2011 IEEE International Symposium on Information Theory Proceedings
Mixed Strategy Nash Equilibrium in Two-User
Resource Allocation Games
Jie Gao, Sergiy A. Vorobyov, and Hai Jiang
Dept. of Electrical and Computer Engineering, University of Alberta, Edmonton, AB, T6G 2V4 Canada
Emails: {jgao3, vorobyov, hai.jiang}@ece.ualberta.ca
Abstract—The problem of power allocation/channel selection in
two-user games is considered. Unlike most of the game theoretic
studies on resource allocation problems which consider pure
strategies, this work investigates mixed strategies and mixed
strategy Nash equilibrium (MSNE) that enables users to adopt
more subtle strategies to improve their utilities. The necessary
and sufficient conditions for the existence/uniqueness of MSNE
are derived, first in a two-channel case and then in a more
practical N channel case. In the two-channel game, the MSNE
which maximizes the utilities of both users is found. In the N channel game, a channel selection algorithm for the users, the
outputs of which can be used to check the existence/uniqueness
of MSNE, is proposed.
I. I NTRODUCTION
Resource allocation among multiple users is a common
problem in communications, for example, in cognitive radio (CR) networks [1]. In the literature, noncooperative game
theoretic approaches are vastly adopted to model and solve
such resource allocation problems [1]- [3]. However, most
game theoretic studies on resource allocation focus on pure
strategies. Nevertheless, there are strong motivations to investigate mixed strategies as an extension. Mixed strategies
introduce deliberate randomness into the decision of a player
such that the player can use more subtle strategies in the
competition with other players. In consequence, the utilities
obtained due to applying mixed strategies can be potentially
improved for the users. Introducing mixed strategies is also
instrumental for capturing the stochastic regularities of equilibria and players’ strategies in noncooperative games [4].
There are a few works on games with mixed strategies in
the literature. In [2], a two-user channel selection game is
considered. Each user in the game assigns different probabilities to different power levels that it uses to communicate on
each channel. The total power of each user is, however, not
constrained. A transmitter’s channel selection game is studied
in [3] and the equilibrium of the game is derived. However, the
problem does not involve power allocation. More importantly,
both of the above works consider games in which the users’
strategies are presented by discrete probability mass functions.
The objective of this work is to investigate mixed strategies and mixed strategy Nash equilibrium (MSNE)1 in noncooperative resource allocation games in which the users’
strategies are represented by continuous probability distributions with discrete distributions as special cases.
1 MSNE
includes pure strategy Nash equilibrium as a special case.
978-1-4577-0594-6/11/$26.00 ©2011 IEEE
II. A T WO -U SER T WO -C HANNEL S YSTEM M ODEL
Consider first a system of two users, i.e., two transmitterreceiver pairs, sharing two channels. The total transmission
power for user i is limited by Pimax while each user is
assumed to be able to communicate on either one or both of
the channels. The users interfere with each other when they
transmit on the same channel. The channel state information
is known at both transmitter and receiver sides for both users
and both users use Gaussian codebooks. Each user here is a
player which seeks to maximize its own utility defined as its
sum information rate on the two channels.
The following two assumptions are made. The received
signal power on each channel at the receiver of each user
is assumed to be low as compared to the sum of interference
and noise power on that channel.2 Under this assumption, the
utility of user i can be approximated by3
ui (ti , tj ) =
b1ii ti
b2ii (1 − ti )
+
, ∀i
σ12 + b1ji tj
σ22 + b2ji (1 − tj )
(1)
where the approximation log(1 + x) ≈ log(e) · x is used
and the constant multiplier log(e) is neglected, ti ∈ [0, 1]
denotes the portion of Pimax that user i allocates on channel
1, σk2 is the noise power on channel k, bkii = Pimax |hkii |2 and
bkij = Pimax |hkij |2 , and hkii and hkij are the channel gain of the
kth channel from the transmitter of user i to the receiver of
user i and from the transmitter of user i to the receiver of
user j, respectively. It is also assumed that the noise power
at the receivers can not be neglected as compared to the
interference power. It is well-known that a noncooperative
approach and the Nash equilibrium (NE) resulted from it
can be inefficient. In [5], it is proved that the cooperative
approach tends to be more efficient, in terms of providing a
larger rate region, than that of the noncooperative approach if
interference is the major impairment to the data transmission
of the users. However, the cooperative approach becomes less
efficient as the ratio of interference to noise power decreases.
