TESTING HYPOTHESES IN UNIVERSAL MODELS

TESTING HYPOTHESES IN UNIVERSAL MODELS

Eva Fiˇ serov´ a

Department of Mathematical Analysis and Applied Mathematics Faculty of Science, Palack´y University

Tomkova 40, 779 00 Olomouc, Czech Republic e-mail: fiserova@inf.upol.cz

Abstract

A linear regression model, when a design matrix has not full column rank and a covariance matrix is singular, is considered. The problem of testing hypotheses on mean value parameters is studied. Conditions when a hypothesis can be tested or when need not be tested are given.

Explicit forms of test statistics based on residual sums of squares are presented.

Key words: universal linear model, unbiased estimator, tests hypotheses.

2000 Mathematics Subject Classification62J05, 62F03, 62F10.

1. Introduction

Let a linear regression model be under consideration. Generally, no assump- tions on the rank of design and covariance matrices are given. When testing linear hypotheses on mean value parameters in universal (singular) models, three typical situations can occur; either a hypothesis cannot be tested, or a hypothesis need not be tested, since it is automatically true, or a hypothesis can be tested.

The aim of the paper is to investigate possible situations which can occur when testing hypothesis in universal models and to find proper test statistics based on residual sums of squares.

∗Supported by the Council of Czech Government MSM 6 198 959 214.

2. Notations and auxiliary statements

Let A be an m × n matrix. Let M(A) = {Au : u ∈ R

} ⊂ R

and Ker (A) = {u : u ∈ R

, Au = 0} ⊂ R

denote the column space and the null space of the matrix A, respectively. Let W be an m × m symmetric positive semidefinite matrix such that M(A) ⊂ M(W). Then P

= A(A

WA)

A

W denotes a projector on M(A) in the W-seminorm.

The symbol M

means I − P

. If W = I (identity matrix), symbols P

and M

are used. The W-seminorm of x, x ∈ R

, is given by kxk

= √

x

Wx. Symbols A

and A

mean the g-inverse and the Moore- Penrose inverse of the matrix A, respectively.

Let N be an n × n symmetric positive semidefinite matrix. The symbol A

denotes the minimum N-seminorm g-inverse of the matrix A, i.e., the matrix A

satisfies equations

(1) AA

A = A, NA

A = A



A



N.

One of representation of the matrix A

is

A

=

 



N

A

(AN

A

)

if M(A

) ⊂ M(N),

(N + A

A)

A

[A(N + A

A)

A

]

otherwise.

In more detail cf. [4].

Lemma 2.1. Let M(B) ⊂ M(A) and M(B

) ⊂ M(C). Then (2) (A − BC

B

)

= A

+ A

B (C − B

A

B)

B

A

.

P roof. It is an obvious consequence of Rohde theorem (cf., e.g., [1], p. 446,

Lemma 10.1.40).

3. Universal model The universal linear model is considered in the form

(3) Y ∼ N



Xβ, Σ 

, β ∈ R

,

where Y is an n-dimensional normally distributed random vector, Xβ is the mean value of Y and Σ its covariance matrix. X is a given matrix of the type n × k and Σ is a given n × n symmetric positive semidefinite matrix.

The best linear unbiased estimator (BLUE) of the function Xβ in the universal model (3) is (cf. [4], p. 148)

(4) Xβ d = X[(X

)

]

Y = XD

X

T

Y with the covariance matrix

Var  Xβ d 

= XD

X

T

ΣT

XD

X

= X(D

− I)X

, where

T = Σ + XX

, D = X

T

X.

Let a null hypothesis

H

: h + Hβ = 0, h ∈ M(H),

where H is a given h × k matrix and h is a given h-dimensional vector, be tested in the universal model (3) against an alternative hypothesis

H

: h + Hβ = ξ 6= 0.

If the hypothesis is taken into account as constraints on the parameter β,the estimator of Xβ can be determined in the following way. Let β

be any solution of the equation h + Hβ = 0. Then the parameter β, β ∈ {u : h + Hu = 0}, can be expressed by the help of a new parameter γ

β = β

+ K

γ, γ ∈ R

,

where K

is a k × [k − rank(H)] matrix such that M(K

) = Ker(H).

