# Evolutionary Cooperative Spectrum Sensing Game - Signals and ...

Evolutionary Cooperative Spectrum Sensing Game - Signals and ...

WANG et al.: EVOLUTIONARY COOPERATIVE SPECTRUM SENSING GAME: HOW TO COLLABORATE? 899

User 1

User 2

User 3

0.2

0.15

0.1

0.05

0

1

0.8

0.6

0.4

0.2

1

0.8

0.6

0.4

0.2

Probability of cooperation

20 40 60 80 100 120 140 160 180 200

number of iterations

Probability of cooperation

20 40 60 80 100 120 140 160 180 200

number of iterations

Probability of cooperation

20 40 60 80 100 120 140 160 180 200

number of iterations

Fig. 5: Behavior dynamics of a heterogeneous 3-user sensing

game.

with M t =(1− x) K−1 U 0 (1 − τ). ByusingL’Hôpital’s rule,

we know that lim x→0 ¯UC (x)− ¯U U D (x) = lim 0 x→0 K [−Kτ(1−

x) K−1 +K(1−x) K−1 −Kx(K−1)(1−x) K−2 ]=U 0 (1−τ) >

0. Thus, x =0is not a solution to equation ¯U C (x)− ¯U D (x) =

0, and the solution must satisfy τ(1−x) K +Kx(1−x) K−1 −

τ =0.

B. Proof of Proposition 1

Summing x ai in (17) over a i yields

˙x C +˙x D = ε[x C ¯U(C, xD )+x D ¯U(D, xC )−(x C +x D ) ¯U(x)].

(39)

Since ¯U(x) =x C ¯U(C, xD )+x D ¯U(D, xC ), and initially a

user chooses x C + x D =1, (39) is reduced to ˙x C +˙x D =0.

Therefore, x C (t) +x D (t) = 1 holds at any t during the

dynamic process. A similar conclusion also holds in an

asymmetric game.

total throughput (Mbps)

2.5

2

1.5

1

0.5

0

comparison of ESS and full cooperation

1 2 3 4 5 6 7 8

ESS

Full

C. Proof of Proposition 2

Substituting (17) into (25), we get

˙x ai ¯U(ai ,x −ai )

a i∈A

= ∑

ε ¯U(a i ,x −ai )[ ¯U(a i ,x −ai ) − ¯U(x)]x ai

a i∈A

= ε ∑

a i∈A

[ ∑

x ai ¯U 2 (a i ,x −ai ) − ε

a i∈A

(40)

] 2.

x ai ¯U(ai ,x −ai )

According to Jensen’s inequality, we know (40) is nonnegative,

which completes the proof. In addition, we can show

(25) also holds for a game with heterogeneous players in a

similar way.

Fig. 6: Comparison of ESS and full cooperation.

VI. APPENDIX

A. Proof of Equation (24)

Subtracting ¯U D (x) from ¯U C (x) we get

¯U C(x) − ¯U D(x)

( )

K−1

∑ K − 1

=

x j (1 − x) K−1−j [U C(j +1)− U D(j)]

j

j=0

( )

K−1

∑ K − 1

=

x j (1 − x) K−1−j [U 0(1 −

τ

j

j +1 ) − U0]+Mt

j=0

( )

K−1

∑ K − 1

= − U 0τ

x j (1 − x) K−1−j 1

j

j +1 + Mt

j=1

= − τU0

K−1

xK

= − τU0

xK

K!

(j +1)!(K − j − 1)! xj+1 (1 − x) K−j−1 + M t

j=1

( )

K∑ K

x j (1 − x) K−j + M t

j

j=2

= τU0

xK [(1 − x)K + Kx(1 − x) K−1 − 1] + M t

[ ]

= U0 τ(1 − x) K + Kx(1 − x) K−1 − τ

,

K

x

(38)

D. Proof of Theorem 1

From the simplified dynamics (23), we know that the sign

of ˙x C (t) is determined by the sign of ¯U C (x) − ¯U D (x), given

x ∈ (0, 1) and ε>0. ¯U C (x) and ¯U D (x) are simplified as the

following

¯U C (x) =U 0 − U 0 (1 − x) K−1 τ

K−1

( ) K − 1

− U 0 x j (1 − x) K−j−1 τ

j

j +1 , (41)

j=1

¯U D (x) =U 0 − U 0 (1 − x) K−1 .

Furthermore, the difference ¯U C (x) − ¯U D (x) is calculated in

Appendix VI-A as

¯U C (x) − ¯U D (x) = U [

0 τ(1 − x) K + Kx(1 − x) K−1 ]

− τ

.

K

x

(42)

According to different values of parameter τ, weprovethe

theorem in three different cases.

Case I (τ = 1): from (41) we know ¯U C (x) < ¯U D (x),

dx

dt < 0, and the replicator dynamics converge to x∗ =0.

Case II (τ = 0): from (41) we have ¯U C (x) > ¯U D (x),

dx

dt > 0, and the replicator dynamics converge to x∗ =1.

Case III (0

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