Therefore, the noncooperative approach used in this work is
more appropriate and efficient when the second assumption
applies.
2 Similar cases are considered in [6] and [7], while received signal-tointerference-plus-noise ratios are directly chosen as users’ utilities in [8]
instead of the information rates.
3 Note that we neglect the condition i ̸= j for brevity and assume it as
default when applies.
2632
In our game, a mixed strategy of user i is represented by
the probability distribution of ti . Denote the mixed strategy of
user i as fi (ti ), which is assumed to be continuous in general
while a discrete distribution is included as a special case. A
combination of strategies {f1 (t1 ), f2 (t2 )} is called a strategy
profile. An MSNE is a strategy profile {f1⋆ (t1 ), f2⋆ (t2 )} that
satisfies [9]
Eti ,tj {ui (ti , tj )|fi (ti )=fi⋆ (ti ),fj (tj )=fj⋆ (tj ) }
= max Eti ,tj {ui (ti , tj )|fi (ti ),fj (tj )=fj⋆ (tj ) }, ∀i
fi (ti )
(2)
where E{·} denotes expectation. It can be proved that
{f1⋆ (t1 ), f2⋆ (t2 )} satisfies (2) iff the conditions
Etj {ui (ti , tj )} = ci , ∀ti ∈ Si⋆ ,
′
Etj {ui (ti , tj )} ≥ Etj {ui (ti , tj )}, ∀ti ∈
(3)
Si⋆ ,
′
∀ti ∈
/
Si⋆
(4)
are satisfied for all i given that fj (tj ) = fj⋆ (tj ), where
Si⋆ = {ti | ti ∈ [0, 1], fi⋆ (ti ) ̸= 0} is defined as the support of
fi⋆ (ti ) and ci is a constant. The following is a brief illustration
of the necessity. If (3) is not satisfied or, equivalently, if there
exist t1i , t2i ∈ Si⋆ such that Etj {ui (t2i , tj )} > Etj {ui (t1i , tj )},
then ui (ti , tj ) can be increased by transferring the probability density assigned on t1i to t2i . If (4) is not satisfied
/ Si⋆ such
or, equivalently, if there exist t3i ∈ Si⋆ and t4i ∈
4
3
that Etj {ui (ti , tj )} > Etj {ui (ti , tj )}, then ui (ti , tj ) can be
increased by transferring the probability density assigned on
t3i to t4i . The illustration for the sufficiency is straightforward
and is neglected here due to the space limitations.
III. MSNE
IN A
T WO -U SER T WO -C HANNEL G AME
The following result on the existence and uniqueness of
MSNE in the considered game is of interest.
Theorem 1: The considered two-user game has either a
unique or infinitely many MSNE. The necessary and sufficient
condition for the latter is
σ12 + b1ji
b1ii
σ12
≤
≤
, ∀i.
σ22 + b2ji
b2ii
σ22
(5)
Proof: First we prove the necessity of (5). The expectation of
(1) with respect to tj can be found as
∫
b2ii
Etj {ui (ti , tj )} =
f (tj ) dtj
σ22 + b2ji (1 − tj )
Sj
∫
+ ti κj (tj )f (tj ) dtj
(6)
Since κj (tj ) is a decreasing function of tj on [0, 1] with
κj (0) < 0 1
(9)
,
κ
(1)
>
0
j
b1ii σ12 +b1ji ,
2
b
σ1
ii
b2
ii
(
κj (tj ) =
b1ii
b2ii
−
σ12 + b1ji tj
σ22 + b2ji (1 − tj )
.
