Biomedical Engineering Reference
In-Depth Information
and this
D
can be rewritten as
K
1
2
T
D
=
1
(
x
k
− ¯
x
k
)
ʓ
(
x
k
− ¯
x
k
)
+
.
(4.21)
k
=
The derivation of the above equation is presented in Sect.
4.10.1
, and
on the right-
hand side of Eq. (
4.21
)isgiveninEq.(
4.97
).
Using the results in Eq. (
4.21
), let us compute the integral on the right-hand side
of Eq. (
4.19
). First, we have
]
exp
d
x
exp [
K
k
=
1
(
1
2
T
−
D
] d
x
=
exp [
−
−
x
k
− ¯
x
k
)
ʓ
(
x
k
− ¯
x
k
)
.
(4.22)
Considering the Gaussian with its mean
x
k
and precision
¯
ʓ
, the integral on the
right-hand side is computed such that,
exp
d
x
k
=
(
N
/
2
1
2
(
2
ˀ)
T
−
x
k
− ¯
x
k
)
ʓ
(
x
k
− ¯
x
k
)
.
1
/
2
|
ʓ
|
Substituting the above results into Eqs. (
4.22
) and (
4.19
), we have
M
/
2
K
2
]
(
K
1
/
2
N
/
2
|
ʦ
|
2
ˀ)
p
(
y
|
ʱ
)
=
exp [
−
.
(4.23)
N
/
2
1
/
2
(
2
ˀ)
ˀ
|
ʓ
|
Taking the logarithm of both sides, and omitting constant terms, which are unrelated
to the arguments, we get
K
1
2
log
1
2
log
M
2
log
p
(
y
|
ʱ
)
=
−
|
ʓ
|+
|
ʦ
|+
log
ʲ
−
.
(4.24)
Let us define
ʣ
y
such that
ʣ
y
=
ʲ
−
1
I
ʦ
−
1
H
T
+
H
.
(4.25)
Using the formula in Eq. (C.95) in the Appendix, the relationship
|
ʦ
||
ʲ
−
1
I
ʦ
−
1
H
T
|=|
ʲ
−
1
I
H
T
H
+
H
||
ʦ
+
ʲ
|
(4.26)
holds. Thus, substituting Eqs. (
4.25
) and (
4.13
) into the above equation, we get
|
ʦ
||
ʣ
y
|=|
ʲ
−
1
I
||
ʓ
|
,
(4.27)
and
log
|
ʣ
y
|=
log
|
ʓ
|−
M
log
ʲ
−
log
|
ʦ
|
.
(4.28)
