m507b-hw1soln-s08

M507b-hw1soln-s08 - 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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 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 ....
View Full Document

This note was uploaded on 06/23/2008 for the course MATH 507A taught by Professor Alexander during the Spring '08 term at USC.

Page1 / 3

M507b-hw1soln-s08 - 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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online