The new model without constraints is in the form

Y ∼ N

(Xβ

+ XK

γ, Σ ) , γ ∈ R

.

Hence the BLUE of Xβ in the universal model (3) respecting the null hypothesis is

d d

Xβ = Xβ

+ \ XK

γ

= Xβ

+ XK

h

(K

X

)

i

(Y − Xβ

).

Since

M (M

) = Ker(H), M (XK

) = M (XM

) , and

P {Y − Xβ

∈ M(Σ)} = 1, it holds that

XK

h

(K

X

)

i

(Y − Xβ

) = XM

h

(M

X

)

i

(Y − Xβ

) and thus

(5) Xβ d d = Xβ

+ XM

h

(M

X

)

i

(Y − Xβ

).

The symbol b means the estimator in the universal model (3) and bb means the estimator in the universal model (3) respecting the null hypothesis.

4. Testing linear hypotheses Here approach of χ

-tests based on residual sums of squares

R

= 

Y − d Xβ 

h Var 

Y − d Xβ i



Y − d Xβ 

,

R

=



Y − Xβ d d



 Var



Y − Xβ d d





Y − Xβ d d



is used (cf. [3], p. 153-157, the first and the second theorems of the least squares theory).

Lemma 4.1. Let in the universal model (3) the null hypothesis be considered.

(i) Matrices Σ

, (Σ + XM

X

)

and (Σ + XX

)

can be chosen as a g-inverses of the matrix Var(Y − d Xβ).

(ii) Matrices Σ

and (Σ + XM

X

)

can be chosen as a g-inverses of the matrix Var(Y − Xβ). d d

P roof. Obviously covariance matrices of residual vectors are Var 

Y − d Xβ



= Σ − X h

(X

)

i

Σ, Var



Y − Xβ d d



= Σ − XM

h

(M

X

)

i

Σ.

Let the matrix (Σ + XM

X

)

be chosen. Then



Σ − X h

(X

)

i

Σ



(Σ + XM

X

)

n

Σ − Σ(X

)

X

o

=



I − X h

(X

)

i



Σ(Σ + XM

X

)

Σ + XM

X

− XM

X

× n

I − (X

)

X

o

=



I − X h

(X

)

i

 Σ n

I − (X

)

X

o

= Var 

Y − d Xβ  .

The other statements can be proved in an analogous way.

Theorem 4.2. Let in the universal model (3) the null hypothesis be consid- ered. Let M(H

) ∩ M(X

) = {0}. Then R

− R

= 0, i.e, the hypothesis cannot be tested by the help of the statistic R

− R

.

P roof. If M(H

) ∩ M(X

) = {0}, then M(XM

) = M(X), since

rank(X) + rank(H) = rank X H

!

= rank(XM

) + rank(H)

(cf. [4], p. 137). Thus Y − d Xβ = n

I − X h

(X

)

i

o Y = n

I − X h

(X

)

i

o

(Y − Xβ

)

= n

I − XM

h

(M

X

)

i

o

(Y − Xβ

) = Y − Xβ d d

and with respect to Lemma 4.1 we obtain R

− R

= 0.

The last theorem implies that those rows of the matrix H, which cannot be obtained from rows of the matrix X by a linear combination, cannot be used in the hypothesis. Therefore in the following text M(H

) ⊂ M(X

) is assumed. Moreover, the assumption M(H

) ⊂ M(X

) implies that the vector function Hβ is unbiasedly estimable in the universal model (3) as

Hβ d = H h

(X

)

i

Y = HD

X

T

Y ,

Var  Hβ d 

= HD

X

T

ΣT

XD

H

= H(D

− I)H

.

Before a theorem on a test of a linear hypothesis some auxiliary statements must be proved.

Lemma 4.3. Let in the universal model (3) the null hypothesis be considered.

(i) As a minimum Σ-seminorm g-inverse (X

)

of the matrix X

also matrices (X

)

H0X0)

and (X

)

can be chosen.

