Let $B_t$ be a standard Brownian motion. Does there with probability one exist a sequence of partitions $\{t_{k, n} : k = 0, 1, \dots, k_n\}$ $$0 = t_{0, n} < t_{1, n} < \dots < t_{k_n, n} = 1,$$with$$\lim_{n \to \infty} \max\{t_{j, n} - t_{j - 1, n} : j = 1, \dots, k_n\} = 0$$and$$\liminf_{n \to \infty} \sum_{j=1}^{k_n} (B(t_{j, n}) - B(t_{j - 1 n}))^2 > 1?$$
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2 Answers
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If the sequence of partitions is fixed/deterministic, the answer is no by an old result of Paul Lévy (P. Levy , Theorie de l'addition des variables aleatoires, Paris, 1937).
If one allows the partitions to depend on the element of the probability space than the quantity will get as large as $2\log \log n$ almost surely where the partitions are restricted to be $1/n$ separated. This is closely related to the law of the iterated logarithm for Brownian motion and was proved by S. J. Taylor in 1972 (S. J. Taylor, Exact asymptotic estimates of Brownian path variation, Duke Math. J. Volume 39, Number 2 (1972), 219-241.)
The introduction to Taylor's paper is a good source for a discussion of this topic including Lévy's result.
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Yes. See Exercise 1.13(a) of Mörters and Peres, Brownian motion.
answered Oct 7, 2015 at 20:50
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$\begingroup$ I should point out the difference in our answers depends on one's interpretation of the question. In exercise 1.13(a) the sequence of partitions are random (that is they vary with each element of the probability space). In this case Taylor's result gives the exact asymptotic dependence. If one uses a deterministic sequence of partitions, which is how I read the question, the answer is 'no' based on Lévy's work. $\endgroup$ Commented Oct 7, 2015 at 21:04
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1$\begingroup$ @MarkLewko "Does there with probability one exist a sequence" seems to be asking whether $P($there exists a sequence$)=1$, though, right? $\endgroup$ Commented Oct 7, 2015 at 21:05
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$\begingroup$ It still seems a bit unclear but it's probably just me. In any event I have reworded my answer. $\endgroup$ Commented Oct 7, 2015 at 21:28