By Peres Y.

Those notes list lectures I gave on the information division, college of California, Berkeley in Spring 1998. i'm thankful to the scholars who attended the direction and wrote the 1st draft of the notes: Diego Garcia, Yoram Gat, Diogo A. Gomes, Charles Holton, Frederic Latremoliere, Wei Li, Ben Morris, Jason Schweinsberg, Balint Virag, Ye Xia and Xiaowen Zhou. The draft was once edited by means of Balint Virag, Elchanan Mossel, Serban Nacu and Yimin Xiao. I thank Pertti Mattila for the invitation to lecture in this fabric on the joint summer season university in Jyvaskyla, August 1999.

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**Extra info for An invitation to sample paths of Brownian motion**

**Sample text**

To see the equivalence, note that l is a cut line if and only if the Brownian motion, projected onto a line perpendicular to l, has a point of increase. It was proved by Bass and Burdzy (1997) that Brownian motion almost surely does not has cut lines. It is still unknown whether a Gaussian random walk in the plane will have cut lines. Burdzy (1989) showed that Brownian motion in the plane almost surely does have cut points, which are points B(t0 ) such that the Brownian motion path with the point B(t0 ) removed is disconnected.

Thus, P(Fn (j) ∩ In (k)) ≥ P(Fj (j)) p2k−j pn−k ≥ p2k−j P(Fj (j)) P (Sj is minimal among Sj , . . , Sn) , since pn−k ≥ pn−j . Here the two events on the right are independent, and their intersection is precisely Fn (j). Consequently P(Fn (j) ∩ In (k)) ≥ p2k−j P(Fn (j)) . Decomposing the event In (k) according to the first point of increase gives n n pk pn−k k=0 n k P(In (k)) = = k=0 j=0 k=0 n/2 P(Fn (j) ∩ In (k)) j+ n/2 ≥ n/2 p2k−j P(Fn (j)) j=0 k=j ≥ n/2 P(Fn (j)) j=0 p2i . 3) i=0 This yields an upper bound on the probability that {Sj }nj=0 has a point of increase by time n/2; but this random walk has a point of increase at time k if and only if the “reversed” walk {Sn − Sn−i }ni=0 has a point of increase at time n − k.

By the law of large numbers λn /n will converge to 1, and the corollary follows. 7 (Wald’s Lemma for Brownian Motion). Let τ be a stopping time for Brownian motion such that E[τ ] < ∞, then E[B(τ )] = 0. Sketch of Proof. Let Xi be independent and have the distribution of B(τ ). , then it would follow that E[B(τ )] = 0. s. ). Define τn inductively by stopping the Brownian motion {B(t) − B(τn−1 )}t≥τn−1 at the stopping time τ . s. , and therefore, limn→∞ n i=1 n Xi = 15. SKOROKHOD’S REPRESENTATION 39 15.