The Wideband Slope of Interference Channels: The Large Bandwidth Case
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The Wideband Slope of Interference Channels:The Large Bandwidth Case
Minqi Shen, Anders Høst-Madsen
Abstract
It is well known that minimum received energy per bit E b N (cid:12)(cid:12)(cid:12) min in the interference channel is − . dB as ifthere were no interference. Thus, the best way to mitigate interference is to operate the interference channel in thelow- SNR regime. However, when the SNR is small but non-zero, E b N (cid:12)(cid:12)(cid:12) min alone does not characterize performance.Verdu introduced the wideband slope S to characterize the performance in this regime. We show that a widebandslope of S S , no interference = is achievable. This result is similar to recent results on degrees of freedom in thehigh SNR regime, and we use a type of interference alignment using delays to obtain the result. We also show thatin many cases the wideband slope is upper bounded by S S , no interference ≤ for large number of users K . Index Terms
Interference channels, wideband slope, interference alignment.
I. I
NTRODUCTION
Recently there has been much interest in interference channels [1], [2], [3]. In [4] it was shown that in thehigh-SNR regime, it is possible to achieve K/ degrees of freedom in a K -user interference channel (half of the K degrees of freedom if there were no interference). The basic idea is to align interference from all K − undesiredusers in half the signal space, and then receive the desired signal in the other half space without interference, an ideapioneered by [5]. The paper [4] has inspired a large body of research on interference alignment in the high-SNRregime, for example [6], [7], [8], [9], [10], [11].In this paper we consider the interference channel in the low-SNR regime, where explicitly SNR , PBN , (1) P is the input power, and B is the system bandwidth. While the work in [4] and follow-up work shows impressivelythat much can be done to mitigate the effect of interference in the high-SNR regime, one could argue that the bestway to mitigate the effect of interference is to avoid the high-SNR regime and instead operate in the low-SNR The authors are with the Department of Electrical Engineering, University of Hawaii Manoa, Honolulu, HI 96822 (e-mail:{minqi,ahm}@hawaii.edu. This work was supported in part by NSF grants CCF 0729152 and CCF 1017823. This paper was presented inpart at the 48th annual Allerton Conference on Communication, Control and Computing, October 2010 (Urbana-Champaign, IL).
June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 2 regime, when possible. It is well-known (e.g., [12]) that in a point-to-point channel the received minimum energyper bit E b N (cid:12)(cid:12)(cid:12) min = − . dB is achieved as the spectral efficiency (bits/s/Hz) R → . It is also known from [12] thatthis energy is unchanged in the presence of interference. Thus, in this limit the effect of interference is completelyeliminated. However, as Verdu pointed out in [13], in practical systems the spectral efficiency must be non-zero,though it might still be small. One way to characterize the effect of this is through the wideband slope . The widebandslope is defined by S , lim EbN ↓ EbN R (cid:16) E b N (cid:17)
10 log E b N −
10 log E b N (cid:12)(cid:12)(cid:12) min
10 log , (2)where R (cid:16) E b N (cid:17) is the spectral efficiency as a function of E b N . The wideband slope essentially represents a secondorder approximation in the low power regime of the spectral efficiency as a function of SNR, or first orderapproximation of the spectral efficiency as a function of E b N . For example, we can write R ≈ S
10 log (cid:18)
10 log E b N −
10 log E b N (cid:12)(cid:12)(cid:12)(cid:12) min (cid:19)
10 log E b N ≈
10 log E b N (cid:12)(cid:12)(cid:12)(cid:12) min + R S
10 log Examples in [13] show that this is a good approximation for many channels up to fairly high spectral efficiencies,e.g., 1 bit/s/Hz. Further, [13] shows that E b N (cid:12)(cid:12)(cid:12) min and S can be determined by the first and second order Taylorexpansion coefficients of R (SNR) at SNR = 0 , namely E b N (cid:12)(cid:12)(cid:12)(cid:12) min = log e R (0) (3) S = − (cid:16) ˙ R (0) (cid:17) ¨ R (0) (4)where ˙ R (0) = dR (SNR) d SNR (cid:12)(cid:12)(cid:12)
SNR=0 and ¨ R (0) = d R (SNR) d SNR (cid:12)(cid:12)(cid:12) SNR=0 .The reference point for wideband slope is the point-to-point AWGN (additive white Gaussian noise) channel,which has a wideband slope of 2. The wideband slope also characterizes the bandwidth required to transmit ata given rate (in the low-
SNR regime). For example, if the wideband slope is decreased from 2 to 1, twice thebandwidth is required for transmitting at a given rate.The wideband slope for interference channels was considered for the 2-user channel in [14] (a generalizationto QPSK can be found in [15]). They showed that TDMA (time-division multiple access) is not efficient in thelow-
SNR regime. In Section III we will extend the results of [14]. However, the main focus of the paper is the K -user channel, and in particular how interference alignment as in [4] can be used in the low- SNR regime.Traditional interference alignment as in [4] does not work in the low-
SNR regime. The results in [4] depend ontime or frequency selectivity of the channel. However, to achieve the minimum energy per bit in a non-flat channel,all data needs to be transmitted on the strongest channel only – which means that the wideband slope is poor (e.g., K for a K -user interference channel if only the strongest user transmits). On the other hand, delay differencesbetween different paths can be effectively used. Delay differences for interference alignment was also considered in June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 3 [16], [17], [18]. However, delay is a more natural fit for the low-
SNR regime. Namely, as the bandwidth B → ∞ even the smallest delay will eventually be magnified to the point of being much larger than the symbol duration.Therefore, delays can be efficiently manipulated and used for high bandwidth.In this paper we will prove that interference alignment using delays can be used to achieve half the widebandslope of an interference-free channel, similar to losing half the degrees of freedom in the high-SNR regime. We willalso show that generally it is difficult to obtain a larger wideband slope. The fact that wideband slope is reducedby only half means that near single-user performance can be obtained in the low-power regime. For example, ifit is desired to transmit at R = 0 . spectral efficiency, in the interference-free channel this requires . extraenergy over the minimum energy per bit for R = 0 . With interference, . , e.g., . extra energy is neededto overcome interference, independently of the number of users .II. S YSTEM M ODEL
We consider a scalar complex K -user interference channel with Gaussian noise with line-of-sight (LOS) propa-gation. There are K transmitters, numbered 1 to K , and K receivers, also numbered 1 to K . Transmitter j needs totransmit a message to receiver j , and receiver j has no need for messages from transmitter i, i = j . All transmittersand receivers have one antenna. As the specifics of the wireless model affect the results, we will discuss in moredetails the physical modeling of the system. The transmitters and receivers are placed in a two or three dimensionalspace, where the distance from transmitter i to receiver j is denoted d ji . Consistent with the LOS model, we assumethe wireless signal propagates directly from transmitter j to receiver i , and the delay in signal arrival is thereforedetermined by d ji .While the LOS model is particular, it does apply directly to some real systems, for example fractionated spacecraft[19]. An extension of results to multipath may be possible, but far from straightforward. Therefore, to obtain aconcise mathematical theory we restrict attention to the LOS model.Consider at first a single transmitter-receiver pair, i and j . Let the complex discrete-time transmitted signal oftransmitter i be x i [ n ] and the corresponding baseband (continuous-time) signal be x i ( t ) with (two-sided) bandwidth B . Specifically, to satisfy a strict band limit we must have x i ( t ) = X n x i [ n ] sinc ( Bt − n ) . This is modulated with the carrier signal c ( t ) = exp ι ( ω ( t − ς i )) , where ω is the carrier frequency and ς i is thedelay (phase offset) in the oscillator at transmitter i (and ι = √− ). The real part is transmitted, s i ( t ) = ℜ { exp ι ( ω ( t − ς i )) x i ( t ) } = cos( ω ( t − ς i )) ℜ{ x i ( t ) } − sin( ω ( t − ς i )) ℑ{ x i ( t ) } . The received signal at receiver j is r j ( t ) = A ji ℜ{ exp ι ( ω ( t − ς i − τ ji )) x i ( t − τ ji ) } + ˜ z j ( t ) τ ji = d ji c June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 4 where A ji is an attenuation factor, c is the speed of light, and ˜ z j ( t ) is white Gaussian noise with power spectraldensity N . This is modulated to baseband by multiplying with exp( − ιω ( t − υ j )) , where υ j is the delay in theoscillator at receiver j , and using a lowpass filter, resulting in the baseband signal y j ( t ) = A ji exp( ιω ( ς i + τ ji − υ j )) x i ( t − τ ji ) + z j ( t ) . This expression is valid on the assumption that ω > B . Here z j ( t ) is white Gaussian noise filtered to a bandwidth B .Return now to the interference channel. When all users transmit, the received signal at receiver j is y j ( t ) = A jj exp( ιω ( ς j + τ jj − υ j )) x j ( t − τ jj ) + X i = j A ji exp( ιω ( ς i + τ ji − υ j )) x i ( t − τ ji ) + z j ( t ) . (5)This is sampled at the Nyquist frequency f s = B (as B is the two-sided bandwidth). Let n ji = (cid:4) τ ji B + (cid:5) (6) δ ji = τ ji B − (cid:4) τ ji B + (cid:5) (7)where ⌊ x ⌋ is the largest integer smaller than or equal to x . Without loss of generality we can assume that thereceived signal at receiver j is sampled symbol-synchronous with the desired signal. Then the discrete-time modelis y j [ n ] = A jj exp( ιω ( ς j + τ jj − υ j )) x j [ n − n jj ] + X i = j A ji exp( ιω ( ς i + τ ji − υ j ))˜ x i [ n − n ji ] + z j [ n ] (8)where z j [ n ] is as sequence of i.i.d circularly symmetric random variables, z j [ n ] ∼ N (0 , BN ) , and ˜ x i [ n ] = ∞ X m = −∞ x i [ m ]sinc( n − m + δ ji ) . (9)We will also occasionally make the dependency on the fractional delay explicit as follows ˜ x i [ n, δ ji ] = ∞ X m = −∞ x i [ m ]sinc( n − m + δ ji ) . (10)By the Shannon sampling theorem, this discrete-time model is equivalent with the original continuous-time model.Results do not change if we normalize the time at each receiver so that n jj = 0 . And as the carrier frequency islarge, the phases exp( ιω ( ς i + τ ji − υ j )) can be reasonably modeled as independent uniform random variables θ ji over the unit circle. We therefore arrive at the following expression for the received signal y j [ n ] = A jj exp ιθ jj x i [ n ] + X i = j A ji exp ιθ ji ˜ x i [ n − n ji ] + z j [ n ]= C jj x j [ n ] + X i = j C ji ˜ x i [ n − n ji ] + z j [ n ] (11)where C ji = A ji exp ιθ ji .Notice that this model makes no assumptions on or approximations of modulation, e.g., it does not assumerectangular waveforms. Transmission in our model is strictly bandlimited to a bandwidth B , as opposed to [16]. June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 5
