On the Sum-Rate Capacity of Poisson MISO Multiple Access Channels

In this paper, we analyze the sum-rate capacity of two-user Poisson multiple access channels (MAC), when the receiver is equipped with single antenna. We first characterize the sum-rate capacity of the non-symmetric Poisson MAC when each transmitter has a single antenna. While the sum-rate capacity of the symmetric Poisson MAC with single antenna at each transmitter has been characterized in the literature, the special property exploited in the existing method for the symmetric case does not hold for the non-symmetric channel anymore. We obtain the optimal input that achieves the sum-rate capacity by solving a non-convex optimization problem. We show that, for certain channel parameters, it is optimal for a single user to transmit to achieve the sum-rate capacity. This is in sharp contrast to the Gaussian MAC, in which both users must transmit, either simultaneously or at different times, in order to achieve the sum-rate capacity. We then characterize the sum-rate capacity of the Poisson multiple-input single-output (MISO) MAC with multiple antennas at each transmitter and single antenna at the receiver. By converting a non-convex optimization problem with a large number of variables into a non-convex optimization problem with two variables, we show that the sum-rate capacity of the Poisson MISO MAC with multiple transmit antennas is equivalent to a properly constructed Poisson MAC with a single antenna at each transmitter.

optical (FSO) and visible light communication (VLC). The point-to-point single-user Poisson channel has been investigated from various perspectives, including single antenna [3], multiple antennas [4], fading channels [5], [6], in continuoustime [3], [7]- [9] and discrete-time [10]- [12]. On the other hand, Poisson channels with multiple users are not that well understood. Among limited existing work, [13], [14] focus on the Poisson broadcast channel, [15] studies the Poisson multiple-access channel (MAC), [16] considers the optimization of the capacity region of Poisson MAC with respect to different power constraints, and [17] investigates the Poisson channel with side information at the transmitter. Furthermore, [18], [19] discuss the discrete-time Poisson channel and show that sum-capacity achieving distributions of the Poisson MAC under peak amplitude constraints are discrete with a finite number of mass points. [20] discusses a discrete memoryless Poisson MAC with noiseless feedback.
Of particular relevance to our study is [15], which thoroughly investigates the continuous-time Poisson MAC with each user equipped with single antenna. [15] shows that the approximation of the complex continuous-time continuousinput discrete-output Poisson MAC by a discrete-time binaryinput binary-output MAC does not result in a loss in terms of the capacity region. [15] determines the sum-rate capacity of the symmetric Poisson MAC, in which the channel gains and power constraints for all users are identical under the maximum power constraint. Furthermore, it characterizes the boundary points on the capacity region of the symmetric MAC under maximum power constraint and analyzes the maximum-throughput under peak-power and average power constraints.
In this paper, we first study the single antenna nonsymmetric Poisson MAC, in which the channel gains and power constraints at the two users are not necessarily the same. We refer to such a channel as Poisson SISO-MAC. This scenario naturally arises in multiuser optical communications when the transmitters have different distances to the receiver or have different transmission powers. Unfortunately, the method used in [15] to characterize the sum-rate capacity for the symmetric case does not apply to the non-symmetric case anymore. In particular, the method in [15] exploits the property that the objective function involved is a Schur concave function for the symmetric Poisson MAC, which greatly simplifies the analysis. However, in the non-symmetric channel, the objective function is not symmetric, and hence is not Schur concave anymore. As a result, we resort to a different approach from the one used in [15] to study the 0018-9448 © 2017 IEEE. Personal use is permitted, but republication/redistribution requires IEEE permission.
sum-rate capacity. More specifically, we show that characterizing the sum-rate capacity is equivalent to solving a nonconvex optimization problem. We show that there are at most four possible candidates for the optimal solution to this optimization problem with two candidate solutions corresponding to the cases when only one user transmits. We further show that, for some channel parameters, it is indeed optimal to allow only one user to transmit in order to achieve the sumrate capacity under the maximum power constraint. This is in sharp contrast to the Gaussian MAC with an average power constraint, in which it is always optimal for both users to transmit, either simultaneously or at different time, to achieve the sum-rate capacity. We also identify conditions under which it is optimal for both users to transmit in order to achieve the sum-rate capacity. We then extend the study to Poisson MAC with multiple antennas at each transmitter and one antenna at the receiver. We refer to this as Poisson MISO-MAC. Similarly to the Poisson SISO-MAC, the complex continuous-time continuous-input discrete-output Poisson MAC can be converted to a discrete-time binary-input binary-output Poisson MAC. However, the resulting problem is much more challenging than that of the Poisson SISO-MAC. In particular, to characterize the sum-rate capacity, we need to solve a non-convex optimization problem with 2 J 1 + 2 J 2 variables, in which J n is the number of antennas at user n. Despite this challenge, we show that the optimal value obtained from this optimization problem with a large number of variables is the same as that of an optimization problem with only 2 variables. Furthermore, this reduced dimension optimization problem is equivalent to a problem arising in the Poisson SISO-MAC with properly chosen parameters. As the result, characterizing the sum-rate capacity of the Poisson MISO-MAC is equivalent to characterizing the sum-rate capacity of a Poisson SISO-MAC. Hence, the techniques and asymptotic analysis developed in the SISO-MAC case can be used for the MISO-MAC case. There are two major steps in our proof. In the first step, we show that the original optimization problem with 2 J 1 +2 J 2 variables can be converted to a non-convex optimization problem with J 1 + J 2 variables by showing and exploiting the fact that, at the optimality, if the antenna with a smaller duty cycle is on, then the antenna with a larger duty cycle is also on. In the second step, we show that the optimization problem with J 1 + J 2 variables obtained in step 1 can be converted to an optimization problem with only 2 variables. The key ingredient in this step is to show that, at the optimality, all antennas at each transmitter have to be simultaneously on or off.
A natural next step is to characterize the sum-rate capacity of the scenario when there are also multiple antennas at the receiver. Another natural extension is to characterize the set of boundary points of the full capacity region. As will be discussed in sequel, there are significant technical challenges in extending our methods to these cases. Hence, new techniques are required for these extensions. One possible direction to pursue is to explore the relationship between the derivative of mutual information and mean squared error as discussed in [21] and [22]. The remainder of the paper is organized as follows. Section II describes the model under consideration. Section III analyzes the Poisson SISO-MAC. Section IV analyzes the Poisson MISO-MAC. Section V discusses challenges for certain extensions. Numerical analysis is presented in Section VI and concluding remarks are presented in Section VII.

