lecture10 notes

V c c n i1 proof n n 1 1 2p sup i xi c i xi

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Unformatted text preview: � i=1 � � �� n � n �1 � � � 1� � � � � ≤ PX � � I (Xi ∈ CX ) − I (Xi ∈ CX )� ≥ t/2�∃CX �n � � n i=1 i=1 �n � � � � n �1 � � � 1� � � � � PX � sup � I (Xi ∈ C ) − I (Xi ∈ C )� ≥ t/2�∃CX . � � n i=1 C ∈C � n i=1 Since indicators are 0, 1-valued, ⎛ ⎞ �n � ⎜ ⎟ �� � ⎟ 1⎜ � ⎜ sup � 1 I⎜ I (Xi ∈ C ) − P (C )� ≥ t⎟ � ⎟ � 2 ⎝C ∈C � n i=1 ⎠ �� � � ∃CX �n � � � n �1 � � � 1� � � � � I (Xi ∈ C ) − I (Xi ∈ C )� ≥ t/2�∃CX · I (∃CX ) sup � � � n i=1 C ∈C � n i=1 �n � � � n �1 � � 1� � � � ≤ PX,X � sup � I (Xi ∈ C ) − I (Xi ∈ C )� ≥ t/2 . � n i=1 C ∈C � n i=1 � ≤ PX � Now, take expectation with respect to Xi ’s to obtain � � � � n �1 � � � � PX sup � I (Xi ∈ C ) − P (C )� ≥ t � C ∈C � n i=1 � � � � n n �1 � � 1� � � � ≤ 2 · PX,X � sup � I (Xi ∈ C ) − I (Xi ∈ C )� ≥ t/2 . � n C ∈C � n i=1 i=1 � Theorem 10.1. If V C (C ) = V , then � � � � � �V n �1 � � nt2 2en...
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