In order to satisfy (3) in this game, it is necessary that
∫
κj (tj )fj (tj ) dtj = 0.
tj ∈Sj
(7)
(8)
b2
ii
>
2
σ2
tj ∈S̄j
In this case, (6) can be rewritten as
∫
b2ii
f¯j (tj ) dtj
Etj {ui (ti , tj )} =
2
2
σ2 + bji (1 − tj )
(11)
S̄j
which satisfies (3) for user i. Moreover, condition (4) is inherently satisfied if user j uses f¯j (tj ) because Etj {ui (ti , tj )}
does not depend on fi (ti ) on [0,1]. Since there are infinitely
many different ϵ1j and ϵ2j , which satisfy ϵ1j > 0, ϵ2j > 0
and [t0j − ϵ1j , t0j + ϵ2j ] ⊆ [0, 1], there must be infinitely many
distributions f¯j (tj ) which satisfy (10). Denote the set of all
¯ f . Since it is the same case for user j,
such f¯j (tj ) as ∆
j
¯ f and ∆
¯ f both have infinitely
it can be concluded that ∆
1
2
many elements if (5) is satisfied. Moreover, any strategy profile
¯ f and f¯2 (t2 ) ∈ ∆
¯f
{f¯1 (t1 ), f¯2 (t2 )} that satisfies f¯1 (t1 ) ∈ ∆
1
2
constitutes an MSNE. Therefore, the game has infinitely many
MSNE upon the satisfaction of (5).
According to Theorem 1, there could be infinitely many
MSNE in the considered two-user game, which can lead to
different utilities for the users. Therefore, it is also of interest
to investigate the most efficient MSNE.
Theorem 2: In the case when there exist infinitely many
MSNE in the considered game, the one MSNE among all which
maximizes the utilities for both users is
f˜j (tj ) = ξj δ(tj ) + (1 − ξj )δ(tj − 1), ∀j
(12)
where δ(·) is the Dirac delta function and
ξj =
)
2 +b2
σ2
ji
there is no strategy fj (tj ) that satisfies (8) if (5) is not satisfied.
It can be shown that there exists only one MSNE which is
actually a pure strategy NE if (5) is not satisfied.
Now we prove the sufficiency of (5). If (5) is satisfied for
user i, then there exists a point t0j ∈ (0, 1) such that κj (t0j ) = 0
and κj (tj ) > 0, ∀tj < t0j ; κj (tj ) < 0, ∀tj > t0j . It can be
proved that for any given ϵ1j > 0 and ϵ2j > 0 such that S̄j =
[t0j − ϵ1j , t0j + ϵ2j ] ⊆ [0, 1], there exists at least one distribution
f¯j (tj ) defined on S̄j which satisfies
∫
κj (tj )f¯j (tj ) dtj = 0.
(10)
Sj
where
<
b2ii
σ22
b2ii
σ22
−
−
b1ii
σ12 +b1ji
b1ii
σ12 +b1ji
+
b1ii
σ12
−
b2ii
σ22 +b2ji
, ∀j.
(13)
Proof: Assume that the most efficient MSNE is {f˜1 (t1 ),
˜
f2 (t2 )} and the support of f˜i (ti ) is S̃i . From Theorem 1, it
˜
∫can be seen that fj (tj ) is the distribution which maximizes
κ̂
(t
)f
(t
)
j dtj among all distributions subject to (8),
tj ∈Sj j j
where
b2ii
κ̂j (tj ) = 2
(14)
2
σ2 + bji (1 − tj )
2633
is a strictly convex and increasing function on [0,1]. Denote
κ̌j (tj ) = κj (tj ) + κ̂j (tj ), then
κ̌j (tj ) =
b1ii
σ12 + b1ji tj
TABLE I
A LGORITHM
(15)
tj ∈Sj
Therefore, ∫among all distributions,∫ the distribution f˜j (tj )
maximizes Sj κ̂j (tj )f (tj ) dtj and Sj κ̌j (tj )f (tj ) dtj simultaneously subject to the condition
∫
∫
˜
κ̂j (tj )fj (tj ) dtj =
κ̌j (tj )f˜j (tj ) dtj .