(ii) As a minimum Σ-seminorm g-inverse (M

X

)

of the matrix M

X

also the matrix (M

X

)

H0X0)

can be chosen.

P roof. (i) Both matrices (X

)

H0X0)

and (X

)

are g-inverses of the matrix X

. Thus it suffices to prove the symmetry of matrices

Σ(X

)

H0X0)

X

and Σ(X

)

X

. The matrix

(Σ + XM

X

)(X

)

H0X0)

X

is symmetric with respect to definition of the matrix (X

)

H0X0)

. Since

(Σ + XM

X

)(X

)

H0X0)

X

= Σ(X

)

H0X0)

X

+ XM

X

, the matrix Σ(X

)

H0X0)

X

is symmetric too.

Analogously for the matrix Σ(X

)

X

. (ii) It can be proved in the same way as (i).

Lemma 4.4. Let in the universal model (3) the null hypothesis be under consideration. Let M(X) ⊂ M(Σ + XM

X

) and M(H

) ⊂ M(X

).

Then one choice of the g-inverse of the matrix Var( d Hβ + h) is n

H 

X

(Σ + XM

X

)

X 

H

o

.

P roof. The covariance matrix of the random vector d Hβ + h is Var 

Hβ d 

= H 

X

(Σ + XM

X

)

X 

X

(Σ + XM

X

)

Σ

× (Σ + XM

X

)

X 

X

(Σ + XM

X

)

X 

H

,

since the assumption M(X) ⊂ M(Σ + XM

X

) implies that one version of the minimum Σ-seminorm g-inverse of the matrix X

is

X

m(Σ)

= (Σ + XM

X

)

X 

X

(Σ + XM

X

)

X 

.

The last term of the expression for Var( d Hβ) is the matrix H

and thus it is sufficient to prove the equality

H

n H 

X

(Σ + XM

X

)

X 

H

o

Var  Hβ d 

= H

. It holds that

H

n H 

X

(Σ + XM

X

)

X 

H

o

Var  Hβ d 

= H

n H 

X

(Σ + XM

X

)

X 

H

o

H 

X

(Σ + XM

X

)

X 

× X

(Σ + XM

X

)

Σ + XM

X

− XM

X

× (Σ + XM

X

)

X 

X

(Σ + XM

X

)

X 

H

= H

n H 

X

(Σ + XM

X

)

X 

H

o

H 

X

(Σ + XM

X

)

X 

× 

X

(Σ + XM

X

)

X  

X

(Σ + XM

X

)

X 

H

= H

.

The following theorem deals with a test of a linear hypothesis under the special condition M(X) ⊂ M(Σ + XM

X

).

Theorem 4.5. Let in the universal model (3) the null hypothesis be

considered. If M(H

) ⊂ M(X

) and M(X) ⊂ M(Σ + XM

X

), the test

statistic is

R

− R

= 

Hβ d + h 

n H 

X

(Σ + XM

X

)

X 

H

o



Hβ d + h  . The statistic R

− R

has the central chi-squared distribution when the null hypothesis is true and the noncentral chi-squared distribution when the null hypothesis is not true; the parameter of noncentrality is

δ = (Hβ

+ h)

 H

h

X

Σ + XM

X

X

i

H



(Hβ

+ h)

where β

is an actual value of the parameter β. Degrees of freedom are f = rank(H).

P roof. According to Lemma 4.3 the matrix (M

X

)

H0X0)

is one choice of the minimum Σ-seminorm g-inverse of the matrix M

X

. Thus using relations

h

(M

X

)

H0X0)

i

= h

M

X

Σ + XM

X

XM

i

M

X

Σ + XM

X

, h

M

X

Σ + XM

X

XM

i

= h

X

Σ + XM

X

X i

− h

X

Σ + XM

X

X i

H

×

 H h

X

Σ + XM

X

X i

H



H h

X

Σ + XM

X

X i

,

(X

)

= (X

)

H0X0)

and

M(H

) ⊂ M(X

) ⇒ H h

(X

)

