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2019年-应用多元统计分析课后习题答案高惠璇(第三章部分习题解答)-PPT精选文档_图文





3-1 XNn(,2In), A ,rk(A)=r(rn),
A
01,r0 (ri)
2


3



4


3-2 XNn(,2In), ABn.
AB 0 ,XAXXBX.
rk(A)=r. An,,
A=diag(1,...,r 0,..,0) YXYNn(,2In),
r
X A X ()Y A Y A iY i2 i 1 5


XBX=YB Y= YHY
H=B XBX Yr+1...,YnH 0 XAX XBX
6


rk(A)=r.
r=n,ABO,B Onn, XAXXBX
r=0,A0, . 0rn.An, ,
7


i0A(i=1,...,r).



r

ABODrH11O DrH12O . Dr,H11OrrH12Or(n-r) .
HH21O(n-r)r .
8


HB
YXY Nn(,2In), r X A X ()Y A Y A iY i2 i 1 Y r1
XB X YB YH Y (Y r1, ,Y n)H 2 2Y n
Y1...,Yr ,Yr+1 ,...,Yn XAXXBX.
9


3-3 XNp(,),0,ABp, (X-)A(X-)(X-)B(X-)
AB0pp.

(

1
2

12 1)



10


"1.6"
1 11 1
CDO 2A22B2 O ABO
11



3-4 Wishart(4)T2(5).

4 Wishart:X() Np(0,) (

1,...,n)


1211

12 r 22pr

W n 1X ()X ( ) W W 1 21 1W W 1 2 2 2p r r~ W p(n , )



12



:
X()X X(((( 1 2))))prr, X X (( 2 (( ))1))~~N N p rr((00 ,, 12)1)2,,



X
np

xij

X(1)| X(2) , nr n(pr)



W X X X X ( ( 1 2 ) ) X X ( ( 1 1 ) ) X X ( ( 1 2 ) ) X X ( (2 2 ) ) W W 1 21 1 W W 1 2 2 2 ,
W 1 1 X ( 1 ) X ( 1 )W ,2 2 X ( 2 ) X ( 2 )

13


3.1.4
W 11 X (1 )X (1 )n(X (( 1 )))X (( 1 ))~ W r(n , 1)1; 1
W 2 2X (2 )X (2 )n(X (( 2 ) ))X (( 2 ) )~ W p r(n , 2)2 1

12 =O ,1,2,...,n,
.W11W22.

X((1))X((2))

14


5 ,T2 .
:X() (1,...,n) p Np(,), XAxX ,1
T x 2n(n1 )X ()A x 1(X)
~T2(p,n1 ).
Y (i)C X(i)d(i1,...,n)
Cppdp1
Y (i)~ N p (C d ,C C )(i 1 ,2 ,.n ) ..,
15



Y CXd, n

y Cd

Ay (Y(i) Y)(Y(i) Y)

i1 n

C[ (X(i) X)(X(i) X)]CCAxC

i1

Ty2 n(n1)(Yy)Ay1(Yy)

n(n1)(X)C[CxA C]1C(X)

n(n1)(X)Ax1(X)Tx2



T

2 x



T

2 y

16



3-5 pNp(,)

H0:=0(=0)

.
:XNp(

,

P66=0
0)( 00),X()( =1,...,n)

(np)pX.

m 0L ( a0 ,x 0 )m L ( a, x 0 )

|2 1 0|n /2e x 1 2 p n 1 (X ()0 ) 0 1 (X ()0 )

|2 1 0|n /2e x 1 2 tp [ r0 1 n 1 (X ()0 )X ( ()0 )]
17


|2 10|n/2exp1 2t[r0 1A 0]
L (X , 0)m L a (x , 0)
|2 1 0|n /2ex 1 2 p n 1(X () X ) 0 1 (X () X ) |2 1 0|n /2e x 1 2 p t[ r0 1 n 1 (X () X )X (() X )]
|2 10|n/2ex p1 2t[r0 1A]
18


m 0L ( a0 ,x 0 )m L ( a, x 0 )
extp[r1 20 1A]t[r1 20 1A0]
e x t[1 2 rp 0 1 A ] t[1 2 r 0 1 (A n (X 0 )X (0 )]
ex p n 2t[rX (0)0 1(X0) ]
expn 2(X0)0 1(X0)
19


ln n 2(X0)0 1(X0)
def
2 lnn (X 0) 0 1(X 0)
X H ~ 0 N p(0 ,1 n 0 ),n (X 0 )H ~ 0 N p(0 , 0 )
3"2.1"
2ln~2(p).
20


