Performance Evaluation of Image Analysis Methods 401
For the probability of under-detection P
ud
, because ∆
1
> 0 and ∆
2
< 0, ∆
may be negative (when σ
2
0
is small or moderate) or positive (when σ
2
0
is very
large). When ∆ < 0, P
ud
has the same functional form as in Eq. (12.62 ) but
with different values of the s ame parameters. When ∆ > 0, from Eqs. (1 2.5),
(12.15), and (12.14) and using Eq. (12.60), we have
P
ud
=
Z
∆
−∞
h(z)dz =
Z
0
−∞
h(z)dz +
Z
∆
0
h(z)dz =
γ
m
1
−m
0
(γ + γ
−1
)
m
0
+m
1
+
1
m
0
X
l=0
(
m
1
m
0
+m
1
−l
)[γ(1 + γ
2
)
l
+ γ
−1
(1 + γ
−2
)
l
(1 − e
1
)], (12.64)
where
γ =
r
K
1
ˆσ
1
K
0
ˆσ
0
a
nd e
1
=
l
X
j=0
(
K
1
ˆσ
1
2
∆)
j
j!
e
−
K
1
ˆσ
1
2
∆
. (
12.65)
Thus, Eq. (12.6 2) finally becomes
P
ov
=
m
0
X
l=0
(
m
1
m
0
+m
1
−l
)
γ
m
1
−m
0
+l+1
e
0
(γ + γ
−1
)
m
1
+m
0
−l+
1
. (12.66)
Eq. (12.64), when ∆ ≥ 0, becomes
P
ud
=
m
0
X
l=0
(
m
1
m
0
+m
1
−l
)
γ
m
1
−m
0
+l+1
(1 + γ
−2l−2
(1 − e
1
))
(γ + γ
−1
)
m
0
+m
1
−l+
1
, (12.67)
and when ∆ < 0, becomes
P
ud
=
m
0
X
l=0
(
m
1
m
0
+m
1
−l
)
γ
m
1
−m
0
+l+1
e
0
(γ + γ
−1
)
m
1
+m
0
−l+
1
e
0
. (12.68)
Because J >> K
0
and J >> K
1
,
J−K
0
2
− 1 ≃
J−K
1
2
− 1, that is, m
0
≃ m
1
(Eq. (12.61)). Let m
0
≃ m
1
∆
= m, Eqs. (1 2.60) and (12.61) b ecome Eqs. (12 .16)
and (12.17), and E qs. (12 .66), (12.68), and (12.67) become Eqs. (12.18) and
(12.19), respectively.
A.4 Special C ases
When γ = 1, ∆
2
in Eq. (12.14) becomes z ero, which leads to ∆ = ∆
1
. Thus,
Eq. (12.15) becomes
P
e
= P (Z < ∆) = P (Z < ∆
1
), (12.69)
and Eqs. (12.18) and (12.19) become
P
ov
=
m
X
l=0
(
m
2m−l
)e
0
2
2m−l+1
and P
ud
=
m
X
l=0
(
m
2m−l
)(2 − e
1
)
2
2m−l+1
. (12.70)
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