A. Approaching the Low-
SNR
Regime: Large B Case and Small B Case
What is interesting is that there are two distinct ways to approach the low-
SNR regime, which have very differentimpacts on the performance of the interference channel defined by (11). Although approaching the low-
SNR regimeby letting B → ∞ is emphasized in previous papers, it is not the only way. As can be noted from the definitionof SNR (1), SNR approaches zero if either B → ∞ or P → . Consider a point-to-point AWGN channel withspectral efficiency R = log (cid:18) PBN (cid:19) . The low-
SNR results are based on a Taylor series of log(1 + x ) , as also seen by (3-4); therefore as long as SNR =
PBN → low-power results such as minimum energy per bit and wideband slope are valid. The key isthat the spectral efficiency R → , not that B → ∞ . For the interference channel (11), on the other hand, differentresults are obtained depending on how the low- SNR regime is approached.In the first approach, let B → ∞ while P is fixed and finite. We call this approach the large bandwidth case .In this case, the propagation delay is large compared with the symbol duration, i.e., as B → ∞ , n ji can becomearbitrarily large even for very small τ ji .In the second approach, let P → while B is fixed and finite. We further assume that the propagation delay ismuch smaller than the symbol duration, i.e., τ ij B ≪ . Under this assumption, n ji = 0 and δ ji ≈ . This approachis called the small bandwidth case .The large bandwidth case is the topic of this paper; the small bandwidth case will be considered in a later paper. B. Performance criteria
In [14] the whole slope region of the interference region in the 2-user case was analyzed. However, for more thantwo users it is complicated to compare complete slope regions, and we are therefore looking at a single quantityto characterize performance. We denote the power constraint for each user P i and the spectral efficiency R i ; wefurther set SNR i = P i N B . We consider two different constraints • The equal power constraint . In this case we maximize the sum spectral efficiency R s = R + R + · · · R K under the constraint P = P = · · · = P K , i.e., SNR = SNR = · · · = SNR K . We want to characterize thewideband slope of the sum spectral efficiency R s . • The equal rate constraint . In this case we minimize the total power per Hz
SNR = SNR +SNR + · · · +SNR K under the constraint R = R = R = · · · = R K . We want to characterize the wideband slope of the sumspectral efficiency R s = K · R .The equal power constraint could correspond to a scenario where each user needs to consume energy at the samerate, e.g., so that batteries last the same for all users. The equal rate constraint could correspond to a scenario wherewe want to minimize total system energy consumption. Each constraint can be easily generalized to unbalancedcases, e.g. µ SNR = µ SNR = · · · = µ K SNR K , but we only consider the balanced case here to keep resultsconcise. June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 6
As performance measure we use ∆ S = S S , no interference . The quantity S , no inteference is the wideband slope of a K -user interference channel where all interference linksare nulled, | C ij | = 0 , i = j , but the direct links C ii are unchanged. We can interpret ∆ S as the loss in widebandslope due to interference, or equivalently (∆ S ) − as (approximately) the additional bandwidth required to overcomeinterference. Alternatively, if we define ∆ E b = 10 log E b N −
10 log E b N as the extra energy required to operate at a spectral efficiency R > , we have ∆ E b ≈ (∆ S ) − ∆ E b, no interferencefor small increases in spectral efficiency. Thus, (∆ S ) − also measures the amount of energy needed to overcomeinterference. C. E b N (cid:12)(cid:12)(cid:12) min of the Interference Channel The papers [13] and [20] show that the minimum energy per bit of the interference channel is equal to that of aninterference-free channel, achievable by Treating Interference as Noise (TIN) and TDMA. The following theoremgives the transmitted E b N (cid:12)(cid:12)(cid:12) min under the two different constraints. Theorem 1.
The minimum energy per bit of the interference channel defined by (5) is E b N = P (cid:16) | C jj | − (cid:17) K log e (12) under the equal rate constraint; and E b N = K log e P Kj =1 | C jj | (13) under the equal power constraint. The best known achievable rate for the interference channel is the Han-Kobayashi region [21]. For the Gaussianinterference channel, in particular the idea of transmitting common messages has been shown to be powerful [22].However, the common message does have a higher E b N in the low power limit than the minimum, and thereforedoes not improve the wideband slope. To make fair comparison any bound imposed on the wideband slope musthave the correct E b N (cid:12)(cid:12)(cid:12) min . We emphasize this requirement in the following remark. Remark . For a bound on rate to be useful as a bound on wideband slope, it needs to have the correct E b N (cid:12)(cid:12)(cid:12) min given in Theorem 1. June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 7
III. T HE SER C ASE
We will start by analyzing the 2-user case as this is instrumental for the K -user case. As we have discussed inSection II-A, the essential difference between large bandwidth case and small bandwidth case is that they impactthe behavior of propagation delays differently. It turns out that all results in the 2-user case are independent ofdelay, thus independent of how the low- SNR regime is approached. This indicates that the capacity region in the2-user case could be independent of delay in general, but we have not been able to prove so.
A. Achievable Schemes
First we will outline the strategies that can be used for the achievable rate. In order to use these to inner bound thethe sum slope, as mentioned in Remark 2, they must have the correct E b N (cid:12)(cid:12)(cid:12) min , and that only leaves three strategies1) Interference decoding .If | C ji | > | C ii | , user j can decode the message from user i , and the capacity region of the interferencechannel is equivalent to the capacity region of the multi-access channel formed by transmitter i , transmitter j , and receiver j , which is R j ≤ log (cid:16) | C jj | SNR j (cid:17) (14) R i ≤ log (cid:16) | C ii | SNR i (cid:17) (15) R i + R j ≤ log (cid:16) | C ji | SNR i + | C jj | SNR j (cid:17) . (16)In the low- SNR regime, as
SNR → , there always exists some real number ǫ > such that if SNR j , SNR i <ǫ the sum slope outer bound given by the summation of (14) and (15) is less than the sum slope outer boundgiven by (16) because | C ji | > | C ii | . Therefore, (16) can be discarded and the multi-access bound is equivalentto the rectangular capacity region of a channel with no interference. Thus, interference does not affect widebandslope in this case.2) Treating interference as noise (TIN).The transmitters use i.i.d Gaussian code books, and the receivers treat the interference as part of the backgroundnoise. Notice that delay does not affect the distribution of interference as ˜ x i [ n − n ji ] has same distributionas x i [ n ] . The achievable ( R , R ) is R ≤ log | C | SNR | C | SNR ! (17) R ≤ log | C | SNR | C | SNR ! . (18)3) TDMA .In time-division multiple access the transmitters use orthogonal time slots. Because of the delay differences,users have to insert buffers with no transmission around each TDMA frame, so that they are orthogonal atboth users. However, the length of these buffers is finite, so as the code length converges towards infinity (as
June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 8 required by capacity analysis), the effect of these buffers on spectral efficiency will converge towards zero.TDMA therefore achieves the following spectral efficiency also in the case of delays, R ≤
12 log (cid:16) | C | SNR (cid:17) (19) R ≤
12 log (cid:16) | C | SNR (cid:17) . (20)The achivable sum slope can easily be straightforwardly calculated from these equations using (4). The expressionsare too complex to give much insight, so we will only state them for later reference for a canonical 2-user channelwith symmetric interference link gains. Theorem 3.
Consider a 2-user interference channel where | C jj | = 1 and | C ji | = a, i = j . The sum slope isinner bounded by S ≥ a > < a < a a ≤ (21) under both the equal power constraint and the equal rate constraint.B. Outer Bounds In this section, we will state some sum slope outer bounds, and discuss the so-called noisy interference channel,where the exact sum slope is known.The following theorem generalizes Theorem 2 from [23] to channels with delay.
Theorem 4 (Kramer’s bound) . Suppose that | C | < | C | . Then R ≤ log (cid:16) | C | SNR (cid:17) (22) R ≤ log | C | SNR + | C | SNR | C | | C | R + 1 − | C | | C | (23) R + R ≤ log (cid:16) | C | SNR (cid:17) (24) + log | C | SNR | C | SNR ! (25) independent of delay.Proof: Put C = 0 to enlarge the capacity region. Now assume that, different from the system model (11),receiver 2 also samples the received signal synchronously with the transmitted signal of user 1 . A Z-channel with According to the sampling theorem, this does not change the capacity region.