(17)
tj ∈S̃j
′
tj ∈S̃j
tj ∈S̃j /{tj }
′
′
+(1 − tj )f˜j (tj )κ̂j (0) + tj f˜j (tj )κ̂j (1) (18)
∫
κ̌j (tj )f˜j (tj ) dtj <
κ̌j (tj )f˜j (tj ) dtj
′
′
tj ∈S̃j
CHANNEL GAME
tj ∈S̃j
First, we prove that S̃j ⊆ {0, 1}, ∀i. Assume that there exists
′
′
′
tj such that∫ 0 < tj < 1, tj ∈ S̃j and f˜j (tj ) defined on S̃j
maximizes Sj κ̂j (tj )f (tj ) dtj among all possible fj (tj ) and
satisfies (17). Since both κ̂j (tj ) and κ̌j (tj ) are strictly convex,
we can write that
∫
∫
κ̂j (tj )f˜j (tj ) dtj
κ̂j (tj )f˜j (tj ) dtj <
∫
N
2
2
1. Let νi1 = [b1ii /σ12 , . . . , bN
= [b1ii /(σ12 +
ii /σN ] and νi
2
N
b1ji ), . . . , bN
ii /(σN + bji )] for each user i. Let ∆i=1 = ∆i=2 =
{1, . . . , N }. Initialize d = 1.
2. For user i = 1, let k = ∆i=1 (d) where ∆i=1 (d) is the dth element of
the set ∆i=1 . Check if the inequalities ν11 (k) > ν12 (l), ∀l ∈ ∆i=1 ̸= k
are all satisfied. If not, let tk1 = 0, ν22 (k) = bk22 /σk2 , remove ∆i=1 (d)
from ∆i=1 and set d = d − 1. Check if d < L(∆i=1 ) where L(·) denotes
the cardinality of a set. If yes, set d = d+1 and repeat the above procedure
in Step 2. If no, set d = 1 and proceed to Step 3.
3. For user i = 2, let k = ∆i=2 (d) and check if the inequalities ν21 (k) >
ν22 (l), ∀l ∈ ∆i=2 ̸= k are all satisfied. If not, let tk2 = 0, ν12 (k) =
bk11 /σk2 , remove ∆i=2 (d) from ∆i=2 and set d = d − 1. Check if d <
L(∆i=2 ), If yes, set d = d+1 and repeat the above procedure in Step 3. If
not, and no element was deleted from ∆i=2 in this step, proceed to Step 4;
otherwise set d = 1 and return to the beginning of Step 2.
4. Output ∆i=1 and ∆i=2 .
and κ̌j (tj ) is a strictly convex and decreasing function on
[0,1]. Then (8) can be rewritten as
∫
∫
κ̂j (tj )fj (tj ) dtj =
κ̌j (tj )fj (tj ) dtj .
(16)
tj ∈Sj
FOR CHANNEL SELECTION IN TWO - USER
′
tj ∈S̃j /{tj }
′
IV. E XTENSION TO A T WO - USER N - CHANNEL G AME
The two-user N -channel case is more general yet complicated. In the N -channel case, the utility of user i extends as
∑N −1 k
N
−1
∑
bN
bkii tki
ii (1−
k=1 ti )
ui (ti , tj ) =
+
∑N −1 , ∀i (22)
2 + b k tk
2
N
σ
σN +bji (1− k=1 tkj )
ji j
k=1 k
−1
where ti = [t1i , . . . , tN
] and tki ∈ [0, 1] is the portion of
i
∑N −1
max
Pi that user i allocates on channel k subject to k=1 tki ∈
[0, 1]. Conditions (3)-(4) extend accordingly as
′
′
′
+(1 − tj )f˜j (tj )κ̌j (0) + tj f˜j (tj )κ̌j (1). (19)
The above inequalities imply that both the left-hand side and
′
the right-hand side of (17) can be increased by setting f˜j (tj ) =
′
′
0 and transferring the probability densities (1 − tj )f˜j (tj ) and
′
′
tj f˜j (tj ) to tj = 0 and tj = 1, respectively. Let t ∈ [0, 1] and
denote the increases on the left-hand sides of (18) and (19)
′
due to transferring the probability densities (1 − t)f˜j (tj ) and
′
tf˜j (tj ) to tj = 0 and tj = 1 as δ̂j (t) and δ̌j (t), respectively.