H0X0)

i

Xβ

= Hβ

= −h the estimator d Xβ d given by (5) can be rewritten as

d d

Xβ = d Xβ + k, where

k = −X 

X

(Σ + XM

X

)

X 

H

n

H 

X

(Σ + XM

X

)

X 

H

o

× 

Hβ d + h  . Then

R

− R

=



Y − Xβ d d



Σ + XM

X



Y − d d Xβ



− 

Y − d Xβ



Σ + XM

X



Y − d Xβ



= 

Y − d Xβ − k 

Σ + XM

X



Y − d Xβ − k 

− 

Y − d Xβ 

Σ + XM

X



Y − d Xβ 

= −2k

Σ + XM

X



Y − d Xβ 

+ k

Σ + XM

X

k.

It is easy to show that

k

(Σ + XM

X

)



Y − d Xβ 

= 0 and

k

(Σ + XM

X

)

k

= 

Hβ d + h 

n

H[X

(Σ + XM

X

)

X]

H

o



Hβ d + h 

.

Further f = rank h

Var  Hβ d i

= rank n H 

X

(Σ + XM

X

)

X 

X

(Σ + XM

X

)

× (Σ + XM

X

− XM

X

)(Σ + XM

X

)

× X 

X

(Σ + XM

X

)

X 

H

o

= rank n H 

X

(Σ + XM

X

)

X 

H

− HM

H

o

= rank(H).

The rest of the proof is obvious.

The condition M(X) ⊂ M(Σ + XM

X

), which enable us to utter the statement on testing linear hypotheses in the classical form, is an obstacle in a general solution of the problem. Therefore we shall investigate a situation in which we assume M(H

) ⊂ M(X

), however M(X) ⊂ M(Σ+XM

X

), is not assumed.

Theorem 4.6. Let in the universal model (3) the null hypothesis be consid- ered. Let M(H

) ⊂ M(X

). Then the BLUE of the vector 

H

 β is

\ \ X H

!

β = Xβ d Hβ d

!

− X

H

!

(D

− I)H



H(D

− I)H





Hβ d + h  .

The expression  \ \

H



β is invariant with respect to the choice of the g-inverse.

P roof. The universal model (3) with the null hypothesis can be written in the form

Y

−h

!

∼ N

"

X H

!

β, Σ, 0 0, 0

!#

.

Then the sought estimator is

\ \ X H

!

β = X H

! h

(X

, H

)

m

Σ, 0 0, 0



i

Y

−h

! .

Further h

(X

, H

)

m

Σ, 0 0, 0



i

=

"

(X

, H

) T, XH

HX

, HH

!

X H

!#

(X

, H

) T, XH

HX

, HH

!

=

"

(X

, H

) Q, R S, U

! X H

!#

(X

, H

) Q, R S, U

! ,

where (cf. Rohde theorem, e.g., in [1], p. 446, Lemma 10.1.40) Q = T

+ T

XH

[H(I − D)H

]

HX

T

, R = −T

XH

[H(I − D)H

]

,

S = −[H(I − D)H

]

HX

T

, U = [H(I − D)H

]

.

Then the expression

(X

, H

) Q, R S, U

!

can be rewritten as

X

T

− (I − D)H

[H(I − D)H

]

HX

T

; (I − D)H

[H(I − D)H

]

and

"

(X

, H

) Q, R S, U

! X H

!#

= n

D + (I − D)H

[H(I − D)H

]

H (I − D) o

. Further, with respect to formula (2) we have

n

D + (I − D)H

[H(I − D)H

]

H (I − D) o

= D

−D

(I−D)H

h

H (I−D)H

+H(I−D)D

(I−D)H

i

H (I−D)D

and using DD

H

= H

we obtain

H (I − D)H

+ H(I − D)D

(I − D)H

= HH

+ HD

H

− 2HDD

H

= H(D

− I)H

. Now the expression for the BLUE of the vector 

H



β is obvious.

The statement on an arbitrary choice of a g-inverse is a consequence of the relationship

P

( Y

−b

!

∈ M

"

Σ, 0 0, 0

!

