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inequality is given by
2
n
2
n
σ
α n + 1 σ
σ n
2
+
1
+
1
(6.81)
ζ )
l 1
2
(
1
ζ
u 1
2
ζ
u 2
From inequality (6.40) it follows that
2
2
2 ( 1 ζ )
u 3
b
γ
OLS
1 ζ
α n + 1
=
(6.82)
For ζ 0 (OLS), the bounds l 1and u 3 tend to finite values, while u 1and
u 2 tend to infinity. It follows that
2
2
2
2
2 ( 1 ζ ) b
b
α n + 1
= γ OLS
(6.83)
2
From (6.81) we also derive
2
n
σ
+
1
γ OLS
(6.84)
2
For
ζ 0 . 5, the following strictest
inequalities derived from the previous
theory are true:
2
n + 1
2 ζ
l 3
2
n + 1
σ
σ
0 . 5 ζ ζ +
α n + 1
1 ζ )
u 2
2
(
(6.85)
2
n
σ
σ n
2
ζ + ζ 1
+
1
α n + 1
ζ
l 3
2
ζ
u 3
For ζ 1 (DLS) the bounds l 3and u 3 tend to finite values. Hence, the bound
for the DLS smallest eigenvalue is given by
2
n
σ
α n + 1 σ n
2
+
1
(6.86)
2
which shows that the bound (2.117) is also valid for i = n + 1. In other words,
the bound (2.117) is valid for each finite eigenvalue of K .
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