Database Reference
In-Depth Information
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
<α
i
,β
i
,γ
i
<
1
,α
i
,β
i
,γ
i
∈ R
.
Search WWH ::
Custom Search