Bagging, Boosting and Ransac - LASA

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MACHINE LEARNING - 20138Proof of convergenceHypothesis: The average will converge to something meaningful• Assumptions• Y (1) ,...,Y (m) are iid• E(Y) = y (E(Y) is an unbiased estimator of y)• Expected ErrorE((Y y) 2 )=E((Y E(Y )) 2 = 2 (Y )• With AggregationZ = 1 mXY (i) E(Z) = 1 mmi=1mXE(Y (i) )= 1 mi=1mXy = yi=1E((Z y) 2 )=E((Z E(Z)) 2 = 2 (Z) = 2⇣ 1mXm! i=1= 1 Xm 2m 2 (Y (i) )= 1 1mX2 (Y (i) ) = 1 m mmi=1i=1Y (i)⌘2 (Y )infinite observations = zero error : we have our underlying estimator!
MACHINE LEARNING - 20139In layman termsyZYThe expected error (variance) of Y is larger than ZThe variance of Z shrinks with m
Page 1 and 2: MACHINE LEARNING - 2013MACHINE LEAR
Page 3 and 4: 3MACHINE LEARNING - 2013The Main Id
Page 5 and 6: MACHINE LEARNING - 20135The Main Id
Page 7: MACHINE LEARNING - 20137Aggregation
Page 11 and 12: MACHINE LEARNING - 2013Peculiaritie
Page 13 and 14: MACHINE LEARNING - 201313Adaptive B
Page 15 and 16: MACHINE LEARNING - 201315Weak Learn
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Page 19 and 20: MACHINE LEARNING - 2013AdaBoostRins
Page 21 and 22: MACHINE LEARNING - 201321Boosting I
Page 23 and 24: MACHINE LEARNING - 2013Boosting In
Page 25 and 26: MACHINE LEARNING - 201325A thousand
Page 27 and 28: MACHINE LEARNING - 201327A celebrat
Page 29 and 30: MACHINE LEARNING - 201329Some simpl
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Page 33 and 34: MACHINE LEARNING - 201333Cascades o
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Page 37 and 38: MACHINE LEARNING - 201337RANSAC in
Page 39 and 40: MACHINE LEARNING - 201339Some examp
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