June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 9 delay is formed: y ′ [ n ] = C x [ n ] + z [ n ] y [ n ] = C ˜ x [ n − n ] + C x [ n ] + z [ n ] (26)where ˜ x [ n ] is defined by (9).Next, we show that the capacity region of (26) is independent of delay. The channel (a) and (b) illustrated inFigure 1 have identical capacity regions because p ( ˇ y | x , ˜ x ) and p ( y | x , ˜ x ) have the same distribution. ˇ z [ n ] isi.i.d Gaussian noise independent of z [ n ] and the input signals, with power (cid:16) − | C | | C | (cid:17) N B . Because | C | | C | < ,such ˇ z [ n ] is guaranteed to exist. The argument is identical to (a)~(c) of Figure 6 in [24]. Details are skipped here.Figure 1: Channels with Equivalent Capacity RegionThe channel ( b ) has the form y ′ [ n ] = C x [ n ] + z [ n ] (27) ˇ y [ n ] = C ˜ x [ n − n ] + C C y ′ [ n ] + ˇ z [ n ] . (28)Using Fano’s inequality as usual, we can now bound the capacity of this channel by nR − nǫ n ≤ h (ˇ y n ) − h (ˇ y n | ˜ x ) (29) ≤ h (ˇ y n ) − h (ˇ y n | w ) (30) = h (ˇ y n ) − h (cid:18) C C y ′ n + ˇ z n | w (cid:19) , (31) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 10 where w is the message sent by transmitter 2. The step (29) to (28) is from the data processing inequality, as ˜ x is a function of the transmitted codeword x , which is a function of w . The second term in (31) is independentof delay, and can be lower bounded by the entropy power inequality [25]. The first term can be upper boundedby the delay-free case. Therefore, the capacity region of (26) is identical to that of the channel without delay. Thepapers [24] and [23] show that the capacity region of delay-free channel can be derived from an equivalent degradedbroadcast channel. Given Theorem 1 in [26] its rate region has upper bound (22). Finally, it is easy to see that thecapacity region of (11) is contained within that of the Z-channel. The equation (25) is a restatement of (47) in [23].We use Theorem 4 to obtain a sum slope outer bound under the equal power constraint as follows, Corollary 5.
Suppose that | C | < | C | . Under the equal power constraint, the wideband slope for the sum ratehas outer bound S ≤ | C | + | C | ) | C | | C | + | C | + | C | (32) ∆ S ≤ | C | | C | | C | + | C | + 1 , (33) independent of delay.Proof: This result can be easily shown combining (25) and the formulas (3) and (4).Results similar to Theorem 4 and Corollary 5 can be obtained for | C | < | C | case by interchanging the indices ′ ′ and ′ ′ .For the equal-rate constraint if only one interference link is weak, bound (22) does not have the correct E b N (cid:12)(cid:12)(cid:12) min and therefore cannot be used for bounding the wideband slope by Remark 2 . If both interference links are weak,we have following corollary.
Corollary 6.
Suppose that | C | < | C | and | C | < | C | . Under the equal rate constraint, the wideband slopefor the sum rate is upper bounded by S ≤ · (cid:0) | C | + | C | (cid:1) − | C | | C | | C | | C | ! · (cid:0) | C | + | C | + | C | − | C | | C | ! + | C | − | C | | C | !! − ∆ S ≤ (cid:0) | C | + | C | (cid:1) − | C | | C | | C | | C | ! · (cid:0) | C | + | C | + | C | − | C | | C | ! + | C | − | C | | C | !! − independent of delay. June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 11
Proof: (22) gives | C | SNR + | C | SNR ≥ R | C | | C | R − | C | | C | + 1 ! − (34) | C | SNR + | C | SNR ≥ R | C | | C | R − | C | | C | + 1 ! − . (35)Under the equal rate constraint, R = R = R s , and our objective is to minimize SNR + SNR . We constructthe following optimization problem min SNR + SNR s . t . A SNR SNR ≥ b P j ≥ where A = | C | | C | | C | | C | and b = Rs / (cid:16) | C | | C | Rs / − | C | | C | + 1 (cid:17) − Rs / (cid:16) | C | | C | Rs / − | C | | C | + 1 (cid:17) − . Using simple linear pro-gramming principles, one optimal solution can be found at the vertex of the feasible region. That is, SNR + SNR | min =SNR o + SNR o where SNR o SNR o = A − b > . We solve this simple linear system and get SNR o = | C | − · RsK (cid:16) RsK − (cid:17) | C | | C | (cid:16) − | C | | C | (cid:17) + (cid:16) − | C | | C | (cid:17) (cid:16) RsK − (cid:17) − | C | | C | | C | | C | (36) SNR o = | C | − · RsK (cid:16) RsK − (cid:17) | C | | C | (cid:16) − | C | | C | (cid:17) + (cid:16) − | C | | C | (cid:17) (cid:16) RsK − (cid:17) − | C | | C | | C | | C | . (37)Now we have the expression of sum power as a function of sum rate. The following formulas are equivalent to (3)and (4). E b N = d SNR ( R ) dR (cid:12)(cid:12)(cid:12)(cid:12) R =0 (38) S = 2 d SNR( R ) dR (cid:12)(cid:12)(cid:12) R =0 d SNR( R ) dR (cid:12)(cid:12)(cid:12) R =0 log 2 . (39)They can be proved using a technique similar to (140)~(144) in [13]. Details are skipped here. Combining (36),(37) and (39), we have E b N = (cid:16) | C | − + | C | − (cid:17) e S = 4 · (cid:0) | C | + | C | (cid:1) − | C | | C | | C | | C | ! · (cid:0) | C | + | C | + | C | − | C | | C | ! + | C | − | C | | C | !! − June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 12
For the 2-user interference channel without delay, [1], [3] and [2] show that there exists a class of channelswhose optimal sum spectral efficiency can be achieved by i.i.d. Gaussian inputs and treating interference as noise.This class of channel is one of the few where the exact capacity is known, and consequently also the exact sumslope.We here extend these results to channels with delay.
Theorem 7.
For a 2-user interference channel defined by (11), if there exist complex numbers ρ , ρ and positivereal numbers σ , σ such that, | ρ | ≤ σ ≤ − | ρ | σ (40) | ρ | ≤ σ ≤ − | ρ | σ (41) C = ρ C | C | SNR + 1 (42) C = ρ C | C | SNR + 1 , (43) then the optimal sum capacity R + R ≤ log (cid:18) | C | SNR | C | SNR (cid:19) + log (cid:18) | C | SNR | C | SNR (cid:19) (44) is achievable by i.i.d. Gaussian input and treating interference as noise at the receivers. Further, (40) ~ (43) aresatisfied as long as s | C | | C | (cid:0) | C | SNR (cid:1) + s | C | | C | (cid:0) | C | SNR (cid:1) ≤ . (45) . Proof: Please see Appendix A.Theorem 7 is identical to the case where there is no delay, which is discussed in [27, Theorem 6]. We now useTheorem 7 to derive the exact sum slope,
Corollary 8.
Consider the 2-user interference channel defined by (11). Under the equal power constraint, if thechannel coefficients satisfy s | C | | C | + s | C | | C | < , (46) then i.i.d. Gaussian inputs and treating interference as noise achieve the optimal sum slope S , which is S = 2 (cid:0) | C | + | C | (cid:1) | C | + | C | + 2 ( | C | | C | + | C | | C | ) (47) ∆ S = 1 + 2 (cid:0) | C | | C | + | C | | C | (cid:1) | C | + | C | . (48) Proof:
Under the equal power constraint where
SNR i = SNR s there must exist some ǫ > , such that if SNR < ǫ then (45) can be satisfied. Because the low-
SNR regime is approached as
SNR → , this gives (46).Given (44), under the equal power constraint the sum rate achieved by treating interference as noise is June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 13 R s ≤ log | C | | C | ! + log | C | | C | ! . (49)Combining (49) with (3) and (4) we have (47).Figure 2 illustrates the sum slope region of a 2-user interference channel with unit direct link gain, and symmetriccross link gain, that is, | C | = | C | = 1 , and | C | = | C | = a . In this figure, the inner bound is given byTheorem 3: the inner bounds labeled “Strong Int.”, “Achievable, TDMA”, and “Achievable TIN” are representedby the first, the second and the last line in (21), respectively. The outer bound is given by (32). Given Corollary 8,if a ≤ , treating interference as noise achieves optimal the sum slope, i.e., the inner bound is tight, which is alsoindicated on the figure.The focal point here is the point a = 1 . Just above that, the effect of interference is completely eliminated. Justbelow that, interference is at its worst. One could wonder if, for the K -user case, the former fact could be usedeffectively. It turns out that is not the case. In Section IV-B, we will show that in a K -user interference channelwhen K is large, with high probability each user will form an 2-user weak interference pair, where a is just below1, with some other user.Figure 2: Sum slope versus | C || C | . In the legend, TIN stands for treating interference as noise.IV. T HE K - USER CASE
A. Achievable Scheme and Inner Bound
For the 2-user case, the achievable rates are unaffected by delay, as seen in Section III-A. However, the K -usercase is very different from the 2-user case, just as for the high SNR case considered in [4]. Similar to the high June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 14
SNR case, we can obtain a significant increase in rate by using a variation of interference alignment. The type ofinterference alignment used in [4] based on time or frequency selectivity does not work in the low-SNR regime;however, propagation delays can be used. Specifically, we show that for any set of delays τ ji , i, j = 1 , · · · , K that are linearly independent over the rational numbers, there exist arbitrarily large B that can make the directpropagation delays τ ii arbitrarily close to an even integer while the cross-delays τ ji , j = i are close to some oddinteger. As a result, if we let each user use even time slots in the discrete time baseband channel model, then atthe receiver, the desired message and the interference signal are almost orthogonal in the time domain. Therefore,the interference channel can achieve ∆ S = .The idea of interference alignment over time domain is also used in [16], [17], and [18]. However, delay is muchmore efficient when we let B → ∞ , as n ji = ⌊ τ ji B ⌋ can become arbitrarily large. We therefore do not need touse the approximation δ ji ≈ as in [16] or large K as in [17]. The paper [18] mainly discusses how to design analgorithm to place K users in an N dimensional space, N > , so that interference alignment can be realized. Inour work, N ≥ and user locations are given. Our method for interference alignment works for any user location(with probability 1 for a continuous distribution of user locations).In order to state the results, we need to refine the definition of wideband slope (2) as E b N = lim inf B → + ∞ PR · N B (50) S , lim sup EbN ↓ EbN R (cid:16) E b N (cid:17)
10 log E b N −
10 log E b N min
10 log . (51)Notice that if the limit exists, (51) is identical to (2), so this is not a new definition, but a widening of the applicabilityof the wideband slope. We will see through an example in Section IV-A2 that this generalized definition has anoperational meaning, as does the wideband slope in [13].For comparison purposes, we will list the results for the interference-free case, i.e., C ji = 0 for i = j as follows(directly obtained from [13, Theorem 9]): S , no interference = 2 (cid:16)P j | C jj | (cid:17) P j | C jj | equal power constraint; S , no interference = 2 K equal rate constraint .
