On Layered Erasure Interference Channels without CSI at Transmitters

0.52 MB·

by  Yan Zhu

1
On Layered Erasure Interference Channels without CSI at Transmitters
Yan Zhu Aerohive Network Inc.
330 Gibraltar Dr, Sunnyvale, CA 94089
Email: [e-mail]
Abstract
Thispaperstudiesalayerederasuremodelfortwo-userinterferencechannels,whichcanbeviewedasasimplified
version of Gaussian fading interference channel. It is assumed that channel state information (CSI) is only available
6
1 atreceiversbutnotattransmitters.Undersuchassumption,anouterboundisderivedforthecapacityregionofsuch
0 interference channel. The new outer bound is tight in many circumstances. For the remaining open cases, the outer
2
bound extends previous results in [1].
n
a
J
4 I. INTRODUCTION
2
Since the breakthrough of Gaussian interference channel by Etkin, Tse and Wang [2], the study of interference

]
T networking has made plenty of progress [3]. However, most works still focus on situations where channel state
I
. information (CSI) is static and known to both transmitters (CSIT) and receivers (CSIR). These results usually valid
s
c
undersituationswhereCSIvariatesslowlyandsystemshaveefficientsoundingandfeedbackmechanismstoupdate
[

global CSI timely. For communications experiencing fast channel fading, we usually do not have such luxury to
1
v
have (accurate and timely) CSI at transmitters.
3
2 Inthispaper,weinvestigatealayerederasuremodeloftwo-userinterferencechannel(IC),whichsharesthesame
4
6 spirit as deterministic model used in [4], except that the transmit binary vectors are erased randomly. We assume
0
. that the erasure levels (which model the fading states or CSI) are known at the receivers but not at transmitters.
1
0 In particular, we derive an outer bound for general two-user layered erasure interference channels with no CSI at
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1 transmitters. The obtained new bound is generally tight for many important cases but whether it is tight for all
:
v situations is still open in this paper. Comparing with previous results in [1], this paper can be viewed as a fully
i
X
extension to multi-layer situations.
r
a Theremainingpaperisorganizedasfollowing.Innextsection,thechannelmodelisformallydescribedandsome
notation is introduced to assist the further discussion. In Section III, our main finding is presented and followed
by several remarks to clarify the new outer bound. We continue our discussion by investigating some non-trivial
situations in Section IV. Limited by the space, we only present a brief proof of the outer bound in Section V.
Finally, we conclude this paper in Section VI.
II. MODELANDNOTATIONS
Consider a layered erasure interference channel as shown in Fig. 1. At each time i, transmitters 1 and 2 emit
binary-vector signals W[i] and X[i], respectively, which take value in Fq. Only a certain top portion of each
2
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W Y
N
11
Tx1 Rx1
X Z
Rx2
Tx2
Fig. 1: Layered Erasure Interference Channel Model.
vector signal reaches the two receivers, and the remaining part is erased randomly. Mathematically, let s denote
a q × q matrix whose elements are all 0 except that s = 1 for k = 1,...,q − 1. It is easy to see that
k+1,k
s[x ,x ,...,x ]T = [0,x ,...,x ]T, and sq−n[x ,x ,...,x ]T = [0,...,0,x ,...,x ]T, which is equivalent
1 2 q 1 q−1 1 2 q 1 n
to a zero-padding downward shift of the vector so that only its first n elements are left. With these notations, at
each time i, the two received signals, Y[i] and Z[i], can be written as,
Y[i]=sq−N11[i]W[i]⊕sq−N21[i]X[i] (1a)
Z[i]=sq−N12[i]W[i]⊕sq−N22[i]X[i], (1b)
respectively, where for t,r = 1,2, each integer random process {N [i]} models the channel fading process from
tr
transmitter t to receiver r. We assume that the four fading processes are independent of each other and each of
them is independent and identically distributed (i.i.d) over time (so that the channel is memoryless). In this paper,
we study situations where receiver 1 knows realization of (N [i],N [i]) and receiver 2 knows realization of
11 21
(N [i],N [i]) at each time i, but no channel state information (CSI) is available at both transmitters except for
22 12
the statistical law of those fading processes.