II. SYSTEM MODEL
In this section, we introduce the model considered in this paper. As shown in the Fig. 1, we consider the continuous-time two-user Poisson MISO-MAC with two users communicating with a single antenna receiver. For transmitter n, it is equipped with J n transmit antennas. Let X nj (t) be the input of the j th transmitter from n th user and Y (t) be the doubly-stochastic Poisson process observed at the receiver antenna. The inputoutput relationship can be described as: in which S nj is the channel response between the j th antenna of the n th user to the receiver, λ is the dark current at receiver antenna, and P(·) is the nonlinear transformation converting the light strength to the doubly-stochastic Poisson process that records the timing and number of photon's arrivals. In particular, for any time interval [t, t + τ ], the probability that there are k photons arriving at the receiver is We consider the peak power constraint, i.e., the transmitted signal X nj (t) must satisfy the following constraint: where A nj is the maximum power allowed by the j th antenna of the n th transmitter. We use μ nj to denote the duty cycle of each transmitting antenna, i.e., μ nj is the percentage of time at which the j th antenna of the n th user is on. We use µ to denote the vector of all μ nj s. Throughout the paper, we use the following notation: It is easy to check that 0 < α(x) < 1 for x ≥ 0. Our goal is to characterize the sum-rate capacity of this Poisson MAC.

III. SISO-MAC ANALYSIS
In this section, we focus on the special case in which each transmitter has only one antenna, i.e., J 1 = 1 and J 2 = 1. Hence for the sake of convenience, we drop the subscript j in this section. The techniques developed in this section will be used in the more complicated setup considered in Section IV.

A. Optimality Conditions
It has been shown in [15] that the continuous-time continuous-input discrete-output Poisson MAC can be converted to a much simpler discrete-time binary-input binaryoutput MAC. In particular, the input waveform can be limited to be piecewise constant waveforms with two levels 0 or A n for the n th transmitter. Let μ n be the duty cycle of the n th transmitter (i.e., μ n is the fraction of the time that transmitter n is on). It has been shown in [15] that the sum-rate capacity is given by in which The optimization problem (8) has been solved by [15] for the symmetric case with S 1 A 1 = S 2 A 2 . In particular, [15] shows that the objective function I X 1 ,X 2 ;Y (μ 1 , μ 2 ) is a Schur concave function when S 1 A 1 = S 2 A 2 . As the result, if (μ 1 ,μ 2 ) is the optimal solution to (8) for the symmetric case,μ 1 must be equal toμ 2 . Hence, the problem can be converted into a one-dimensional optimization problem, which can be solved easily.
However, the situation for the non-symmetric case is different. In particular, if is not a Schur concave function anymore. This can be observed from the fact that a Schur concave function must be a symmetric function (see page 258 of [23]), while I X 1 ,X 2 ;Y (μ 1 , μ 2 ) is not a symmetric function when S 1 A 1 = S 2 A 2 . Therefore, the techniques developed in [15] for the symmetric case cannot be extended to the non-symmetric case. Furthermore, for general values of S n A n and λ, I X 1 ,X 2 ;Y (μ 1 , μ 2 ) is not necessarily a concave function of (μ 1 , μ 2 ) (see the proof in Appendix A). Hence, (8) is a non-convex optimization problem in general.
In the following, we solve this non-convex optimization problem. We start with the necessary KKT conditions (since the problem is not convex, these conditions are not sufficient for optimality). For convenience, we write I X 1 ,X 2 ;Y = I and hence the corresponding Lagrangian equation is given by: The optimal solution (μ 1 ,μ 2 ) must satisfy the following KKT constraints: and In order to further analyze the above KKT conditions, we need to consider 16 cases corresponding to different combinations of active constraints (i.e., whether η i = 0 or not for i = 1, · · · , 4). For example, if η 1 = 0, η 2 = 0, η 3 = 0, η 4 = 0, then the above KKT conditions can be simplified to from which we obtain Since max I (μ 1 , 0) > max I (μ 1 , 1), (12) is clearly not an optimal solution. Using similar arguments, we can show that among these 16 possible combinations, 13 constraint combinations result in non-optimal solutions. We are then left with the following three possible scenarios: This scenario corresponds to the case where both users are active. From (10) and (11), we can see that both ∂ I are nonlinear functions of (μ 1 , μ 2 ). Hence, there can be multiple (μ 1 , μ 2 ) pairs satisfying (13) and (14) simultaneously. However, we now show that there are in fact at most 2 possible (μ 1 , μ 2 ) pairs that satisfy (13) and (14) simultaneously.
We have just converted (13) and (14) into equivalent forms: Hence, (μ 1 , μ 2 ) pairs where f (μ 1 ) and g(μ 1 ) intersect with each other satisfy (13) and (14) simultaneously. As f (μ 1 ) is a linear function of μ 1 , while g(μ 1 ) is a strictly convex function of μ 1 , they can have at most two intersecting points as shown in Fig. 2. Therefore, there can be at most two pairs of (μ 1 , μ 2 ) that satisfy both conditions simultaneously. Let these solutions be (μ 1 ,μ 2 ) and (μ 1 ,μ 2 ). We then need to check whether 1] or not. If yes, we keep it. If not, then for the presentation convenience, we replace it with (0, 0). We do the same for (μ 1 ,μ 2 ).
Solving the corresponding KKT conditions, we obtain From the property of α(·), we know that 0 ≤μ 1 ≤ 1, and hence (μ 1 , 0) is a valid input. This scenario corresponds to the case where only user 1 is active.
In summary, we have the following theorem. Theorem 1: The sum-rate capacity of the Poisson MAC is It is interesting to note that unlike the Gaussian MAC with an average power constraint, it can be optimal to allow only one user to transmit in order to achieve the sum-rate capacity in the Poisson MAC with a maximum power constraint. For example, when solution for (20) and (21) in the desired range of 0 ≤ μ 1 ≤ 1 and 0 ≤ μ 2 ≤ 1, because (20) and (21) do not intersect (as shown in Fig. 3). Hence, for such a set of parameters, it is optimal to allow only one user (in this case, user 2) to transmit to achieve the sum-rate capacity.
On the other hand, there are scenarios under which it is optimal for both users to transmit. For example, when S 1 A 1 = 10, S 2 A 2 = 15, λ = 0.5, it is easy to check that it is optimal for both users to transmit in order to achieve the sum-rate capacity. Please refer to Fig. 7 in Section VI for an illustration of optimal operating regions for varying values of Motivated by these observations, we further analyze (24) to characterize conditions under which it is optimal to either allow one user or two users to transmit.