1) The Achievable Sum Slope:
In the following, we will precisely specify the interference alignment scheme weuse to obtain the achievable sum slope.
Definition 9 (Delay-based interference alignment) . Fix the transmission bandwidth B ≤ B .1) At transmitter j • Use a codebook generated from independent Gaussians according to x j [ n ] ∼ N (cid:16) , P j B o (cid:17) n = 2 k n = 2 k + 1 k ∈ Z . (52) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 15 • Generate the baseband transmitted signal according to x j ( t ) = + ∞ X n = −∞ x j [ n ] sinc ( B o ( t − nT )) . Notice that this signal has bandwidth B ≤ B .2) At receiver j : • Sample the received continuous time baseband signal with rate B o and symbol synchronize with x j [ n ] ; • Discard y j [2 m + 1] , m ∈ Z • Decode the desired message from y j [2 m ] , m ∈ Z , with typical decoding while treating any remaininginterference as noise.Since we only use every other time slot for transmission in (52) we can at most achieve a wideband slope ∆ S = . In the following we will show that it is possible to choose the transmission bandwidth B so that mostof the interference lines up in the discarded time slots y j [2 m + 1] , m ∈ Z , which in turn means that we can actuallyachieve ∆ S = .The concept of linearly independent over rational number will be introduced first. Definition 10 ([28]) . A set of real numbers θ = { θ , θ , · · · , θ n } are linearly independent over rational number if P a i θ i = 0 only if a i = 0 for all a i ∈ Z . Lemma 11. If τ ji , i, j ∈ { , · · · , K } , i = j are linearly independent over the rational numbers, then for any δ > , there exist an arbitrarily large real number B , such that | τ ji B − k ji − | ≤ δ (53) for some integers k ji , j, i ∈ { , · · · , K } . The proof of Lemma 11 is based on the following fundamental approximation results in number theory.
Theorem 12 ([28, Theorem 7.9, First Form of Kronecker’s Theorem]) . If α , · · · , α n are arbitrary real numbers,if θ , · · · , θ n are linearly independent real numbers over the rational numbers, and if ǫ > is arbitrary, then thereexists an real number t and integers h , · · · , h n such that | tθ i − h i − α i | < ǫ (54) ∀ i ∈ { , , . . . , n } We also have
Lemma 13. [28, Exercise 7.7, page 160]Under the hypotheses of Theorem 12, if
T > is given, there exists areal number t > T satisfying the n inequalities (54). Now let us prove Lemma 11.
June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 16
Proof of Lemma 11:
Let α , · · · , α n = 0 . , ǫ = δ . According to Theorem 12, there exist arbitrarily large realnumber ˆ B and some integers n ji , i, j ∈ { , · · · , K } such that (cid:12)(cid:12)(cid:12) τ ji ˆ B − k ji − . (cid:12)(cid:12)(cid:12) ≤ δ . Let B = 2 ˆ B , we have | τ ji B − k ji − | ≤ δ. Combining the inequality above with Lemma 13, Lemma 11 is proved.Lemma 11 shows that using this transmission scheme, the desired signal is almost orthogonal with the interferencesignal in time domain. However, there is always some interference leaking into the signal time slots. We need toshow that as the fractional delay (7) δ ji → , the power of this interference become negligible. For this we needthe following lemma, Lemma 14.
Under the assumptions x j [2 m ] are i.i.d. Gaussian random variable with distribution N (0 , P j ) and x j [2 m + 1] = 0 for all j and m , E [˜ x i [ n , δ ji ]˜ x ∗ i [ n , δ ji ]] is a continuous function of δ ji which satisfies lim δ ji ↓ E [˜ x ∗ i [ n , δ ji ]˜ x i [ n , δ ji ]] = P i if n = n = 2 k, for some integer k o.w. Proof:
Please see Appendix B.Equipped with the interference alignment scheme in definition 9, Lemma 11, and Lemma 14, we proceed toshow the main results on the achievable sum slope of the K -user interference channel. Theorem 15.
Suppose that the set of delays τ ji , i, j ∈ { , · · · , K } , i = j are linearly independent over therational numbers. Then the following wideband slope is achievable S = (cid:16)P j | C jj | (cid:17) P j | C jj | equal power constraint; S = K equal rate constraint . Under both constraints, ∆ S = 12 is achievable.Proof: Assume that the system uses the transmission scheme proposed in Definition 9. Let ǫ j ( B ) = K X i = j (cid:12)(cid:12)(cid:12) E h (˜ x i [2 n ]) i(cid:12)(cid:12)(cid:12) denote the power of the leaked interference. June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 17
The best rate with this scheme is clearly achieved if the leaked interference power is zero; in that case the channelis an interference-free channel where half the symbols are not used. We can therefore conclude ∆ S ≤ (55)On the other hand, taking into account the leaked interference, the achievable rate at receiver j is R j = 12 log | C jj | P j BN ǫ j ( B ) BN ! (56) = | C jj | P j BN − (cid:16) ǫ j ( B ) | C jj | P j + | C jj | P j (cid:17) (cid:18) BN (cid:19) + o (cid:18) BN (cid:19) ! . (57)The wideband slope is a continuous function of the coefficients in the first two terms in the Taylor series of R j in B . According to Lemma 11 for any δ > there exists some B δ and a set of integers k ji such that n ji = 2 k ji + 1 ,i.e., the integer part of the delay is an odd number, and the fractional part of the delay satisfies | δ ji | ≤ δ. From Lemma 11 and Lemma 14, we can then conclude that there exists a sequence of real numbers { B o , B o , · · · } , B o ( k +1) > B ok , so that k → ∞ and ǫ j ( B ok ) → for all j = 1 , · · · , K . This means that ∆ S = is a limit point,and together with (55) this shows that ∆ S = is the limit superior.The Theorem has the following corollary. Corollary 16.
Suppose that all transmitters and receivers have independent positions and each node position hasa continuous distribution. Then the propagation delays τ ji , i, j ∈ { , · · · , K } , are linearly independent over therational numbers with probability one, and ∆ S = 12 is achievable. So in practice ∆ S = is achievable, since transmitters and receivers can never be positioned accurately ina grid; there is always some nano-scale inaccuracy (dither) in positions, at the fundamental level due to quantummechanics!
2) Practical Implementation and Simulation Results:
In this section we will show that the interference alignmentideas of the previous section can be used in a practical system, and show some simulation results. This will alsomake it clear why the modified definition (51) is needed.We can see that one key question concerning the transmission scheme defined by Definition 9 is: how to find B o ? Here we propose an algorithm, stated in the following proposition. Proposition 17.
Assume that both the transmitters and the receivers have perfect channel knowledge.Initialize B to be any positive integer. Proceed with the following while loop:While ( ∃ i = j : | τ ji B − k ji − | > δ ) { Increase B by 1 , i.e., B = B + 1 . } June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 18 If τ ji are linearly independent over the rational numbers the algorithm terminates after a finite number ofiterations. The output B of the algorithm satisfies ∀ i = j : | τ ji B − k ji − | ≤ δ, which can therefore can be chosen as B o . Lemma 18 guarantees that the searching algorithm defined in the proposition above terminates. The proof isalmost identical to the proof of Lemma 11. However, the essential difference is that while Lemma 11 only showsthe existence of B satisfying (53) over the set of positive real numbers R + , the results in this section ensure thatsuch B can be found even if we restrict B to be integer. Lemma 18. If τ ji are linearly independent over the rational numbers, then for any δ > , there exist an integer B , such that | τ ji B − k ji − | ≤ δ (58) for some integers k ji . Further, B can be made arbitrarily large. The proof of Lemma 18 is based on the second form of Kronecker’s theorem[28, Theorem 7.10, Second Formof Kronecker’s Theorem], which shows that Theorem 12 still holds even if we require t to be an integer. Detailsare skipped here.We can see that the brute force algorithm of searching through all integer B is guaranteed to find good operatingbandwidths. Fig. 3b shows the performance of the proposed achievable scheme when the system operating at asequence of B δ , δ = 0 . . However, designing more efficient B o -searching algorithm could be a subject of furtherresearch.In the simulation, we consider a 3-user channel with symmetric channel gain: | C jj | = 1 , | C ji | = 0 . . Noticethat for channels with symmetric link gains, equal power and equal rate constraints are equivalent. The delays τ ji are chosen such that they are linearly independent over the rational numbers.Fig. 3a shows the simulation results of the case where bandwidth B increases continuously. The system per-formance shows a noticeable oscillating behavior. This phenomenon can be explained as followed. At receiver j the interference caused by user i is an increasing function of δ ji ; and δ ji is a periodic function of B , oscillatingbetween 0 and 1. It can be proved that the cumulative effect of leaked interferences from all other users has same(almost) periodic behavior. The proof is similar to that of Lemmas 72 and 13; details are skipped here.Fig. 3a also shows why we need the modified definition (51). In this case the limit (2) does not exist; onedefinition of a limit of a function is that for any sequence x n → x , f ( x n ) → f ( x ) . In Fig. 3a the points alongthe upper envelope and the points along the low envelope, for example, give different slope. However, the lim sup always exists. One sequence that achieves the lim sup is shown in Fig. 3b. What is important to notice that thenew definition of the wideband slope is still operational as in [13]. That is, it is possible to choose some finite June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 19 bandwidth where the performance is close to the wideband approximation. But different from [13] is not enoughto use a bandwidth that is sufficiently large. It also has to be chosen very carefully. (a) Performance for arbitrary bandwidth. (b) Peak points of performance.