For remaining discussion, we need following notations. Suppose X ∈ Fq is an arbitrary random vector. We
2
use X to denote its j-th element and Xk to denote its sub-vector [X ,...,X ]T. For the special case where
j j j k
subscript j = 1, we often use Xk instead of Xk. We often use lower-case letters to denote realizations of their
1
corresponding random vectors or random variables. For example, x = [x ,x ,...,x ]T should be interpreted as
1 2 q
a particular realization of the random vector X. For any sequence of random vectors {X[i]}, let (X)i2 denote
i1
thesubsequenceX[i ],...,X[i ].Consequently,(Xk)i2 denotesthesubsequenceXk[i ],Xk[i +1],...,Xk[i ].
1 2 j i1 j 1 j 1 j 2
In summary, the indices outside the parentheses always refer to time while inside ones refer to element(s) of
the corresponding vectors. Binary addition (⊕) between two vectors with different lengths is aligned at the least
significant bits: if n ≥ n , then define Xn1 ⊕Wn2 = [X ,...,X ,X ⊕W ,...,X ⊕W ]T.
1 2 1 n1−n2 n1−n2+1 1 n1 n2
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Since we only consider memoryless channels in this paper, we often surpress the time index i to ease the notations.
For example, P(N ≥l) is equivalent to P(N [i]≥l), distribution of N [i] is referred as distribution of N ,
11 11 11 11
et al.With the convention introduced above, channel model (1) can be rewritten as
Y =WN11 ⊕XN21 (2a)
Z =WN12 ⊕XN22. (2b)
III. MAINRESULTS
Our main findings are summarized in Theorem 1, which needs a few of more definitions. Define following three
regions within first orthant of R2 as
R1a := (cid:92) (cid:8)(R1,R2)∈R2≥0(cid:12)(cid:12)R1+ωR2 ≤B1a(ω)(cid:9) (3a)
ω∈[0,1]
R1b := (cid:92) (cid:8)(R1,R2)∈R2≥0(cid:12)(cid:12)R1+ωR2 ≤B1b(ω)(cid:9) (3b)
ω∈[0,1]
R1c := (cid:92) (cid:8)(R1,R2)∈R2≥0(cid:12)(cid:12)(1+µ)R1+ωR2 ≤B1c(ω,µ)(cid:9) (3c)
µ∈[0,ω]
ω∈[0,1]
where
q
(cid:88)
B1a(ω)=EN11+ωE[N21−N11]++ [ωβ1(l)−α1(l)]+, (4a)
l=1
q q
B1b(ω)≤(1−ω)EN11+ωEN21+(cid:88)[ωγ1(l)−α1(l)]++ω(cid:88)max(cid:0)P(N11−N21 ≥l),P(N12 ≥l)(cid:1),
l=1 l=1
(4b)
q q
(cid:88) (cid:88)
B1c(ω,µ)=EN11+ωE[N21−N11]++ [ωγ1(l)−α1(l)]++ max(µP(N11 ≥l),ωP(N12 ≥l)), (4c)
l=1 l=1
and
α (l)=P(N ≥l)−P(N −N ≥l), (5a)
1 21 21 11
(cid:2) (cid:3)+
β (l)= P(N ≥l)−P(N −N ≥l) (5b)
1 22 21 11
(cid:2) (cid:3)+
γ (l)= P(N −N ≥l)−P(N −N ≥l) . (5c)
1 22 12 21 11
By swapping subscripts 1 and 2 in (5), we can define α (l), β (l), and γ (l) accordingly. For example, α (l) :=
2 2 2 2
P(N ≥l)−P(N −N ≥l). In turn, we can define B and R for j ∈ {a,b,c} by exchanging subscripts
12 12 22 2j 2j
1 with 2 in (3) and (4).
Theorem 1: The capacity region C of the two-user interference channel (1) is contained in R ∩R ∩R ∩
1a 1b 1c
R ∩R ∩R .
2a 2b 2c
To help understand the outer bound, we make following remarks.
