lecture11 notes

Hence there exists cx such that 18465 n i xi c i1

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Unformatted text preview: never P (C ) ≥ n . Indeed, P (C ) ≥ n since i=1 I (Xi ∈ C ) ≥ nP (C ) ≥ 1. Otherwise P ( i=1 I (Xi ∈ C ) = 0) = �n n � / i=1 P (Xi ∈ C ) = (1 − P (C )) can b e as close to 0 as we want. Similarly to the proof of the previous lecture, let � � �n 1 P (C ) − n i=1 I (Xi ∈ C ) � (Xi ) ∈ sup ≥t . P (C ) C 24 Lecture 11 Optimistic VC inequality. Hence, there exists CX such that 18.465 �n I (Xi ∈ C ) � i=1 ≥ t. P (C ) P (C ) − 1 n P (CX ) − 1 n Exercise 1. Show that if and �n I (Xi ∈ CX ) � i=1 ≥t P (CX ) n 1� � I (Xi ∈ CX ) ≥ P (CX ) , n i=1 then 1 n �n i=1 � I (Xi ∈ CX ) − 1 n �n n Hint: use the fact that φ(s) = s−a √ s I (Xi ∈ CX ) t ≥√ . � 2 i=1 I (Xi ∈ CX ) i=1 �� n 1 �n 1 I (Xi ∈ CX ) + n √ s = s − √s is increasing in s. i=1 From the above exercise it follo...
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This note was uploaded on 01/23/2014 for the course MATH 18.465 taught by Professor Panchenko during the Spring '07 term at MIT.

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