Computational Statistics Handbook with MATLAB phần 2 pot
Computational Statistics Handbook with MATLAB phần 2 pot
... θ()ln
σ
2
L θ()
1
σ 2
x
i
µ–()
2
2σ
2
–
exp
i 1=
n
∏
1
2 σ
2
n 2
1
2
2
x
i
µ–()
2
i 1=
n
∑
–
exp==
L θ()[]ln
1
2 σ
2
n
2
1
2
2
x
i
µ–()
2
i 1=
n
∑
–
expln+ln=
© ... estimators
VX() EX
2
[]EX[]()
2
–
t
λ
2
==
λ
t λEX[]=
EX
2
[]EX[]()
2
–
λEX[]
λ
2
=
λ
λ
EX[]
EX
2
[]EX[]()
2
–
=
λ
t
EX[]()
2
EX
2...
Computational Statistics Handbook with MATLAB phần 4 potx
... parallel axis.
Clustering in Both Dimensions
x2
x1
Clustering in x
1
x2
x1
© 20 02 by Chapman & Hall/CRC
21 2 Computational Statistics Handbook with M
ATLAB
M = 1000;
alpha = 0.05;
% Get the ... vectors.
for j = 1:d /2
a (2* (j-1)+1) = cof*sin(th(j)*t);
a (2* j) = cof*cos(th(j)*t);
b (2* (j-1)+1) = cof*cos(th(j)*t);
© 20 02 by Chapman & Hall/CRC
170 Computational Stat...
Computational Statistics Handbook with MATLAB phần 5 pot
... uncertainty associ-
B α 2 ⋅θ
ˆ
*b
B 1 α 2 –()⋅
q
ˆ
α 2
q
ˆ
1 α 2 –
1.03 1.45,()
θ
ˆ
θ
ˆ
θ
ˆ
.
θ
BC
a
© 20 02 by Chapman & Hall/CRC
25 0 Computational Statistics Handbook with M
ATLAB
Example ... 15 20
4
6
8
10
12
14
5 10 15 20
4
6
8
10
12
14
5 10 15 20
4
6
8
10
12
14
5 10 15 20
4
6
8
10
12
14
ρ
ˆ
0. 82=
T
i–()
JT() T
i
n⁄
i 1=
n
∑
=
)
SE
ˆ
JackP
T()
1
nn 1...
Computational Statistics Handbook with MATLAB phần 8 potx
... 1000;
X(1 :25 0,1) = unifrnd(0,1 ,25 0,1);
X(1 :25 0 ,2) = unifrnd(0.5,1 ,25 0,1);
y(1 :25 0) = 2+ sqrt (2) *randn(1 ,25 0);
X (25 1:500,1) = unifrnd(-1,0 ,25 0,1);
X (25 1:500 ,2) = unifrnd(-0.5,1 ,25 0,1);
y (25 1:500) ... measure as follows
(10 .29 )
x
1
x
1
x
2
R
α
T() Rt() αT+=
)
© 20 02 by Chapman & Hall/CRC
420 Computational Statistics Handook with M
ATLAB
available in...
Computational Statistics Handbook with MATLAB phần 3 pps
... the surface with the one shown in
Figure 5.18.
−4 2 0 2 4
−4
2
0
2
4
2
0
2
2
0
2
0
0.05
0.1
© 20 02 by Chapman & Hall/CRC
148 Computational Statistics Handbook with M
ATLAB
at 2- D scatterplots ... It is created using the contourf function.
−3 2 −1 0 1 2 3
−3
2
−1
0
1
2
3
−6
−4
2
2
0
0
0
2
2
4
6
8
−3 2 −1 0 1 2 3
−3
2
−1
0
1
2
3
© 20 02 by...
Computational Statistics Handbook with MATLAB phần 6 doc
... rougher.
−4 2 0 2 4
0
0 .2
0.4
0.6
0.8
h = 0.11
−4 2 0 2 4
0
0 .2
0.4
0.6
0.8
h = 0 .21
−4 2 0 2 4
0
0 .2
0.4
0.6
0.8
h = 0. 42
−4 2 0 2 4
0
0 .2
0.4
0.6
0.8
h = 0.84
n 100=
µ
K
0=
0 σ
K
2
∞.<<
h
Ker
*
RK()
nσ
k
4
Rf″()
... class 2
−6 −4 2 0 2 4 6 8
0
0.05
0.1
0.15
0 .2
0 .25
Feature − x
P(x | ω
2
) * P(ω
2
)
P(x | ω
1
) * P(ω
1
)
ω
1
()
ω
2
().
©...
Computational Statistics Handbook with MATLAB phần 7 pdf
... corresponding to features
190 143 52
174 131 50
21 8 126 49
130 131 51
138 127 52
211 129 49
X
1
X
2
X
3
© 20 02 by Chapman & Hall/CRC
348 Computational Statistics Handbook with M
ATLAB
% This shows ... =
0 1.0000 1.0000 2. 828 4 3.6056
1.0000 0 1.41 42 3.6056 4.4 721
1.0000 1.41 42 0 3.6056 4 .24 26
2. 828 4 3.6056 3.6056 0 1.0000
3.6056 4.4 721 4 .24 26 1.0000 0...
Computational Statistics Handbook with MATLAB phần 9 docx
... the function csintkern with .
500 1000 1500 20 00 25 00
1000
1500
20 00
25 00
3000
3500
4000
h 22 0=
© 20 02 by Chapman & Hall/CRC
486 Computational Statistics Handbook with M
ATLAB
evidence for ... process with specified intensity.
−5 −4 −3 2 −1 0 1 2 3 4 5
−5
−4
−3
2
−1
0
1
2
3
4
5
Homogeneous Poisson Process, λ = 2
λ
ˆ
2. 05=
© 20 02 by Chapman & Ha...
Computational Statistics Handbook with MATLAB phần 10 ppsx
... 1=
n
∑
2 e()log+=
φ
2
h
γ
γ,αβ,=
h
γ
1.06n
15⁄–
z
i
γ
z
j
γ
n⁄
j 1=
n
∑
–
2
n 1–()⁄
i 1=
n
∑
1
2
=
On
2
().
PI
M
αβ,()
1
12
κ
30
2
3κ
21
2
3κ
12
2
κ
03
2
1
4
κ
40
2
4κ
31
2
6κ
22
2
4κ
13
2
κ
04
2
++++
()++++
=
κ
30
n
n ... κ
40
2
4κ
31
2
6κ
22
2
4κ
13
2
κ
04
2
++++
()++++
=
κ
30
n
n 1–()n 2 ()
z
i
α
()
3
i 1=...