Week3Student2009

# Week3Student2009 - Week 3 Lecture 4 Model Let Y 1 Y 2 Y n...

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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 ) Â¡ 2 dx Â¡ M o . We have shown there is a kernel estimator b f n such that sup f 2F Z E Â¢ b f n ( x ) & f ( x ) Â£ 2 Â¡ 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 Â¢ b f n ( x ) & f ( x ) Â£ 2 Â¢ 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 Â£ F such that inf b f n sup f 2F 1 Z E Â¢ b f n ( x ) & f ( x ) Â£ 2 Â¢ 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 Â¢ b f n ( x ) & f ( x ) Â£ 2 Â¢ 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 Â¤ h m n G Â¤ x & x n;j h n Â¥ , & 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 Â£ F : Â¥ R G 2 > : Â¥ R G = 0 ) R f n;& ( x ) = 1 . Â¥ j G j is bounded which implies f n;& ( x ) Â¢ as n ! 1 . 1 & G ( s ) ( Â¡ 1) = G ( s ) (1) = 0 , s = 1 ;:::;m , and Z h G ( m ) ( x ) i 2 Â¢ M ) Z h f ( m ) n;& ( x ) i 2 Â¢ h n r n Z h G ( m ) ( x ) i 2 = Z h G ( m ) ( x ) i 2 Â¢ M . If G ( y ) satis&es the property above, so is cG ( y ) with c small....
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