Remark 1: Each region R , where k ∈ {1,2} and j ∈ {a,b,c}, is a convex region characterized by a set of
kj
weighted sum-rate bounds, which is associated with each B . Although the intersections are done among infinite
kj
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numbers of ω and µ in (3), it is not difficult to see that only finite numbers of them are necessary. Therefore, each
R is a polytope. In the remaining discussion, we refer to the weighted bound of form R +ωR ≤ B (ω) as
kj 1 2 1a
bound B (ω), and (1+µ)R +ωR ≤B (ω,µ) as bound B (ω,µ), et al., for each fixed ω and µ. Furthermore,
1a 1 2 1c 1c
on the boundary of each region R (inside the first orthant), R has larger weight than R , which corresponds to
1j 1 2
a situation where rate of user 1 is preferred to that of user 2. By symmetry, similar interpretation can be made to
each region R , where user 2 is the preferred one.
2j
Remark 2: Bysettingeach{N [i]},t,r =1,2,toaconstant,sayn ,thenewouterboundrecoversthecapacity
tr tr
region of its deterministic counterpart, which consists of all positive rate pairs satisfying [4], [5]:
R ≤n r =1,2 (6a)
r rr
R +R ≤max(n ,n )+(n −n )+ (6b)
1 2 11 21 22 21
R +R ≤max(n ,n )+(n −n )+ (6c)
1 2 22 12 11 12
R +R ≤max((n −n )+,n )+max((n −n )+,n ) (6d)
1 2 11 21 12 22 12 21
2R +R ≤max(n ,n )+(n −n )++max((n −n )+,n ) (6e)
1 2 11 12 11 21 22 12 21
R +2R ≤max(n ,n )+(n −n )++max((n −n )+,n ). (6f)
1 2 22 21 22 12 11 21 12
We leave the detailed proof in Appendix A. Table I is a brief summary of the proof and it also highlights the
relation between each region R , k = 1,2 and j ∈ {a,b,c}, and each constraint in (6). We will look into more
kj
sophisticated examples in Section IV.
Theorem1 Constraintin(6)
B1a(ω)andB2a(ω)withω=0 (6a)
B1a(ω)withω=1 (6b)
B2a(ω)withω=1 (6c)
B1b(ω)orB2b(ω)withω=1 (6d)
B1c(ω,µ)withω=1,µ=1 (6e)
B2c(ω,µ)withω=1,µ=1 (6f)
TABLE I: A summary of how to obtain capacity region of the deterministic IC via Theorem 1. It also shows how
each bound B , for k =1,2 and j ={a,b,c}, relates to each constraint of (6).
kj
Remark 3: Regions R and R are actually capacity regions of Z-IC with N ≡ 0 and N ≡ 0, respec-
1a 2a 21 12
tively [6]. The remaining four are new.
IV. MOREDISCUSSION:LAYEREDERASURECASES
In this section, we will investigate Theorem 1 under several special situations, which include cases where
Theorem 1 is tight as well as some open cases. Inspired by classification done in [1] for q = 1, we define
following three cases:
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1) Stochastically strong interference: ∀l∈{1,...,q},
P(N ≥l)≥P(N ≥l)
12 11
P(N ≥l)≥P(N ≥l);
21 22
2) Stochastically weak interference: ∀l∈{1,...,q},
P(N −N ≥l)≥P(N ≥l)
11 21 12
P(N −N ≥l)≥P(N ≥l);
22 12 21
3) Stochastically moderate interference: ∀l∈{1,...,q},
P(N ≥l)>P(N ≥l)>P(N −N ≥l) (9a)
11 12 11 21
P(N ≥l)>P(N ≥l)>P(N −N ≥l). (9b)
22 21 22 12
The first case can be interpreted as following: from the viewpoint of any given layer l, signal reaches to the
undesiredusermoreoftenthanthedesiredone,whichissimilartothestronginterferencechannelsinusualsense[7].
Both of the other two cases implies that P(N ≥l) ≤ P(N ≥l) and P(N ≥l) ≤ P(N ≥l) for any l.
12 11 21 22
Therefore, they can be interpreted as cases where signal reaches to the desired user more often than the undesired
one from the viewpoint of each layer. Hence, they both fall into weak interference category in usual sense.
