Ergodic Fading Z-Interference Channels Without State Information at Transmitters

IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 5, MAY 2011 2627

Ergodic Fading Z-Interference Channels WithoutState Information at TransmittersYan Zhu, Student Member, IEEE, and Dongning Guo, Member, IEEE

Abstract—This paper studies the capacity region of a two-userergodic interference channel with fading, where only one user issubject to interference from the other user, and the channel stateinformation (CSI) is only available at the receivers. A layered era-sure model with arbitrary fading statistics is studied first, whosecapacity region is completely determined as a polygon. Each domi-nant rate pair can be regarded as the outcome of a tradeoff betweenthe rate of the interference-free user and the rate loss its interfer-ence causes the other user. Using insights from the layered erasuremodel, inner and outer bounds of the capacity region are providedfor fading Gaussian Z-interference channels. The gap between theinner and outer bounds is no more than 12.8 bits per channel useper user, regardless of the signal-to-noise ratio (SNR) and fadingstatistics.

Index Terms—Capacity region, channel state information (CSI),deterministic model, fading, incremental channel, interferencechannel, layered erasure model.

I. INTRODUCTION

T HE capacity region of Gaussian interference channels(ICs), comprised of one or more cross links, remains

open despite of decades of research. Etkin, Tse, and Wang[1] recently made an important progress by characterizingthe capacity region of the two-user Gaussian IC to within asmall constant number of bits (see also [2]). Since then, severalnew results have been obtained, including the sum capacityin special interference regimes (e.g., [3] and [4]), the degreesof freedom of Gaussian ICs [5]–[8] and MIMO Gaussian ICs[9]–[15]. The capacity of fading ICs has also been studied,e.g., in [16] and [17], where the focus has been on scenarioswhere channel state information (CSI) is fully available at thetransmitters, as well as at the receivers.

This work studies fading interference channels where the in-stantaneous channel state is available at the receivers but notat the transmitters. This is the case in many practical systemswhere the channel state can only be measured by the receiverswhich cannot inform the transmitters of the state accurately in atimely manner through a feedback link. Specifically, this paper

Manuscript received November 05, 2009; revised September 27, 2010; ac-cepted December 30, 2010. Date of current version April 20, 2011. This workwas supported by the NSF by Grant CCF-0644344 and by DARPA by GrantW911NF-07-1-0028. This work was presented in part at the IEEE InformationTheory Workshop, Cairo, Egypt, January 2010, and at the IEEE InternationalSymposium on Information Theory, Austin, TX, June 2010.

The authors are with the Department of Electrical Engineering and ComputerScience, Northwestern University, Evanston, IL 60208 USA.

Communicated by A. S. Avestimehr, Associate Editor for the special issue on"Interference Networks".

Color versions of one or more of the figures in this paper are available onlineat http://ieeexplore.ieee.org.

Digital Object Identifier 10.1109/TIT.2011.2120090

assumes independent (fast) fading over time, where the fadingstatistics are known to the transmitters. The result could alsoapply to some situations where the transmitters are given an es-timate of the channel state over a coding block, but cannot trackits instantaneous variations. The fading statistics are differentthan in the work of Raja, Prabhakaran, and Viswanath [18] oncompound interference channels, where the channel state (froma finite set) is unknown to the transmitters but remains staticover the course of a codeword. The key issue therein is to find acoding scheme which is essentially compatible with all channelstates. The results of [18] and [19] are applicable to (slow) blockfading interference channels with no CSI at transmitters. Thecurrent paper, however, investigates the ergodic case where thecode is designed to perform over a typical realization of thetime-varying fading process.

To make progress, this paper considers interference chan-nels with two single-antenna users where the interference isone-sided, i.e., only one user is subject to interference from theother user. The capacity region of such a channel, also known as

-interference channel (Z-IC), is open in general even withoutfading. Such an interference model is suitable if one of the re-ceivers is within the range of both transmitters, while the otherreceiver is out of the range of the interfering transmitter. Onesimple scenario for this case is a line network of four nodeswhere the two transmitters and their corresponding re-ceivers are interconnected by the links described usingarrows as .

Like a number of recent works (e.g., [7], [20]–[23]), thispaper makes use of the deterministic model approach to obtaininsights for general interference models. Despite its simplicity,the deterministic model captures two most important phe-nomena of wireless communication, namely, the broadcastnature of wireless transmission and the superposition of mul-tiple signals at the receiver. Furthermore, a fading wirelesschannel can be simplified to a time-varying version of the de-terministic model, called the layered erasure model, where thestate of a link corresponds to the number of most significant bitsnot obliterated by noise. The capacity region of such a modelfor two-user fading broadcast channel has been established byTse and Yates, who then adapt the achievable scheme to obtaina constant-gap characterization of the capacity region of thecorresponding fading Gaussian broadcast channel [24].

The main contributions of this paper include:• Determining the exact capacity region of the layered era-

sure Z-IC with arbitrary fading statistics;• Deriving a new outer bound for the capacity region of the

fading Gaussian Z-IC using insights from the converse re-sult for the layered erasure model;

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2628 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 57, NO. 5, MAY 2011

• Developing a specific coding scheme which achieves a rateregion to within a gap of 12.8 bits per channel use per userfrom the outer bound, regardless of the fading statistics,signal-to-noise ratios (SNR) and interference-to-noise ra-tios.

Also related to this work are [25]–[27]. The capacity resultswith various assumptions of strong or weak interference [25]are special cases of the general result here. The outer bound de-veloped in [25] using the techniques developed in [28] is looserthan the bounds developed in this paper. Reference [26] studiesIC with Rayleigh fast fading. However, the approaches in [26],as well as in a more recent work [27], do not lead to as sharp acharacterization on fading Z-IC as is obtained in this paper.

The remainder of this paper is organized as follows. The Z-ICmodels are described in Section II. The main results are summa-rized in Section III. In order to make the development more ac-cessible, the capacity region for single-layer erasure channel isestablished first in Section IV, and the development for the gen-eral case is relegated to Section V. The result for the Gaussianfading model is found in Section VI. Conclusion is drawn inSection VII.

