本研究探討核密度函數估計量之偏量性質的改善方法, 分成六個章節；第一章導論, 描述在估計母體密度函數f的無母數方法中, 使用核密度函數估計量的優點；第二章提岀改良式核密度函數估計量f ̂_D；第三章為探究改良式核密度函數估計量之理論性質；第四章針對兩筆實際資料, 分別是紐約州水牛城1910-1972一共62年的降雪量資料, 以及320位男性胸痛病人血漿總膽固醇濃度資料, 以改良式核密度函數估計量進行密度函數估計；第五章為模擬研究, 針對改良式核密度函數估計量f ̂_D進行一系列基於模擬資料的研究, 並與傳統f ̂ 做比較；第六章為本研究結論；第七章為理論證明。

Kernel density estimator is a nonparametric method to recover the true probability density functions, It is a useful nonprametric smoothing method as it uses no assumption about the true probability function. Unfortunately, optimal efficient kernel density estimators are biased which hinders the development of methods for further statistical inference. In this dissertation, we apply the method of Cheng et al. (2018), which serves to reduce the bias of kernel regression function estimator, to reduce the bias of classical kernel density estimates. The proposed method does not inflate the variance. It needs not to use higher-order kernel function whose theoretical advantages do not transfer to reasonably sized samples. The theoretical properties of the proposed estimator are provided two real-data applications are demonstrated. Simulations are carried out for comparisons with regular kernel density estimator. The results reveals that the biased corrected kernel density estimator enjoys noticeably smaller mean integrated square errors.

章節目錄 I
圖表目錄 II
第一章 導論 1
第二章 改良式核密度函數估計量 5
第三章 理論性質  8
第四章 基於實際資料之f ̂_D與舉隅  10
第五章 模擬研究  13
第六章 結論  23
第七章 定理證明  24
參考文獻  27

圖1-1 f(x)=[exp(-x^2/2)]/√2π 的直方圖與折線圖估計量  2
圖1-2 f(x)=[exp(-x^2/2)]/√2π 的核估計量與折線圖估計量  3
圖1-3 f與f ̂ (h=0.25、1.75)  4
圖4-1 基於水牛城降雪資料之核密度函數估計量  11
圖4-2 基於血漿總膽固醇濃度資料之核密度函數估計量  12
圖5-1 0.5N(-1.5,0.5^2)+0.5N(1.5,0.5^2)  15
圖5-2 3/10 N(-3,2^2)+ 4/10 N(2,1^2)+ 3/10 N(10,4^2)  16
圖5-3 2/3 N(0,1)+ 1/3 N(0,0.1^2)  17
圖5-4 1/2 N(0,1/10^2)+ 1/10 N(-1,1/10^2)+ 
      1/10 N(-0.5,1/10^2)+ 1/10 N(0,1/10^2)+
      1/10 N(0.5,1/10^2)+ 1/10 N(1,1/10^2)   18

表5-1 四種混合常態分布  14
表5-2 IMSE之模擬研究的混合常態分布  20
表5-3 (n=100,h_d=C_d h_0, C_d=1+(4-d)/10,
      d=1,2,..,7)  21
表5-4 (n=50,h_d=C_d h_0, C_d=1+(4-d)/10,
      d=1,2,..,7)  21
表5-5 (n=100,h_d=C_d h_0, C_d=1+(6-d)/10,
      d=1,2,..,11)  22
表5-6 (n=50,h_d=C_d h_0, C_d=1+(6-d)/10,
      d=1,2,..,11)  22

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