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Unformatted text preview: 6.262: Discrete Stochastic Processes Lecture 24  5/5/10 • Sequential decision theory (when to de cide?) – Cherno ff bound and tilted rv’s – Wald’s identity – Bound on threshold crossing probability – Sequential decisions • Martingales – Isolation of past – Submartingales and supermartingales – Stopped martingales and Wald identity 1 Cherno ff bound review Assume { X n ; n ≥ 1 } is IID sequence with X < and E e Xr < ∞ for r ≥ . Let γ ( r ) = ln( E e Xr ) . Then for a > X Pr { S n ≥ na } ≤ exp { n [ γ ( r ) ra ] } for r > where S n = X 1 + · · · + X n . = = = = = = slope = X r o γ ( r o ) r * slope = γ ( r o ) = a γ ( r o ) r o a r Optimizing r = r o 2 Tilted probabilities Assume X is discrete with PMF p X ( x ) . For any given r , define the tilted PMF q X,r ( x ) = p X ( x ) exp[ rx γ ( r )] , where the tilted X is the same mapping from sample points to real values as X , but the prob ability is di ff erent, tilted to favor large values for r > . = = = = =2 1 1 2 p X ( x )2 1 1 2 = = = = = q X,r ( x ) x q X,r ( x ) = ∑ x p X ( x ) e xr E e Xr = 1 3 Assume X n ; n ≥ 1 are IID and also define them to be IID in this new tilted measure. Then q X n , r ( x 1 , . . . , x n ) = p X n ( x 1 , . . . , x n ) exp[ n i =1 rx i n γ ( r )] Sum over all x 1 , . . . , x n that have a common sum s n . Then q S n ,r ( s n ) = p S n ( s n ) exp[ rs n n γ ( r ) Thus a sum of tilted IID rv’s are related to the untilted sum in the same way as a single tilted rv is related to the untilted rv. This is a key to large deviation analysis. 4 The mean of X , tilted by r , is E r [ X ] = x x p X ( x ) exp[ rx γ ( r ) = ∑ x p X ( x ) exp( rx ) E [exp( rX )] = γ ( r ) Now consider Pr { S n ≥ na } again, using tilted probabilities with r chosen so that γ ( r ) = a . Pr { S n ≥ na } = s n ≥ na p S n ( s n ) = s n ≥ na q S n ,r ( s n ) exp[ rs n + n γ ( r )] ∼ exp[ rna + n γ ( r )] since S n with the tilted measure obeys the CLT and is centered on na . Thus the Cherno ff bound is exponentially tight. 5 Thm: (Wald) Let γ ( r ) = ln( E [exp( rX )]) exist over ( r < < r + ) . Let J be trial at which S n first crosses α > or β < . Then E [exp( rS J J γ ( r )] = 1 for r ∈ ( r , r + ) Pf: q X n ,r ( x 1 , . . . , x n ) = p X n ( x 1 , . . . , x n ) exp[ rs n n γ ( r )] q J,r ( n ) = x n : { J = n } p X n ( x n ) exp[ rs n n γ ( r )] = E [exp( rS n n γ ( r ))  J = n ] p J ( n ) Summing both sides over n completes the proof....
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This note was uploaded on 10/21/2010 for the course EE 5581 taught by Professor Moon,j during the Spring '08 term at Minnesota.
 Spring '08
 Moon,J

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