′
′
If δ̂j (tj ) = δ̌j (tj ), then (17) is still satisfied after the above
transferring of probability densities. Note that κ̂j (tj ) is strictly
increasing and κ̌j (tj ) is strictly decreasing on [0, 1]. Therefore,
′
′
′
′
if δ̂j (tj ) > δ̌j (tj ), then there exist ϵ > 0 and ṫj ∈ [tj − ϵ, tj )
′
such that ṫj ∈ (0, tj ) and δ̂j (ṫj ) = δ̌j (ṫj ) > 0. Similarly, if
′
′
′
′
′
′
δ̂j (tj ) < δ̌j (tj ), then there exist ϵ > 0 and ẗj ∈ (tj , tj + ϵ ]
′
such that ẗj ∈ (tj , 1) and δ̂j (ẗj ) = δ̌j (ẗj ) > 0. In any
of the above three cases, (17) can be satisfied and at the
same time both sides of (17) can be increased. Thus, f˜j (tj )
′
defined on any S̃j that includes
∫ tj ∈ (0, 1) cannot be the
distribution which maximizes Sj κ̂j (tj )f (tj ) dtj subject to
(17). Therefore, S̃j ⊆ {0, 1}. It is the same for S̃i .
Second, assume that f˜j (tj ) = ξj δ(tj ) + (1 − ξj )δ(tj − 1)
where 0 ≤ ξj ≤ 1. Then
∫
b2
b2
κ̂j (tj )f˜j (tj ) dtj = ξj 2 ii 2 + (1 − ξj ) ii2 (20)
σ2 + bji
σ2
tj ∈S̃j
∫
1
1
b
b
κ̌j (tj )f˜j (tj ) dtj = ξj ii2 + (1 − ξj ) 2 ii 1 . (21)
σ
σ
tj ∈S̃j
1
1 + bji
Using the condition (17), ξj can be derived as in (13).
Etj {ui (ti , tj )} = ci , ∀ti ∈ Si⋆ ,
′
(23)
′
Etj {ui (ti , tj )} ≥ Etj {ui (ti , tj )}, ∀ti ∈ Si⋆ , ∀ti ∈
/ Si⋆(24)
with the mixed strategy of user i now represented by the joint
distribution of tki , ∀k ∈ {1, . . . ,N − 1} and denoted as fi (ti ).
The existence/uniqueness of MSNE in the N -channel game
can be derived based on the outputs of the algorithm in
Table I.4
Theorem 3. The following properties hold for the proposed
algorithm in Table I.
i) The algorithm converges to the same result regardless of
the ordering of users or channels.
ii) Denote Γi = {k ∈ ∆i , k ∈
/ ∆j }, then L(Γi ) ≤ 1, ∀i at
the output of the algorithm.
iii) MSNE is unique in the game iff L(∆i=1 ) = 1 or
L(∆i=2 ) = 1. Otherwise, infinitely many MSNE exist.5
Proof: Denote ΩN = {1, . . . , N } as the set of all channels
and define Φ0i = {k ∈ ΩN |∃l ∈ ΩN ̸= k : νi1 (k) ≤ νi2 (l)}. If
Φ0i ̸= Ø, the first iteration of Step 2 of the algorithm deletes
Φ01 from ∆i=1 and increases ν22 (k) to bk22 /σk2 , ∀k ∈ Φ01 . In the
first iteration of Step 3, in consequence, the set of channels
4 This algorithm provides a method of checking the existence and uniqueness of MSNE rather than a mechanism of reaching MSNE. The latter
functionality is another topic not included in this work due to the page limit.
5 There is a trivial exception. If ∆ = {k }, ∆ = {k , k }, and ν 1 (k ) =
1
1 2
2
i
j
j
νj2 (k1 ) where k1 , k2 ∈ {1, . . . , N }, infinitely many MSNE exist. However,
in this case, all other MSNE generate smaller utilities for user i (and the
same utility for user j) than the MSNE which is achievable and unique when
∆i = {k1 } and ∆j = {k2 }. Therefore, we consider the former case as
equivalent to the latter one.