In this section, we will show that the new outer bound actually coincides with the capacity regions for the first
two cases. However, for the moderate interference case, whether Theorem 1 is tight still remains open. Note that
general layered erasure channel can be much more complicated so that it can be none of these three cases. We will
conclude this section with some discussion about general cases.
A. Stochastically Strong Interference
For strong interference, it is well known that the capacity region is the same as capacity region of compound
multi-access channel at receivers 1 and 2 [7], i.e.,
 (cid:12) 
(R1,R2)∈R2≥0(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) R1+RR2i ≤≤ EmNaxrr(cid:0)Erm=ax1(N,211,N21),Emax(N22,N12)(cid:1) . (10)
From Theorem 1, we see that all summation terms vanish in B , k = 1,2 and j ∈ {a,b,c} under stochastically
kj
strong interference assumption. Therefore, it is not difficult to see that the outer bound becomes R ∩R , which
1a 2a
coincides with the region defined by (10).
B. Stochastically Weak Interference
Theorem 2: For interference channel which satisfies condition (8) ∀ l ∈ {1,...,q}, the capacity region is
characterized by Theorem 1. In particular, it coincides with R ∩R and sum-capacity is given by
1b 2b
Csum =E[N22−N12]++E[N11−N21]+. (11)
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Proof: We start with the converse, which makes construction of achievable schemes more intuitive. Since (8)
holds for any l, we can simplify B (ω) as
1b
q q
(cid:88) (cid:88)
B1b(ω)=(1−ω)EN11+ωEN21+ [ωγ1(l)−α1(l)]++ω P(N11−N21 ≥l)
l=1 l=1
q
(cid:88)
=(1−ω)EN11+ωEN21+ [ωγ1(l)−α1(l)]++ωE[N11−N21]+
l=1
q
(cid:88)
=EN11+ωE[N21−N11]++ [ωγ1(l)−α1(l)]+. (12)
l=1
With (8) and (5b), we have β (l)≥γ (l). Therefore B (ω)≤B (ω) ∀ω ∈[0,1], which implies R ⊂R . For
1 1 1b 1a 1b 1a
B (ω,µ), we have
1c
q
(cid:88)
B1c(ω,µ)=EN11+ωE[N21−N11]++ [ωγ1(l)−α1(l)]+
l=1
q
(cid:88)(cid:2) (cid:0) (cid:1) (cid:3)
+ max µP(N ≥l),ωP(N ≥l) −µP(N ≥l)+µP(N ≥l)
11 12 11 11
l=1
q
(cid:88)
=(1+µ)EN11+ωE[N21−N11]++ [ωγ1(l)−α1(l)]+
l=1
q
(cid:88)
+ [ωP(N ≥l)−µP(N ≥l)]+. (13)
12 11
l=1
Therefore, bound B (ω,µ) can be rewritten as
1c
ω ω (cid:88)q (cid:20) ω 1 (cid:21)+
R1+ 1+µR2 ≤EN11+ 1+µE[N21−N11]++ 1+µγ1(l)− 1+µα1(l)
l=1
q (cid:20) (cid:21)+
(cid:88) ω µ
+ P(N ≥l)− P(N ≥l) . (14)
1+µ 12 1+µ 11
l=1
Comparing with (12), we see that the right-hand side (RHS) of (14) is greater or equals to B ( ω ) for each
1b 1+µ
(cid:84)
fixed ω and µ. Therefore, R ⊂ R . Hence, we have R ⊂ R R . By symmetry, we can argue that
1b 1c 1b 1a 1c
R ⊂ R ∩R . Therefore, the outer bound in Theorem 1 equals to R ∩R under the stochastically weak
2b 2a 2c 1b 2b
interference assumption.
Let ω =1 in (12), we obtain
q
R1+R2 ≤Emax(cid:0)N11,N21(cid:1)+(cid:88)[γ1(l)−α1(l)]+. (15)
l=1
With (5c) and (5a), we have
(cid:2) (cid:3)+
γ (l)−α (l)= P(N −N ≥l)−P(N −N ≥l) −P(N ≥l)+P(N −N ≥l)
1 1 22 12 21 11 21 21 11
(cid:0) (cid:1)
=max P(N −N ≥l),P(N −N ≥l) −P(N ≥l).