II. MODELS AND NOTATION

Consider an interference channel with two pairs of transmit-ters and receivers, where the message of transmitter 1 is in-tended to receiver 1, and the message of transmitter 2 is intendedto receiver 2. It is assumed that the interference is one-sidedfrom transmitter 2 to receiver 1, so that the direct link of user 2is free of interference.

A. The Gaussian Model

Let , , , and denote the transmitted and received sig-nals of user 1 and user 2, respectively. The input-output rela-tionship over time interval is described as

(1a)

(1b)

where and denote the channel gain andphase of the two direct links, respectively, denotesthe gain and phase of the cross link from transmitter 2 to re-ceiver 1, and noise and consist of independent circu-larly symmetric complex Gaussian (CSCG) variables with unitvariance. The channel is depicted by Fig. 1(a). Let the averagepower of each transmitted codeword be constrained by 1. Thestate can be regarded as the instantaneous SNR of the cor-responding link. Since the CSI is not available at transmitters,the input signals are independent of the channel state. It is as-sumed that each link’s fading process is indepen-dent and identically distributed (i.i.d.) over time ,and that the amplitude process and phase process of each fadingprocess are also mutually independent. Thus all variables,

, with and , are mutually in-dependent. Furthermore, it is fair to assume the phasesto be uniformly distributed on . We often drop the timeindex when referring to a sample from an i.i.d. process, e.g.,is identically distributed as .

Fig. 1. Z-interference channels with fading. (a) A Gaussian model. (b) A lay-ered erasure model.

B. The Layered Erasure Model

In the spirit of the deterministic models introduced in [29],the layered erasure channel model for the Z-IC is depicted byFig. 1(b) and described as follows. Let the signals emitted bytransmitters 1 and 2 at the th time interval be denoted byand respectively, which take values in , the -vectorspace of binary Galois field . Let denote a matrixwith for all and all other ele-ments being 0, so thatrepresents a single shift, and denotes a downward shiftof the elements of the vector with its least significant bitsdropped out and zeros padded from the top of the vector. Thereceived signals at time interval are then expressed as

(2a)

(2b)

where , and take values inand represent the fading states of the three physical links, re-spectively. Let , and be mutually inde-pendent, and let each of the three processes be i.i.d. over time(so that the channel is memoryless).

We introduce the following notation for the layered era-sure model for convenience. For , let denote itsth element and denote . For a vector

process , we use to denote the sequence, and use to denote the sequence

. The indexes outside the parentheses al-ways refer to time. If or , the index can be omitted,e.g., and . To ease notation,we also use to denote with stricken out, i.e.,

ZHU AND GUO: ERGODIC FADING Z-INTERFERENCE CHANNELS 2629

Fig. 2. Single-layer Erasure Model. � , � , and � are the erasure probabilitiesof corresponding links.

. Binary addition of vectors ofdifferent lengths is aligned at the least significant bits; e.g., if

, then

Since the channel described by (2) is memoryless, we often sup-press the time index to describe the model as

(3a)

(3b)

Furthermore, the distribution of an i.i.d. sequence of randomvariables is often represented by the variable with the time indexsuppressed, e.g., is identically distributed as .

In both the layered erasure model and the Gaussian model, itis assumed that the fading state of each link is known only tothe receiver of that link.

III. MAIN RESULTS AND SPECIAL CASES

Throughout the paper, all information units are bits and alllogarithms are of base 2.

Theorem 1: The capacity region of the fading Gaussian Z-in-terference channel (1) is contained in the following region:

(4)where

(5)

and

(6)

Furthermore, regardless of the fading statistics, the outer boundcan be achieved to within a gap of at most 12.8 bits per channeluse per user.

It is instructive to investigate the following special cases ofTheorem 1.

1) Consider the case of stochastically strong interference,where for every . Thisimplies that for every andevery . Therefore, (4) reduces to

Since dominates stochastically, the region is infact within the capacity region of the multiaccess channel(MAC) formed by the two transmitters and receiver 1.Hence, this region can be exactly achieved by letting bothusers use i.i.d. Gaussian codebooks and allowing receiver1 to decode the message of user 2. This result has beenessentially established in [30] as a special case.

2) Consider the case of stochastically weak interference,where for every . It is notdifficult to verify that for all . Letting

in (4), we have the sum-rate constraint

(7)

Although the achievability of the outer bound given by (4)is not clear in general, the bound (7) on the sum rate can beachieved as follows: Let both users use Gaussian signalingand let user 2 transmit at the full rate ; re-ceiver 1 decodes its own message by treating interferenceas noise.

3) Consider the case of static channel, where ,, and are deterministic numbers known to all

transmitters and receivers. The channel is a special case ofeither the stochastically strong IC (if ) or stochas-tically weak IC (if ). In fact, the outer bound (4)becomes

(8)

which is the outer bound for the Z-IC without fading es-tablished in [31]. This bound can be achieved to within onebit [1]. Here (4) reduces to (8), because

. Therefore, the third constraint can be simpli-fied as

for all . It is easy to see that the boundaries ofthese constraints all pass through a point on the right sideof line . Therefore, the tightest constraintis the one with .


Fig. 3. Capacity region for single-layer erasure channel with different cases drawn by solid lines. (a) Very strong interference � � � � . (b) Strong but not verystrong interference � � � � � � . (c) Weak interference � � � .

Theorem 2: The capacity region of the layered erasure Z-in-terference channel (2) is

(9)where, for every

(10)

and

(11)

The parameter represents the probability that the th signif-icant bit of the signal of user 2 is not erased by the cross channeland interferes with a certain bit level of user 1. The parameter

represents the probability that the th significant bit of user 2is received by user 2, but is not received by user 1 due to erasureor interferes with a certain bit level of user 1.

The following special cases of Theorem 2 are similar to thoseof Theorem 1, except that the exact capacity region is alwaysknown.

