• Keine Ergebnisse gefunden

Eigen Problems and the Singular V alue Deomposition

De-omposition

Anymatrix

X ∈ R n × p

denesalinearmap

R p → R n

via

w 7→ Xw

. Thesingular

valuedeompositionof

X

isarepresentationofthis mapintermsoforthonormal basis vetors forboth

R p

and

R n

suhthat the mapdened by

X

isassimple as

possible.

Proposition A.7 (SingularValue Deomposition). For any matrix

X ∈ R n × p

, there is an orthonormal basis

u 1 , . . . u p

of

R p

and a set of orthonormal vetors

v 1 , . . . , v p ∈ R n

suh that

Xu i = σ i v i , σ i ≥ 0 .

The quantities

σ i

are alled the singular values of

X

and are numbered in

de-reasingorder. In matrix notation, we have

X = V Σ U t

(A.8)

with

V t V = I p

and

U t U = I p .

It follows immediatelythatthe rankof

X

equalsthe numberofnonzero singular

values (ounted with multipliities). We note that

V

is a basis of the olumn

spae of

X

and

U

is a basis of the row spae of

X

. We an extend the vetors

v i

toan orthonormalbasis of

R n

.

Denition A.8. A vetor

u ∈ R p \ { 0 }

is alled an eigenvetor of a quadrati

matrix

A ∈ R p × p

if there is a salar

λ ∈ R

suh that

Au = λu

. We all

λ

an

eigenvalue of

A

.

The eigendeomposition of

A

is a representation of the form

A = U Λ U 1 .

For some matries, there is no eigendeomposition. If

A

is however symmetri,

we have anorthogonal eigendeomposition

A = U Λ U t , U t U = I p .

The eigenvetors of a symmetri matrix an be omputed with the help of the

so-alledpowermethod.

Algorithm A.9 (Power method). For a symmetri matrix

A

and an initial

vetor

b 0

, the power method omputes iteratively

e b k+1 = Ab k

matrix multipliation

b k+1 = 1

k

e

b k+1 k e b k+1

normalization

The power algorithm onverges to the eigenvetor

u

for whih the

orrespond-ingeigenvalue has the greatest absolutevalue, if this eigenvalue is dominant (in

absoluteterms)and ifthe startingvetor

b 0

isnot orthogonalonthe eigenvetor

u

.

A.4 Projetions

Letusonsider ageneralHibertspae

V

. Forasubspae

U

andanyvetor

v ∈ V

,

we denethe following optimizationproblem:

arg

min k v − u k ,

subjet to

u ∈ U .

As we assume that

V

is a Hilbert spae, the solution exists if

U

is a losed

sub-spae. We all the unique solution the (orthogonal) projetion of

v

onto

U

and

denoteit by

P U v

.

If

U

is nite-dimensional, we an give a short representation of the projetion operator. Denote by

U = (u 1 , . . . , u k )

any set of vetors that generate the

subspae

U

. For any other set

V = (v 1 , . . . , v l )

of vetors we dene the

k × l

matrix

h U , V i = ( h u i , v j i ) .

Furthermore,we dene the (symboli) multipliationof

U

with a vetor

α ∈ R k

as

U α = X k

i=1

α i u i .

The projetion mapis then

P U v = U ( h U , U i ) h U , v i .

(A.9)

Wenow listsome properties of projetion operators.

Proposition A.10. Denote by

P U

the projetion onto the subspae

U

.

1.

P U

isa symmetri map.

2. The projetion operator is idempotent,

P U 2 ≡ P U

.

3. If thespae

U

thatis orthogonal on

U

isalosed subspae, then

(Id V − P )

is the projetion onto that spae.

4. If

V

is nite-dimensional and

P U

an be represented by a matrix

P

, then

trae

(P ) = dim U

.

In Chapter 4, we need the rst derivative of a projetion operator. We now

present this result. Let us assume that both vetors

v = v(y), z = z(y) ∈ R n

depend ona vetor

y

. The projetion of

z

onto

v

isdened as (see (A.9))

P v z = v v t v − 1

v t z .