22 12 21 11 21
With condition (8), we have P(N −N ≥l)≥P(N ≥l)≥P(N −N ≥l). Therefore,
22 12 21 21 11
γ (l)−α (l)=P(N −N ≥l)−P(N ≥l). (16)
1 1 22 12 21
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From condition (8) and equation (16), we observe that that γ (l)−α (l)≥0 for any l. Substitute (16) into (15),
1 1
we have R +R ≤C .
1 2 sum
By symmetry, we have shown that B (1) = B (1) = C . In fact, it is not difficult to see that boundaries
1b 2b sum
of R1b and R2b intersects inside the first orthant at point (R1,R2)=(E[N11−N21]+ , E[N22−N12]+). Indeed,
this point can be achieved by treating interference as noise at both receivers. We illustrate R and R in Fig. 2
1b 2b
by shadowing them with blue and green, respectively. To show the overlapped region is achievale, it is sufficient
to construct a coding scheme for each extreme point. Let us focus on one of them, say, point A in Fig. 2.
(12)holdssomeinsightsaboutthecodingschemes.Notethat(cid:0)EN11,E[N21−N11]+(cid:1)isalwaysontheboundary
ofR1b,whichmarksby(cid:70)inFig2.Ifuser1anduser2transmittheirinformationatrateEN11andE[N21−N11]+,
respectively,user1’srateisachievablebydecodingandcancelinguser2’smessagecompletely.Onetheotherhand,
user 2 should not have issue to decode its message at rate of E[N21−N11]+ by treating user 1’s signal as noise,
becauseE[N21−N11]+ ≤E[N22−N12]+ byweakinterferenceassumption.Ifuser2wouldliketosendathigher
rate, it faces a tradeoff by generating more interference, which, consequently, will reduce user 1’s rate. This type of
tradeoff is captured by the summation term in (12). Therefore an achievable scheme for point A can be constructed
as following. Both users generate their codebooks according to the distribution of q-size random vector whose
elements are all i.i.d. Ber(1/2) random variables. Here, Ber(p) denotes Bernoulli distribution with probability p
taking value 1 and probability 1−p taking value 0. To finalize the coding scheme, we only need to determine the
allocation of layers for private and common messages, respectively, in the spirit of Han-kobayashi (HK) scheme.
Suppose that the right boundary segment of point A is on a line of form R +ω R = const. Let us define
1 A 2
two subsets of {1,...,q}: L := {l|ω γ (l) ≥ α (l)} and L := {l|ω γ (l) < α (l)}. Thus, we can write the
p A 1 1 c A 1 1
coordinate of point A as
 
(cid:88) (cid:88)
EN11− α1(l), E[N21−N11]++ γ1(l). (17)
l∈Lp l∈Lp
In terms of coding scheme, we allocate all layers of user 1 for private message. For user 2, we allocate layers in
L to carry private message and layers in L to carry common message. At receiving sides, receiver 1 decodes and
p c
removes common message of user 2 before decoding its own message. By evaluating the corresponding mutual
(cid:80)
information, one can verify that user 1 can achieve transmit rate at EN11− l∈Lpα1(l). On the other hand, we
have
(cid:88)
E[N21−N11]++ γ1(l)
l∈Lp
q
(cid:88) (cid:88)
=E[N21−N11]++ γ1(l)− γ1(l)
l=1 l∈Lc
q
(cid:88) (cid:0) (cid:1) (cid:88)
= max P(N −N ≥l),P(N −N ≥l) − γ (l)
22 12 21 11 1
l=1 l∈Lc
q
(cid:88) (cid:88)
= P(N −N ≥l)− γ (l) (18)
22 12 1
l=1 l∈Lc
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≥E[N22−N12]+ (19)
where (18) is due to weak interference assumption (8). By (19), we see that user 2 can decode its own message at
rate E[N21−N11]++(cid:80)l∈Lpγ1(l) by treating interference from user 1 as noise.