Figure 3: Achievable spectral efficiency versus E b N . The straight line shows the performance approximated to firstorder by the wideband slope. B. Outer bounds
In Section IV-A we have seen that ∆ S = can be achieved. Is this the best possible? Clearly no. In the 2-userchannel the interference alignment scheme proposed in Definition 9 reduces to TDMA. As we have seen in SectionIII, interference decoding and treating interference as noise can be better than TDMA. For K > case, it is alsonot difficult to construct examples where ∆ S > is achievable. However, in this section we will show that forlarge K this happens rarely.Let us first define two concepts: (1 − ǫ ) -interference pair and weak (1 − ǫ ) -interference pair. Definition 19.
We say that users i and j form an (1 − ǫ ) -interference pair if − ǫ ≤ | C ji | | C ii | , | C ij | | C jj | < , and form a weak (1 − ǫ ) -interference pair if − ǫ ≤ | C ji | | C ii | < or − ǫ ≤ | C ij | | C jj | < . In section IV-B1, we will show that as the number of users K → ∞ , the event { user j, ∀ j ∈ { , · · · , K } , forms a (1 − ǫ ) -interference pair with at least one other user } June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 20 happens with high probability. Consider Fig. 2: when two users form an (1 − ǫ ) -interference pair, they operate inthe point just below 1 in the figure, where Kramer’s bound bounds each user’s wideband slope by 1. This resultsin ∆ S ≤ + δ, ∀ δ > under the equal rate constraint.Similarly, in section IV-B2, we will show that as K → ∞ , the event (cid:26) K users form K disjoint weak (1 − ǫ ) − interference pairs (cid:27) happens with high probability, which gives ∆ S ≤ + δ, ∀ δ > under the equal power constraint if the distributionof C ji satisfies some additional conditions.The outer bounds in this section are proven under the assumption that the channel coefficients C ji for all i, j ∈ { , · · · , K } are i.i.d. random variables. However, this is not a necessary condition, only a convenientcondition to simplify proofs; later in the section we will comment more on this.
1) The Equal Rate Constraint:
First consider the equal rate constraint. We assume that the channel coefficients C ji are i.i.d. and E h | C ii | − i < ∞ ; if the latter assumption were not satisfied, lim K →∞ K P Ki =1 P i = ∞ even inthe interference-free case (see (12)), so the energy per bit and wideband slope would not be well-defined for large K (see also the comment at the top of page 1325 in [13] about Rayleigh fading).Let F | C ii | be the CDF of | C ij | ; this defines a probability measure on the real numbers through µ F (( a, b ]) = F | C ii | ( b ) − F | C ii | ( b ) (this is true for any random variable) For ∀ ǫ, ˆ ǫ > , define two sets R ˆ ǫ = { x ∈ R : F | C ii | ( x ) − F | C ii | ((1 − ǫ ) x ) < ˆ ǫ } ; D ˆ ǫ = n all i ∈ { , . . . K } : | C ii | ∈ R ˆ ǫ o . The following lemma shows that as the number of users K → ∞ a user in D c ˆ ǫ , with high probability forms a (1 − ǫ ) -interference pair with at least one other user. Lemma 20.
Given ∀ ǫ, ˆ ǫ > , denote B ǫ, ˆ ǫ = { i ∈ D c ˆ ǫ : user i does not form an (1 − ǫ ) -inteference pair with any other user } Then lim K →∞ Pr ( B ǫ, ˆ ǫ = ∅ ) = 1 . Proof:
Please see Appendix C.On the other hand users in D ˆ ǫ do not necessarily form (1 − ǫ ) -pairs. The following lemmas are used to showthat the probability of the set D ˆ ǫ is small, Lemma 21.
Given any infinite sequence ˆ ǫ n > satisfying ˆ ǫ n > ˆ ǫ n +1 and ˆ ǫ n → , the corresponding sequence of R ˆ ǫ n satisfies1) R ˆ ǫ n +1 ⊆ R ˆ ǫ n ; June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 21 µ F ( R ˆ ǫ n ) → .Proof: Please see Appendix D
Lemma 22.
Let X be a positive random variable with E [ X ] < ∞ , and let µ X be the measure induced by theCDF of X . Let G i ⊂ R be a sequence of measurable sets with G i +1 ⊆ G i and lim i →∞ µ X ( G i ) = 0 . Define X i = X X ∈ G i X / ∈ G i . Then lim i →∞ E [ X i ] = 0 . Proof:
Please see Appendix E.Our main result is stated in the following theorem.
Theorem 23.
Suppose that the channel coefficient C ij are i.i.d.. Under the equal rate constraint ∀ δ > K →∞ Pr (cid:18) ∆ S ≤
12 + δ (cid:19) = 1 . (59) Proof:
We discuss users in the set D ˆ ǫ and those in the set D c ˆ ǫ separately.First, let us look at user j , j ∈ D c ˆ ǫ . We assume that each user j ∈ D c ˆ ǫ forms a (1 − ǫ ) -interference pair withsome user i ( j ) . Given Lemma 20, this happens with high probability. Consider a single (1 − ǫ ) -interference pair ( j, i ( j ) ) . We can get an upper bound on the spectral efficiency, by eliminating all interference links except the linksbetween users j and i ( j ) , so that the received signal is y j = C jj x j + C ji ( j ) x i ( j ) + z j y i ( j ) = C i ( j ) j x j + C i ( j ) i ( j ) x i ( j ) + z i ( j ) . Let (cid:12)(cid:12)(cid:12) C ji ( j ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) C i ( j ) i ( j ) (cid:12)(cid:12)(cid:12) = 1 − ǫ ji ( j ) , (cid:12)(cid:12)(cid:12) C i ( j ) j (cid:12)(cid:12)(cid:12) | C jj | = 1 − ǫ i ( j ) j . Since (cid:8) y j , y i ( j ) (cid:9) is a (1 − ǫ ) -interference pair, we have ≤ ǫ ji ( j ) , ǫ i ( j ) j < ǫ. (60)Applying (36) and (37) to (cid:8) y j , y i ( j ) (cid:9) , we have the optimum solution SNR i ( j ) o = (cid:12)(cid:12) C i ( j ) i ( j ) (cid:12)(cid:12) − · RsK (cid:16) RsK − (cid:17) (cid:0) − ǫ i ( j ) j (cid:1) ǫ ji ( j ) + ǫ i ( j ) j (cid:16) RsK − (cid:17) − (cid:0) − ǫ ji ( j ) (cid:1) (cid:0) − ǫ i ( j ) j (cid:1) (61) SNR jo = | C jj | − · RsK (cid:16) RsK − (cid:17) (cid:0) − ǫ ji ( j ) (cid:1) ǫ i ( j ) j + ǫ ji ( j ) (cid:16) RsK − (cid:17) − (cid:0) − ǫ ji ( j ) (cid:1) (cid:0) − ǫ i ( j ) j (cid:1) . (62)And SNR i ( j ) + SNR j ≥ SNR i ( j ) o + SNR jo . Notice that the RHS of (63) and (64) are monotonically decreasingfunction of either ǫ ji ( j ) or ǫ i ( j ) j . Thus, given the condition (60), we can relax (63) and (64) by substituting ǫ ji ( j ) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 22 and ǫ i ( j ) j by ǫ , SNR i ( j ) o ≥ (cid:12)(cid:12) C i ( j ) i ( j ) (cid:12)(cid:12) − · RsK (cid:16) (1 − ǫ ) 2 RsK + ǫ (cid:17) − − ǫ (63) SNR jo ≥ | C jj | − · RsK (cid:16) (1 − ǫ ) 2 RsK + ǫ (cid:17) − − ǫ . (64)Thus SNR jo = 2 RsK (cid:16) (1 − ǫ ) 2 RsK + ǫ (cid:17) − ǫ | C jj | − , if j ∈ D c ˆ ǫ . (65)Second, for user k , k ∈ D ˆ ǫ , we treat them as being interference-free. In this case, we have SNR k ≥ (cid:16) RsK − (cid:17) | C kk | − , if k ∈ D ˆ ǫ . (66)Combining (64) and (66), the minimum sum power required for an equal rate system with sum spectral efficiency R s is lower bounded by SNR s ≥ RsK (cid:16) (1 − ǫ ) 2 RsK + ǫ (cid:17) − ǫ X j ∈ D c ˆ ǫ | C jj | − + (cid:16) RsK − (cid:17) X k ∈ D ˆ ǫ | C kk | − . (67)Using (39) on (67) we get E b N = P (cid:16) | C jj | − (cid:17) K log 2 (68) ∆ S = (2 − ǫ )(4 − ǫ ) (1 − θ ) + (2 − ǫ ) θ (69)where θ , P k ∈ D ˆ ǫ | C kk | − P Kj =1 | C jj | − = K P k ∈ D ˆ ǫ | C kk | − K P Kj =1 | C jj | − . Notice that the outer bound converges to the correct E b N (cid:12)(cid:12)(cid:12) min , and (69) can therefore be used as an outer bound onthe slope.Now, we want to show that ∀ ǫ > , θ can be made arbitrarily small. Define random variable H j, ˆ ǫ as H j, ˆ ǫ = | C jj | − j ∈ D ˆ ǫ j / ∈ D ˆ ǫ . Given the fact that H j, ˆ ǫ and H i, ˆ ǫ , i = j are independent, and P k ∈ D ˆ ǫ | C kk | − = P Kj =1 H j, ˆ ǫ , we can apply the lawof large number to θ , which gives P lim K →∞ θ = E ( H j, ˆ ǫ ) E (cid:16) | C jj | − (cid:17) = 1 . (70) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 23
Combining Lemma 21 and Lemma 22, we have lim ˆ ǫ ↓ E ( H j, ˆ ǫ ) = 0 . (71)This proves (59) explicitly as follows. For any δ > we can choose ǫ, θ > sufficiently small to make (69) lessthan + δ . We can choose ˆ ǫ > sufficiently small to make E ( H j, ˆ ǫ ) E ( | C jj | − ) smaller than θ . Finally we can choose K large enough to make P k ∈ D ˆ ǫ | C kk | − P Kj =1 | C jj | − smaller than θ with high probability and Pr ( B ǫ, ˆ ǫ = ∅ ) close to 1.