For any funtion

f

that depends on

y

, we use

df = df (y)

as a shortut. Using

proposition A.5,wehave

d ( P v z) = d

v v t v 1

v t z

= (dv) v t v 1

v t z + v

d v t v 1

v t z + v v t v 1

d v t z

= (dv) v t v 1

v t z − v v t v 1

d v t v

v t v 1

v t z +v v t v 1

d v t

z + v t dz

= (dv) v t v 1

v t z − v v t v 1

d v t

v v t v 1

v t z

− v v t v 1

v t dv v t v 1

v t z + v v t v 1

d v t

z + v v t v 1

v t dz

= (dv) v t v 1

v t z − v v t v 1

d v t

P v z − P v dv v t v 1

v t z +v v t v 1

d v t

z + P v dz

= (dv) v t v 1

v t z − v v t v 1

d (v) t P v z − P v dv v t v 1

v t z +v v t v 1

d (v) t z + P v dz

Using(A.2), this isequivalent tothe following. For all

h ∈ R n

,

(d ( P v z)) h = ((dv) h) v t v 1

v t z − v v t v 1

(d (v) h) t P v z

−P v (dvh) v t v 1

v t z + v v t v 1

(d (v) h) t z + P v dzh .

This an be further simplied by fatoring out the expression

(v t v) 1

and

rear-rangingsome terms. We obtain

(d ( P v z)) h = 1 v t v

v t z ((dv) h) − vz t P v t ((dv) h) − v t z P v ((dv) h) + vz t ((dv) h)

+ P v dzh

= 1

v t v

v t z − vz t P v − v t z P v + vz t

((dv) h) + P v dzh .

Finally,we use the denition of the rst derivative A.1 and obtain the following

result.

Proposition A.11. The rst derivate of the projetion operator is

∂ P v z

∂y = 1

v t v

vz t (I − P v ) + v t z (I − P v ) ∂v

∂y + P z ∂z

∂y .

A.5 The Moore-Penrose Inverse

The ontents of this setion an befound e.g. in Kokelkorn (2000). If amatrix

A

is not quadrati oris not of full rank, we have tond a suitablesurrogate for

itsinverse. In this work,we use the Moore-Penrose inverse.

Proposition A.12 (Moore-PenroseInverse). For any matrix

A ∈ R p × l

, there is

a unique matrix

A ∈ R l × p

suh that

A = AA A , A = A AA , AA t

= AA , A A t

= A A .

Proposition A.13. If

A

is a symmetri matrix with eigendeomposition

A = U Λ U t ,

the Moore-Penrose inverse of

A

is dened in the followingway. Set

Λ

ij =

 

 

 

 

0 i 6 = j

1

λ i i = j

and

λ i 6 = 0 0 i = j

and

λ i = 0 .

Then

A = U Λ U t .

Proof. It follows readily fromthe deniton of

Λ

that

ΛΛ = Λ Λ =

diag

( 1, . . . , 1

| {z }

rk

(A) −

times

, 0, . . . , 0) .

This impliesthat

Λ

isindeedthe Moore-Penrose inverse of

Λ

,asthe properties in proposition A.12are fullled. It follows that

AU Λ U t A = U ΛΛ Λ U t = U Λ U t = A

and

U Λ U t AU Λ U t = U Λ ΛΛ U t = U Λ U t .

Finally,we remark that the matrix

AU Λ U t = U Λ U t A = U

diag

(1, . . . , 1, 0, . . . , 0)U t

issymmetri.

Proposition A.14. The system of linear equations

Ax = b

has asolution if and onlyif

x = A b

isa solution. Any solution of these linear

equations has the form

x = x + I − A A v

for any vetor

v

. The two omponents of

x

are orthogonal.

Results of the Simulation Study

We display the results of the simulation study that is desribed in Setion 7.3.

Thefollowingtables show theMSE-RATIO for

β b

aswellasfor

y b

. In additionto

theMSE-RATIO,wedisplaytheoptimalnumberofomponentsforeahmethod.

Itisinterestingtosee thatthetwoquantitiesarethesamealmostallofthetimes.

ollinearity no no no med. med. med. high high high

stnr 1 3 7 1 3 7 1 2 7

1 0.833 0.861 0.676 0.958 1.000 0.993 1.000 0.999 1.000

2 0.980 0.976 0.975 0.995 0.938 0.864 0.847 0.965 0.866

3 1.000 0.993 1.001 0.969 0.960 0.993 0.954 0.980 0.967

4 1.000 1.001 0.999 0.988 1.000 1.002 0.997 0.993 0.992

m opt P LS

2 5 2 2 4 3 1 2 5

m opt T RN

2 5 2 2 4 3 1 2 5

Table B.1: MSE-RATIOof

β b

for

p = 5

. The rsttworows display the settingof

the parameters. The rows entitled 1-4 display the MSE ratio for the respetive

number of omponents.