1) The case of stochastically strong interference, wherefor every . The

capacity region (9) reduces to

which is established in [30].2) The case of stochastically weak interference, where

for every . Thesum capacity is

3) The case of pure deterministic model, where ,, and . The capacity region (9) becomes

which is implied by [28].The rate region (4) and the capacity region (9) are each sur-

rounded by a collection of simple affine constraints. This resultsfrom the tradeoff between the rate of user 2 and the loss in therate of user 1 due to interference from user 2, depending on thesignaling of the users.

In Theorem 2, setting in the second constraint in (9)yields the single-user bound for the rate ofuser 1. This does not apply to Theorem 1 because of the extraconstant 1 in the third constraint of (4).

We first prove Theorem 2 in the special case of a single layermodel in Section IV to illustrate some of the key ideas and tech-niques, so that the proof for the general layered erasure modelin Section V becomes easier to follow. Insights developed fromtreating the layered erasure model are then adapted to proveTheorem 1 for the fading Gaussian IC in Section VI.

IV. PROOF FOR THE SINGLE-LAYER ERASURE MODEL

Throughout this section, a single layer is assumed, i.e.,. Denote the erasure probability of the link labeled with

as , and let for notational convenience. Themodel is depicted in Fig. 2. Evidently, is the probability thatthe input symbol actually traverses the corresponding link. Theregion defined in Theorem 2 with is illustrated in Fig. 3for all possible configurations of the parameters. The region isprecisely described in the following proposition.

Proposition 1: Let . In case of strong interference, i.e.,, the capacity region of the single-layer IC depicted by

Fig. 2 is the pentagon with boundary constraints ,, and

(12)


which reduces to a rectangle when . In case of weakinterference, i.e., , the capacity region is the pentagonwith boundary constraints , , and

(13)

where and .

Lemma 1: The region defined in (9) is identical to the pent-agon described in Proposition 1.

Proof of Lemma 1: By (10) and (11)

and

If , then , so that the second constraint in (9)reduces to for all . Forevery , the upper bound passes the point , hence thetightest of such bounds is the one with , i.e., (12).

If, on the other hand, , then , sothat the second constraint in (9) becomes

for all . For every , the upper boundbecomes , the tightest of which is (13)achieved at . For every , the upperbound becomes . All of these boundsas well as (13) pass the point . Because ,the tightest of these bounds is still (13), which describes thedominant face of the region .

In the remainder of this section, the region described inProposition 1 is first shown to be achievable, and then amatching converse is established.

A. Proof of Achievability of Proposition 1

Since and are scalars and or 1 isknown to the receiver, we can replace by andby in model (3). In each subfigure of Fig. 3, we shadowthe capacity region of the following multiple access channel:

(14)

which is the pentagon region enclosed by the axes, the lines, , and . Note that if

an achievable rate pair for channel (2) falls into theMAC capacity region, then the message from transmitter 2 canbe decoded at receiver 1. With these in mind, we investigate theachievability for all two possible cases.

In case of strong interference ( ), is contained inthe MAC capacity region [as shown in Fig. 3(a) and (b)]. Let

denote the Bernoulli distribution which puts prob-ability masses of and at values 1 and 0, respectively. Anyrate pair in can be achieved by using inputsand letting receiver 1 decode messages from both transmitters.

In case of weak interference ( ), it suffices to showthat the two corner points and , which aremarked with star and square in Fig. 3(c), respectively, areachievable. Because the point is also a corner pointof the MAC channel capacity region, it can be achieved. Toachieve the second point, both users can use random codebooksgenerated according to distribution. Let thecode rate of user 2 be . If the fading state ,then ; for all other realizations of , isindependent of . Therefore, from receiver 1’s viewpoint, thisis equivalent to an erasure channel with erasure probability

. Thus the rate is achievable by user 1, whichshows that the corner point is also achievable in thiscase.

B. Proof of Converse of Proposition 1

Every achievable rate pair must satisfyand . Therefore, it suffices to show that (12) must besatisfied in the strong interference case, whereas (13) must besatisfied in the weak interference case.

Because the two decoders operate separately, the capacityregion of any IC depends only on the marginal distributionof the channel outputs conditioned on the inputs, but not onthe joint conditional distribution [32]. It is assumed in theremainder of this section that the fading states andare “aligned” such that forevery , whereas the state variables remain independentof . Hence, if the realization of the weaker onebetween and is equal to 1, then the realization ofthe stronger one must also be equal to 1. This does not changethe capacity region. Neither does the alignment change and

defined in Theorem 2.Consider first the strong interference case, . For no-

tational simplicity, let so thatdenotes all fading states

from time 1 to time . By Fano’s inequality, must satisfy

(15)

for some as , where (15) follows from thatis maximized by setting both and

to be i.i.d sequences. Also due to Fano’s in-equality, must satisfy

(16)

By assumption, , so that , thus. Comparing (15)

and (16) yields (12) as .


Next, consider the weak interference case, . Letbe an i.i.d. se-

quence independent of all channel inputs and channel states,and let . Fano’s inequality requires that

(17)

(18)

(19)

where (17) follows from data processing theorem and (18) is dueto the fact that is identically distributed as . Breakingdown the mutual information in (19) in another way, we obtain

(20)

Also by Fano’s inequality and by supplying user 2 with sideinformation

(21)

The upper bounds (20) and (21) can be understood as fol-lows: The rate pair can be achieved by letting user 1decode and cancel the signal of user 2 as shown in Section IV-B.By choosing the signaling , user 2 can improve its own rateby no greater than , where the in-terference causes user 1 to lose its rate in the amount of atleast . In the following, we precisely

quantify this tradeoff over all choices of the signal . To-ward this goal, we introduce a new Marton-style expansion inthe following, which reduces some conditional mutual informa-tion of vectors to that of scalar variables.

Lemma 2 (A Marton-Style Expansion of Mutual Information):Consider uses of a memoryless channel described by randomtransformation . Let and denote the inde-

pendent input and state sequences, respectively. Then for any

(22)

Lemma 2 is proved in Appendix A. To use Lemma 2, Let ,, and in (22) be , , , and, respectively. The assumption in Lemma 2 holds. Also, by

Lemma 2 and the fact that

for , we have

(23)

To bound (23), we need the following lemma.