R
2
EN22 B
A
EN11 R1
Fig. 2: Sketch of capcity region of a weak interference channel. Regions with blue and green shadow are R and
1b
R ,respectively.Eachofthemisapolytopeandtheirintersectionisthecapacityregion.PointBistheintersection
2b
between boundaries of the two regions and it achieves the sum-capacity.
C. Stochastically Moderate Interference
Whether the outer bound given by Theorem 1 is also tight or not for moderate interference is not clear. But with
condition (9), we can simply the bound further:
Proposition 1: For moderate interference channel, which satisfies condition (9) ∀l, we have
q
(cid:88)
B1a(ω)=EN11+ωE[N21−N11]++ [ωβ1(l)−α1(l)] (20a)
l=1
(cid:0) (cid:1)
B1b(ω)≤(1−ω)EN11+ω EN21+EN12 (20b)
q
(cid:88)
B1c(ω,µ)=EN11+ωE[N21−N11]++ max(µP(N11 ≥l),ωP(N12 ≥l)), (20c)
l=1
and B , B and B are in similar forms with subscripts 1 and 2 exchanged, accordingly.
2a 2b 2c
The proof is similar as reverse part of Theorem 2, except that with condition (9) we have γ (l) < α (l), for
k k
k =1,2 and ∀l∈{1,...,q}.
With q = 1, Proposition 1 does not improve results for sum-rate obtained in [1]. However, it does provide a
slightlybetterouterboundforthewholecapacityregion.Toseethis,wesimplifythenotationforsymmetricsingle-
layered case, i.e. q =1 as following. P(N =1)=P(N =1)=p and P(N =1)=P(N =1)=p . The
11 22 d 21 12 c
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moderate interference assumption (9) becomes p > p > pd , or, equivalently, p p > p −p > 0. With these
d c 1+pd d c d c
notations, we have
α (1)=α (1)=p p
1 2 d c
β (1)=β (1)=p −p +p p
1 2 d c d c
(cid:16) (cid:17) (cid:16) (cid:17)
For k =1,2, consider B (0), B α1(1) , B (1), and B 1, pc accordingly. We obtain an outer bound as
ka ka β1(1) kb kc pd
 (cid:12) 
(R1,R2)∈R2≥0(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Rpd1−+ppdcR−pppdcdp1p+−dd+ppppccdcRRR+ppdcp111p+dd+++ppccRRRRRR22222i ((((((≤≤≤≤≤≤fadecb)))))) pppp2ppdddddc++++ipppp=ddddpp−−++dd1pppppp,ddcccc++pp2ppccccpp((dd11ppcc−−ppccpp((dd11))−−ppdd))  (21)
Note that conditions (a)–(d) are shown in [1] and none of them is redundant when p p > p −p > 0. We
d c d c
will show that in some condition (e) is stronger than (b). In fact, we observe that lines R + pdpc R =
1 pd−pc+pdpc 2
p + pdpc p (1−p ) and R + pd R = p + pd p (1−p ) intersect at (p ,p (1−p )). Regarding
d pd−pc+pdpc c d 1 pd+pc 2 d pd+pc c d d c d
the slopes of these two lines, pd > pdpc under the condition of p −p > p2. In other word, when
pd+pc pd−(1−pd)pc d c c
p −p > p2, (b) becomes redundant to (e). Note that p −p > p2 can happen under moderate interference
d c c d c c
assumption.
D. General Cases
One can evaluate Theorem 1 under all kinds of different conditions beyond what we have discussed above. For
example, one can mixed conditions of weak, moderate, strong interference between the two users. Moreover, in
above three situations, we require (7), (8), or (9) hold for all l. One can also come up a case where conditions of
(7), (8), and (9) are mixed across different layers. Therefore, structures of general layered erasure channels could
be very complicated. However, we believe the limitation of Theorem 1 roots on our limited understanding of the
moderate interference with q =1. In particular, we conjecture that
Conjecture 1: Theorem 1 characterizes the capacity region of general layered erasure channels, which satisfy
condition that there does not exist l∈{1,...,q} such that either (9a) or (9b) holds.