2634
not satisfying the inequalities ν21 (k) > ν22 (l), ∀l ∈ ∆i=2 ̸= k
for user 2 can be potentially extended to Φ12 = Φ02 + Φ̄12 , where
Φ̄12 denotes the extra set of channels which do not satisfy the
above inequalities due to the deletion of Φ01 from ∆i=1 in
Step 2. The deletion of Φ12 from ∆i=2 in Step 3 could break
the inequalities of ν11 (k) > ν12 (l), ∀l ∈ ∆i=1 ̸= k on certain
channels in ∆i=1 (which has been updated in Step 2) and the
process potentially repeats as Steps 2 and 3 iterate. Denote
Φ̄qi , q ≥ 1 as the set of channels which do not satisfy the
aforementioned inequalities for user i due to the deletion of
Φ̄q−1
(if q ̸= 1) or Φ0j (if q = 1) from ∆j in the preceding
j
step. Note that Φ̄qi = Ø if Φ̄q−1
= Ø. According to the
j
definition of Φ0i and Φ̄qi , it follows that Φ21 = Φ̄11 + Φ̄21 ,
and iteratively Φqi = Φ̄q−1
+ Φ̄qi , q = 2, 4, . . . , q max for
i
i = 1 and q = 1, 3, . . . , q max + 1 for i = 2. Here
q max = min{r|r ∈ {0, 2, 4, . . . , N }, Φr+1
= Ø}.
2
Proof of i) At any iteration of the algorithm, if k̂ ∈ Φqi ,
then ∃p ̸= k̂ ∈ ΩN such that νi2 (p) ≥ νi2 (k̂). Otherwise,
there exists l such that νi1 (k̂) ≤ νi2 (l) and νi2 (l) < νi2 (k̂). In
consequence, it leads to the situation when νi2 (k̂) > νi1 (k̂)
which is impossible. Thus, deleting any k̂ ∈ Φqi will not
change maxk∈∆i νi2 (k). Therefore, the result of checking the
inequalities νi1 (k) > νi2 (l), ∀l ∈ ∆i ̸= k for any other channel,
i.e. for k ̸= k̂, will not be affected. In conclusion, the ordering
of channels is irrelevant to the result of the algorithm.
Now consider the ordering of users. When the algorithm starts from user 1, the max
sequences of
deletions are
q max
for user 1
Φ01 , Φ̄11 + Φ̄21 , Φ̄31 + Φ̄41 , . . . , Φ̄q1 max−1 + Φ̄max
1
+ Φ̄q2 +1 for user 2
and Φ02 + Φ̄12 , Φ̄22 + Φ̄32 , . . . , Φ̄q2
through
all max
iterations of max
Steps 2 and 3, respectively. Here
max
Φ̄q2
+ Φ̄q2 +1 = Φq2 +1 = Ø according to the definitions of Φqi and q max . In this case, the outputs of the
algorithm are ∆1i=1 = ΩN − ∪q Φq1 , q = 2, 4, . . . , q max
and ∆1i=2 = ΩN − ∪q Φq2 , q = 1, 3, . . . , q max + 1. If the
ordering of users is changed or, equivalently, if the algorithm
starts from user 2, the sequences
of max
deletions
change to
max
max
+ Φ̄q2 ,max
Φ̄q2 +1 for user
Φ02 , Φ̄12 + Φ̄22 , Φ̄32 + Φ̄42 , . . . , Φ̄q2 −1max
+ Φ̄q1 max+1 for user 1,
Φ̄31 , . . . , Φ̄q1
2 and Φ01 + Φ̄11 , Φ̄21 +max
q
+1
respectively.
Here Φ̄1 max = Ø max
since Φ̄q2max = Ø, while
q
q
+1
q max
+ Φ̄q2 +1 = Ø. Note
= Φ̄2
Φ̄2
= Ø because Φ2
q
that Φ0i , ∀i and Φ̄i , ∀i, ∀q keep unchanged regardless of the
ordering of users according to their definitions. The outputs of
the algorithm are then ∆2i=2 = ΩN −∪q Φq2 , q = 2, 4, . . . , q max
and ∆2i=1 = ΩN − ∪q Φq1 , q = 1, 3, . . . , q max + 1. Using
max
max
the facts that Φq2 +1 = Ø and Φ̄q1 +1 max
= Ø, it can be
s=q
shown that ∆1i=1 = ∆2i=1 =
Ω
−
∪
Φ̄s1 − Φ01 and
N
s=1
s=q max −1 s
1
2
0
∆i=2 = ∆i=2 = ΩN − ∪s=1
Φ̄2 − Φ2 if q max ≥ 2 and
1
2
0
∆i=1 = ∆i=1 = ΩN − Φ1 and ∆1i=2 = ∆2i=2 = ΩN if
q max = 0. Therefore, the ordering of users is irrelevant.
