Bayes and EB benchmarkig for SAE. Dissertation 2012.pdf

# Next we return to 521 and show that e ˆ θ eb w ˆ

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Next, we return to ( 5.2.1 ) and show that E [( ¯ ˆ θ EB w ¯ ˆ θ B w ) 2 ] = o ( m 1 ). Consider E [( ¯ ˆ θ EB w ¯ ˆ θ B w ) 2 ] = summationdisplay i w 2 i E bracketleftBig ( ˆ θ EB i ˆ θ B i ) 2 bracketrightBig + 2 m 1 summationdisplay i =1 m summationdisplay j = i +1 w i w j E bracketleftBig ( ˆ θ EB i ˆ θ B i )( ˆ θ EB j ˆ θ B j ) bracketrightBig = 2 m 1 summationdisplay i =1 m summationdisplay j = i +1 w i w j E bracketleftBig ( ˆ θ EB i ˆ θ B i )( ˆ θ EB j ˆ θ B j ) bracketrightBig + o ( m 1 ) (5.2.3) since i w 2 i E [( ˆ θ EB i ˆ θ B i ) 2 ] = o ( m 1 ). The latter holds because E [( ˆ θ EB i ˆ θ B i ) 2 ] = g 2 i ( σ 2 u ) + g 3 i ( σ 2 u ) = O ( m 1 ), max 1 i m w i = O ( m 1 ), and i w i = 1. Thus, it suffices to show E bracketleftBig ( ˆ θ EB i ˆ θ B i )( ˆ θ EB j ˆ θ B j ) bracketrightBig = o ( m 1 ) for all i negationslash = j , and we do so by expanding ˆ θ EB i about ˆ θ B i . For simplicity of notation, denote ˆ θ B i ∂σ 2 u = ˆ θ B i ( σ 2 u ) ∂σ 2 u and 2 ˆ θ B i ( σ 2 u ) 2 = 2 ˆ θ B i ( σ 2 u ) ( σ 2 u ) 2 . This results in ˆ θ EB i ˆ θ B i = ˆ θ B i ∂σ 2 u σ 2 u σ 2 u ) + 1 2 2 ˆ θ B i ( σ 2 u ) 2 σ 2 u σ 2 u ) 2 for some σ 2 u between σ 2 u and ˆ σ 2 u . The expansion of ˆ θ EB j about ˆ θ B j is similar. We now consider E [( ˆ θ EB i ˆ θ B i )( ˆ θ EB j ˆ θ B j )] for i negationslash = j . Notice that E [( ˆ θ EB i ˆ θ B i )( ˆ θ EB j ˆ θ B j )] = E bracketleftBigg ˆ θ B i ∂σ 2 u ˆ θ B j ∂σ 2 u σ 2 u σ 2 u ) 2 bracketrightBigg + 1 2 E bracketleftBigg ˆ θ B i ∂σ 2 u 2 ˆ θ B j ( σ 2 u ) 2 σ 2 u σ 2 u ) 3 bracketrightBigg + 1 2 E bracketleftBigg 2 ˆ θ B i ( σ 2 u ) 2 ˆ θ B j ∂σ 2 u σ 2 u σ 2 u ) 3 bracketrightBigg + 1 4 E bracketleftBigg 2 ˆ θ B i ( ∂σ 2 u ) 2 2 ˆ θ B j 2 ( σ 2 u ) 2 σ 2 u σ 2 u ) 4 bracketrightBigg := R 0 + R 1 + R 2 + R 3 . In R 1 , consider that E bracketleftBigg ˆ θ B i ∂σ 2 u 2 ˆ θ B j ( σ 2 u ) 2 σ 2 u σ 2 u ) 3 bracketrightBigg = E bracketleftBigg ˆ θ B i ∂σ 2 u 2 ˆ θ B j ( σ 2 u ) 2 σ 2 u σ 2 u ) 3 I σ 2 u > 0) bracketrightBigg E bracketleftBigg ˆ θ B i ∂σ 2 u 2 ˆ θ B j ( σ 2 u ) 2 ( σ 2 u ) 3 I σ 2 u 0) bracketrightBigg . (5.2.4) 60

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Observe that E bracketleftBigg ˆ θ B i ∂σ 2 u 2 ˆ θ B j ( σ 2 u ) 2 ( σ 2 u ) 3 I σ 2 u 0) bracketrightBigg σ 6 u E 1 4 braceleftBigg ˆ θ B i ∂σ 2 u bracerightBigg 4 E 1 4 braceleftBigg 2 ˆ θ B j ( σ 2 u ) 2 bracerightBigg 4 P 1 2 σ 2 u 0) σ 6 u E 1 4 braceleftBigg ˆ θ B i ∂σ 2 u bracerightBigg 4 E 1 4 sup σ 2 u 0 braceleftBigg 2 ˆ θ B j ( σ 2 u ) 2 bracerightBigg 4 P 1 2 σ 2 u 0) = o ( m r ) for all r > 0 by Lemma 2 (ii) and 3 , which we have proved in Appendix A, and since P σ 2 u
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• Spring '16
• Yessi
• The Land, Estimation theory, Mean squared error, Bayes estimator, Empirical Bayes method

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