B. Single-User or Two-User Transmission?
In this subsection, we present conditions under which it is optimal for a single-user to transmit and conditions under which it is optimal for two-user transmission.
We first focus on the optimality of single-user transmission. As discussed above, the solution for two-user transmission is characterized by the intersections of (20) and (21). The following simple proposition characterize the conditions under which (20) and (21) do not have an intersection in the desired region [0, 1] × [0, 1] and hence two-user transmission is not optimal.
Proof: It suffices to argue that two-user transmission is not optimal under the assumption of the proposition. This happens when (20) and (21) do not intersect. Therefore we prove that Here, (a) follows from the linearity of f (·), (b) follows from the assumption g(0) < f (0) and g(1) < f (1), and (c) follows from the strict convexity of g(·). Proposition 2 gives sufficient conditions for single-user transmission to be optimal. Even if the conditions in Proposition 2 are not satisfied, which implies there are solutions for two-user transmission, it is still possible for single-user transmission to be optimal if the corresponding rate achieved is the largest. In the following, we will show that if one of the S i A i s is sufficiently large, then it is optimal for one user to transmit. As the roles of users are symmetric, we restrict our analysis to S 2 A 2 → ∞ as an example. We show that as S 2 A 2 → ∞, the conditions in Proposition 2 are satisfied and hence f (μ 1 ) and g(μ 1 ) do not intersect.
Lemma 3: The functions f (μ 1 ) and g(μ 1 ) have the following properties: Therefore, (20) and (21) do not intersect as S 2 A 2 → ∞. Proof: Please refer to Appendix B. Lemma 3 is illustrated in Fig. 4. As discussed in Scenario 1 of Section III-A, (20) and (21) do not intersect in our region of interest, and hence we replace (μ 1 , as the optimal solution, i.e., only user 2 transmitting achieves the sum-rate capacity. Now, we discuss the conditions under which it is optimal for both users to transmit. In particular, the following proposition characterizes conditions under which single-user transmission is not optimal.
Proposition 4: User 1 transmitting alone is not optimal if in which W is defined in (17). Similarly, user 2 transmitting alone is not optimal if Furthermore, if both conditions above are satisfied, it is optimal for both users to be active. Proof: To prove this proposition, we will find out conditions under which a single-user transmission can be eliminated as a candidate for optimality. To eliminate (0, μ * 2 ), which is obtained in (23), as a candidate for the optimal solution, we check whether ∂ I is larger than 0 or not. If it is larger than 0, then we know that (0, μ * 2 ) cannot be the optimal solution. By replacing the value of log( Hence the condition for ∂ I Following similar arguments, we can conclude that (μ 1 , 0), which is obtained in (22), is not optimal whenμ 1 > γ 1 , Ifμ 1 > γ 1 and μ * 2 > γ 2 , two-user transmission is the optimal solution.

