Information Technology Reference
In-Depth Information
C = [ A ; α b ].If b R ( A ) ,thenrank ( A ) = n + 1. The rank n approximation of
C ,say C , is found such that the Frobenius norm of the correction is minimal.
The vector in ker( C ) scaled as x T , 1 T yields a solution of the weighted
TLS problem. This formulation has the drawback of requiring an infinite value
of α for the DLS problem and holds only asymptotically for the OLS problem.
5.2 GENERALIZED TOTAL LEAST SQUARES
(GeTLS EXIN) APPROACH
As
illustrated
in
Proposition
33,
in
the
unidimensional
EIV
model,
under
the
assumption
that
the
rows
of
[
A
;
b ]
(i.e.,
a ij ,
b i ,
i
=
1,
...
, m ,
j
, n ) are i.i.d. with common zero-mean vector and common covariance
matrix of the form
=
1,
...
2
ν
I n + 1 , the TLS method is able to compute strongly
consistent estimates of the unknown parameters. In many cases of practical
interest, the assumption above is not realistic. For instance, the relative errors
= σ
b i may be i.i.d. random variables with zero-mean value
and then the previous assumption no longer holds. The generalized GeTLS
EXIN deals with the case in which the errors a ij are i.i.d. with zero mean
and common variance σ a and the errors b i are also i.i.d. with zero mean
and common variance σ b = σ a . This formulation is justified because b and A
represent different physical quantities and therefore are often subject to different
measurement methods with different accuracy. Solving GeTLS [24,31,35,156],
implies finding the vector x minimizing
a ij
/
a ij
and
b i
/
2
F
2
F
ζ A
+ ( 1 ζ ) b
with
0 ζ 1
(5.2)
where
2
a
ζ
1 ζ
= σ
(5.3)
2
b
σ
Recalling the OLS, TLS, and DLS formulations:
m b
b 2
b
OLS :
min
b
subject to
R
(
A
)
(5.4a)
[ A ; b ] [ A ;
b ] F
b R (
A )
TLS :
min
subject to
(5.4b)
[ A
b ]
m × ( n + 1 )
;
n A
A F
A )
DLS :
min
A
subject to
b
R
(
(5.4c)
m
×
the formulation (5.2) comprises these problems. Indeed,
ζ = 0 results in the
a
OLS formulation because σ
= 0, whereas ζ = 0 . 5 results in the TLS formula-
a
2
b
2
ν
tion because σ
= σ
= σ
,and ζ = 1 results in the DLS formulation because
2
b
σ
= 0. Equation (5.3) defines the problem for intermediate values of ζ :inan
Search WWH ::




Custom Search