ollinearity no no no med. med. med. high high high

stnr 1 3 7 1 3 7 1 2 7

1 0.775 0.780 0.570 0.919 1.000 0.970 1.004 0.995 0.999

2 0.978 0.972 0.9697 0.994 0.882 0.786 0.828 0.951 0.823

3 1.001 0.990 1.001 0.969 0.967 0.992 0.960 0.977 0.973

4 1.000 1.001 0.999 0.990 1.000 1.001 0.997 0.996 0.993

m opt P LS

3 5 3 2 4 4 1 2 5

m opt T RN

2 5 2 2 4 4 1 2 3

Table B.2: MSE-RATIOof

y b

for

p = 5

.

ollinearity no no no med. med. med. high high high

stnr 1 3 7 1 3 7 1 2 7

1 0.929 0.963 0.972 0.98 0.998 0.989 1.000 1.000 1.000

2 0.938 0.959 0.977 0.922 0.91 0.978 0.789 0.793 0.792

3 0.907 0.952 0.981 0.875 0.91 0.945 0.849 0.843 0.849

4 0.905 0.933 0.971 0.879 0.913 0.912 0.857 0.864 0.868

5 0.901 0.942 0.954 0.879 0.924 0.898 0.870 0.883 0.879

6 0.898 0.942 0.945 0.878 0.915 0.891 0.882 0.890 0.893

7 0.892 0.926 0.949 0.887 0.906 0.891 0.891 0.895 0.898

8 0.899 0.926 0.956 0.892 0.904 0.895 0.897 0.897 0.903

9 0.908 0.933 0.955 0.897 0.910 0.895 0.903 0.902 0.904

10 0.913 0.938 0.951 0.900 0.916 0.898 0.902 0.899 0.901

11 0.913 0.937 0.947 0.902 0.919 0.907 0.906 0.901 0.902

12 0.917 0.931 0.944 0.909 0.919 0.917 0.908 0.904 0.906

13 0.924 0.932 0.946 0.919 0.92 0.925 0.913 0.907 0.914

14 0.933 0.939 0.946 0.927 0.917 0.936 0.921 0.911 0.922

15 0.94 0.945 0.95 0.933 0.916 0.936 0.928 0.916 0.931

16 0.949 0.945 0.951 0.935 0.918 0.941 0.938 0.922 0.936

17 0.956 0.945 0.954 0.939 0.922 0.945 0.944 0.926 0.936

18 0.961 0.944 0.959 0.943 0.931 0.95 0.946 0.930 0.935

19 0.968 0.946 0.964 0.946 0.934 0.958 0.953 0.939 0.936

20 0.973 0.951 0.973 0.949 0.935 0.962 0.961 0.947 0.939

21 0.977 0.958 0.977 0.954 0.936 0.966 0.968 0.955 0.943

22 0.98 0.965 0.981 0.961 0.94 0.973 0.972 0.962 0.948

23 0.984 0.97 0.984 0.968 0.945 0.98 0.976 0.967 0.950

24 0.987 0.976 0.988 0.975 0.948 0.983 0.98 0.970 0.953

25 0.989 0.98 0.99 0.978 0.953 0.987 0.981 0.973 0.959

26 0.992 0.985 0.993 0.982 0.959 0.991 0.984 0.977 0.966

27 0.994 0.989 0.996 0.986 0.966 0.992 0.987 0.981 0.975

28 0.995 0.991 0.997 0.988 0.973 0.994 0.99 0.985 0.984

29 0.996 0.993 0.998 0.99 0.978 0.995 0.993 0.988 0.988

30 0.997 0.994 0.999 0.992 0.982 0.996 0.995 0.991 0.99

31 0.998 0.995 0.999 0.994 0.985 0.997 0.996 0.992 0.993

32 0.998 0.996 0.999 0.996 0.99 0.998 0.996 0.994 0.995

33 0.999 0.997 1.000 0.996 0.991 0.999 0.997 0.995 0.996

34 0.999 0.998 1.000 0.997 0.993 0.999 0.998 0.996 0.997

35 0.999 0.999 1.000 0.999 0.994 0.999 0.998 0.997 0.998

36 1.000 1.000 1.000 0.999 0.996 0.999 0.999 0.998 0.998

37 1.000 1.000 1.000 0.999 0.998 1.000 0.999 0.998 0.999

38 1.000 1.000 1.000 1.000 0.999 1.000 0.999 0.999 0.999

39 1.000 1.000 1.000 1.000 0.999 1.000 1.000 0.999 1.000

m opt P LS

1 3 5 1 2 2 1 1 1

m opt T RN

1 3 5 1 2 2 1 1 1

Table B.3: MSE-RATIO of

β b

for

p = 40

.

