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Unformatted text preview: Week 3 Lecture 4 Model : Let Y 1 ; Y 2 , . . . , Y n i.i.d. on [ & 1 ; 1] with density f 2 F , F = n f; R & f ( m ) ( x ) ¡ 2 dx ¡ M o . We have shown there is a kernel estimator b f n such that sup f 2F Z E ¢ b f n ( x ) & f ( x ) £ 2 ¡ Cn & 2 m= (2 m +1) . Question : Is this rate optimal? Can you improve this rate? Answer : No. We will show inf b f n sup f 2F Z E ¢ b f n ( x ) & f ( x ) £ 2 ¢ C 1 n & 2 m= (2 m +1) , for some C 1 > . Because it is hard to analyze the whole ball F , we will construct a simpler ball F 1 £ F such that inf b f n sup f 2F 1 Z E ¢ b f n ( x ) & f ( x ) £ 2 ¢ C 1 n & 2 m= (2 m +1) . Remark: For higher dimensional case, we may show inf b f n sup f 2F Z E ¢ b f n ( x ) & f ( x ) £ 2 ¢ C 1 n & 2 m= (2 m + d ) , for some C 1 > . Parameter space F 1 : For h n = n & 1 = (2 m +1) and r n = 1 =h n , let x n; 1 < x n; 2 < ::: < x n;r n with x n;i = & 1 + (2 i & 1) h n . The meshwidth is 2 h n . For a &xed probability density f and a &xed G with support on ( & 1 ; 1) , de&ne F 1 = 8 < : f n;& ( x ) : f n;& ( x ) = f ( x ) + r n X j =1 & j ¤ h m n G ¤ x & x n;j h n ¥ , & 2 f ; 1 g r n 9 = ; : Let f ( x ) = 1 2 I [ & 1 ; 1] ( x ) . The function G needs to satisfy some properties such that F 1 £ F : ¥ R G 2 > : ¥ R G = 0 ) R f n;& ( x ) = 1 . ¥ j G j is bounded which implies f n;& ( x ) ¢ as n ! 1 . 1 & G ( s ) ( ¡ 1) = G ( s ) (1) = 0 , s = 1 ;:::;m , and Z h G ( m ) ( x ) i 2 ¢ M ) Z h f ( m ) n;& ( x ) i 2 ¢ h n r n Z h G ( m ) ( x ) i 2 = Z h G ( m ) ( x ) i 2 ¢ M . If G ( y ) satis&es the property above, so is cG ( y ) with c small....
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This note was uploaded on 11/06/2009 for the course STAT 680 at Yale.
 '09
 HarrisonH.Zhou

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