400 Statistics of Medical Imaging
1 ≤ l ≤ K
1
) with a few pixels is cre ated. This newly created r egion has a
large var iance (but not very large) and a very small weight (ˆπ
l
≃ 0) such that
ˆπ
l
ˆσ
2
l
is very small. Other re gions (say, the k-th region, 1 ≤ k ≤ K
1
and k 6= l)
are essentially the same as the original regions. Thus ˆσ
2
1
=
P
K
1
k=1
ˆπ
k
ˆσ
2
k
≃
P
K
1
k=1, k6=l
ˆπ
k
ˆσ
2
k
≃
P
K
0
k=1
ˆπ
k
ˆσ
2
k
= ˆσ
2
0
, or ˆσ
1
is slightly larger than ˆσ
0
. As a
result, K
1
ˆσ
1
> K
0
ˆσ
0
. In the case of under-detection (K
1
< K
0
), some regions
(say, the i-th and j-th regions, 1 ≤ i, j ≤ K
0
) are merged into one region
(say, the l-th region, 1 ≤ l ≤ K
1
). This newly merged region has a very large
variance and a big weight (ˆπ
l
= ˆπ
i
+ ˆπ
j
) such that ˆπ
l
ˆσ
2
l
≫ ˆπ
i
ˆσ
2
i
+ ˆπ
j
ˆσ
2
j
. Other
regions (say, the k-th region, 1 ≤ k ≤ K
1
and k 6= l) remain almost the same
as the original regions. Thus, ˆσ
2
1
=
P
K
1
k=1
ˆπ
k
ˆσ
2
k
≫
P
K
0
k=1
ˆπ
k
ˆσ
2
k
= ˆσ
2
0
. As a
result, K
1
ˆσ
1
> K
0
ˆσ
0
(even though K
1
< K
0
). So , in both of the above cases,
∆
2
= 2J(
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
) ≤ 0
. The equality occurs if and only if K
1
ˆσ
1
= K
0
ˆσ
0
.
Simulations also showed that when σ
0
increases, ˆσ
0
and ˆσ
1
increase. Therefore,
∆
2
will monotonically increase.
For (3), we also set |K
1
− K
0
| = 1. From the discussion in (2) above, we
know that ∆
2
= 2J(
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
) ≤ 0
, but due to ˆσ
1
≃ ˆσ
0
(K
1
> K
0
)
and ˆσ
1
≫ ˆσ
0
(K
1
< K
0
), |
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
| is
close to zero when K
1
> K
0
and |
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
| is
quite different from zero when K
1
< K
0
. Ther efore,
|∆
2
(K
1
> K
0
)| < |∆
2
(K
1
< K
0
)|.
For (4), in the case of over-detectio n (K
1
> K
0
), because ∆
1
< 0 a nd
∆
2
< 0, we have ∆ < 0. In the case of under-detection (K
1
< K
0
), due to
|∆
1
| ∝ ln J whe n ∆
1
≥ 0 and |∆
2
| ∝ J when ∆
2
< 0; therefore, for the not
very large σ
0
, ∆ = ∆
1
+ ∆
2
< 0. For very large σ
0
, |
1
K
1
ˆσ
1
−
1
K
0
ˆσ
0
| is
very
small so that ∆
2
becomes very small. As a result, ∆ = ∆
1
+ ∆
2
> 0.
A.3 Probabilities P
ov
and P
ud
This section proves Eq. (12.18) and Eq. (12.19).
Proof.
For the pr obability of over-detection P
ov
, because ∆ < 0, from Eqs. (12.5),
(12.15), and (12.14) and using Eq. (12.60), we have
P
ov
=
Z
∆
−∞
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
e
0
], (12.62)
where
γ =
r
K
1
ˆσ
1
K
0
ˆσ
0
a
nd e
0
=
l
X
j=0
(−
K
0
ˆσ
0
2
∆)
j
j!
e
K
0
ˆσ
0
2
∆
. (
12.63)
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