In many cases, an NE results from learning and evolution processes of all the game participants. Therefore, it is fundamental to predict and characterize the set of such points from the system design perspective of wireless networks. In the rest of the paper, we focus on characterizing the set of NEs. The following questions are addressed one by one.
Does an NE exist in our game?
Is the NE unique or there exist multiple NE points?
How to reach an NE if it exists?
How does the system perform at NE?
Throughout this section we investigate the existence and uniqueness of a Nash equilibrium.
It is known that in general an NE point does not necessarily exist. In the following theorem we establish the existence of a Nash equilibrium in our game.
A Nash equilibrium exists in game .
Since is convex, closed, and bounded for each ; is continuous in both and ; and is concave in for any set , at least one Nash equilibrium point exists for [12, 22].
Once existence is established, it is natural to consider the characterization of the equilibrium set. The uniqueness of an equilibrium is a rare but desirable property, if we wish to predict the network behavior. In fact, many game problems have more than one NE . As an example of games with infinite NEs, we could consider a special case of our game , namely, the symmetric waterfilling game  where the channel coefficients are assumed to be symmetric. Then, in general, our game does not have a unique NE. But with the assumption of independent and identically distributed (i.i.d.) continuous entries in , we will show that the probability of having a unique NE is equal to .
For any player , given all other players' strategy profile , the best-response power strategy can be found by solving the following maximization problem:
which is a convex optimization problem, since the objective function is concave in and the constraint set is convex. Therefore, the Karush-Kuhn-Tucker (KKT) conditions for optimization are sufficient and necessary for the optimality . The KKT conditions are derived from the Lagrangian for each player ,
and are given by
where for all and for all are dual variables associated with the power constraint and transmit power positivity, respectively. The solution to (11)–(13) is known as waterfilling :
where and satisfies
In order to analyze the equilibrium set, we establish necessary and sufficient conditions for a point being an NE in the game
A power strategy profile is a Nash equilibrium of the game if and only if each player's power is the single-player waterfilling result while treating other players' signals as noise. The corresponding necessary and sufficient conditions are:
The proof can be found in Appendix A.
From (16), it is easy to verify that necessarily , since and for all and for all . Also, from (17), we have
This equation implies that, at the NE, all APs transmit at their maximum power by conveniently distributing the power over all the orthogonal channels.
However, it is still difficult to find an analytical solution from (16)–(18), since the system consisting of (14) and (15) is nonlinear. To simplify this problem, we could consider linear equations instead of nonlinear ones. The following lemma provides a key step in this direction.
For any realization of channel matrix , there exist unique values of the Lagrange dual variables and for any Nash equilibrium of the game . Furthermore, there is a unique vector such that any vector corresponding to a Nash equilibrium satisfies
The proof can be found in Appendix B.
Now, let be the following matrix:
where is the th column of , is the identity matrix, and is the zero vector of length . Let be the following vector of length :
Then, (19) and (20) can be written in the form of linear matrix equation
Define the following sets:
and denote by and their cardinalities. From (18), if an index we must have . Without loss of generality, we assume that for . Let be the matrix formed from the first rows and first columns of , is formed from the first elements of , and is formed from the first elements of . Then, any NE solution must satisfy
Let be the matrix formed from the columns of that correspond to the elements of . Similarly, let be the vector of length with entries such that (same order as they were in ). Then, any NE solution satisfies
For any realization of a random channel gain matrix with i.i.d. continuous entries, if , the probability that is equal to .
() If and , the probability that is equal to .
() If , the probability that is equal to .
The proofs of Lemmas 2 and 3 can be found in Appendices C and D, respectively.
Based on Lemmas 1, 2, and 3, we derive the following theorem.
For any realization of a random channel gain matrix with statistically independent continuous entries, the probability that a unique Nash equilibrium exists in the game is equal to .
The proof can be found in Appendix E.
Thus, from Theorems 1 and 3, we have established the existence and uniqueness of NE in our game .