V. PROOFOFTHEOREM1
A. Preliminary Lemmas
The proof relies on following two lemmas.
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Lemma 1: Consider n uses of a memoryless channel described by random transformation P . Let Xn
Y,Z,T|X,S
and Sn denote the independent input and state sequences, respectively. Then for any µ ≥µ ≥0,
2 1
µ I(Xn;Zn|Tn,Sn)−µ I(Xn;Yn|Tn,Sn)
1 2
n
≤(cid:88)(cid:104)µ I(cid:0)X ;Z |Zi−1,Yn ,Tn,Sn(cid:1)−µ I(cid:0)X ;Y |Zi−1,Yn ,Tn,Sn(cid:1)(cid:105). (22)
1 i i i+1 2 i i i+1
i=1
ItisaMarton-styleexpansionofmutualinformation,whichessentiallyconvertsamulti-lettermutualinformation
difference into a single-letter one. For the proof, please refer to [6, Appendix A] 1.
Lemma 2: Suppose that (W)n and (X)n are two independent arbitrary random processes taking value in Fq.
2
Let (N )n and (N )n be two i.i.d fading processes taking value in {0,...,q}. Then we have
1 2
(cid:16) (cid:12) (cid:17) (cid:16) (cid:12) (cid:17)
I (W)n,(X)n;(WN1 ⊕XN2)n(cid:12)(cid:12)(N1)n,(N2)n ≤nEN1+I (X)n;(X(N2−N1)+)n(cid:12)(cid:12)(N1)n,(N2)n (23)
Proof:
(cid:16) (cid:12) (cid:17)
I (W)n,(X)n;(WN1 ⊕XN2)n(cid:12)(N )n,(N )n
(cid:12) 1 2
(cid:16) (cid:12) (cid:17)
=H (WN1 ⊕XN2)n(cid:12)(N )n,(N )n
(cid:12) 1 2
≤H(cid:16)(WN1 ⊕W(cid:102)N1 ⊕XN2)n(cid:12)(cid:12)(cid:12)(N1)n,(N2)n(cid:17)
where (W(cid:102))n be an i.i.d. random sequence independent of all other random processes and each element of W(cid:102)[i]
is an independent Ber(1/2) random variable. Note that (W(cid:102) ⊕W)n has same distribution as W(cid:102) and it is also
independent of (X)n. Let W[i]=W(cid:102)[i]⊕W[i]. Then
(cid:16) (cid:12) (cid:17)
I (W)n,(X)n;(WN1 ⊕XN2)n(cid:12)(N )n,(N )n
(cid:12) 1 2
(cid:16) (cid:12) (cid:17)
≤H (WN1 ⊕XN2)n(cid:12)(N )n,(N )n
(cid:12) 1 2
(cid:16) (cid:12) (cid:17)
=I (W)n,(X)n;(WN1 ⊕XN2)n(cid:12)(N )n,(N )n
(cid:12) 1 2
(cid:16) (cid:12) (cid:17)
=EN1+I (X)n;(WN1 ⊕XN2)n(cid:12)(cid:12)(N1)n,(N2)n (24)
(cid:16) (cid:12) (cid:17)
=EN1+I (X)n;(X(N2−N1)+)n,(WN1 ⊕XN1)n(cid:12)(cid:12)(N1)n,(N2)n . (25)
where (24) is due to chain rule and (25) leads us to (23) by noting that (WN1 ⊕XN1)n and (X(N2−N1)+)n are
independent of each other given ((N )n,(N )n).
1 2
Basically, Lemma 2 claims that for multi-access layered erasure channel, we can fix one input as (W(cid:102))n without
reducing the sum-rate.
B. Proof
First, by letting N [i]≡0, we obtained a z-interference channel, whose capacity region can serve as a natural
12
outer bound. Therefore, bounds B (ω) and B (ω) are direct consequence of [6, Theorem 2]. Thus, it is sufficient
1a 2a
to show the remaining four bounds. By symmetry, we only need to show bounds B and B , respectively.
1b 1c
1In[6],thelemmaisinaformofµ2=1andµ1∈[0,1].Lemma1hereisatrivialextension.
January26,2016 DRAFT