第五章数据分析（梅长林）习题

第五章题
1题51
解：假定两总体服正态分布协方差矩阵误判损失相先验概率例分配通SAS计算先验概率表：
Class Level Information
group
Variable
Name
Frequency
Weight
Proportion
Prior
Probability
G1
G1
6
60000
0428571
0428571
G2
G2
8
80000
0571429
0571429
：
计算：

计算总体协防差距矩阵S：
Pooled WithinClass Covariance Matrix DF 12
Variable
x1
x2
x1
1081944444
0310902778
x2
0310902778
0174756944

：
计算广义方距离函数：

计算验概率：

回代判结果
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
1
G1
G1

09387
00613
2
G1
G1

09303
00697
3
G1
G1

09999
00001
4
G1
G2
*
04207
05793
5
G1
G1

09893
00107
6
G1
G1

10000
00000
7
G2
G2

00007
09993
8
G2
G2

00026
09974
9
G2
G2

00008
09992
10
G2
G2

00586
09414
11
G2
G2

00350
09650
12
G2
G2

00006
09994
13
G2
G2

00038
09962
14
G2
G2

00012
09988
见误判回代估计：

交叉确认法定义广义方距离：

逐剔 交叉判验概率式计算：

通SAS计算表示结果发现样属G14号误判G2误判率交叉确认估计
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
1
G1
G1

09060
00940
2
G1
G1

07641
02359
3
G1
G1

10000
00000
4
G1
G2
*
01950
08050
5
G1
G1

09743
00257
6
G1
G1

10000
00000
7
G2
G2

00012
09988
8
G2
G2

00051
09949
9
G2
G2

00014
09986
10
G2
G2

00713
09287
11
G2
G2

00422
09578
12
G2
G2

00009
09991
13
G2
G2

00059
09941
14
G2
G2

00022
09978


中121138

验概率p：0048709
题53
解：（1）先验概率相假设前提建立矩离判线性判函数利SASproc discrim程首先计算总体协方差矩阵表：
Pooled WithinClass Covariance Matrix DF 25
Variable
x1
x2
x3
x4
x5
x6
x7
x8
x1
225705591
091513311
034259974
06084399
09576508
08929719
00539445
02192724
x2
09151331
252318255
03390873
25515272
50966371
078571637
00835586
437529806
x3
034259974
033908734
330063123
142276017
178692343
040208409
00676655
00732213
x4
06084399
255152726
142276017
607845863
578100857
232039331
03205116
048605897
x5
09576508
509663714
178692343
578100857
815854743
344983429
01096651
008904743
x6
08929719
078571637
040208409
232039331
344983429
416657066
02236278
087862549
x7
00539445
008355869
00676655
03205116
01096651
02236278
026009291
00767347
x8
02192724
437529806
00732213
048605897
008904743
087862549
00767347
251054423
总体马氏方距离见表：
Generalized Squared Distance to group
From group
G1
G2
G1
0
2461468
G2
2461468
0
线性判函数：


训练样回判法判结果表：
Error Count Estimates for group

G1
G2
Total
Rate
00000
00000
00000
Priors
05000
05000

训练样交叉确认判结果：
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
17
G1
G2
*
04501
05499
19
G1
G2
*
00920
09080


Error Count Estimates for group

G1
G2
Total
Rate
01000
00000
00500
Priors
05000
05000


（2）假设两总体服正态分布先验概率例分配误判损失相两总体协方差矩阵相条件进行Bayes判分析通SAS discrim程结果：
Error Count Estimates for group

G1
G2
Total
Rate
00000
00000
00000
Priors
07407
02593

交叉确认判结果：
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
19
G1
G2
*
02246
07754
25
G2
G1
*
05282
04718

Error Count Estimates for group

G1
G2
Total
Rate
00500
01429
00741
Priors
07407
02593

先验概率例分配假设前提利SASproc discrim程进行Bays判分析时总体训练样单独估计总体协方差矩阵训练样回判交叉确认结果：
回判结果：
Error Count Estimates for group

G1
G2
Total
Rate
00000
00000
00000
Priors
07407
02593

交叉确认判结果：
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
21
G2
G1
*
10000
00000
22
G2
G1
*
10000
00000
23
G2
G1
*
10000
00000
24
G2
G1
*
10000
00000
25
G2
G1
*
10000
00000
26
G2
G1
*
10000
00000
27
G2
G1
*
10000
00000

Error Count Estimates for group

G1
G2
Total
Rate
00000
10000
02593
Priors
07407
02593

（3）假设前提采判方法判样判结果：
1距离判分析西藏海广东判结果：
Posterior Probability of Membership in group
Obs
Classified into group
G1
G2
1
G2

00000
10000
2
G2

00000
10000
3
G2

00000
10000
2协方差矩阵相前提Bayes西藏海广东判结果：
Posterior Probability of Membership in group
Obs
Classified into group
G1
G2
1
G2

00000
10000
2
G2

00000
10000
3
G2

00000
10000
3协方差矩阵相前提Bayes西藏海广东判结果：
Posterior Probability of Membership in group
Obs
Classified into group
G1
G2
1
G1

10000
00000
2
G1

10000
00000
3
G1

10000
00000
3题54
解：（1）假设两总体服正态分布两总体协方差矩阵相先验概率相条件进行Bayes判分析通SAS discrim程结果：
首先线性判函数：

回代误判结果：
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
9
G1
G2
*
03401
06599
29
G2
G1
*
08571
01429
计算结果发现第9号样误判G229号样误判G1误判率634

Error Count Estimates for group

G1
G2
Total
Rate
00833
00435
00634
Priors
05000
05000

交叉确认判结果：计算发现总四样判错分9282935号样品累计误判率1069
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
9
G1
G2
*
00973
09027
28
G2
G1
*
06130
03870
29
G2
G1
*
09643
00357
35
G2
G1
*
08470
01530

Error Count Estimates for group

G1
G2
Total
Rate
00833
01304
01069
Priors
05000
05000


（1）假设两总体服正态分布两总体协方差矩阵相先验概率例分配误判损失相条件进行Bayes判分析通SAS discrim程结果：
首先线性判函数：
Linear Discriminant Function for group
Variable
G1
G2
Constant
9991796
9541991
x1
3035060
2987680
x2
015214
015210
x3
078868
022662
x4
195176
139528
x5
058964
006490
x6
10810195
8533735
x7
031156
025957
回代误判结果
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
9
G1
G2
*
02119
07881
29
G2
G1
*
07579
02421

Error Count Estimates for group

G1
G2
Total
Rate
00833
00435
00571
Priors
03429
06571

交叉确认误判结果：
Posterior Probability of Membership in group
Obs
From group
Classified into group
G1
G2
5
G1
G2
*
03436
06564
9
G1
G2
*
00532
09468
11
G1
G2
*
04052
05948
12
G1
G2
*
03519
06481
29
G2
G1
*
09338
00662
35
G2
G1
*
07428
02572

Error Count Estimates for group

G1
G2
Total
Rate
03333
00870
01714
Priors
03429
06571


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