This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ∞ ) = 1. Let V ( t ) = exp {2 μX ( t ) } , which is a martingale. As T a (b ) is stopping time, we have (2.3) E [ V ( T a (b ) )] = E [ V (0)] = 1 Hence (2.4) 1 = P ( X ( T a (b ) ) = a ) exp {2 μa } + P ( X ( T a (b ) ) =b ) exp {2 μ (b ) } . And (2.5) P ( T a < Tb < ∞ X (0) = x ) = P ( X ( T a (b ) ) = a  X (0) = x ) = P ( X ( T ( ax )(bx ) ) = ax  X (0) = 0) . 1 2XIXI WANG, CITED FROM THE BOOK “APPLIED STOCHASTIC PROCESS” BY YUANLIE LIN, TSINGHUA UNIVERSITY PRESS Therefore P ( T a < Tb < ∞ X (0) = x ) = 1exp { 2 μ ( b + x ) } exp {2 μ ( ax ) } exp { 2 μ ( b + x ) } = exp { 2 μb } exp {2 μx } exp { 2 μb } exp {2 μa } (2.6) The problem gets proved....
View
Full
Document
This note was uploaded on 10/02/2009 for the course STAT 87528 taught by Professor Pitman,jim during the Spring '09 term at Berkeley.
 Spring '09
 Pitman,Jim

Click to edit the document details