3-6 ()
XNp( , )( 0),X() ( 1,...,n)(np) pX ( 1,..., p).C kp(k<p),rank(C)=k,rk. H0:C r.
Y ()C(X )( 1 ,2 , ,n )
Y()( 1,...,n) kY
Y ( )~ N k ( C ,C C ) ;y C , y C C .
21



H 0:C rH 0 : y r
k

.3.2y = CC ,:

F

nk

H0
T2 ~F(k,nk)

(n1)k

T 2(n 1 )n(Yr)A y 1(Yr).

(n 1 )n(C Xr)CC A 1(C Xr).

n

A (X(i)X)(X(i)X).

i1

22



3-7 XNp() (0), X() (1,...,n)(np)p
XXA. (1,...,p).H0:1=2=...=p ,H1:1,2,...,p .

C11

1 0

0 1



0 0





,

1 0 0 1(p1)p

H0C=0p-1,H1C 0p-1

H0 .

H0:12 p, H1:1,2,,p .
23



H 0 :C 0 ,H 1 :C 0 ,

3-6H0



F

n(p1)

H0
T2 ~ F(p1,np1),

(n1)(p1)

T2(n1)n(CX)[CC A]1CX

~T2(n1,p1).H (0 )
(:3-6kp-1)
24



3-8 :(X1),
(X2)(X3)641. X ()(1,...,n)X=(X1,X2,X3). XN3(,)3.5 ()(H0, ).
()

. 1: 2: 3 6 :4 :1 C 0



C10

0 1

6423,

: 1 6 , 2 4

3 1

3 1







1 2



63 43



0 0

.

25


C 2 10 36 0 , C 0 23 14 0
H0 H 0:C0 ,H 1:C0 ,
3-6
F n2T2H ~ 0F(2,n2) 2(n1)
T 2 (n 1 )n (C X )[XC A ] 1 C X .
(p=3,n=6) T2=47.1434, F=18.8574, p=0.009195<=0.05,
H0,.
26


3-9 pNp(,) H0:=0 .
:XNp( , ),X()( =1,...,n)
pX.
m L a (x , 0)m ,L a (,x )
^X ,
L(X,0)|2 1 0|n/2ex p 1 2n 1(X()X)0 1(X()X)
27



|2 1 0|n /2e x 1 2 p t[ r0 1 n 1 (X () X )X (() X )]

(2)n2p|0|n 2et r1 20 1A

L(X,1A )mL a(x ,)

n

,

np

np

2 ne2

|A|n 2(2)n2p n2
e

n
|A| 2

28



m | 0 |L a n2(etrnx ,p 0)12 m 0,1 A L a (,x )

n2

n
|A| 2

e

np





e n



2

etr





1 2



1 0

A



|



0

1

A

n
|2

3.2.1,n, 2ln~2p(p1).
2 29



3-10 pNp((1),)Np ((2),) H0 (1)=(2).

:

X (i) ( )

(1,...,ni)XNp((i),)

(i=1,2),,>0.H0

mL (a, x ) ma L (x (1 ), (2 ), )

, 0



n i

(1 ), (2 ), 0

A i(X ( ( i) ) X (i))X ( ( ( i) ) X (i))(i 1 ,2 )n n 1 n 2

1



X (i)n 1 i n i1X (( i)) (i 1 ,2 ) ,X1 ni 2 1 n i1X (( i)),

30



2 nk

T

(X((ij))X)(X((ij))X)

i1 j1

2

2

A i n i(X (i)X )X ((i)X )A B

i 1

i 1

A=A1+A2.B.