ollinearity no no no med. med. med. high high high

stnr 1 3 7 1 3 7 1 2 7

1 0.781 0.797 0.791 0.877 0.983 0.924 1.013 1.004 1.001

2 0.870 0.857 0.853 0.785 0.702 0.868 0.673 0.684 0.680

3 0.853 0.899 0.914 0.776 0.818 0.853 0.778 0.772 0.778

4 0.874 0.891 0.896 0.818 0.836 0.838 0.81 0.818 0.822

5 0.889 0.92 0.893 0.839 0.891 0.846 0.835 0.855 0.856

6 0.898 0.942 0.921 0.844 0.884 0.859 0.862 0.881 0.881

7 0.897 0.938 0.929 0.876 0.902 0.88 0.886 0.898 0.898

8 0.923 0.941 0.943 0.886 0.898 0.896 0.9 0.906 0.914

9 0.924 0.944 0.960 0.904 0.916 0.901 0.915 0.92 0.917

10 0.935 0.958 0.961 0.913 0.93 0.915 0.915 0.914 0.921

11 0.943 0.967 0.959 0.922 0.937 0.916 0.924 0.92 0.927

12 0.954 0.967 0.958 0.929 0.942 0.938 0.932 0.926 0.931

13 0.959 0.967 0.965 0.941 0.95 0.942 0.939 0.933 0.937

14 0.961 0.961 0.966 0.948 0.949 0.954 0.947 0.942 0.942

15 0.97 0.969 0.977 0.954 0.953 0.96 0.953 0.948 0.949

16 0.975 0.971 0.976 0.964 0.962 0.967 0.961 0.954 0.957

17 0.979 0.976 0.983 0.968 0.962 0.974 0.967 0.957 0.957

18 0.982 0.981 0.985 0.972 0.966 0.979 0.968 0.960 0.966

19 0.986 0.985 0.988 0.976 0.969 0.980 0.974 0.965 0.970

20 0.989 0.987 0.991 0.977 0.970 0.983 0.979 0.972 0.974

21 0.991 0.99 0.992 0.980 0.973 0.985 0.984 0.977 0.978

22 0.993 0.99 0.994 0.984 0.979 0.988 0.988 0.982 0.981

23 0.995 0.992 0.996 0.987 0.98 0.991 0.990 0.986 0.983

24 0.996 0.993 0.997 0.989 0.982 0.993 0.992 0.987 0.984

25 0.996 0.995 0.997 0.99 0.983 0.994 0.993 0.989 0.985

26 0.997 0.996 0.998 0.992 0.986 0.996 0.994 0.991 0.987

27 0.998 0.997 0.999 0.994 0.990 0.997 0.995 0.993 0.989

28 0.999 0.997 0.999 0.995 0.991 0.998 0.996 0.994 0.991

29 0.999 0.998 0.999 0.996 0.992 0.998 0.997 0.996 0.994

30 0.999 0.999 1.000 0.997 0.993 0.999 0.998 0.997 0.994

31 0.999 0.999 1.000 0.998 0.994 0.999 0.998 0.997 0.996

32 0.999 0.999 1.000 0.998 0.995 0.999 0.999 0.998 0.997

33 1.000 0.999 1.000 0.998 0.996 0.999 0.999 0.998 0.998

34 1.000 0.999 1.000 0.999 0.997 1.000 0.999 0.999 0.998

35 1.000 1.000 1.000 0.999 0.998 1.000 0.999 0.999 0.999

36 1.000 1.000 1.000 0.999 0.998 1.000 1.000 0.999 0.999

37 1.000 1.000 1.000 1.000 0.998 1.000 1.000 0.999 0.999

38 1.000 1.000 1.000 1.000 0.999 1.000 1.000 0.999 1.000

39 1.000 1.000 1.000 1.000 0.999 1.000 1.000 1.000 1.000

m opt P LS

1 3 9 1 1 2 1 1 1

m opt T RN

1 3 4 1 2 2 1 1 1

Table B.4: MSE-RATIO of

y b

for

p = 40

.