almost surely. Intersecting these countably many probability-one events, all negative integer levels are reached on the same path. Their hitting times must become arbitrarily large, because a continuous path is bounded on every compact time interval. Hence the path cannot tend to positive infinity. Applying the same argument to positive levels also gives ${\mathrm{limsup}}_{t\to \infty }{B}_t=+\infty$ almost surely.

True. The negative-level hitting times from the preceding calculation are almost surely finite for every negative integer, and occur at unbounded times. ThereforeThis is the almost-sure negation in the original PDF, not another assertion of convergence. The stronger oscillation statement $\mathrm{liminf}{B}_t=-\infty$ and $\mathrm{limsup}{B}_t=+\infty$ rules out that convergence path by path.

True. The unit-time increments $Z_k=B_k-B_{k-1}$ are independent and identically distributed random variables with standard normal distribution, hence integrable with mean zero. The strong law of large numbers gives $B_n/n\to 0$ almost surely at integer times.

To control the intervals between integers, put $M_n={\sup }_{0\le u\le 1}|B_{n+u}-B_n|$. Stationary increments and the Doob L2 maximal inequality give $\mathbb{E}{M}_n^2\le 4$. Thus the Markov inequality yieldsThe Borel-Cantelli lemma, followed by a countable intersection over positive rational $\epsilon$, implies $M_n/n\to 0$ almost surely. For $n\le t\le n+1$,Both terms tend to zero on the same probability-one event, proving $\boxed{B_t/t\to 0\text{ almost surely}}$ for continuous time as well.

False. The preceding interpolation argument proves that the event $\{B_t/t\to 0\}$ has probability one. Its complementary event, which is the nonconvergence assertion printed here, therefore has probability zero. The normalization by $t$ dominates both the integer-time increments and the fluctuations within each unit interval.

False. For every deterministic $t>0$, Brownian scaling gives $B_t/\sqrt{t}\sim N(0,1)$. In particular, for any $\epsilon >0$,independently of $t$. Almost-sure convergence would imply convergence in probability, but these fixed positive tail probabilities rule out convergence in probability to zero. The square-root normalization retains fluctuations of order one.

True. The preceding distributional argument alone shows that almost-sure convergence to zero cannot hold with probability one. To establish almost-sure nonconvergence, rather than merely failure of a probability-one convergence claim, use the Kolmogorov zero-one law.

Let $A=\{B_n/\sqrt{n}\to 0\}$. For every fixed $m$, subtracting $B_m/\sqrt{n}\to 0$ shows thatThus $A$ belongs to the tail sigma-algebra of the independent unit-time increments; it is a tail event. The Kolmogorov zero-one law gives $\mathbb{P}(A)\in \{0,1\}$. If its probability were one, then $B_n/\sqrt{n}\to 0$ in probability, contrary to its standard normal distribution at every $n$. Therefore $\mathbb{P}(A)=0$. Continuous-time convergence would imply integer-time convergence, soThe more general Brownian fluctuations exceed the square-root scale result is consistent with this direct tail-event proof. No independence of the normalized values themselves has been assumed.