C. Special Case: Symmetric Channel
In this section we show that the results obtained in Section III-A can recover the results obtained in [15] for the symmetric case. We show this using the following three steps.
Step 3: We show that there is a unique pair (μ 1 , μ 2 ) that satisfies (20) and (21) of scenario 1. To prove the uniqueness of the solution, as illustrated in Fig. 5 is a strictly convex function while μ 2 = μ 1 is a linear function, f (μ 1 ) and g(μ 1 ) have a single intersecting point in the range 0 ≤ μ 1 ≤ 1.
Proof: Please refer to Appendix C. Hence it can be concluded that if S 1 A 1 = S 2 A 2 , then there is a unique solution to the problem and at optimalityμ 1 =μ 2 . This result is consistent with the one shown in [15].

IV. MISO-MAC ANALYSIS
In this section, we extend the analysis to the case when the transmitters are equipped with more than one antennas.

A. Sum-Rate Capacity of MISO-MAC
Similarly to the single-antenna case studied in Section III, the continuous-time continuous-input discrete-output Poisson MAC can be converted to discrete-time binary-input binaryoutput MAC. In particular, the input waveform of each antenna can be limited to be piecewise constant waveforms with two levels 0 or A nj for the j th antenna of the n th transmitter, which can be established by following analysis in [15] and [24]. At the end of Section IV-A, we also provide another intuitive way to understand the optimality of two-level signaling. Depending on the on-off states of each antenna of user n, there are 2 J n states at user n. In the following, we use i n ∈ [1, · · · , 2 J n ] to index each of these 2 J n states at user n. We will use P n (i n ) to denote the probability that user n lies in state i n and p n [P n (1), · · · , P n (2 J n )] to denote the vector of probabilities of states at user n. We will use the binary variable b nj (i n ) to indicate whether the j th antenna of the n th user is on or off at state i n , i.e., b nj (i n ) = 1 if the j th antenna of the n th user is on for state i n and is 0 otherwise. The sum-rate achievable using (p 1 , p 2 ) is given by It is easy to see that  To characterize the sum-rate capacity, we need to solve the following optimization problem: Problem (P1) is a complex non-convex optimization problem with a large number of variables. In particular, the number of variables 2 J 1 + 2 J 2 increases exponentially with the number of antennas. The main result of this section is the following theorem.
Theorem 6: Solving problem (P1) is equivalent to solving the following problem with where Remark 7: Compared with (P1), there are only 2 variables in (P2). Although (P2) is still a non-convex optimization problem, it has the same form as the problem (P0) solved in Section III-A and hence all techniques and results there (e.g., the analysis on whether single-user transmission is optimal or not) can be applied here. Intuitively, this theorem says that the sum capacity of a MISO-MAC (with channel gains (S n1 , S n2 ) and power constraints (A n1 , A n2 ) for each transmitter n) is the same as the sum capacity of a SISO-MAC (with channel gain S n1 A n1 + S n2 A n2 and power constraint 1 for each transmitter n).
The proof of Theorem 6 has the following two major steps. In Step-1, we prove the following proposition that simplifies the optimization problem from (p 1 , p 2 ) to µ. This proposition shows that, at the optimality, instead of being a function of (p 1 , p 2 ), the objective function can be simplified to a function of µ. As the result, the dimension of the problem is reduced from 2 J 1 + 2 J 2 to J 1 + J 2 . The central issue addressed here is that, for a given µ, there are infinite number of combinations of (p 1 , p 2 ) that satisfy (27). The main idea is to show that, for any user n, if the antenna with a smaller duty cycle is on, then the antenna with a larger duty cycle is also on at the optimality. As the result, at the optimality, the value of (p 1 , p 2 ) is determined by µ. Detailed proof of this proposition can be found in Section IV-B. For the example shown in Figure 6, assuming μ 21 ≥ μ 22 , there are four initial states shown in Fig. 6 (a): only the antenna with the larger duty cycle is on, both of the antennas are on, only the antenna with the smaller duty cycle is on and none of the antennas is on. We argue that a state with only the antenna having smaller duty cycle to be on is not optimal. Hence, at the optimality, we have the scenario shown in Fig. 6 (b). In Step-2, we show the following proposition that characterizes the optimal value of µ.
Proposition 9: At the optimality, for each user, all antennas must have the same duty cycle and they must be on or off simultaneously.
This proposition shows that, at the optimality, the antennas of each user must have the same duty cycle (i.e., μ n1 = · · · = μ n J n μ n ) and are aligned. Hence, the dimension of the problem is further reduced from J 1 + J 2 to 2. The main idea of this step is to show that, at the optimality, all antennas of user n are either simultaneously on or off. Hence, from receiver's perspective, transmitter n can be viewed as a single antenna with power constraint 1 and channel gain Step 2 is illustrated Fig. 6 (b) and Fig. 6 (c). The proof can be found in Section IV-C.
In order to prove Theorem 6, Propositions 8 and 9 need to be proved. In the following subsections, we prove these propositions in detail. For the MISO case, the results can be proved in the continuous time domain as follows. For any codeword for transmitter 1 that sends One can easily see that x 1 j satisfies the peak power constraint of the j th antenna, and that the new codeword induces the same output distribution as the original one. Hence there is no loss of optimality in letting all transmit antennas send the same waveform. As the result, the MISO-MAC can be converted to a SISO-MAC. Since in SISO-MAC, letting each antenna to send two-level signal is optimal, this also shows that letting each antenna sending two-level signal is optimal in the multiple transmit antenna case. We note that, however, the proofs presented below are more general and can be used to handle more sophisticated scenarios when the receiver also has multiple antennas, as discussed in Section V, or when we have average power constraint at each transmit antenna.