Proof of ii) According to the algorithm and the definition
of Γi , tkj = 0, ∀k ∈ Γi . In the algorithm, tkj = 0 occurs
together with setting νi2 (k) = bkii /σk2 at all times. Thus,
νi2 (k) = νi1 (k), ∀k ∈ Γi . Meanwhile, the inequalities νi1 (k) >
νi2 (l), ∀l ∈ ∆i ̸= k must be satisfied ∀k ∈ ∆i for user i
at the output of the algorithm. If L(Γi ) ≥ 2, then there exist
ˆl, ˇl (ˆl ̸= ˇl) such that the inequalities ν 1 (ˆl) > ν 2 (ˇl) = ν 1 (ˇl) and
i
i
i
νi1 (ˇl) > νi2 (ˆl) = νi1 (ˆl) are satisfied at the same time, which is
impossible. Thus L(Γi ) ≤ 1.
Proof of iii) It can be shown that the channel indexes
removed from ∆i = 1 and ∆i = 2 in Steps 2 and 3 correspond
to the channels which must not be used for users 1 and 2,
respectively, in any MSNE. It can also be shown that there
exists one MSNE, in which both users end up allocating
Pimax on one channel in the output ∆i , if L(∆i=1 ) = 1 or
L(∆i=2 ) = 1. Given the above two facts, it follows that a
unique MSNE exists if L(∆i=1 ) = 1 or L(∆i=2 ) = 1. It
proves the sufficiency of the uniqueness condition of MSNE
and the necessity of the condition for the existence of infinitely many MSNE at the same time. Now assume that
L(∆i ) > 1, ∀i. Denote Li = L(∆i ), Ti = ∆i (Li ) and
˜ i = {∆i (1), . . . , ∆i (Li − 1)}. Then Et {ui (ti , tj )} at the
∆
j
output of the algorithm, denoted as Etj {ui }, can be written
as

∫ ∑

Etj{ui }= 
Sj
bkii tki
2
σ +bkji tkj
˜i k
k∈∆
bTiii (1−
+
tki )
˜i
k∈∆
σT2 i +bTjii (ζ−
∫∑
∫
=
tki ιkfj (tj )dtj +
˜i
Sj k∈∆
∑
Sj
∑
˜i
k∈∆
Ti
bii
σT2 i+bTjii (ζ−


fj (tj )dtj
tkj )
∑
˜i
k∈∆
tkj )
fj (tj )dtj
(25)
∑
where ζ = 1− k∈Γj tkj is the total power that user j allocates
on the channels represented by the indexes in ∆i and
ιk =
bkii
bTiii
−
∑ k .
σk2 + bkji tkj σT2 i +bTjii (ζ −
tj )
(26)
˜i
k∈∆
In
to satisfy (23), it is required in this
∫ order
∑
∑ gamek kthat
k k
t
ι
f
(t
)
=
0.
The
minimum
of
˜i i
˜ i ti ι as
j j
k∈∆
k∈∆
Sj
k
k k
˜
˜
a function of tj , ∀k ∈ ∆i is min{ti ι , ∀k ∈ ∆i |tkj =1 }. Denote
Υ = {k | k ∈ ∆i ∩ ∆j }, then Υ is nonempty given that
L(∆i ) > 1 as assumed, and L(Γi ) ≤ 1 as proved in statement
˜ i |tk =1 } < 0 if
ii. It can be shown that min{tki ιk , ∀k ∈ ∆
j
k
k
k
˜ i.