Lemma 3: Assume alignment of and . Letbe any collection of random variables independent of

. Let be a randomvariable independent of . Then

(24)

(25)

In Lemma 3, is the probability that is seen throughchannel but not through and is the probabilitythat is seen through channel but not through .

Proof: There is no uncertainty about conditional onunless . Hence (24) is established

To see (25), write

Since and are aligned,and . Also,


by assumption of. Hence the proof of Lemma 3.

With Lemmas 2 and 3 established, we prove the converseresult next. Define

(26)

For every , we apply Lemma 3 to (23) withreplaced by , which is independent of , to obtain

(27)

By (23) and (27)

(28)

(29)

where (28) is due to the facts that conditioning reduces entropyand when , and (29) is by the chain rule. Werewrite (29) as

(30)Comparing (20), (21), and (30), we have

Sending yields (13).

V. PROOF FOR THE GENERAL LAYERED ERASURE MODEL

In this section, we prove Theorem 2 in full generality. We firstestablish the converse using similar techniques as developed inSection IV-C, after which the achievable scheme becomes intu-itive.

A. Proof of the Converse Part of Theorem 2

Recall that in Section IV-C, the converse result for the single-layer model is proved separately for the strong interference caseand weak interference case. With multiple layers, the Z-IC cannot be classified into those two cases only (e.g., in caseand takes values of 0 or 2, the channel is neither of weakor strong interference). Nonetheless, in layer , the signal isseen more often either through the direct link or the cross link.The converse proof is essentially by adapting the converse proofof the single-layer model to each of the layers.

Similarly as in Section IV, we can assume arbitrary joint dis-tribution of the fading coefficients and as long as themarginals remain the same. Throughout the proof of the con-verse of Theorem 2 (Section V-A), it is assumed that

is independent of , and and arealigned as follows. Let denote the cu-mulative distribution function of random variable . We define

and let

(31)

for , where are i.i.d. and uniformly dis-tributed on . Evidently, for , are i.i.d. andidentically distributed as . The alignment is that a larger re-alization of implies a larger realization of , and viceversa. For the special case where and only take valuesof 0 or 1 as in the single-layer model, we see that (31) is equiv-alent to , as is assumed inSection IV-C. It is important to note that , and the region

defined in Theorem 2 remain the same after the alignment be-cause they do not depend on the joint distribution of and .

The first constraint in (9) is trivial by the cut-set bound. Wefocus on proving the second constraint for . Let theelements of , i.e., , and , bei.i.d. random variables, and let be inde-pendent of all channel inputs and states. For notational conve-

nience, let .As in the single-layer case, replacing with does not

reduce the rate of user 1. By Fano’s inequality

(32)

(33)

where (32) is because is identically distributed as . ByFano’s inequality and providing side information to receiver2, we have

(34)


By property of memoryless channels, the first term on the right-hand side (RHS) of (34) can be upper bounded

(35)

By (34) and (35), we have

(36)By (33) and (36), which generalize and admit similar interpre-

tation as their respective single-layer counterparts (20) and (21),we have the following weighted bound for every :

(37)

By

for , and by applying Lemma 2 to the memory-less channel with as the input, as the state, and

as the output, we can further upper bound thetwo remaining entropy terms on the RHS of (37) as

(38)

The following generalization of Lemma 3 is needed.

Lemma 4: Suppose and are aligned according to (31).Let be a collection of random variables independent of

. Let be an arbitrary random vector in inde-pendent of . Then

(39)

(40)

where and are given by (10) and (11), respectively.An interpretation for Lemma 4, which is proved in

Appendix B, is as follows. Suppose that we can observe

through three channels: , , and .By (10), is the probability that layer of can be seen

in but not through the channel . Wheneverthis event happens, the amount of entropy isaccumulated (via the chain rule). Hence, (39) follows. Similarinterpretation can be obtained for (40) by noting that is theprobability that layer of can be seen in but not through

the channel under the assumption of alignmentbetween and .

Applying Lemma 4 for to both entropy terms in(38) with expressed as

(41)

we have

(42)

Therefore, substituting (42) into (37) and noting that as, we have established the converse part in Theorem 2.

B. The Achievability Scheme: A General Description

We first show that the boundary of region defined by (9)is piece-wise linear. In other words, although (9) consists oflinear constraints corresponding to every , only afinite number of them are dominant. Once all the corner pointson the boundary are identified, their achievability implies theachievability of the region by the time-sharing argument. InSection V-C, we revisit the achievability of the single-layermodel, and the difficulty of generalizing to the multiple-layercase is discussed. The formal proof of the achievability isrelegated to Section V-D.

Assume ; otherwise, the capacity region istrivial. The region bounded by the second constraint in (9) canbe viewed as where

(43)and . Let us order

as ,and let the corresponding permutation be referred to as so that

, . In addition, let .

Lemma 5: Except for the constraints, all other

are redundant, i.e.

(44)


Proof: For every , is a half plane to theleft side of a straight line. For , the boundary of

and the boundary of intersect at

(45)

which is denoted by from now on. This is easy to check byplugging (45) into the constraint (43) for andto get equalities. In addition, let be the intersection pointof the boundary of and the boundary of , whosecoordinate is also given by (45) by changing the subscript to. Define intervals for , and

. For every , it is not difficult to seethat for all . Furthermore, the boundaryof also passes point . Thus we see that the constraint

is redundant to and (or for ).Hence the proof of the proposition.

In general, the overall boundary of the second constraint in(9), henceforth referred to by , is piece-wisely linear, and itscorner points are . Point is always on the upperleft side of . In some special cases, the two points may coin-cide. Region is the region enclosed by , line , andthe two axes. Line can intersect with curve at var-ious positions. This point of intersection achieves the maximumsum rate, and shall be referred to as the max-point. To prove theachievability of the region, it suffices to show that the max-pointand all corner points on below it are achievable.