2) The Equal Power Constraint:
We now consider the equal power constraint. Assume that the number of users K is an even integer, K = 2 M . For ∀ ǫ > , we define the event A ǫ , { K users can form M disjoint weak (1 − ǫ ) − pairs } , and denote the indices of users belonging to the same weak (1 − ǫ ) -pairs as { m , m } .Let the channel coefficients C ij , i, j = 1 , . . . , K be random variables with a distribution that could depend on K . We consider the following property of this sequence of distributions Property 1. ∀ ǫ : Pr ( A ǫ ) → as K → ∞ . Proposition 24.
If the channel gains C ij are i.i.d (independent of K ) with continuous distribution, Property 1 issatisfied.Proof: Please see Appendix F.
Theorem 25.
If property 1 is satisfied and the direct channel gains C jj are i.i.d with finite 4th order moments,then under the equal power constraint ∀ δ > K →∞ Pr ∆ S ≤ E [ | C jj | ]) E [ | C jj | ] + 1 + δ = 1 . (72) Proof:
For the equal power constraint where
SNR j = SNR s K , if property 1 is satisfied, then for K = 2 M users, M disjoint weak (1 − ǫ ) -pairs { m , m } , m = 1 , · · · , M can be formed with high probability, and we will assumethis is the case. Applying Kramer’s bound Theorem 4 on each pair, we have R m + R m ≤ min log (cid:18) | C m m | SNR s K (cid:19) + log | C m m | s / K | C m m | s K ! , log (cid:18) | C m m | SNR s K (cid:19) + log | C m m | s / K | C m m | s K !! (73) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 24 in nats / s . For each weak (1 − ǫ ) -pair, (73) gives d ( R m + R m ) dP s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 = | C m m | + | C m m | K − d ( R m + R m ) d SNR s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 ≥ | C m m | + | C m m | + 2 min n | C m m | | C m m | , | C m m | | C m m | o K = ≥ | C m m | + | C m m | + 2 min n (1 − ǫ ) | C m m | | C m m | , (1 − ǫ ) | C m m | | C m m | o K ≥ | C m m | + | C m m | + 2 (1 − ǫ ) | C m m | | C m m | K since the M pairs are disjoint and using the linearity of derivatives, we have dR s d SNR s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 = M X m =1 d ( R m + R m ) d SNR s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 = P Kj =1 | C jj | K − d R s d SNR s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 = M X m =1 − d ( R m + R m ) d SNR s (cid:12)(cid:12)(cid:12)(cid:12) P s =0 ! ≥ P Mm =1 (cid:16) | C m m | + | C m m | + 2 (1 − ǫ ) | C m m | | C m m | (cid:17) K therefore E b N = K log e P Kj =1 | C jj | S ≤ (cid:16)P Kj =1 | C jj | (cid:17) P Mm =1 (cid:16) | C m m | + | C m m | + 2 (1 − ǫ ) | C m m | | C m m | (cid:17) = 2 K (cid:16) K P Kj =1 | C jj | (cid:17) K P Kj =1 | C jj | + (1 − ǫ ) M P Mm =1 | C m m | | C m m | . (74)Now K K X j =1 | C jj | P → E h | C jj | i K K X j =1 | C jj | P → E h | C jj | i M M X m =1 | C m m | | C m m | P → E h | C jj | i as K → ∞ , where P → stands for convergence in probability since all random variables are positive and the moments June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 25 are assumed to exist. Using standard rules for convergence of transformation, we then obtain ∀ ǫ > K →∞ Pr ∆ S ≤ − ǫ ) ( E [ | C jj | ]) E [ | C jj | ] + 1 = 1 from (74). Equivalently, ∀ ǫ > K →∞ Pr ∆ S ≤ E [ | C jj | ]) E [ | C jj | ] + 1 + δ ǫ = 1 where δ ǫ = ǫ ( E [ | C jj | ]) E [ | C jj | ](1 − ǫ ) (cid:18) ( E [ | C jj | ]) E [ | C jj | ] (cid:19) + (2 − ǫ ) ( E [ | C jj | ]) E [ | C jj | ] + 1 Notice that ∀ δ , ∃ ǫ > such that δ ǫ < δ . Therefore, we obtain (72).It is perhaps illustrative to write (72) in terms of central moments, E [ | C jj | ]) E [ | C jj | ] + 1 = 1 µ − σ µ + σ µ +6 µ σ +4 γ µσ + γ σ + 1 (75) µ = E [ X ] σ = var[ X ] γ = E [( X − µ ) ] σ γ = E [( X − µ ) ] σ . It can be seen that for σ ≪ µ , i.e., a nearly constant distribution, (75) is close to .We will discuss some implications of these theorems. For the equal rate constraint, (59) essentially states that thewideband slope is bounded by of that of no interference for large K . Since this is also achievable by Theorem15, this is indeed the wideband slope, and delay-based interference alignment is optimum. The bound for the equalpower constraint is slightly weaker, but is still close to for some distributions.Theorem 23 and 25 have been proven under an i.i.d. assumption on all channel coefficients. This can seemrestrictive and not that realistic in a line of sight model. However, the i.i.d. assumption is not essential. In Theorem23 it is used to prove that every user has at least one other user with which it forms an (1 − ǫ ) -pair with highprobability. This might be true under many other model assumptions. It is also used to invoke the law of largenumbers, which has a wide range of generalizations. In Theorem 25 the i.i.d. assumption is used to prove that usersform disjoint weak (1 − ǫ ) -pairs, and again for invoking the law of large numbers.What can be concluded is that for small special examples it is possible to find a better wideband slope byoptimizing a combination of interference alignment, interference decoding, and treating interference as noise.However, it probably does not pay off to try to find a general algorithm for optimizing wideband slope: comparingthe achievable sum slope given by Theorem 15 and the upper bounds provided by Theorem 23 and 25, we can June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 26 see that as the number of users K grows large, the gap between the upper bounds and the inner bounds achievedby the interference alignment scheme defined by Definition 9 becomes arbitrarily small. Furthermore, finding suchschemes are hard based on our experimentation.Another interesting observation is that the outer bounds do not depend on delay, only on the channel gains. Thus,the outer bound depends on the macroscopic location of transmitters and receivers (e.g., if gain is proportional to d αji for some α > ), while the inner bounds depend on the microscopic location (i.e., fractional delay differences). Thisalso means that the outer bounds apply to general scalar interference channels, not only LOS channels. However,for non-LOS channels better outer bounds can be proven, which is the subject of a later paper (for initial results,see [29]) V. C ONCLUSIONS
In this paper we have shown that by using interference alignment with delay differences, a wideband slope ofhalf of the interference-free case is achievable. We have also shown that, mostly, it is the best achievable. What itmeans is that near single-user performance can be achieved in the interference channel in the low-
SNR regime. Onesurprising conclusion is that orthogonalizing interference is (near) optimum in the low-power regime. It is not toosurprising that this is optimum in the high-SNR regime [4], since that regime is interference limited. But since thelow-power regime is also noise-limited, one could have expected that orthogonalizing interference is sub-optimum.That is indeed the case for a 2-user channel. But for a K -user channel, orthogonalizing is near optimum, as shownby Theorem 25.A number of questions remain open. What if the bandwidth remains fixed, but the transmission rate approacheszero (e.g., in a sensor network)? This case is more complicated, and will be covered in a later paper. How canthe delay based interference alignment be implemented in practical systems? As we have seen in section IV-A2,the achievable spectral efficiency is very dependent on choosing the right symbol rate, so this touches on issues ofchannel knowledge and estimation, and how to optimize symbol rate in a given spectral efficiency region, as wellas up to what spectral efficiencies the wideband slope provides a good approximation.R EFERENCES[1] V. Sreekanth Annapureddy and V. Veeravalli, “Sum capacity of the gaussian interference channel in the low interference regime,” in
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June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 28 A PPENDIX AP ROOF OF T HEOREM
Lemma 26.
Let X n = { X , X , · · · , X n } be a sequence of random variables satisfying power constraint n P ni =1 cov ( X i ) ≤ SNR . Let X nG = { X G , X G , · · · , X nG } be a sequence of i.i.d. Gaussian random variable, X G ∼ N (0 , SNR) . Let Z n and Z n be two sequence of i.i.d. random variables with distributions Z ∼ N (cid:0) , σ (cid:1) and Z ∼ N (cid:0) , σ (cid:1) .Then we have the following inequality h ( X n + Z n ) − h ( X n + Z n + Z n ) ≤ h ( X G + Z ) − h ( X G + Z + Z ) . Lemma 27.
