Sparse Bayesian (Champagne) Algorithm - Electromagnetic Brain Imaging - page 72

where A is a real symmetric matrix and m k is a column vector, and represents

remaining terms. The first term on the right-hand side is rewritten as

1
2 x k

K K

1
2

T

x k

1 (

x k − m k ) A ( x k − m k ) = Ax k − Am k +··· .

k = k = 1

Comparing the right-hand side of the equation above with that of Eq. ( 4.95 ), we get

H T H A = ʦ + ʲ ,

H T H − 1

H T y k namely m k = ʲ H T y k . Am k = ʲ ʦ + ʲ

Comparing the equations above with Eqs. ( 4.13 ) and ( 4.14 ), we have the relation-
ships A = ʓ and m k = ¯ x k ,giving

K

1
2

T

=

x k − ¯

ʓ (

x k − ¯ x k ) + . D

1 (

x k ) (4.96)

k =

This equation shows that D reaches theminimumat x k = ¯ x k , and theminimumvalue

is equal to .Thevalueof is obtained by substituting x k = ¯ x k into Eq. ( 4.20 ),

such that

K K

= 2

1
2

2

x k ʦ ¯

1

y k − H ¯ x k +

1 ¯

x k . (4.97)

k = k =

4.10.2 Derivation of Eq. ( 4.29 )

The derivation starts from in Eq. ( 4.97 ). On the right-hand side of this equation,

the k th terms containing x k are
2

y k

x k

1
2 ¯

x k = 2

1
2 ¯

2

x k ʦ ¯ x k H T y k + ¯ x k H T H x k ʦ ¯ y k − H ¯ x k + y k − 2 ¯ ¯ + x k

x k

1
2

y k x k ʲ H T y k + ¯ x k ( ʦ + ʲ H T H = ʲ y k − 2 ¯ ) ¯

x k

1
2

y k x k ʲ H T y k + ¯ x k ʓ ¯ = ʲ y k − 2 ¯ . (4.98)

H T y k = ʓ ¯

Using the relationship ʲ x k ,wehave

ʲ x k ʲ x k

1
2

1
2

y k x k ʓ ¯ x k ʓ ¯ y k x k ʓ ¯ y k − 2 ¯ x k + ¯ = y k − ¯

ʲ

1
2

y k y k − (ʲ ʓ − 1 H T y k )

T

ʓ (ʲ ʓ − 1 H T y k ) =