The achievability scheme is a special case of that of the well-known Han and Kobayashi (HK) scheme for general interfer-ence channels [33], with only user 2’s message split into privateand common messages. Specifically, we identify which layersof user 2 are used for sending the private message and whichlayers are used for sending the common message. In the proofof the converse result, the term is interpretedas a tradeoff of the rates of the two users. For given , inorder to maximize , layer should be used to encodethe common message of user 2 if , because there is nonegative impact on the weighted sum rate; if , layershould be used to encode private message of user 2.1 In partic-ular, to achieve corner point , we should use layers infor the private message and those in for the commonmessage, as long as the rate of user 2 is not saturated, i.e., isbelow the line .

C. The Achievability of the Single-Layer Case Revisited

Let us apply the HK scheme to the single-layer model (seethe region depicted in Fig. 4). The technique will then be gener-alized to the case of multiple layers. In the case of weak interfer-ence, and hence . For any , user2 sends only common message so that the rate pair

1This is in contrast to the layered erasure broadcast channel problem studiedin [24], where the role of the weighting parameter in a weighted sum-rate char-acterization of the capacity region is interpreted as a preference between the twousers.

Fig. 4. An illustration of the capacity region of the layered erasure channel. Theregion is generally enclosed by the two axes, line� � � and a piece-wiselylinear curve �. The max-point is marked by a square.

is achieved; for any , user 2 sends only privatemessage, so that the rate pair is achieved. In the caseof strong interference, and . For any ,user 2 always sends common message. With the constraint onthe direct link of user 2, i.e., , we can achieveif , as shown in Fig. 3(a) and (b).

In general, the max-point cannot be achieved using timesharing between the corner points because the corner pointto its upper left is outside of the capacity region. To overcomethe difficulty, let us revisit the achievability of the max-point inthe single-layer case.

1) The very strong interference case, : The max-point is as shown in Fig. 3(a). With both users usingi.i.d. signaling, the max-point is achievedby having user 1 carry out successive decoding.

2) The strong but not very strong interference case,: The max-point

[see Fig. 3(b)] can be achieved by using rate split-ting [34] as follows. User 2 achieves the rateusing i.i.d. codebook. User 1 isviewed as two virtual users with two i.i.d. codebooks,whose distributions follow and

, respectively, where. Let the transmitted codeword be

with for ,which is an i.i.d. Bernoulli(1/2) sequence because

. Receiver1 can first decode the message encoded in at rate


and then decode the message of user 2 as long as the ratedoes not exceed

and finally it can decode the message encoded in at rate

The rate for user 1 is and itis easy to verify that . Letbe such that . As varies from 0 to , thepoint moves from to .In fact, every point in the capacity region on the boundaryof , including the max-point, can beachieved by rate splitting without time-sharing.

3) The weak interference case, : The max-pointis [see Fig. 3(c)]. Let user 1 use i.i.d.

codebook. We split the rate ofuser 2 and let and

be the two distributions usedto generate the two i.i.d. random codebooks. Let the trans-mitted codeword be with ,

. Receiver 1 can decode the common messageat rate

(46)

After removing interference generated by , receiver 1 candecode its own message at rate

At receiver 2, the common message of rate givenby (46) can be decoded because

After removing the common message, receiver 2 can de-code its private message at rate

It is easy to verify that andsatisfy

By varying from 0 to 1, every point on the line segmentconnecting and can be achieved by ratesplitting.

D. Proof of the Achievability Part of Theorem 2

We generalize the preceding techniques for designing rate-splitting codes to prove the achievability result for channels withmultiple layers.

1) The Corner Points : For , con-sider the achievability of the corner point . Let user 1 generatea random codebook of rate

(47)

and let user 2 generate two codebooks: one is for private mes-sage at rate

(48)

the other for common message at rate

(49)

where for every . For no-tational convenience, for a random vector and a subset

, we denote as a -dimensional vectorwhose th element is if and is equal to 0 otherwise. Thecodeword of user 1 is transmitted as . The codeword forthe private message of user 2 is transmitted as over thelayers in , whereas the codeword for the common messageis transmitted as over the remaining layers. All code-books consist of i.i.d. entries.


By (48) and (49)

(50)

From (47) and (50), the codebooks carry exactly the rate pair(45) at . Therefore, to show achievability of point is equiv-alent to show that the rate triple is achievable.Indeed, receiver 1 can decode the common message, because

Intuitively, for given and , the signal inlayer contributes to the mutual information if andonly if . After canceling the interference caused bythe common message, receiver 1 can decode its own message,because

(51)

where (51) can be regarded as a consequence of Lemma 4. Re-ceiver 2 can decode its private message because

Receiver 2 can also decode the common message

where the inequality is because .2) The Max-Point: We establish the achievability of the max-

point in all three possible cases depending on its position.

Case 1: The max-point is below . In this case,, so that the region (9) becomes rectangular (cf.

the single layer model in the case of in Section V-C,where the capacity region is shown in Fig. 3(a)). It suffices toshow that is achievable.

Let user 1 and user 2 each generate a random codebook withi.i.d. entries, with rate and , respec-tively. Receiver 2 can decode its own message because

Then receiver 1 can decode the message of user 2, because

After canceling the interference, receiver 1 can decode its ownmessage, because

Case 2: The max-point is between and , i.e., the inter-section of line and the curve is on the boundaryof for some . Point is not achievable,because it is above the line . Therefore, we can notachieve this max-point using time sharing between pointsand . However, the rate-splitting approach for achievingthe rate pair in Section V-C is applicable here. Recall the per-mutation defined in Section V-C. The basic idea is to transmit aprivate message over layers and a common mes-sage over layers , where rate splitting is appliedover layer .

Let user 1 generate a random codebook with i.i.d.entries. Define set

and set . Let user 2 encode itscommon message onto , and encode its private mes-sage onto , where and are two binary signals.The transmitted signal then consists of , and

. The codebook for the common messageconsists of random independent entries, where the elementsin of are andfor some . The codebook for the private message isgenerated similarly, but with .We note that

(52)


For fixed , the common message can be decoded at receiver1 as long as its rate does not exceed

where (52) is used to reach the last equality. Once the interfer-ence caused by the common message is removed, receiver 1 candecode its own message at rate

(53)

(54)

where (53) follows by Lemma 4.Choose the value of such that the common message of rate

is decodable at receiver 2. Once the common message iscanceled, then the following private rate is achievable:

It suffices to show that we can makecoincide with the max-point by choosing some . Note that

(55)

Multiplying (55) with and adding with (54), and noting that, we have

which coincides with the boundary of . As variesfrom 0 to 1, point traverses the line segmentconnecting and . There must exist some such that

coincides with the max-point.