B. Proof of Proposition 8
In this subsection, we prove Proposition 8 by characterizing the optimal value of (p 1 , p 2 ) for any given µ. Hence, in this subsection, µ is fixed. More specifically, we show that, at the optimality in the MISO-MAC, if the antenna with the smaller duty cycle is on, then the other antenna should also be on.
From (26), it is clear that to optimize over (p 1 , p 2 ) for a given µ, we only need to focus on equation (34) shown at the bottom of the page.
We focus on finding the optimal values of p 2 first. To facilitate the understanding, we list the labeling of states of user 2 and the corresponding values of b 2 j s in Table I.
Using the definition of ζ function in (6), we can easily check that which is simultaneously true for any value of i 1 . As and d(·, ·) are defined in (34), as shown at the bottom of this page, hence, for any given µ, maximizing H 1 (i 1 ) is a linear programming problem, for which we have the following (assuming μ 21 ≥ μ 22 , the other case being similar): 1) As d(i 1 , 4) is the largest, P 2 (4) should be as large as possible. Therefore, we assign P 2 (4) = μ 22 . 2) As μ 22 has been all used, we should set P 2 (2) = 0. 3) As d(i 1 , 3) > d(i 1 , 1), we assign the remaining part of μ 21 to state (i 1 , 3) and hence P 2 (3) = μ 21 − μ 22 . 4) For the last state related to the term d(i 1 , 1), allot the remaining probability. Hence P 2 (1) = 1 − μ 21 . This assignment implies that if the antenna with a smaller duty cycle is on, the antenna with a larger duty cycle should also be on. This is illustrated in Fig. 6 (b). Note that the above arguments are true for all i 1 s, and hence this assignment maximizes H 1 (i 1 ) for all i 1 simultaneously. Furthermore, this assignment is independent of p 1 .
Notice that the above discussion for J 2 = 2, can be extended for J 2 > 2.
Similarly, by writing then following the same procedure as above, we can calculate the optimal values of p 1 .

C. Proof of Proposition 9
In this subsection, we prove Proposition 9 by solving (P3). For the ease of calculation, we define q 1 = μ 11 − μ 12 , q 2 = μ 12 , q 3 = μ 21 − μ 22 and q 4 = μ 22 and let q = [q 1 , q 2 , q 3 , q 4 ]. Then (35) can be re-written as Correspondingly, (P3) is equivalent to s.t. q k ≥ 0, k = 1, · · · , 4, (42) Similarly to (8), the objective function (41) is not a convex function in general. We use the KKT conditions as necessary conditions to characterize the set of possible candidates for the optimal solution. In the following, we consider only the constraint (42) explicitly. We check (43) and (44) after obtaining the solution. The Langrangian equation for (P4) with constraint (42) is given by The corresponding KKT conditions are: where Now in order to find the set of optimal solutions, we may solve the above KKT conditions (45) and (46). As we can see from the expressions of ∂ I ∂q k s, we need to solve a set of nonlinear equations, which are in general difficult to solve and may have infinite number of solutions. Nevertheless, by exploring the structure of problem, we obtain the following result.
Proposition 10: There are only three possible cases for the solution to Problem (P4): 1) q = (0, α(B 1 /λ), 0, 0), which implies that both antennas of user 1 are active with the same duty cycle α(B 1 /λ) while both antennas of user 2 are off. 2) q = (0, 0, 0, α(B 2 /λ)), which implies that both antennas of user 1 are off while while both antennas of user 2 are active with the same duty cycle α(B 2 /λ). 3) q = (0, μ 1 , 0, μ 2 ), which implies that both antennas of user 1 are active with the same duty cycle μ 1 and both antennas of user 2 are active with the same duty cycle μ 2 . Furthermore, there are only two possible pairs of (μ 1 , μ 2 ) and can be obtained by solving (13) and (14) with S 1 A 1 being replaced with B 1 and S 2 A 2 being replaced with B 2 . Proof: Please refer to Appendix D. Proposition 10 states that the solution to (P4) is the same as the solution to (P2), and hence Theorem 6 is proved.

V. DISCUSSIONS
In this section, we discuss a few interesting open directions along the lines of the results presented in this paper and potential challenges associated with these problems.

A. Boundary Points of the Capacity Region
After characterizing the sum-rate capacity, a natural next step is to characterize all boundary points on the capacity region. Towards this goal, we can follow the same approach developed in this paper to solve the following optimization problem to obtain any boundary rate pair (R 1 , R 2 ) for a given 0 ≤ γ ≤ 1/2 Here, whose expression can be written out as functions of duty cycles of each antenna. Let E 1 be the set of obtained rate pairs (R 1 , R 2 ) by solving (47) with γ varying in [0, 1/2]. Similarly, we can obtain the set E 2 by setting R 1 = I X 1 ;Y |X 2 and R 2 = I X 2 ;Y . Let E = E 1 E 2 . [15] shows that E is a description of the boundary points of the capacity region, if the set However, although one can verify the convexity of R numerically, it turns out to be difficult to analytically verify such convexity even for the symmetric case considered in [15] and further for the non-symmetric case. Such an open problem is interesting and solution to it requires more refined understanding of the structure of the set R.

