By J.-M. Bismut, L. Gross, K. Krickeberg, P. L. Hennequin

**Read or Download Ecole d'Ete de Probabilites de Saint-Flour X - 1980. Proceedings PDF**

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**Additional resources for Ecole d'Ete de Probabilites de Saint-Flour X - 1980. Proceedings**

**Sample text**

Ii) If EX = 0 and EX 2 < ∞, then τ can be chosen to have finite mean. Only part (ii) of the theorem is useful. Proof. (i) Pick X according to its distribution. Define τ = min{t : B(t) = X}. s. s (ii) Let X have distribution ν on R. , ν({0}) = 0. For, suppose ν({0}) > 0. Write ν = ν({0})δ0 + (1 − ν({0})˜ ν , where the distribution ν˜ has no mass on {0}. Let stopping time τ˜ be the solution of the problem for the distribution ν˜. The solution for the distribution ν is, τ= τ˜ with probability 1 − ν({0}) 0 with probability ν({0}).

For the lower bound, fix q > 1. In order to use the Borel-Cantelli lemma in the other direction, we need to create a sequence of independent events. 5 for large x: P(Z > x) ≥ ce−x x 2 /2 . Using this estimate we get P(Dn) = P Z ≥ and therefore ≥c q n − q n−1 n P(Dn ) e− log log(q −q ) ce− log(n log q) c ≥ > n n−1 n log n 2log log(q − q ) 2 log(n log q) n ψ(q n − q n−1 ) n−1 = ∞. Thus for infinitely many n B(q n ) ≥ B(q n−1 ) + ψ(q n − q n−1 ) ≥ −2ψ(q n−1 ) + ψ(q n − q n−1 ) where the second inequality follows from applying the previously proven upper bound to −B(q n−1 ).

Also, ∞ ∞ ∞ µ(Sk (x))2kα ≤ C k=1 |22−k |β 2kα = C k=1 2k(α−β), k=1 2β where C = 2 C. Since β > α, we have ∞ Eα (µ) ≤ C 2k(α−β) < ∞, k=1 which proves the theorem. Definition. The α-capacity of a set K, denoted Capα (K), is inf Eα (µ) µ −1 , where the infimum is over all Borel probability measures supported on K. If Eα(µ) = ∞ for all such µ, then we say Capα (K) = 0. 3 (McKean, 1955). Let B denote Brownian motion in Rd . Let A ⊂ [0, ∞) be a closed set such that dimH (A) ≤ d/2. Then, almost surely dimH B(A) = 2 dimH (A).