+ X

H

!

(X

, H

)

#)

= 1.

Theorem 4.7. Let in the universal model (3) the null hypothesis be consid- ered. If M(H

) ⊂ M(X

), then the test statistic is

R

− R

= 

Hβ d + h 



H(D

− I)H





Hβ d + h  ,

where H(D

− I)H

= Var(d Hβ + h).

The statistic R

− R

has the central chi-squared distribution when the null hypothesis is true and the noncentral chi-squared distribution when the null hypothesis is not true; the parameter of noncentrality is

δ = (Hβ

+ h)



H(D

− I)H



(Hβ

+ h) ,

where β

is an actual value of the parameter β. The degrees of freedom are rank 

H(D

− I)H

 .

P roof. With respect to Lemma 4.1 as a g-inverse of both matrices Var



Y − Xβ d d



and Var 

Y − d Xβ 

the matrix W

= (Σ + XM

X

)

can be chosen and at the same time the quadratic forms



Y − Xβ d d



 Var



Y − Xβ d d





Y − Xβ d d



and 

Y − d Xβ 

h Var 

Y − d Xβ i



Y − d Xβ  are invariant with respect to the choice of the g-inverse. Let

d d

Xβ = d Xβ − Xa, where

a = (D

− I)H



H(D

− I)H





Hβ d + h  . Then



Y − Xβ d d



W



Y − Xβ d d



= 

Y − d Xβ



W



Y − d Xβ



− 2 

Y − d Xβ



W

Xa + a

X

W

Xa.

In the following consideration the matrix W

will be used instead of W

. Thus we can proceed in simpler way.

At first the validity of the equality (Y − d Xβ)

W

Xa = 0 will be proved.

Using the relationship (2), the identity

(Σ + XM

X

)

= (T − XP

X

)

can be rewritten as

T − XP

X

= T

+ T

XP

I − P

X

T

XP

P

X

T

. Then we obtain



Y − d Xβ



W

Xa = Y

I − T

XD

X

× h

T

+ T

XP

I − P

X

T

XP

P

X

T

i

Xa = 0, since

I − T

XD

X

T

X = 0.

Further a

X

W

Xa

= 

Hβ d + h 



H(D

− I)H



H(D

− I)X

Σ + XM

X

× X(D

− I)H



H(D

− I)H





Hβ d + h 

= 

Hβ d + h 



H(D

− P

)H



H(D

− P

)X

Σ + XM

X

× X(D

− P

)H



H(D

− P

)H





Hβ d + h 

= 

Hβ d + h 



H(D

− I)H





Hβ d + h 

= 

Hβ d + h 



H(D

− I)H





Hβ d + h 

,

since

X

Σ + XM

X

X

= X



T

+ T

XP

(I − P

DP

)

P

X

T

 X

= D + DP

(I − P

DP

)

P

D = (D

− P

)

. Finally

Var 

Hβ d + h 

= HD

X

T

ΣT

XD

H

= H(D

− I)H

. The other statements are obvious.

Theorem 4.8. Let in the universal model (3) the null hypothesis be consid- ered. If M(H

) ⊂ M(X

M

), then the hypothesis need not be tested since in this case P{d Hβ + h = 0} = 1.

P roof. This statement is implied by the fact that Var( d Hβ) = 0. It is a consequence of the relationship

M(H

) ⊂ M(X

M

) ⇒ ∃E : H

= X

M

E, which implies

Var  Hβ d 

= Var



E

M

X h

(X

)

i

Y



= E

M

X h

(X

)

i

Σ(X

)

X

M

E

= E

M

X h

(X

)

i

ΣM

E = 0.

If the null hypothesis is not true, the test statistic R

−R

has the noncentral chi-squared distribution with f degrees of freedom and the parameter of noncentrality is

δ = (Hβ

+ h)

h Var 

Hβ d + h i

(Hβ

+ h) ,

where β

is a true value of the parameter β. The power of the test at the point ξ is

p (ξ) = P 

R

− R

≥ χ

(0, 1 − α)| H

: Hβ + h = ξ 6= 0 .