Case 3: The max-point is above , i.e., the intersection ofline and curve is on the boundary of . Thisis similar to the single-layer model with , whosecapacity region is shown in Fig. 3(b). The approach here is sim-ilar as in Section V-C. The basic idea is to split the message ofuser 2 into a common message and a private message as in case2; and regard user 1 as two virtual users.

Let user 2 transmit its private message usingand transmit its common message using . Definerandom vectors , where the elements ofare i.i.d. , and the elements of are i.i.d

, where . We split the user1 into two virtual users, with codewords and , respectively.The transmitted codeword consists of ,

.Let receiver 1 decode and cancel first, then decode and

cancel , and finally decode . For fixed , following ratetriple is achievable by receiver 1:

At receiver 2, the following private rate is achievable:

As long as is chosen such that the rate of the common messagesatisfies , the common message canbe decoded at both receivers. It then suffices to show that thereexists such that

, coincides with the max-point.By the chain rule

(56)


Thus, is on the boundary of . Note thatincreases as increases because larger indicates larger

part of user 1’s signal is removed before decoding the commonmessage at receiver 1. Let , then

where the last inequality follows by the facton . Since both and

are continuous functions, there must exist a such thatis the max-point, which falls on the line

segment between and . Thiscompletes the proof of Theorem 2.

VI. FADING GAUSSIAN Z-INTERFERENCE CHANNEL

In this section, we study the fading Gaussian Z-interferencechannel model described in Section II and prove Theorem 1.A key technique for deriving the outer bound is to convert thechannel to a “layered” model through the use of “incrementalchannels,” whereas the achievability result is established usinga “layered” signaling. In either case, insights obtained in thestudy of the layered erasure model in Section V are instruc-tive. Since the phases of each channel state is known and can becompensated by the receiver, we assume for

without loss of generality, throughout Section VI.The phases remain uniformly distributed and i.i.d.

A. The Outer Bound for the Gaussian Model

The capacity region depends only on the marginal distribu-tions of the channel outputs and conditioned on the channelinputs and states. Throughout Section VI-A, we make additionalassumptions on the model without changing the capacity region.First, we assume that the gains of the direct link for user 2 andthe cross link are aligned, i.e., for every , the SNRsare driven by the same random variable :

where are i.i.d. and uniformly distributed on . Thestates remain independent of . It is important tonote that the region remains the same, because the bounds in(4) are invariant to the dependence of and introduced here.Secondly, we assume that the additive noises at the two receiversare also aligned. For ease of description, let ,

, be independent continuous CSCG processes,each of which is of independent increments withand , i.e., is a complex-valued Brownianmotion (Wiener process). Without changing the capacity region,throughout Section VI-A, we revise model (1) to following:

(57a)

(57b)

Fig. 5. An illustration of noise alignment via incremental channel. The signal� is corrupted by a circularly symmetric standard complex Brownian motion.The observations � , � , and � are generated by taking the corrupted signal outat time �� , �� , and �� , respectively.

where and are i.i.d. uniform on .See Fig. 5 for an illustration. In case where the realizationof the channel state is such that , the additive noise

is interpreted as a CSGC noise with unitvariance (i.e., the limit of ). Since and

are both unit CSCG random variables, channelmodel (57) is no different than (1) except that the additiveGaussian noises at the two receivers are samples from the sameWiener process, and are hence aligned.

Let and for conve-nience. The notation for the time index in this section is dif-ferent than that used for the layered erasure model in Sections IVand V: Here refers to a signal at time interval , andrefers to the signal over time intervals, . More-over, let and

.By Fano’s inequality, the rates of the two users must satisfy

(58)

(59)

with as . Let us introduce auxiliary signalsand as follows:

where we have implicitly defined as a unit CSCG randomvariable, which is proportional to the increment of the Brownianmotion, and is hence independent of and the additive noise

.By the chain rule

(60)


Hence, for , we have

(61)

where (61) is because is maximizedby letting both and be i.i.d. unit CSGC sequences. By[35, Corollary 2], setting the distribution of the input to unitCSCG incurs no more than bit of loss in .Therefore, by (60)

(62)

Substituting (62) into (61) upper bounds the rate of user 1

(63)

where (63) is because is a Markov chain.The rate of user 2 can be upper bounded by providing side

information to receiver 2, so that (59) is relaxed to

(64)

Combining (63) and (64), we have

(65)

Bound (65) admits a similar interpretation as (37).

Applying Lemma 2 to the memoryless channel of transitionprobability , we have

(66)

We need the following result, which is a parallel of Lemma 4.

Lemma 6: Suppose a collection of random variables isindependent of . Then

(67)

(68)

where and are given in (5) and (6), respectively, and

stands for minimum mean-square error of estimating givenits observation in Gaussian noise conditioned on .

The proof of Lemma 6 is given in Appendix C and it has asimilar interpretation as the one for Lemma 4.

Define

Applying Lemma 6 to the summation term in (66), we have

(69)

(70)

where, in (69), has a mixture distribution of and we usethe fact that is a concave function of the input dis-tribution [36], and (70) is because the MMSE is upper boundedby for . Therefore, substituting (70) into(66) and sending , we have established the third con-straint in (4).

B. The Inner Bound: A Constant Gap Result

A coding scheme is developed next to achieve a rate re-gion within a constant gap to the outer bound developed inSection VI-A. The gap applies to all SNRs and fading statistics,and at high SNRs, where the capacity is large, the gap becomes


Fig. 6. An illustration of the capacity inner bound (dashed curve) and outer bound (solid curve).

insignificant. The scheme is inspired by the “layered” signalingintroduced by Tse and Yates for fading broadcast channels [24].