^ X , ^ TA B ,

nn

np

L X T)(2)n2pn2

n
|T| 2

n

e

31



^(1) X(1),^(2) X(2),^A,

n np

L X(1),X(2) A)(2)n2pn2|A|n 2

n

e



||T A||n n//2 2

|

n/2
A| AB | | n/2



2
TABA ni(X(i) X)(X(i) X)

i1

An1n2 (X(1) X(2))(X(1) X(2)) n
32



| T || A n1n2 ( X (1) X (2) )( X (1) X (2) ) | n

A

n1n2 ( X (1) X (2) )



n

n1n2 ( X (1) X (2) )

1

n

| A | [1 n1n2 ( X (1) X (2) ) A1 ( X (1) X (2) )] n

|A|

1

|T| 1n1n2(X(1)X(2))A1(X(1)X(2))

n

33





n1n2(X(1) n

H0
X(2)) ~Np(0,)

A A 1 A 2 ~ W p ( n 2 , )( n , n 1 n 2 )
3.1.5 T2(n2)n1n2(X(1)X(2))A1(X(1)X(2)) n ~T2(p,n2)

| A| 1 , 1 T2 1

|T| 1 1 T2

n2



n2

34


F(n2)p1 T2 np11 p n2 p
H0
~ F(p,np1)
H0
{}{ }{ T2T2}
{ FF }
35


3-11 3.5152(X1)(X2) (X3).X() ( 1,...,6)N3( (1)). Y() (1,...,9)N3 ((2)) .3.5H0:(1) =(2) (=0.05).
:. (p=3,n=6,m=9):
Fnm p 1T2H ~ 0 F (p ,nm p 1 ) (nm 2 )p
36


T 2 (n m 2 )n n m ( m X Y )(A 1 A 2 ) 1 (X Y )

F n m p p 1 n n m ( m X Y )(A 1 A 2 ) 1 (X Y )
: T25.311,F71.498, 2
p0.26930.05
H0.
37



3-12 ABC

3.6.
N3( (i) i)(i123)( =0.05). (1) H0 1 2 3H1 1, 2, 3; (2) H0 (1) (2),H1 (1) (2); (3) H0: (1) (2) (3),H1:ij, (i) (j);
(4) .

:(4)(p=3,k=3)
X ( ( it) )~ N p ((t), )(t, 1 , ,k ;i 1 , ,n t)

0 :1 2 H 1 3 2 30 ,H 1:1,21,32 3 .

H0 m (i),aiiLx((i),ii)maLx((i),)

(i),

38


D001100220033diag)(,
L(X(t),D^)maxL((t);D) (2)n2pD^ n2exp[1tr(D^1A)]
2
39



a1

1
n

0

0

D^ 0 0

a2 2 n
0

a303n1ndia(gA),

k

k n t

A A t (X ((it))X (t))X (((it))X (t))

t 1

t 1i 1

D ^1np ip1 aii,D ^.1na00111

0 a212

0 0,

0 a313

40



p
tr(D ^1A)n aii1aiinp i1

L( X

(t) , D^ )



np
(2 ) 2

D^

n 2

exp[

1

tr(D^ 1A)]

2



(2


)

np 2





1 n



np 2



p i 1

aii



n 2





exp

np 2



np





n
2e



2



p i 1

aii





n 2



41





L(X(t),1 A)maxL((t);)

n

np
(2) 2

1

n 2
A

exp{1tr[(1

A)1A]}

n

2n

(2)n 2 p1 n 2p A n 2ex n p p n n 2p A n 2

n

2 2e

42



H0:





max
( i ) , ii

L(

(i) ,

ii )

max L ( (i) , )

(i) ,





n
2e



np

/

2



p i 1

a ii

n/2

n np / 2 A n / 2

2e

p aii n / 2





i 1
A


n/2







n

2

A
p



n



V 2

aii

i1

43



BoxH0n =-blnV~2(f)



b132 33(333233)6611.01667

f 1[343 12]3

2

i1

V=0.7253, =-blnV=3.2650,

p=0.3525>0.05.

H0( ).

44


3.2.1,n =-2ln~2(f)
f=p+p(p+1)/2-(p+p)=3,
V=0.7253, =0.1240 ,
=-2ln =-nlnV=4.1750,
p=0.2432>0.05. H0( ).
45


3-13 3.3 4.
(1) ?
(2) 2
4.
46