Here χ

(0, 1 − α) is (1 − α)-quantile of the central chi-squared distribution with f degrees of freedom.

The random variable R

− R

∼ χ

(δ) can be approximated by (cf. [2]) χ

(δ) ≈ c

χ

(0),

where

c

= f + 2δ

f + δ , g = (f + δ)

f + 2δ .

Remark 4.9. It is to be pointed out that in practical computing it is nec- essary to be very careful since in some situations some derived formulae can be numerically unstable. For example, small numbers on the main diagonal of the matrix Σ can caused the covariance matrix H(D

− I)H

numerically unstable. In practice it is useful to compute with both expressions

Var 

Hβ d + h 

= H(D

− I)H

,

Var 

Hβ d + h 

= HD

X

T

ΣT

XD

H

and to compare obtained results. If they are different, one can use for example the substitution Σ → kΣ, X → √

kX, where k > 0 is a sufficiently large number, and to compute them once more. This substitution does not influence the result of the original covariance matrix.

Another problem can occur when degrees of freedom are computed.

Here, e.g., expressions rank[(Σ + XX

)

X] or rank[(Σ + XM

X

)

X] can be numerically unstable.

5. Example

Example 5.1. Let a linear part of high-speed lane be under consideration.

One of the safety conditions is that rails are in the line. For the sake of

simplicity let the problem be studied in plane only. An experiment for

the verification that rails are in the straight line can be done, e.g.,

in the following way. Firstly, points X

, i = 1, . . . , 4, are chosen elsewhere on rails. Then other points Z

, Z

, Z

are chosen around rails such that all distances Z

X

and Z

Z

, i, k = 1, 2, 3, i 6= k, j = 1, . . . , 4, can be observed. Finally points X

, Z

are put into proper coordinates system (the map), see Figure 1. Let each distance be measured just once. Let distances Z

X

, i, j = 1, 2, and Z

Z

have been measured in previous experiment by V¨ais¨al¨a interferometer, i.e., the accuracy of measurement is practically σ

= 0. (cf. [5], p. 50). Let other distances be measured by optical range- finder with the accuracy σ = 0.01 m. The problem is to test a hypothesis that all points X

, i = 1, . . . , 4, are located on a straight line.

0 200 400 600 800 1000

300 400 500 600 700 800 900 1000 1100 1200

x (m)

y (m)

Z1

Z₂ Z₃

X1

X2

X3

X4

y1 y₂ y₃ y4

y5 y₆ y7

y8 y9

y10 y₁₁ y12

y13 y14

y15

Figure 1. The design of the experiment

Let the notation

• y = (y

, . . . , y

)

. . . a vector of observed distances Z

X

, Z

Z

, i, k = 1, 2, 3, i 6= k, j = 1, . . . , 4,

• β = (β

, . . . , β

)

. . . a vector of unknown coordinates of points X

= [β

, β

]

, i = 1, . . . , 4, and Z

= [β

, β

]

, i = 5, 6, 7, be used. The mentioned process of measurement can be modelled by

Y ∼ N

(f (β), Σ) , where, e.g.,

f

(β) = p

(β

− β

)

+ (β

− β

)

and Σ is a diagonal matrix given by

Σ = Diag

 

 0, 0, σ

, σ

, 0, 0, σ

, . . . , σ

| {z }

6×

, 0, σ

, σ

 

 . The linear version of the model can be written in the form

Y − f  β



∼ N

(X∆β, Σ) , ∆β = β − β

, where β

are approximate values of the vector β and

X = ∂f(u)

∂u

.

Let straight line p

intersects points X

and X

, i = 1, 2, 3. Straight lines p

can be expressed as

p

: y = a

+ b

x, i = 1, 2, 3, where

a

= β

− β

β

− β

β

− β

, b

= β

− β

β

− β

.

The problem is to test the null hypothesis H

: p

= p

= p

against the alternative hypothesis

H

: ∃i 6= j : p

6= p

, i, j ∈ {1, 2, 3}.

