By Jiruse M., Machek J.

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**Extra info for A Bayesian estimate of the risk of Tick-Borne deseases**

**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.