Digital Signal Processing Reference
In-Depth Information
n
E
˜
)
=
c
(
n
E
[
A
(
n
,
0
)
]
˜
c
(
−
1
)
+
(
1
−
α)
E
[
A
(
n
,
j
+
1
)
]
c
T
(4.60)
j
=
0
At this point we can use the independence assumption on the input regressors. It is
then clear that
E
T
n
B
n
−
j
I
L
−
f
E
[
A
(
n
,
j
+
1
)
]
=
α
(
x
(
i
))
˜
x
(
i
)
=
,
(4.61)
x
i
=
j
+
1
where in the first equality the order of the matrix products is the same as in (
4.58
).
The matrix
B
x
=
E
T
is given by
I
L
−
f
α
(
x
(
j
))
˜
x
(
j
)
E
f
1
/
2
C
x
μ
1
/
2
x
T
B
x
=
α
I
L
−
μ
,
with
C
x
=
(
x
(
n
))
(
n
)
.
(4.62)
Therefore, we can write (
4.60
)as:
n
E
˜
)
=
B
n
−
j
B
n
+
1
x
c
(
n
c
˜
+
(
1
−
α)
c
T
.
(4.63)
−
1
x
j
=
0
This is a matricial equation and we will assume that
B
x
is diagonalizable.
9
We have
then the following lemma:
E
˜
)
<
∞
Lemma 4.2
Equation (
4.63
) will be stable (i.e.,
lim
n
→∞
c
(
n
), for
L.
10
every choice of
˜
c
(
−
1
)
and
c
T
if and only if
eig
i
[
B
x
]
<
1
,
i
=
1
,...,
Proof
As
B
x
is diagonalizable, it exists an invertible matrix
P
such that
B
x
=
P
−
1
, where
P
=
diag
(λ
1
,...,λ
L
)
, and
λ
i
denotes the eigenvalues of
B
x
.Itis
n
P
−
1
. Now, we can write equation (
4.63
)as:
easy to show that
B
x
=
P
n
E
˜
)
=
n
+
1
P
−
1
n
−
j
P
−
1
c
T
c
(
n
P
c
˜
(
−
1
)
+
(
1
−
α)
P
j
=
0
⎛
⎞
n
n
+
1
P
−
1
⎝
n
−
j
⎠
P
−
1
c
T
=
P
c
˜
(
−
1
)
+
(
1
−
α)
P
0
j
=
⎛
⎞
n
j
=
0
n
+
1
P
−
1
⎝
j
⎠
P
−
1
c
T
.
=
P
c
˜
(
−
1
)
+
(
1
−
α)
P
(4.64)
9
If
B
x
is not diagonalizable, we can always find a Jordan decomposition [
35
] for it, and the result
from Lemma
4.2
is still valid [
24
].
10
We use e ig
i
[
A
] to denote the
i
-th eigenvalue of matrix
A
.
