第五章数据分析(梅长林)习题.docx
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1、第五章数据分析(梅长林)习题第五章习题 1.习题5.1 解:假定两总体听从正态分布,且协方差矩阵,误判损失相同又先验概率按比例安排,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: 又计算可得: 有计算的总体协防差距矩阵S为: Pooled Within-Class Covariance Matr
2、ix, DF = 12 Variable x1 x2 x1 1.081944444 -0.310902778 x2 -0.310902778 0.174756944 并且: 计算广义平方距离函数: 并计算后验概率: 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
3、 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: 若根据交叉确认法,定义广义平方距离如下: 逐个剔除, 交叉判别,后验概率按下式计算: 通过SAS计算得到表所示结果。发觉同样也是属于
4、G1的4号被误判为G2,因此误判率的交叉确认估计为 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 9 G2 G2
5、0.0014 0.9986 10 G2 G2 0.0713 0.9287 11 G2 G2 0.0422 0.9578 12 G2 G2 0.0009 0.9991 13 G2 G2 0.0059 0.9941 14 G2 G2 0.0022 0.9978 其中=12.1138, ,又因为,所以, 最终可得后验概率p为:0.048709 习题5.3 解:(1)在并且先验概率相同的的假设前提下,建立矩离判别的线性判别函数。利用SAS的proc discrim过程首先计算得到总体的协方差矩阵,如表: Pooled Within-Class Covariance Matrix, DF = 25 Va
6、riable x1 x2 x3 x4 x5 x6 x7 x8 x1 2.25705591 -0.91513311 0.34259974 -0.6084399 -0.9576508 -0.8929719 -0.0539445 -0.2192724 x2 -0.9151331 25.2318255 -0.3390873 -2.5515272 -5.0966371 0.78571637 -0.0835586 4.37529806 x3 0.34259974 -0.33908734 3.30063123 1.42276017 1.78692343 0.40208409 -0.0676655 -0.07
7、32213 x4 -0.6084399 -2.55152726 1.42276017 6.07845863 5.78100857 2.32039331 -0.3205116 0.48605897 x5 -0.9576508 -5.09663714 1.78692343 5.78100857 8.15854743 3.44983429 -0.1096651 0.08904743 x6 -0.8929719 0.78571637 0.40208409 2.32039331 3.44983429 4.16657066 -0.2236278 0.87862549 x7 -0.0539445 -0.08
8、355869 -0.0676655 -0.3205116 -0.1096651 -0.2236278 0.26009291 -0.0767347 x8 -0.2192724 4.37529806 -0.0732213 0.48605897 0.08904743 0.87862549 -0.0767347 2.51054423 各个总体的马氏平方距离见表: Generalized Squared Distance to group From group G1 G2 G1 0 24.61468 G2 24.61468 0 线性判别函数为: 得到训练样本回判法判别结果如表: Error Count
9、Estimates for group G1 G2 Total Rate 0.0000 0.0000 0.0000 Priors 0.5000 0.5000 训练样本的交叉确认判别结果: Posterior Probability of Membership in group Obs From group Classified into group G1 G2 17 G1 G2 * 0.4501 0.5499 19 G1 G2 * 0.0920 0.9080 Error Count Estimates for group G1 G2 Total Rate 0.1000 0.0000 0.050
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