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By Peres Y.

Those notes checklist lectures I gave on the statistics division, collage of California, Berkeley in Spring 1998. i'm thankful to the scholars who attended the path 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 used to be edited through 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 time institution in Jyvaskyla, August 1999.

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S. , and therefore, limn→∞ n i=1 n Xi = 15. SKOROKHOD’S REPRESENTATION 39 15. Skorokhod’s representation Our goal here is to find a stopping time with finite expected value at which Brownian motion has a given mean 0 distribution. If the distribution is on two points a < 0 < b, then this is easy. Define τa,b = min{t : Bt ∈ {a, b}}. 1) Let τ = τa,b . Then, by Wald’s Lemma, 0 = EBτ = aP(Bτ = a) + bP(Bτ = b). Then, P(Bτ = a) = b , |a| + b P(Bτ = b) = |a| . |a| + b By the corollary of Wald’s Lemma, Eτ = EBτ2 = a2 b b2 |a| + = |a|b.

0≤t≤1 n n n Also, Eψ({B(t)}0≤t≤1) = Eφ( max B(t)). 0≤t≤1 Then, by Donsker’s Theorem, Eφ( max1≤k≤n Sk √ ) −→ Eφ( max B(t)). 3. 1) max{1 ≤ k ≤ n : Sk Sk−1 ≤ 0} =⇒ max{0 ≤ t ≤ 1|B(t) = 0} n The left hand side is the last time between 1 to n, scaled by n, that the random walk crosses 0. The right hand side is the last zero of Brownian motion in [0, 1]. Its distribution can be explicitly calculated. 16. 1), define the function ψ by ψ(f ) = max{t ≤ 1 : f (t) = 0}. ψ is not a ˜ 1]. But it is continuous at every f ∈ C[0, ˜ 1] with the property continuous function on C[0, that f (ψ(f ) − ε, ψ(f ) + ε) contains a neighborhood of 0 for every ε > 0.

D satisfies the Poincare Cone condition, and f is a continuous function on ∂D. 1) x→a,x∈D for each a ∈ ∂D. 52 1. BROWNIAN MOTION Proof. 1. To prove existence, let W be a Brownian motion in Rd and define u(x) = Ex f (Wτ∂D ), where τA = inf{t ≥ 0 : Wt ∈ A} for any Borel set A ⊂ Rd . For a ball B(x, r) ⊂ D, the strong Markov property implies that u(x) = Ex [Ex [f (Wτ∂D )|FτS(x,r) ]] = Ex [u(WτS(x,r) )] = S(x,r) u(y)µr (dy), where µr is the uniform distribution on the sphere S(x, r). 2). 1). Fix z ∈ ∂D, then there is a cone with height h > 0 and angle α > 0 in Dc based at z.

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