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m507b-hw1soln-s08

# m507b-hw1soln-s08 - MATH 507b ASSIGNMENT 1 SOLUTIONS SPRING...

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MATH 507b ASSIGNMENT 1 SOLUTIONS SPRING 2008 Prof. Alexander (4.4) Fix n and define the bounded stopping time N = n min { k : | S ) k | ≥ x } . Since | ξ m | ≤ K we have | S N | ≤ x + K , so ES 2 N ( x + K ) 2 . Also Es 2 N E ( s 2 N 1 { N = n } ) = s 2 n P ( N = n ) s 2 n P (max m n | S m | ≤ x ) . By Exercise 2.6, { S 2 n - s 2 n } is a martingale, so by Theorem 4.1, 0 = E ( S 2 n - s 2 n ) = E ( S 2 N - s 2 N ) ( x + K ) 2 - s 2 n P (max m n | S m | ≤ x ) . Equivalently, P (max m n | S m | ≤ x ) ( x + K ) 2 s 2 n . (4.5) Since ( x + c ) 2 is a convex function of x and { X n } is a martingale, { ( X n + c ) 2 } is a submartingale for all c . Consider c > 0 and λ > 0. By Doob’s Inequality, since EX n = EX 0 = 0 for all n , P (max m n X m λ ) P (max m n ( X m + c ) 2 ( λ + c ) 2 ) (1) E ( X n + c ) 2 ( λ + c ) 2 = EX 2 n + c 2 ( λ + c ) 2 . Let f ( c ) be the fraction on the right side of (1). Since (1) is valid for all c > 0, we can minimize f ( c ) over c : f ( c ) = 0 for c = EX 2 n by easy calculus, so P (max m n X m λ ) f EX 2 n λ = EX 2 n EX 2 n + λ 2 .

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