Straight lines p

and p

are identical if and only if a

= a

and b

= b

, i.e., b

= b

= b ⇔ β

− β

β

− β

= β

− β

β

− β

= b and

a

= a

⇔ β

− β

b = β

− β

b ⇔ β

− β

β

− β

= b.

Analogously for p

, p

and p

, p

. Thus

p

= p

= p

⇔ g(β) = 0, where

g

(β) = (β

− β

)(β

− β

) − (β

− β

)(β

− β

), i = 1, 2, and

g

(β) = (β

− β

)(β

− β

) − (β

− β

)(β

− β

).

Linear version of the null hypothesis can be written as H

: H∆β = 0,

where

H = ∂g(u)

∂u

and the alternative hypothesis as

H

: H∆β = ξ 6= 0.

Let approximate values β

have been chosen as (in meters):

Z

= 100 400

!

, Z

= 800 650

!

, Z

= 250 1100

! ,

X

= 200 660

!

, X

= 320 696

!

, X

= 400 720

!

, X

= 510 753

! .

In this case, terms in linearized model are the following one:

f β⁽⁰⁾

= [ 278.56777, 368.80347, 438.63424, 541.02588, 600.08333, 482.19913, 406.07881, 307.74827, 442.83180, 410.01951, 408.53396, 433.60005, 743.30344, 715.89105, 710.63352 ]⁰,

the design matrix X = (X

, X

), where

X

=



 

 

 

 

 

 

 

 

 

 

 

 

 

 

5.99149, 15.57786, 0, 0, 0, 0, 0

0, 0, 11.45579, 15.41325, 0, 0, 0

0, 0, 0, 0, 14.32419, 15.27913, 0

0, 0, 0, 0, 0, 0, 17.62686

−24.49320, 0.40822, 0, 0, 0, 0, 0

0, 0,−21.85889, 2.09481, 0, 0, 0

0, 0, 0, 0,−19.84974, 3.47370, 0

0, 0, 0, 0, 0, 0,−16.53104

−2.37602,−20.90900, 0, 0, 0, 0, 0

0, 0, 3.45697,−19.95166, 0, 0, 0

0, 0, 0, 0, 7.42125,−18.80050, 0

0, 0, 0, 0, 0, 0, 12.48615

0, 0, 0, 0, 0, 0, 0

0, 0, 0, 0, 0, 0, 0

0, 0, 0, 0, 0, 0, 0



 

 

 

 

 

 

 

 

 

 

 

 

 

 

,

X

=



 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0, −5.99149, −15.57786, 0, 0, 0, 0

0, −11.45579, −15.41325, 0, 0, 0, 0

0, −14.32419, −15.27913, 0, 0, 0, 0

15.17629, −17.62686, −15.17629, 0, 0, 0, 0

0, 0, 0, 24.49320, −0.40822, 0, 0

0, 0, 0, 21.85889, −2.09481, 0, 0

0, 0, 0, 19.84974, −3.47370, 0, 0

5.87137, 0, 0, 16.53104, −5.87137, 0, 0

0, 0, 0, 0, 0, 2.37602, 20.90900

0, 0, 0, 0, 0, −3.45697, 19.95166

0, 0, 0, 0, 0, −7.42125, 18.80050

−16.66421, 0, 0, 0, 0, −12.48615, 16.66421

0, −25.67527, −9.16974, 25.67527, 9.16974, 0, 0 0, −5.60619, −26.16222, 0, 0, 5.60619, 26.16222 0, 0, 0, 20.63193, −16.88067, −20.63193, 16.88067



 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and the matrix H = (H

, 0

), where

H

=



 

24, −80, −60, 200, 36, −120, 0, 0

0, 0, 33, −110, −57, 190, 24, −80 33, −110, −33, 110, −36, 120, 36, −120



  .

Let the simulated data of observed distances be (in meters):

y= [ 278.56778, 368.80346, 438.63423, 541.02588, 600.07120, 482.18593, 406.08812, 307.74839, 442.82535, 410.02757, 408.53628, 433.59015, 743.31683, 715.89395, 710.64831 ]⁰.