第五章数据分析(梅长林)习题VIP免费

第五章习题 1.习题5.1 解：假定两总体服从正态分布，且协方差矩阵21，误判损失相同又先验概率按比例分配，通过SAS计算得到先验概率如表： Class Level Information group Variable Name Frequency Weight Proportion Prior Probability G1 G1 6 6.0000 0.428571 0.428571 G2 G2 8 8.0000 0.571429 0.571429 即：0.4285711 P 0 .571422 P 又计算可得： ( 1 )(2)25.31622.025,2.4161.187xx 有计算的总体协防差距矩阵S为： Pooled Within-Class Covariance Matrix, DF = 12 Variable x1 x2 x1 1.081944444 -0.310902778 x2 -0.310902778 0.174756944 并且：-2.38145ln S 计算广义平方距离函数： 2( )1( )( )()()ln2lnjTjjjjjdpxxxSxxS 并计算后验概率： 222ˆˆ0.5( )0.5( )1ˆ(| )eejkddjkP Gxxx1,2j  回代判别结果如下: Posterior Probability of Membership in group Obs From group Classified into group G1 G2 1 G1 G1 0.9387 0.0613 2 G1 G1 0.9303 0.0697 3 G1 G1 0.9999 0.0001 4 G1 G2 * 0.4207 0.5793 5 G1 G1 0.9893 0.0107 6 G1 G1 1.0000 0.0000 7 G2 G2 0.0007 0.9993 8 G2 G2 0.0026 0.9974 9 G2 G2 0.0008 0.9992 10 G2 G2 0.0586 0.9414 11 G2 G2 0.0350 0.9650 12 G2 G2 0.0006 0.9994 13 G2 G2 0.0038 0.9962 14 G2 G2 0.0012 0.9988 由此可见误判的回代估计： 0.07141/14*rP 若按照交叉确认法，定义广义平方距离如下： 2( )1( )( )( )( )( )( )()()ln2lnjjjTjjxxxxjdpxxxSxxS 逐个剔除, 交叉判别，后验概率按下式计算： 222ˆˆ0.5( )0.5( )1ˆ(| )eejkddjkP Gxxx1,2j  通过SAS计算得到表所示结果。发现同样也是属于 G1 的4 号被误判为 G2，因此误判率的交叉确认估计为*ˆ1/140.0714cp  Posterior Probability of Membership in group Obs From group Classified into group G1 G2 1 G1 G1 0.9060 0.0940 2 G1 G1 0.7641 0.2359 3 G1 G1 1.0000 0.0000 4 G1 G2 * 0.1950 0.8050 5 G1 G1 0.9743 0.0257 6 G1 G1 1.0000 0.0000 7 G2 G2 0.0012 0.9988 8 G2 G2 0.0051 0.9949...