Let X n = { X , X · · · , X n } and Y n = { Y , Y · · · , Y n } be two sequence of random variables. Let ˆ X G , ˇ X G and ˆ Y G , ˇ Y G be random variables satisfying cov ˆ X G ˆ Y G ≤ n n X i =1 cov X i Y i ≤ cov ˇ X G ˇ Y G . (76) Then h ( X n ) ≤ nh (cid:16) ˆ X G (cid:17) ≤ nh (cid:0) ˇ X G (cid:1) h ( Y n | X n ) ≤ nh (cid:16) ˆ Y G (cid:12)(cid:12)(cid:12) ˆ X G (cid:17) ≤ nh (cid:0) ˇ Y G (cid:12)(cid:12) ˇ X G (cid:1) . Proof:
This is a special case of [27, Lemma 2].For the delay-free case where ˜ X i [ n − n ji ] = X i [ n ] , this theorem is identical to the previous results in [30],[1], [2], and a later work [27]. Here we use similar technique as the proof of Theorem 6 in [27] to show that thisresults still hold for channel with non-zero delay.Assume that the channel coefficients and input power constraints satisfy (45). Provide side information S n and S n to receiver 1 and 2 respectively S n = C X n + W n S n = C X n + W n where W j are zero mean i.i.d Gaussian noise. And the joint distribution of W j and Z j is Z j W j ∼ N , ρ j ρ ∗ j σ j , June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 29 ρ j and σ j satisfy (40) to (43). From Fano’s inequality, we have n ( R + R ) ≤ I ( X n ; Y n ) + I ( X n ; Y n ) + o ( n ) ≤ I ( X n ; Y n , S n ) + I ( X n ; Y n , S n ) + o ( n ) ( a ) = h ( S n ) − h ( S n | X n ) + h ( Y n | S n ) − h ( Y n | S n , X n )+ h ( S n ) − h ( S n | X n ) + h ( Y n | S n ) − h ( Y n | S n , X n ) (77) ( b ) = h ( C X n + W ) − h ( W n ) − h (cid:16) C ˜ X n + Z n (cid:12)(cid:12)(cid:12) W n (cid:17) + h (cid:16) C X n + C ˜ X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) + h ( C X n + W n ) − h ( W n ) − h (cid:16) C ˜ X n + Z n (cid:12)(cid:12)(cid:12) W n (cid:17) + h (cid:16) C ˜ X n + C X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) + o ( n ) (78) ( c ) ≤ − nh ( W ) + h ( C X n + W n ) − h ( C X n + Z n | W n )+ h (cid:16) C X n + C ˜ X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) − nh ( W ) + h ( C X n + W ) − h ( C X n + Z n | W n )+ h (cid:16) C ˜ X n + C X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) + o ( n ) (79) ( d ) ≤ − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ h (cid:16) C X n + C ˜ X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ h (cid:16) C ˜ X n + C X n + Z n (cid:12)(cid:12)(cid:12) C X n + W n (cid:17) + o ( n ) (80) ( f ) ≤ − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ nh (cid:16) C X G + C ˜ X G + Z (cid:12)(cid:12)(cid:12) C X G + W (cid:17) − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ nh (cid:16) C ˜ X G + C X G + Z (cid:12)(cid:12)(cid:12) C X G + W (cid:17) + o ( n ) (81) ( g ) ≤ − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ nh ( C X G + C X G + Z | C X G + W ) − nh ( W ) + nh ( C X G + W ) − nh ( C X G + Z | W )+ nh ( C X G + C X G + Z | C X G + W ) + o ( n ) (82)where lim n →∞ o ( n ) /n = 0 , X jG with ’ G ’ subscription means that input at transmitter j is i.i.d. Gaussian, withdistribution X jG ∼ N (0 , SNR j ) . ( a ) is from chain rule. ( c ) holds because both X j and ˜ X j can be obtained from June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 30 sampling the same continuous-time baseband signal X j ( t ) at the Nyquist rate, while Z j and W j are sampled fromwhite Gaussian noise, so that h (cid:16) C ˜ X n + Z n (cid:12)(cid:12)(cid:12) W n (cid:17) = h ( C X n + Z n | W n ) + o ( n ) h (cid:16) C ˜ X n + Z n (cid:12)(cid:12)(cid:12) W n (cid:17) = h ( C X n + Z n | W n ) + o ( n ) because Given (40) and (41), cov ( W n ) ≤ cov ( Z n | W n ) and cov ( W n ) ≤ cov ( Z n | W n ) . Combining Lemma 26and [3, Lemma 3], we have h ( C X n + W n ) − h ( C X n + Z n | W n ) ≤ nh ( C X G + W ) − nh ( C X G + Z | W ) (83)and h ( C X n + W n ) − h ( C X n + Z n | W n ) ≤ nh ( C X G + W ) − nh ( C X G + Z | W ) . (84)Therefore (d) is true.(f) is from Lemma 27, where ˜ X jG are i.i.d. Gaussian random variable satisfying cov (cid:16) ˜ X jG (cid:17) = n Tr (cid:18) ˜ X nj (cid:16) ˜ X nj (cid:17) † (cid:19) .Denote SNR ′ j , n Tr (cid:18) ˜ X nj (cid:16) ˜ X nj (cid:17) † (cid:19) as the power of ˜ X nj . We could see that SNR ′ j ≤ SNR j , because time-shiftingof a signal sampled at the Nyquist rate does not change signal power. Therefore we have (g). This shows that thesum capacity of a channel with delay is outer bounded by that of a channel without delay.We could see that the inequality (g) is independent of the propagation delay. It is identical to the first inequalityin [27, (89)]. Therefore, from this point on, the proof is the same as in the delay-free case.A PPENDIX B June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 31 P ROOF OF L EMMA Proof:
We have
E [˜ x ∗ i [ n , δ ji ]˜ x i [ n , δ ji ]] = E " ∞ X m = −∞ x i [2 m ]sinc( n − m + δ ji ) ! ∗ · ∞ X m = −∞ x i [2 m ]sinc( n − m + δ ji ) ! = ∞ X m = −∞ E h | x i [2 m ] | i sinc( n − m + δ ji )sinc( n − m + δ ji )= ∞ X m =0 E h | x i [2 m ] | i sinc( n − m + δ ji )sinc( n − m + δ ji )+ ∞ X m =1 E h | x i [ − m ] | i sinc( n + 2 m + δ ji )sinc( n + 2 m + δ ji )= 2 P i ∞ X m =0 sinc( n − m + δ ji )sinc( n − m + δ ji )+ ∞ X m =1 sinc( n + 2 m + δ ji )sinc( n + 2 m + δ ji ) ! . (85)Define f m ( δ ji ) and g m ( δ ji ) as f m ( δ ji ) , sinc( n − m + δ ji )sinc( n − m + δ ji ) g m ( δ ji ) , sinc( n + 2 m + δ ji )sinc( n + 2 m + δ ji ) and their partial sums s f,M ( δ ji ) = P Mm =0 f m ( δ ji ) , s f ( δ ji ) , lim M →∞ s f,M ( δ ji ) ; s g,M ( δ ji ) = P Mm =0 g m ( δ ji ) , s g ( δ ji ) , lim M →∞ s g,M ( δ ji ) . Here | f m ( δ ji ) | = (cid:12)(cid:12)(cid:12)(cid:12) sin ( π ( n − m + δ ji )) sin ( π ( n − m + δ ji ))( n − m + δ ji )( n − m + δ ji ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ n − m + δ ji )( n − m + δ ji ) and | g m ( δ ji ) | ≤ n + 2 m + δ ji )( n + 2 m + δ ji ) . Let M f,k , n − m + δ ji )( n − m + δ ji ) , M g,k , n +2 m + δ ji )( n +2 m + δ ji ) . Because P ∞ m =1 1 k is convergent, P ∞ k =0 M f,k and P ∞ k =1 M g,k converge too. Due to Weierstrass’s test for uniform convergence[31], s f,M ( δ ji ) and s g,M ( δ ji ) converge uniformly. And using Theorem 7.11 in [31], we have lim δ ji ↓ lim M →∞ s f,M ( δ ji ) = lim M →∞ lim δ ji ↓ s f,M ( δ ji )lim δ ji ↓ lim M →∞ s g,M ( δ ji ) = lim M →∞ lim δ ji ↓ s g,M ( δ ji ) . June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 32
Thus, (85) becomes lim δ ji ↓ E [˜ x ∗ i [ n , δ ji ]˜ x i [ n , δ ji ]] = 2 P i (cid:18) lim M →∞ lim δ ji ↓ s f,M ( δ ji )+ lim M →∞ lim δ ji ↓ s g,M ( δ ji ) (cid:19) = P i if n = n = 2 k, for some integer k o.w. And given Theorem 7.12 in [31] and the continuity of sinc function, we can conclude that