B. Poisson MIMO-MAC
Another extension is to consider Poisson MIMO-MAC in which the receiver is also equipped with multiple antennas. A natural idea is still to characterize the sum-rate capacity by following the two-step methodology developed in this paper. In particular, in our approach, the first step is to reduce the dimension of the problem by arguing that if the antenna with a smaller duty cycle is on, the other antennas with larger duty cycles must also be on. This step can still be carried out for Poisson MIMO-MAC by following the method presented in this paper. The second step is to characterize the optimal value of the duty cycles. In the case of the Poisson MISO-MAC, we do it by converting the set of KKT conditions related nonlinear equations, to a set of linear equations and a convex equation by substituting the common nonlinear term (i.e., the log term) from one equation to all the others (see, for example, the proof of Case-8 in Appendix D). This enables us to conclude that there are only a finite number of solutions to our problem. However, this procedure requires further exploration for the general MIMO case. In particular, as the nonlinear terms in KKT conditions have different forms, it is difficult to convert nonlinear equations to a set of linear equations so that it is not clear whether there are a finite number of solutions or an infinite number of solutions from the set of nonlinear equations related to KKT conditions. Therefore, the extension to the Poisson MIMO-MAC may require new treatments of the optimization objective function. One possible venue is to explore the relationship between the derivative of the mutual information and the mean squared error as discussed in [21] and [22]. The interesting work on single user MIMO Poisson channel [5], [24], [25] will be also be useful for our study.

VI. NUMERICAL ANALYSIS
In this section, we provide numerical examples to illustrate results obtained in this paper. As shown in the paper, the case of Poisson MISO-MAC can be converted to a Poisson SISO-MAC. Hence, in the following, we provide only example related to the SISO-MAC case. Fig. 7 shows the optimal operating scenarios for different combinations of S 1 A 1 and S 2 A 2 when they range from 0 to 25. In generating this figure, we set λ = 0.5. In Region-I, it is optimal for user 2 to transmit alone. Region-II corresponds to the case in which it is optimal for both users to transmit. In Region-III, it is optimal for user 1 to transmit alone. Fig. 8 illustrates the effect of increasing S 2 A 2 on the optimal value of (μ 1 ,μ 2 ) when S 1 A 1 is constant. In this figure, S 1 A 1 = 12.5. We can see that when S 2 A 2 is small, the optimal valueμ 2 is equal to 0, i.e., it is optimal for user 2 to stay silent. We also observe that once S 2 A 2 starts to increase and has noticeable value compared to S 1 A 1 ,μ 1 starts to decrease whileμ 2 starts to increase. Furthermore, μ 1 and μ 2 intersect with each other, i.e.μ 1 =μ 2 , when S 1 A 1 = S 2 A 2 . This is consistent with the result obtained in [15] for the symmetric case.
VII. CONCLUSION In this paper, by solving non-convex optimization problems, we have characterized the sum-rate capacity for both non-symmetric Poisson SISO-MAC and non-symmetric Poisson MISO-MAC. We have shown that under certain channel conditions, it is optimal for both users to be active and we have also established conditions under which it is optimal for only one user to be active.
There are a few open issues that are interesting to investigate. For example, one can aim to characterize the boundary points of the capacity region. One can also consider the scenario with multiple receiver antennas. These extensions require exploration of new treatment and understanding of the problem as we discuss in Section V. Furthermore, we can investigate the error exponents of Poisson MAC by extending the work in [26]. When the error exponent is considered, even for the non-symmetric Poisson MAC, single user transmit may not be optimal anymore.
In this Appendix, we show that I X 1 ,X 2 ;Y (μ 1 , μ 2 ) is not necessarily concave for general values of S 1 A 1 , S 2 A 2 and λ.
For I X 1 ,X 2 ;Y (μ 1 , μ 2 ) to be concave, 2 I needs to be negative semi-definite. For 2 I to be negative semi-definite, there are two conditions to be satisfied [27]. The first condition is that its first order principle minor must be non-positive.
, this condition holds. The second condition is that the determinant of the Hessian matrix must be non-negative. It is easy to check that The two terms on the right hand side can be dealt separately. First, we show the following.
Proof: Using the definition of ϕ, it is easy to see that ϕ is a strictly increasing function. Let a = λ, b = S 1 A 1 + λ, c = S 2 A 2 + λ and d = S 1 A 1 + S 2 A 2 + λ. then using the mean value theorem, we have: Without loss of generality we can assume that S 1 A 1 < S 2 A 2 , then we will have a < b < c < d. As ϕ is an increasing function and x 1 < x 2 , we have ϕ (x 1 ) < ϕ (x 2 ) and b − a = d − c, then: Hence ϕ(d) + ϕ(a) > ϕ(b) + ϕ(c).

APPENDIX B PROOF OF LEMMA 3
In this section we present the asymptotic analysis of (16) and (19). As the case S 1 A 1 → ∞ is similar to analysis for S 2 A 2 → ∞ due to symmetry, we restrict our analysis to S 2 A 2 → ∞ in this section. We will show that as S 2 A 2 → ∞, f (μ 1 ) and g(μ 1 ) do not intersect. Denoting S 2 A 2 as x, we calculate lim As lim Now for the f (μ 1 ), we notice that lim 1 +x+λ) , (a) follows from the L'hospital rule and (b) follows from multiplying by x 2 (x+λ) x 2 (x+λ) and L'hospital rule.