tj = 1, k ∈ Υ since ι |tkj =1 < 0 and ti > 0, ∀k ∈ ∆
Moreover, if Ti = ∆i (Li ) ∈ Υ, which can always be satisfied
since the elements of ∆i can be ordered in any manner with no
effect on anything else, then it holds that limtkj →0,∀k̸=Ti tki ιk >
˜ i . It follows that the sets Λ1 = {tj |tk = 0, ∀k ∈
0, ∀k∑∈ ∆
/
j
j
∑
k k
2
k
∆j , k∈∆j tkj = 1, and k∈∆
˜ i ti ι < 0} and Λj = {tj |tj =
∑
∑
k k
0, ∀k ∈
/ ∆j , k∈∆j tkj = 1, and k∈∆
˜ i ti ι > 0} are both
nonempty. Then similar to the proof for Theorem 1, it can be
shown that
many fj (tj ), each of which
∫ there
∑ existk infinitely
k
satisfies Sj k∈∆
t
ι
f
(t
)
=
0. Moreover, similar to the
˜i i
j j
proof for Theorem 1, condition (24) is inherently satisfied
upon the satisfaction of (23). The details are omitted due to
the limit of space.
2635
V. S IMULATIONS
Our simulation example illustrates the iterative process
of channel selection described in Table I. Here N = 8,
Pimax = 1, ∀i, and σk2 , ∀k are uniformly generated from
the interval [1, 2]. The real and imaginary parts of hkii and
hkij ∀i, ∀k are generated from zero-mean normal distributions
with variances 1 and 0.25, respectively. The results are
shown in Fig. 1, where the diamonds and squares are generated at coordinates (Re(hk11 ), Im(hk11 ), (|h21 |k )2 ), ∀k and
(Re(hk22 ), Im(hk22 ), (|h12 |k )2 ), ∀k, respectively. The diamond
and square corresponding to the same k are connected by dashdot lines for all k. A diamond/square closer to the corners
corresponds to a channel with higher channel gain for the
corresponding user, while a diamond/square closer to the top
corresponds to a channel with higher gain of interference
from the transmitter of the other user to the receiver of the
corresponding user.
At the end of Step 2/Step 3, the diamonds/squares corresponding to the channel indexes in the updated ∆1 /∆2 are circled in Fig. 1. If the algorithm iterates, the diamonds/squares
corresponding to the channel indexes in the most updated
∆1 /∆2 are circled by circles with a larger radius at the end of
each iteration of Step 2/Step 3. The diamonds/squares with the
maximum number of circles correspond to the channel indexes
in ∆1 and ∆2 at the output of the algorithm. The upper plot of
Fig. 1 shows the case of ∆1 = 1, ∆2 = 1, in which a unique
MSNE exists according to Theorem 3. It can be seen that the
first run of Step 2 selects four channels for user 1 while the
second run further selects one out of the four. The lower plot
of Fig. 1 shows the case of ∆1 = 2, ∆2 = 3, in which two of
the eight channels are shared and infinitely many MSNE exist
according to Theorem 3. Note that the users interfere with
each other only on the channels corresponding to the dash-dot
line with the maximum number of circles at both ends in the
plots. From the figure, it can be seen that the channels selected
in MSNE achieve high channel gains and low interference.
VI. C ONCLUSION
Noncooperative resource allocation games are studied in
mixed strategies. It is shown that applying mixed strategies
can potentially lead to NE which are more efficient than NE
in pure strategies. For two-channel games, the sufficient and
necessary condition for the uniqueness of MSNE is derived.
The game becomes significantly more complicated in the case
of N channels. A channel selection algorithm which simplifies
the game is proposed. Based on the outputs of the algorithm,
the sufficient/necessary condition for the uniqueness of MSNE
in this game is derived. Our simulation results demonstrate
how the proposed algorithm selects channels in the N -channel
game.
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2
(Re(h ),Im(h ),|h | )
11
11
21
(Re(h22),Im(h22),|h12|2)
1
2
2
|h21| /|h12|
2
1.5
0.5
1
0
0
2
1
−1
0
−1
Re(h11)/Re(h22)
−2
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Im(h11)/Im(h22)
(a) At the output of the algorithm ∆1 = 1, ∆2 = 1
(Re(h ),Im(h ),|h |2)
11
11
21
(Re(h22),Im(h22),|h12|2)
|h21|2/|h12|2
3
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1
1
0
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−1
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1
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Im(h11)/Im(h22)
Re(h )/Re(h )
11
22
(b) At the output of the algorithm ∆1 = 2, ∆2 = 3
Fig. 1.
Illustration of the cannel selection algorithm in two examples
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