E [˜ x ∗ i [ n , δ ji ]˜ x i [ n , δ ji ]] isa continuous function of δ ji . A PPENDIX CP ROOF OF L EMMA C i,ǫ be the event that user i does not form an (1 − ǫ ) -pair with any other user;. Then Pr( B ǫ, ˆ ǫ = ∅ ) = Pr K [ i =1 C i,ǫ ! ≤ K X i =1 Pr( C i,ǫ )= K Pr( C ,ǫ ) and Pr( C ,ǫ ) = Pr ∀ j > | C j | | C | or | C j | | C jj | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ ! = Pr ∀ j > | C j | | C | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ ! + Pr ∀ j > | C j | | C jj | / ∈ (1 − ǫ, ! − Pr ∀ j > | C j | | C | and | C j | | C jj | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ ! = (cid:0) − p K − j (cid:1) Pr ∀ j > | C j | | C | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ ! + p K − j where p j = Pr (cid:16) | C j | | C jj | / ∈ (1 − ǫ, (cid:17) ∈ (0 , . Notice that the events | C j | | C jj | / ∈ (1 − ǫ, are independent fordifferent j , and p K − j → . Thus, Pr( C ,ǫ ) → Pr (cid:16) ∀ j > | C j | | C | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ (cid:17) . As the | C jj | June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 33 are independent, Pr ∀ j > | C j | | C | / ∈ (1 − ǫ, and | C | / ∈ R ˆ ǫ ! = Z x/ ∈ R ˆ ǫ K Y j =2 − Z x (1 − ǫ ) x dF | C j | ( u ) ! dF | C | ( x )= Z x/ ∈ R ˆ ǫ (cid:16) − F | C ii | ( x ) + F | C ii | ((1 − ǫ ) x ) (cid:17) K − dF | C ii | ( x ) ≤ (1 − ˆ ǫ ) K − Z x/ ∈ R ˆ ǫ dF | C ii | ( x )= (1 − µ F ( R ˆ ǫ )) (1 − ˆ ǫ ) K − . Thus, P σ ≤ K (1 − µ F ( R ˆ ǫ )) (1 − ˆ ǫ ) K − , and lim K →∞ Pr( B ǫ, ˆ ǫ = ∅ ) = 0 .A PPENDIX DP ROOF OF L EMMA ∀ x ∈ R ˆ ǫ : F | C ii | ( x ) − F | C ii | ((1 − ǫ ) x ) ≤ ˆ ǫ , and given the fact F | C ii | ( x ) − F | C ii | ((1 − ǫ ) x ) < ˆ ǫ n +1 < ˆ ǫ n , it clearly follows that R ˆ ǫ n +1 ⊆ R ˆ ǫ n . Let I ˆ ǫ n ( x ) be the indicator function of R ˆ ǫ n . Using Lebesguedominated convergence we have lim n →∞ Z ∞ I ˆ ǫ n ( x ) dF | C ii | ( x )= Z ∞ lim n →∞ I ˆ ǫ n ( x ) dF | C ii | ( x )= Z ∞ I ( x ) dF | C ii | ( x ) where I ( x ) is the indicator function of the set R = { x ∈ R : F | C ii | ( x ) − F | C ii | ((1 − ǫ ) x ) = 0 } . Since wehave assumed that E h | C ii | − i < ∞ , also Pr( | C ii | = 0) = 0 and clearly µ F ( R ) = 0 , and µ ( R ˆ ǫ n ) → µ ( R ) .A PPENDIX EP ROOF OF L EMMA X be a random variable on the probability space (Ω , F , P ) [ ? ]. Given the definition of X i ,we can conclude that lim i →∞ X i = 0 w.p. 1.Namely, if there is a set B ∈ F with P ( B ) > where lim i →∞ X i = 0 then P ( B ) ≤ µ X ( T ∞ i =1 G i ) whichcontradicts lim i →∞ µ X ( G i ) = 0 .Now X i ≤ X , and therefore by Lebesgue dominated convergence lim i →∞ E [ X i ] = E [ lim i →∞ X i ] = 0 . June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 34 A PPENDIX FP ROOF OF P ROPERTY G K , with K = 2 M vertices u , u , · · · , u n . Vertices u i and u j areconnected by edge E ij if they form a weak (1 − ǫ ) -pair, i.e., (1 − ǫ ) ≤ | C ij | | C jj | ≤ or (1 − ǫ ) ≤ | C ji | | C ii | ≤ . Dividevertices into two disjoint classes V = { u , u , · · · , u M } and V = { u M +1 , u M +2 , · · · , u M } . Now define event ˆ A ǫ = { there exists a perfect matching in the bipartite graph G M, M } . As ˆ A ǫ ⊆ A ǫ , P ( A ǫ ) ≥ P (cid:16) ˆ A ǫ (cid:17) . Thus, if we can show that P (cid:16) ˆ A ǫ (cid:17) = 1 − o (1) as K → ∞ then Property 1 holds.For any bipartite graph, a perfect matching exists if Hall’s condition is satisfied. Theorem 28 (Hall 1935) . Given a bipartite graph G M, M with disjoint vertices class V and V , V S V = V , | V i | = M , whose set of edges is E ( G M, M ) , a perfect matching exists if and only if for every S ⊆ V i , i = 1 or , | N ( S ) | ≥ | S | , where N ( S ) = { y : xy ∈ E ( G M, M ) f or some x ∈ S } . Any bipartite graph that does not have a perfect matching has following properties
Lemma 29.
Suppose G M, M has no isolated vertices and it does not have a perfect matching. Then Hall’s conditionmust be violated by some set A ⊂ V i , i = 1 or . And such set with minimal cardinality satisfies the followingnecessary conditions(i) | N ( A ) | = | A | − ;(ii) ≤ | A | ≤ (cid:6) M (cid:7) (iii) the subgraph of G spanned by A S N ( A ) is connected, and it has at least a − edges;(iv) every vertex in N ( A ) is adjacent to at least two vertices in A ;(v) any subsets of N ( A ) can find a perfect matching in | A | ;Proof: (i), (ii), (iii), and (iv) are proved by Lemma 7.12 in [32], and p.82 of [33]. And (iv) is true because ifthere exists a subset B of N ( A ) that can not find a perfect match, we could just let B be ˆ A , and its neighborsin A be N (cid:16) ˆ A (cid:17) . Then ˆ A violates Hall’s condition, while (cid:12)(cid:12)(cid:12) ˆ A (cid:12)(cid:12)(cid:12) < a . This contradicts the assumption that A is theminimal set violating Hall’s condition.Define the event F a : there is a set A ⊂ V i , i = 1 or , | A | = a . satisfying (i), (ii) and (iii) in Lemma 29. [32]shows that for a graph with no isolated vertex, P ( A ǫ ) = 1 − o (1) is equivalent to P (cid:18)S ⌈ M ⌉ a =2 F a (cid:19) = o (1) . Define F as the event that there exists at least one isolated vertex in G M,M . In our case, we want to show that P ⌈ M ⌉ [ a =2 F a + P ( F ) = o (1) . June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 35
Using the union bound, we have P ( F ) ≤ M X i =1 P ( u i isolated ) ≤ M · P ( u isolated ) ( a ) ≤ M · (1 − p j ) M ( b ) = o (1) where p j , P (cid:16) (1 − ǫ ) ≤ | C j | | C jj | ≤ (cid:17) , j = M +1 , · · · , M . We also define p , P (cid:16) (1 − ǫ ) ≤ | C ij | | C jj | ≤ , or (1 − ǫ ) ≤ | C ji | | C ii | ≤ (cid:17) for later use. (a) holds because the event | C j | | C jj | / ∈ (1 − ǫ, is independent of j . And it is a necessary conditionfor V to be is isolated.Now, let us look into F a for ≤ a ≤ (cid:6) M (cid:7) . Let A ⊂ V , A ⊂ V , and | A | = | A | + 1 = a . Denote P ( A a ) asthe probability that the subgraph of G M,M spanned by A S A satisfies (i), (ii), and (iii) in Lemma 29. We have P ⌈ M ⌉ [ a =2 F a ( d ) ≤ ⌈ M ⌉ X a =2 P ( F a ) ( e ) ≤ ⌈ M ⌉ X a =2 Ma Ma − P ( A a ) . (86) ( d ) is from the union bound; ( e ) is from the union bound, and from that fact that there are Ma choicesfor A with | A | = a , and Ma − more choices for N ( A ) . In [32], [33], the case where edge probabilitiesare i.i.d, whose value is p , is considered. In [32], P ( A a ) is bounded using condition (i), (ii) and (iii), whichgives P ( A a ) ≤ a ( a − a − p a − p a ( n − a +1) . The term p a ( n − a +1) is the probability that the vertices in A donot connect to vertices in V − A . And in [33], condition (iv) instead of (iii) are used, which gives P ( A a ) ≤ a a − p a − p a ( n − a +1) . However, in our case, any two edges having adjacent vertices are dependent. So weuse condition (v). Since for N ( A ) , a perfect match exists, then the subgraph spanned by A S N ( A ) has a − edges that are not adjacent with each other. Thus, P ( A a ) can be bounded by P ( A a ) ≤ P r (condition (i) , (ii) and (iv) are satisfied , (87) vertices in A do not connect to vertices in V − A ) (88) p a − a Y k =2 k p A ¯ A (89) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 36 where p A ¯ A = P | C ij | | C jj | / ∈ [(1 − ǫ ) , , for all u i ∈ A and u j ∈ ( V − A ) ! ≤ Y u j ∈ ( V − A ) P | C ij | | C jj | / ∈ [(1 − ǫ ) , , for all u i ∈ A ! = P | C ji | | C jj | / ∈ [(1 − ǫ ) , , i = 1 , · · · , a j = M + a !! M − a +1 (90)notice that the event | C ij | | C jj | / ∈ [(1 − ǫ ) , , for all u i ∈ A and u j ∈ ( V − A ) is an necessary Substitute | C j | by x j , and | C | by x , denote their CDF by F x ( x ) , and their joint CDF F x ( x ) . Notice that | C ij | are i.i.d.distributed. Then P | C ji | | C jj | / ∈ [(1 − ǫ ) , , i = 1 , · · · , a j = M + a ! = Z A x dF x ( x )= Z ∞ f x M + a ( x M + a ) a Y i =1 Z xixM + a / ∈ [(1 − ǫ ) , f x i ( x i ) dx i ! dx M + a = Z ∞ f x ( x ) Z x xM + a / ∈ [(1 − ǫ ) , f x M +1 ( x M +1 ) dx M +1 ! a dx M + a ( f ) ≤ (cid:18)Z ∞ f x ( x ) dx (cid:19) / (cid:18)Z ∞ g x ( x ) dx (cid:19) a / (91)where g ( x ) , R x xM + a / ∈ [(1 − ǫ ) , f x M +1 ( x M +1 ) dx M +1 . (f) is from Cauchy-Schwartz inequality. Denote q , (cid:18)Z ∞ f x ( x ) dx (cid:19) M − a +1) (cid:18)Z ∞ g x ( x ) dx (cid:19) / q < , and lim M →∞ q = (cid:0)R ∞ g x ( x ) dx (cid:1) / . Notice that this limit value do not depend on the value of M .Now combining (90) and (91), we have P ( A a ) ≤ p a − a Y k =2 k q a ( M − a +1)1 . (92) June 20, 2018 DRAFTEEE TRANSACTIONS ON INFORMATION THEORY 37
Given (86) and (92), P ⌈ M ⌉ [ a =2 F a ≤ ⌈ M ⌉ X a =2 Ma Ma − p a − a Y k =2 k q a ( M − a +1)1 ≤ ⌈ M ⌉ X a =2 (cid:18) eMa (cid:19) a (cid:18) eMa − (cid:19) a − a a − p a − q a ( M − a +1)1 ≤ q ⌈ M ⌉ X a =2 e M ( a − ap q M ! a − ≤ q ⌈ M ⌉ X a =2 (cid:16) e p M q M (cid:17) a − ≤ q M (cid:16) e p M q M (cid:17) = o (1) . This means we can find a perfect matching with high probability, i.e., − o (1) ..