To ease the analysis, we drop the first constraint in (4) andreplace the on the RHS of third constraint in (4) with 1to yield a looser outer bound, denoted by . The third bound at

corresponds to , which is looserthan the first constraint in (4), but within 1 bit.

Similar to the capacity region for the layered erasure model,besides the two axes, the outer bound is enclosed by twocurves, which correspond to the remaining two constraints: Oneis the horizontal line ; the other curve isthe boundary of the region , where

and

(71)

For every , the straight line boundary oftouches the curve at the point

(72)

To verify this, it suffices to show that the point (72) is in aswell as on straight line boundary of . The later is true be-

cause (72) achieves the equality of the constraint . More-over, for every , we have

where the last step is due to the fact thatfor every . Thus the point

for every , which implies that the point is inthe region . Therefore touches at point (72).

Denote points and by and, respectively. See Fig. 6 for an illustration. Generally, the

curve can be divided into three parts: The part on the leftside of is a ray with slope ; the curve between andhas tangent lines with slope steeper than ; the remaining partis a vertically downward ray starting from point . Anotherobservation is that all the extreme points on are contained inthe closure set . However,not all boundary points between and are contained in theset . For example, in case , , are all discrete randomvariables, the outer bound becomes a polyhedron, like thecase of layered erasure model, and consists of only the cornerpoints.

We refer to the intersection of the lineand the curve as the max-point as in the case of the layerederasure model. In the following, we first show that every pointin and below line is achievable withina constant gap, and then deal with the max-point.


1) Points in : It suffices to show that for every ,the rate pair is achievable for some uni-versal constant .

The points in can either be parametrized with as (72)or be asymptotically approached by those can be parametrizedwith . Therefore, it suffices to show the achievability result forthose can be parametrized. For convenience, we define function

where and is equal to 1 if the condition in-side the parentheses is satisfied and equal to 0 otherwise. Givenrandom variable with cumulative distribution function

Since whenever , thecoordinate (72) can be rewritten as

(73a)

(73b)

Let transmitter 1 generate a random codebook with i.i.d. unitCSCG distribution. Let the signaling of user 2 follow the distri-bution of a signal constructed as

(74)

where and and are i.i.d. andequally likely to be . In fact, and correspond toa certain binary expansion of the in-phase and quadrature com-ponents of , respectively. It is easy to check that is uni-formly distributed on a square on the complex plane. Fixto be a constant, for , let

and . For every , let de-note the signal constructed as (74) where the set is replacedby . The transmitted signal of user 2 is the superposition oftwo codewords, which carry the common message and the pri-vate message, respectively, i.e., ,

. The codebooks for the common message and theprivate message are randomly generated using the distributionsof and , respectively.

Signaling like (74) was first proposed in [24] for fading broad-cast channels. Each can be viewed as a signal of

on “layer” . The set contains layers corresponding todefined in (71). The following property of the signaling

is crucial for achieving a sufficiently high rate while limitingthe impact of interference.

Lemma 7: Let be a measurable subset of ,, and . For the random variable

defined by (74) and every nonnegative random variable ,

(75)

and

(76)

where is independent of , and. Furthermore, can be minimized by choosing

, which yields .Lemma 7 is proved in Appendix D. We note that inequality

(75) has been shown in [24]. It is easy to see that the gap sug-gested by Theorem 1 is exactly . In [24], the authors con-jecture that the actual gap in Lemma 7 is much smaller than .If one can improve the gap in Lemma 7 to , then the gap inTheorem 1 can be improved to .

We next analyze the achievable rates using Lemma 7. Recallthat . Receiver 1 first decodes the common mes-sage of user 2 and then can decode its own message at rate

(77)

(78)

(79)

where (77) and (78) are because

are Markov chains.


Now, we compare the rate pair on the outerbound with the achievable rate pair . By (73a)and (78)

(80)

where we have replaced with inthe first mutual information. Applying Lemma 7 to each of thetwo difference terms in (80), we have

User 2 can achieve rate

as long as ,

, and

. The gap between and themaximum achievable is thus no greater than the largerone of

(81)

and

(82)

where the phase is removed in (82) without changing themutual information because and are circularly symmetric.

By (73b) and two uses of Lemma 7, we can upper bound (82)as

(83)

Since is uniform, by [35, Eq. (10)], (81) can be upper boundedby

(84)

Putting (83) and (84) together, we obtain

where stands for the maximum achievable rate using theproposed layered signaling. By minimizing over ,we obtain

(85)

2) Max-Point: We next establish the constant-gap achiev-ability of the max-point. There are three cases, each of whichis in analogy to one case of the layered erasure model inSection V-D.

Case 1: The max-point is below . In this case,, so that the outer bound becomes

a rectangle. The rate pair issimply achievable by using i.i.d. Gaussian codebooks. In fact,for this case, , is exactly thecapacity region.

Case 2: The max-point is between and . If the max-point is in the set , i.e., a corner point, then the preceding“layered” code achieves the bound (85). But it is possible thatthe max-point is between and and it is not in set .In other words, the part of curve can be a line segment withslope steeper than and the max-point is right on this segment.This situation can happen, for example, when , and arediscrete random variables.

Suppose the max-point falls on a line segment on theboundary . Denote the two ends of the line segment by and

, where is below the max-point. Suppose slope of the linesegment is , then ’s coordinate isand the coordinate of is

where . For nota-tional convenience, we define

. Since both and are extremepoints, they belong to . By the preceding analysis, pointcan be achieved up to a gap of 12.8 for each user. Therefore, itis sufficient to show the achievability result for the max-pointbecause points between and the max-point can be shownby time-sharing argument. The main idea is to use layers cor-responding to and for the private and commonmessages, respectively, and divide the layers corresponding to

in a similar manner as for the layered erasure model asdiscussed in Section V-D.

