Performance Evaluation of Image Analysis Methods 375
≃
1
√
2πKˆσ
[J −
1
2
K
X
k=
1
n
k
X
j=n
k−1
+1
(x
j
− ˆµ
k
)
2
ˆσ
2
] − J. (12.10)
Define
K
X
k=1
n
k
X
j=n
k−1
+1
(x
j
− ˆµ
k
)
2
ˆσ
2
= Y
K
, (
12.11)
Eq. (12.10) becomes
ln L
K
(
ˆ
r) ≃
1
√
2πK ˆσ
[J −
1
2
Y
K
] − J
. (12.12)
Because
P
n
k
j=n
k−1
+1
(x
j
−ˆµ
k
)
2
ˆσ
2
has a χ
2
distribution with the degree of free-
dom (n
k
− n
k−1
− 1); Y
K
has a χ
2
distribution with the degree of freedo m
P
K
k=1
(n
k
− n
k−1
− 1) = J − K [45].
Applying Eq. (12.12) to L
K
0
(
ˆ
r) and L
K
1
(
ˆ
r) in Eq. (12.6) with ˆσ = ˆσ
0
(when
K = K
0
) and ˆσ = ˆσ
1
(when K = K
1
), we have
P
e
= P (
1
K
1
ˆσ
1
Y
K
1
−
1
K
0
ˆσ
0
Y
K
0
< (
3
√
2π ln J)(K
0
− K
1
) + 2J(
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
)
), (12.13)
In the following sections and the appendices, σ
0
is the true value and ˆσ
0
is its
estimate. Let
Z =
1
K
1
ˆσ
1
Y
K
1
−
1
K
0
ˆσ
0
Y
K
0
∆
1
=
(3
√
2π ln J)(K
0
− K
1
)
∆
2
= 2J(
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
)
∆
= ∆
1
+ ∆
2
.
(12.14)
Eq. (12.13) becomes
P
e
= P (Z < ∆). (12.15)
Appendix 1 2A shows that the pdf of Z is
h(z) =
Ce
K
0
ˆσ
0
2
z
[
P
m
l=
0
(
l
m
)
(2m−l)!
a
2m−l+1
(−z)
l
] (z < 0)
Ce
−
K
1
ˆσ
1
2
z
[
P
m
l=
0
(
l
m
)
(2m−l)!
a
2m−l+1
(z)
l
] (z ≥ 0),
(12.16)
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