APPENDIX C PROOF OF LEMMA-5
In this section we will prove that when S 1 A 1 = S 2 A 2 , we have g(1) < 1 and g(0) > 0.
By plugging μ 1 = 1 in (19), this is equivalent to show which is true as log(1 + x) < x. Following the similar steps, g(0) > 0 can also be proved.

APPENDIX D PROOF OF PROPOSITION 10
The proof strategy is to analyze different cases corresponding to whether η k s are zero or not. By exploiting the structure of the problem, we will show that, except for three cases, all other cases are not optimal. It will be clear in the sequel, while some cases are easy to handle, it needs significant amount of work to rule out certain cases. Case-1: This implies that none of the users are active. It is clear that I (0, 0, 0, 0) can not the optimal solution. Case-2: These equations imply that user 1 is inactive while at user 2 both transmitting antennas are active with a same duty cycle. These equations lead to from which we solve q 4 : As 0 ≤ α(B 2 /λ) ≤ 1, we obtain a feasible candidate (0, 0, 0,q 4 ). Case-3: These equations imply that user 1 is inactive and at user 2 only one antenna is active. As user 1 is inactive, the scenario is same as a single-user MISO Poisson channel. It is easy to check that, for a single-user MISO Poisson channel, the maximal rate achievable using only a single antenna is less than the maximal rate achievable when both antennas are active, which is Case-2 mentioned above. Hence, Case-3 cannot be the optimal solution. Case-4: This case refers to the scenario when user 1 is inactive and user 2 transmits with both antennas having different duty cycles. Plugging q 1 = 0 and q 2 = 0 into the last two equations leads to the following two equations: which requires: It is easy to check that z(x) α(x)x is a monotonically increasing function. As the result, there does not exist (q 3 , q 4 ) that satisfies these two equations simultaneously as S 21 A 21 < B 2 . Hence, Case-4 is not possible. Case-5: These imply that user 2 is inactive and at user 1 both antennas are active with a same duty cycle. From these equations, we obtain from which we solve q 2 : Hence, the obtained feasible candidate for optimal solution from this case is (0,q 2 , 0, 0).
This case corresponds to the scenario when all of the antennas are active and both antennas at user 1 have same duty cycle and both antennas at user 2 have same duty cycle. By plugging q 1 = 0 and q 3 = 0 into (50) and (51), these two equations have the same form as (13) and (14) (with S 1 A 1 replaced by B 1 and S 2 A 2 replaced by B 2 respectively). Hence, (50)-(51) can be solved in the same manner as (13)- (14). In particular, these two nonlinear equations can be converted into a linear equation and a convex equation, therefore we know that there can be only two such values of q 2 and q 4 that satisfy the equations simultaneously. Lets those values be (0,q 2 , 0,q 4 ) and (0,q 2 , 0,q 4 ). If the solutions lies outside the range of (0, 1)× (0, 1)× (0, 1)× (0, 1), we replace it with (0, 0, 0, 0) for the sake of presentations convenience. Case-7: This case refers to the scenario when both of the antennas at user 1 are active with a same duty cycle but at user 2 only the antenna with the larger duty cycle is active. In Appendix E-A, we show that any sum-rate achieved in this case can also be achieved by the letting both antennas of each user to be simultaneously on or off, which is Case-6. Hence, Case-7 can be ruled out. Case-8: This case corresponds to the scenario when both antennas at the user 2 are active and have different duty cycles but at user 1 both transmitting antennas have the same duty cycle. Following a similar approach as how to obtain (15), we can combine ∂ I ∂q 3 = 0 and ∂ I ∂q 4 = 0 to obtain a linear equation in terms of q 1 and q 2 . By plugging q 1 = 0 to the obtained linear equation, we solve where Now for q 2 to be feasible, we need 0 ≤ q 2 ≤ 1, which requires c 1 and c 2 to have the same sign. To rule out this case, we need the following lemma. Lemma 12: h 1 (x) < 0 for x > 0. Proof: Please see Appendix E-B. Using this lemma, we know c 1 < 0 and c 2 > 0, so q 2 / ∈ [0, 1]. Hence this case not a valid choice. Case-9: In this case user 2 is inactive and at user 1 only the antenna with a larger duty cycle is active. As user 2 is inactive, the scenario is same as the single-user MISO Poisson channel. It is easy to check that, for a single-user MISO Poisson channel, the maximal rate achievable using only a single antenna is less than the maximal rate achievable when both antennas are active, which is Case-5 mentioned above. Hence, Case-9 cannot be the optimal solution. Case-10: This case refers to the scenario when both antennas at user 2 are active with a same duty cycle, while at user 1 only one antenna is active. This case can be ruled out using the same reason as Case-7. Case-11: In this case, only one antenna at both of the users are active. Following similar argument as that in Case-7, we know this case cannot be the optimal solution. Case-12: This case occurs when both antennas at user 