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# N n 1 2 c i xi c 1 i x c 1 n n i1 i xi n i1 1 n n n

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Unformatted text preview: ws that � �n � � 1 1� � � ≤ P(X � ) I (Xi ∈ CX ) ≥ P (CX ) �∃CX � 4 n i=1 ⎛ ⎞ � �n �n � 1 1 � t� i= I (Xi ∈ CX ) − n i=1 I (Xi ∈ CX ) n ≤ P(X � ) ⎝ � � 1 ≥ √ �∃CX ⎠ �n n 1 1 �∈C ) 2� X i=1 I (Xi ∈ CX ) + n i=1 I (Xi n Since indicator is 0, 1-valued, ⎛ ⎞ �n ⎜ ⎟ 1 ⎟ ⎜ P (C ) − n i=1 I (Xi ∈ C ) 1⎜ � I sup ≥ t⎟ ⎟ 4 ⎜C P (C ) ⎠ ⎝ � �� � ∃CX ⎛ ⎞ � � ∈ CX ) − t� i=1 I (Xi ∈ CX ) ≤ P(X � ) ⎝ � � ≥ √ �∃CX ⎠ · I (∃CX ) �n n 1 1 �∈C ) 2� X i=1 I (Xi ∈ CX ) + n i=1 I (Xi n ⎛ ⎞ �n �n 1 1 � I (Xi ∈ C ) − n i=1 I (Xi ∈ C ) t⎠ i=1 n ≤ P(X � ) ⎝sup � � ≥√ . �n n 1 2 C I (Xi ∈ C ) + 1 I (X � ∈ C ) 1 n �n � i=1 I (Xi n i=1 1 n �n n i=1 i Hence, � � �n 1 P (CX ) − n i=1 I (Xi ∈ CX ) 1 � P sup ≥t 4 P (CX ) C ⎛ ⎞ �n �n 1 1 � t⎠ i=1 I (Xi ∈ C ) − n i=1 I (Xi...
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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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