Image Processing Reference
In-Depth Information
Then
μ
=
0 734
.
,
μ
=
04
.,
μ
=
0 464
.
,
μ
=
0 405
.
,
μ
=
0 243
.
a
a
a
a
a
σ
()
1
σ
()
2
σ
()
3
σ
()
4
σ
(()
5
ν
=
0 170
.
,
ν
=
02
.,
ν
=
0 438
.
,
ν
=
0 682
.
,
ν
=
0 528
.
a
a
a
a
a
σ
()
1
σ
()
2
σ
()
3
σ
()
4
σ
(()
5
Suppose
w
= [0.112, 0.236, 0.304, 0.236, 0.112]
T
which is derived from the
normal distribution-based method [16]. A brief idea on the normal distribu-
tion-based method is given:
Let
w
= (
w
1
,
w
2
, …,
w
n
)
T
be a weight vector (
i
= 1, 2, 3, …,
n
), and μ
n
= mean of
the collection which is computed as
1
nn
(
+
1
)
n
+
1
μ
n
=
=
n
2
2
σ
n
is the standard deviation (σ
>
0) which is computed as
n
1
∑
2
σ
=
(
i
−
μ
)
n
n
n
i
=
1
With μ
n
and σ
n
,
w
i
becomes
⎡
⎣
2
⎤
⎦
(
i
−
μ
σ
)
n
n
−
1
⎢
⎢
⎥
⎥
⎡
⎣
2
⎤
⎦
2
(
i
−
μ
σ
)
2
n
n
e
−
⎢
⎢
⎥
⎥
2
2
2
πσ
e
n
w
=
=
i
⎡
⎣
2
⎤
⎦
⎡
⎣
2
⎤
⎦
(
j
−
μ
σ
)
(
j
−
μ
σ
)
n
n
n
n
−
⎢
⎢
⎥
⎥
−
⎢
⎢
⎥
⎥
1
∑
n
∑
n
2
2
2
2
e
e
2
πσ
j
=
1
j
=
1
n
∑
n
2
With
μ
=+
(
n
12
)
/and
σ
=
(
1
/
n
)
(
i
−
μ
)
n
n
n
i
=
1
2
⎡
⎤
⎡
⎣
2
⎤
⎦
(
i
)
−−
+
⎛
⎜
n
1
⎞
⎟
−
μ
σ
n
n
2
⎢
⎢
⎥
⎥
−
i
2
σ
⎢
⎥
n
2
2
2
e
e
⎣
⎦
w
=
=
,
i
=
12,, ,
…
n
i
⎡
⎣
2
⎤
⎦
⎡
2
⎤
(
j
−
μ
σ
)
−−
+
n
1
⎛
⎜
⎞
⎟
2
n
n
⎢
⎥
−
⎢
⎥
j
2
σ
n
∑
n
∑
n
2
2
2
e
⎣
⎦
e
j
=
1
j
=
1
Now, for the earlier problem, we get
51
2
+
1
∑
n
μ
=
=
3
and
σ
=
(
i
−
μ
)
2
n
n
n
n
i
=
1
1
5
13 23 33 43
(
)
=
2
2
2
)
2
2
53 2
=
(
− +−+−+−
)
(
)
(
)
(
+−
(
)
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