The point on the segment can be parametrized as


where

It is easy to see that when varies from 0 to ,moves continuously from to . Since the max-point is be-tween and , it must be for some . Now,let . Let user 1 generate its code-book according to i.i.d. unit CSCG distribution. Let user 2 gen-erate the codebook for private message using the distribution of

and generate the codebook for common message using thedistribution of . Following the exactly same analysis as forthe points in , we see that the max-point can be achieved upto a gap of 12.8 for each user.

Case 3: The max-point is above . In this case, the achiev-ability result for holds by preceding analysis. That is fol-lowing rate pair is achievable and it is at most 12.8 bits awayfrom the critical point for each user:

Since the outer bound between max-point and is a seg-ment with slope , it suffices to investigate the achievabilityof points along the ray with slope and starting from

till the intersection with the other constraint.

As in the case of the layered erasure model, we split user1 into two virtual users [37]. Let and be two inde-pendent CSCG random variables with and

, where . Let user 1 generate thecodebooks for its two virtual users using the distribution ofand , respectively. Receiver 1 first decode , then decode

and finally decode in order. Therefore, the followingthree rates are achievable at receiver 1

Receiver 2 can achieve the private rate

By the chain rule, we have. Thus

is on the ray with slope and starting from. That is the gap between the segment

, , andthe boundary of is at most 12.8 per dimension.

Furthermore, we show next that can bewithin a constant gap to line . Note that

increases as increases, since larger impliesthat a larger part of user 1’s signal is removed before decodingthe common message at receiver 1. Letting

By Lemma 7

(86)

where (86) is due to the fact when-ever . This implies that we can choose so that themax-point can be achieved up to 12.8 for each user. This com-pletes the proof of Theorem 1.

VII. CONCLUSION

This paper derives the first constant-gap result for the capacityregion of the ergodic fading Gaussian Z-interference channelwith channel state information at the receivers but not at thetransmitters. To achieve this, the new outer bound is obtained byinvestigating the tradeoff between the rate of the two users. Theachievability strategy is constructed by artificially layering onetransmit signal. Both the outer and inner bounds are motivatedby the simpler and exact results for the corresponding layerederasure model.

APPENDIX APROOF OF LEMMA 2

For , by Marton’s expansion [39]

(87)

(88)

where (87) is due to the chain rule. Furthermore, using the factorgraph depicted in Fig. 7, we show that the following are twoMarkov chains:

(89)

(90)


Fig. 7. The factor graph for justification of Markov chains (89), (90), and (92).The factor graph summarizes relations among all random variables involved inthe three Markov chains. The key property of factor graph is that conditioned onany cut set of variable nodes, the sets of random variables on the disconnectedsubgraphs are mutually independent. For more details, see [38].

Note that the input is a function of the message, and the channels for to are memoryless.

It is easy to see that conditioned on ,the output is independent of , hence (89). For a similarreason, (90) also holds. Therefore, (88) can be rewritten as

(91)

On the other hand, by the chain rule

Since

(92)

is a Markov chain, which can also be justified using factor graph(Fig. 7), we have

(93)Lemma 2 is established by writing the LHS of (22) as

and applying (91) and (93).

APPENDIX BPROOF OF LEMMA 4

Because signals are aligned at their respective least significantbit in a sum, one can write

(94)

where (94) is due to the chain rule. Hence the proof of (39).Similarly

Furthermore

(95)

where (95) is due to the alignment between and . Hencethe proof of (40).


APPENDIX CPROOF OF LEMMA 6

Since is Markovian, we have

(96)

where (96) is obtained by using the integral representation ofmutual information via MMSE [40]. Hence the proof of (67).

Note that for any realization of , either oris a Markov chain. In the latter case, the mutual

information is zero. Therefore, for every realization of thestates

Thus

Furthermore

(97)

where (97) is due to the alignment between and . Hencethe proof of (68).

APPENDIX DPROOF OF LEMMA 7

Instead of computing the mutual information directly, the au-thors of [24] show that the system with input and outputhas an achievable rate at least the amount of RHS of (75). Hence(75) holds.

To show (76), we apply (75) to the sets and , where,

It can be rewritten as

(98)

Note that has a uniform distribution on a square region incomplex plane. By [35, Eq. (10)], we have

(99)

Comparing (98) and (99), we have established (76).

ACKNOWLEDGMENT

The authors would like to thank the anonymous reviewersand the Associate Editor for their comments which have helpedimprove the presentation of this paper.

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Yan Zhu received the B.E. and M.S. degrees from the Electronic EngineeringDepartment, Tsinghua University, in 2002 and 2005, respectively.

Since 2005, he has been pursuing the Ph.D. degree at Northwestern Uni-versity. His research interests include wireless communications, informationtheory, communication network, and signal processing. His current researchfocuses on the interference problems in wireless systems, including commu-nication scheme design for interference network and fundamental limitations ofinterference channels. In 2010, he joined Broadcom Inc. as a scientist, designstaff.

Mr. Zhu is a corecipient of the 2010 IEEE Marconi Prize Paper Award inWireless Communications.

Dongning Guo (S’97–M’05) received the B.Eng. degree from the University ofScience and Technology of China, the M.Eng. degree from the National Univer-sity of Singapore, and the M.A. and Ph.D. degrees from Princeton University,Princeton, NJ.

He joined the faculty of Northwestern University, Evanston, IL, in 2004,where he is currently an Associate Professor with the Department of ElectricalEngineering and Computer Science. He was an R&D Engineer in the Center forWireless Communications (now the Institute for Infocom Research), Singapore,from 1998 to 1999. He has held visiting positions with Norwegian University ofScience and Technology in summer 2006 and the Chinese University of HongKong in 2010–2011. His research interests are in information theory, commu-nications, and networking.

Dr. Guo is an Associate Editor of the IEEE TRANSACTIONS ON INFORMATION

THEORY. He received the Huber and Suhner Best Student Paper Award in theInternational Zurich Seminar on Broadband Communications in 2000 and is acorecipient of the IEEE Marconi Prize Paper Award in Wireless Communica-tions in 2010. He is also a recipient of the National Science Foundation FacultyEarly Career Development (CAREER) Award in 2007.

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Ergodic Fading Z-Interference Channels Without State Information at Transmitters

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