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Unformatted text preview: I (Xi ∈ C ) − P (C )
≥ t2 /4⎠
�n
�
n i=1
i=1
≤ (by Chebyshev’s Ineq)
= =
since we chose t ≥ � 4
n 2 t2 n
�
i=1 4�
n 2 t2 4E � 1 �n
n i=1 �2
�
I (Xi ∈ C ) − P (C )
t2 �
�
E(I (Xi ∈ C ) − P (C ))(I (Xj ∈ C ) − P (C )) i,j �
E(I (Xi ∈ C ) − P (C ))2 = 4nP (C ) (1 − P (C )
1
1
≤ 2≤
2 t2
n
2
nt 2
n. So,
�
�
�� n
�
�1 �
�
�
�
�
�
�
PX � �
I (Xi ∈ CX ) − P (CX )� ≤ t/2�∃CX ≥ 1/2
�n
�
�
i=1
if t ≥ �
2/n. Assume that the event
�n
�
�1 �
�
�
�
�
I (Xi ∈ CX ) − P (CX )� ≤ t/2
�
�n
�
i=1 occurs. Recall that �n
�
�1 �
�
�
�
I (Xi ∈ CX ) − P (CX )� ≥ t.
�
�n
�
i=1 Hence, it must be that
�n
�
n
�1 �
�
1�
�
�
�
I (Xi ∈ CX ) −
I (Xi ∈ CX )� ≥ t/2.
�
�n
�
n i=1
i=1
21 Lecture 10 Symmetrization. Pessimistic VC inequality. 18.465 We conclude
1
2 �
�
�� n
�
�1 �
�
�
�
�
�
�
≤ PX � �
I (Xi ∈ CX ) − P (CX )� ≤ t/2�∃CX
�n...
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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.
 Spring '07
 Panchenko
 Statistics

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