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Unformatted text preview: Stat 150 Stochastic Processes Spring 2009 Lecture 6: Markov Chains and First Step Analysis II Lecturer: Jim Pitman 1 Further Analysis of Markov Chain In class last time : We found that if h = Ph , then (1) E [ h ( X n +1 )  X ,X 1 ,...,X n ] = h ( X n ) (2) E [ h ( X n +1 )  h ( X ) ,h ( X 1 ) ,...,h ( X n )] = h ( X n ) h ( X n ) is a MC. Why ? This illustrates a conditional form of E [ Y ] = E [ E [ Y  X ]], which can be written more generically as ( ) E [ Y  Z ] = E [ E [ Y  X,Z ]  Z ] To obtain (2), take Y = h ( X n +1 ), Z = ( h ( X ) ,h ( X 1 ) ,...,h ( X n )), X = ( X ,X 1 ,...,X n ). Notice that ( X,Z ) = ( X ,X 1 ,...,X n ,h ( X ) ,h ( X 1 ) ,...,h ( X n )) is the same in formation as ( X ,X 1 ,...,X n ) since Z is a function of X . So E [ h ( X n +1 )  h ( X ) ,h ( X 1 ) ,...,h ( X n )] = E [ E [ h ( X n +1 )  X ,X 1 ,...,X n ,h ( X ) ,h ( X 1 ) ,...,h ( X n )]  h ( X ) ,h ( X 1 ) ,...,h ( X n )] = E [ E [ h ( X n +1 )  X ,X 1 ,...,X n ]  h ( X ) ,h...
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 Spring '09
 Pitman,Jim
 Markov Chains

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