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W i
W i ·
W
i
W
i
W i ·
W i ·
W i ·
Because of cosθ 1 =
, cosθ 2 =
, cosθ 3 =
,wewantto
find the Nash Equilibrium point to make θ 1 = θ 2 = θ 3 ,then cosθ 1 = cosθ 2 = cosθ 3 .
In other words, the Nash Equilibrium point among W i , W i and W
W i
W i ·
W i
W i ·
W i
i
should be the point that has an equal correlation coecient to each of
the vectors. In other words
W i
W
i
W
i
W i ·
W i ·
W i ·
=
=
, then we get:
W i ·
W i
W i ·
W i
W i ·
W i
W i
= W i ·W
= W i ·W
W i ·
.
i
i
W i
W i
W i
ș3
ș 1
ș2
ā
ÿ
Fig. 2. The relationship of W
i
, W
i
, W
i
and W i
Combining (2) and (3), the paper builds the following linear binary simple
equations:
α i + β i + γ i =1
W i ·W i
.
= W i ·W
= W i ·W
i
i
W i
W i
W i
For W i = α i W i + β i W i + γ i W i ,
( α i W i + β i W
i
+ γ i W
i
) ·W i
= ( α i W i + β i W
+ γ i W
i
) ·W
i
= ( α i W i + β i W
+ γ i W
i
) ·W
i
,
i
i
W i
W i
W i
then ( α i W i + β i W i + γ i W i )
· W i ·W
·W
i
i
=( α i W i + β i W i + γ i W i ) · W i ·W i ·W
i
=( α i W i + β i W i + γ i W
· W
i
·W i ·W i
)
.
i
For α i + β i + γ i =1,then
m− 1
j =0
w i,j 2 + m− 1
j =0
w
i,j
W i
W i
+
α i =
) =
2 ;
2 m− 1
j =0
2 + m− 1
j =0
2 + m− 1
j =0
2 (
W i
W i
W i
+
+
w i,j
w i,j
w
i,j
m− 1
j =0
w i,j 2 + m− 1
j =0
w
i,j
W i
W i
+
β i =
) =
2 ;
2 m− 1
j =0
w i,j 2 + m− 1
j =0
w i,j 2 + m− 1
j =0
2 (
W i
W i
W i
+
+
w
i,j
m− 1
j =0
w i,j 2 + m− 1
j =0
w i,j
W i
W i
+
γ i =
) =
2 ;
2 m− 1
j =0
2 + m− 1
j =0
2 + m− 1
j =0
2 (
W i
W i
W i
+
+
w i,j
w i,j
w
i,j
so: 0 i i i < 1 i i i ∈ R .
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