Голяндина Н.Э. Метод Гусеница-SSA: анализ временных рядов
Подождите немного. Документ загружается.
X
IR
L
L ×L
IR
L
IR
K
X
U = [U
1
: . . . : U
L
] V = [V
1
: . . . : V
K
]
d < K V
d+1
, . . . , V
K
U V L × L K × K
U
X, Y ∈ IR
L
(UX, UY ) = (X, Y )
U
IR
L
7→ IR
L
U
−1
= U
T
Λ
X λ
ii
= λ
i
1 ≤ i ≤ d
X = UΛ
1/2
V
T
or Λ
1/2
= U
T
XV.
X U
1
, . . . , U
L
IR
L
V
1
, . . . , V
K
IR
K
L ×K
Λ
1/2
U V
X
i
=
√
λ
i
U
i
V
T
i
X = X
1
+ . . . + X
d
.
X
i
X
i
X
i
X
T
j
= 0
LL
X
T
i
X
j
= 0
KK
i 6= j
X
M
L,K
L×K
LK
X = (x
ij
)
L,K
i,j=1
Y = (y
ij
)
L,K
i,j=1
M
L,K
hX, Yi
M
=
L
X
i=1
K
X
j=1
x
ij
y
ij
.
||X||
2
M
= hX, Xi
M
=
L
X
i=1
K
X
j=1
x
2
ij
dist
M
(X, Y) = ||X −Y||
M
X Y
IR
LK
x
ij
y
ij
M
||X
i
||
2
= λ
i
||X||
2
M
= λ
1
+ . . . + λ
d
λ
i
X
i
=
√
λ
i
U
i
V
T
i
(
√
λ
i
, U
i
, V
i
)
M
k
k
(
√
λ
i
, U
i
, V
i
)
1 min
Y∈M
k
||X −Y||
2
M
=
d
P
i=k+1
λ
i
.
2
Y
0
=
k
X
i=1
p
λ
i
U
i
V
T
i
∈ M
k
,
||X −Y
0
||
2
M
= min
Y∈M
k
||X −Y||
2
M
.
k
k
X = [X
1
: . . . : X
K
]
min
L
K
X
i=1
dist
2
(X
i
, L)
k
L = span(U
i
, . . . , U
k
)
||X −Y
0
||
2
M
||X||
2
M
=
λ
k+1
+ . . . + λ
d
λ
1
+ . . . + λ
d
.
X
I
=
X
i∈I
p
λ
i
U
i
V
T
i
I = {j
1
, . . . , j
k
} ⊂ {1, . . . , d} j
1
> . . . > j
k
k < d
1 −
||X −X
I
||
2
M
||X||
2
M
=
λ
j
1
+ . . . + λ
j
k
λ
1
+ . . . + λ
d
.
j
1
, . . . , j
k
X
1
, . . . , X
K
∈ IR
L
X, P ∈ IR
L
X 6= 0
L
||P || = 1
(X, P )P X
L
P
= span (P ) c = |(X, P )|
c = c(P )
X L
P
c = c(P ) X
span (P )
P L
P
X
1
, . . . , X
K
P
0
||P
0
|| = 1
ν
1
def
=
K
X
i=1
(X
i
, P
0
)
2
= max
P
K
X
i=1
(X
i
, P )
2
,
P ∈ IR
L
||P || = 1
X = [X
1
: . . . :
X
K
] (23)
1 P
0
= U
1
(30) ν
1
= λ
1
.
2 P
0
ν
k
def
=
K
X
i=1
(X
i
, P
0
)
2
= max
P
(k)
K
X
i=1
(X
i
, P )
2
,
(31) P ∈ IR
L
||P || = 1 (P, U
i
) = 0 1 ≤ i < k k ≤ d
P
0
= U
k
(31) ν
k
= λ
k
k > d
ν
k
= 0
U
i
i
X
1
, . . . , X
K
c
j
(U
i
) = (X
j
, U
i
)
X
j
=
d
X
i=1
c
j
(U
i
)U
i
,
c
j
(U
i
) i
X
j
Z
i
= (c
1
(U
i
), . . . , c
K
(U
i
))
T
= X
T
U
i
i
Z
i
=
√
λ
i
V
i
X
X L
K
V
i
V
i
X
√
λ
i
U
i
∗
N = 174
∗
1/ min(L, K) K = N − L + 1 N
L K
L = 84
N = 174 K = 91
w
w
w
w
w
w