2 are active and have different duty cycles while at user 1 only one antenna is active. Following the same steps in Case-8, we obtain in which c 1 = h 1 (λ) and c 3 = −h 1 (S 11 A 11 + λ). Using Lemma 12, we know that q 1 / ∈ [0, 1], hence we may conclude that Case-12 is not a valid case. Case-13: In this case user 2 is inactive while at user 1 both antennas transmit with different duty cycles. By plugging q 3 = 0 and q 4 = 0 into ∂ I ∂q 1 = 0 and ∂ I ∂q 2 = 0, we have that (q 1 , q 2 ) must satisfy the following two equations simultaneously: As mentioned in Case-4, z(x) = α(x)x is a monotonically increasing function. As S 11 A 11 = B 1 , we may conclude that there does not exist such (q 1 , q 2 ) pair and hence this case is not possible. Case-14: This case corresponds to the scenario when at user 1 both antennas are active with different duty cycles and at user 2 both antennas have same duty cycle. Following the same steps in Case-8, we obtain q 4 in (53), as shown at the top of the next page.
However, it is difficult to make any definitive conclusion about q 4 from this form. To rule out this case, we use the following lemma.
Proof: Please see Appendix E-C. Similar to Lemma 12, we can show that h 2 (x) < 0 when x > 0. As the result, q 4 / ∈ [0, 1]. Hence, we know that Case-14 is not a valid choice. Case-15: This case corresponds to the scenario when both antennas at user 1 are active with a different duty cycles while at user 2 only the antenna with the larger duty cycle is active. Following the same steps in Case-8, we obtain the value of q 3 as (54), as shown at the top of this page. Similar to Case-14, it is difficult to directly make any conclusion about the value of q 3 . Following similar steps as in Lemma 13, we can show that in which c 4 = h 2 (λ) and c 6 = −h 2 (S 21 A 21 + λ). Hence, similar to Case-14, we can conclude that q 3 / ∈ [0, 1], and hence this case is not a valid choice. Case-16: This case refers to the scenario when both of the antennas of each user is active and have different duty cycles. Following similar argument in Appendix E-A, we can rule this case out.
In summary, we are left with only three candidates for the optimality, i.e. Case-2, Case-5 and Case-6. Case-2 corresponds to the scenario where only user 2 is active with both antennas are simultaneously on or off with duty cycle α(B 2 /λ) and hence the optimal value of q is (0, 0, 0, α(B 2 /λ)). Case-5 is the scenario where only user 1 is active with both antennas are simultaneously on or off with duty cycle α(B 1 /λ) and therefore q = (0, α(B 1 /λ), 0, 0). Case-6 is the scenario where both users are active with both antennas at user 1 are simultaneously on or off and both antennas at user 2 are also simultaneously on or off and hence q = (0, μ 1 , 0, μ 2 ) where μ 1 and μ 2 are obtained by solving (13) and (14)  In this section we show that any sum-rate achievable for scheme A, where both of the antennas at user 1 are active with a same duty cycle but at user 2 only the antenna with the larger duty cycle is active, can also be achieved by scheme B, where both antennas of each user are simultaneously on or off.
Let p * be the duty cycle used by both antennas of user 1 and x * be the duty cycle used by the antenna with the larger duty cycle of user 2. Then the sum-rate achieved by scheme A is + p * x * ϕ(S 21 A 21 + S 11 A 11 + S 12 A 12 + λ) − ϕ((S 11 A 11 + S 12 A 12 ) p * + S 21 A 21 x * + λ). Now consider scheme B, in which both antennas of user 1 to be simultaneously on-off with duty-cycle p * , for user 2, we let both antennas to be simultaneously on or off with duty cycle x * but with reduced amplitude. In particular, for antenna 1, it uses β 1 A 21 . For antenna 2, it uses β 2 A 22 . We select β 1 and β 2 such that β 1 S 21 A 21 +β 2 S 22 A 22 = S 21 A 21 . It is easy to check that there always exists 0 ≤ β 1 ≤ 1 and 0 ≤ β 2 ≤ 1 such that this relationship holds. Hence, scheme B is a valid scheme. For this scheme, the achievable sum-rate is + p * (1 − x * )ϕ(S 11 A 11 + S 12 A 12 + λ) + p * x * ϕ(β 1 S 21 A 21 + β 2 S 22 A 22 + S 11 A 11 + S 12 A 12 + λ) − ϕ((S 11 A 11 + S 12 A 12 ) p * +(β 1 S 21 A 21 +β 2 S 22 A 22 )x * +λ).
As β 1 S 21 A 21 + β 2 S 22 A 2 2 = S 21 A 21 , we have I A = I B . Therefore, we can conclude that any sum-rate achievable by letting both of the antennas at user 1 to be active with a same duty cycle but letting only one antenna of user 2 to be active can also be achieved by letting both antennas of each user to be simultaneously on or off.

B. Property of Lemma 12
To lighten up the notation, we set a = S 22 A 22 and b = S 21 A 21 . We have Using the fact that for x > 0, we obtain
Recall that we need to solve ∂ I ∂q 1 = 0, By combining ∂ I ∂q 1 = 0 and ∂ I ∂q 2 = 0, we can eliminate the term with log and obtain a linear equation in terms of q 3 and q 4 . By plugging q 3 = 0 to the obtained linear equation, we obtain an alternative form of (53): in which c 4 = h 2 (λ) and c 5 = −h 2 (B 2 + λ) with h 2 (x) = 1 + S 12 A 12 S 11 A 11 ζ(S 11 A 11 , x) − ζ(B 1 , x).