A New Outer Bound and the Noisy-Interference Sum–Rate Capacity for Gaussian Interference Channels

Abstract

A new outer bound on the capacity region of Gaussian interference channels is developed. The bound combines and improves existing genie-aided methods and is shown to give the sum-rate capacity for noisy interference as defined in this paper. Specifically, it is shown that if the channel crosstalk coefficient magnitudes lie below thresholds defined by the power constraints then single-user detection at each receiver is sum-rate optimal, i.e., treating the interference as noise incurs no loss in performance. This is the first capacity result for the Gaussian interference channel with weak to moderate interference. Furthermore, for certain mixed (weak and strong) interference scenarios, the new outer bounds give a corner point of the capacity region.

Full text

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arXiv:0712.1987v2 [cs.IT] 13 Dec 2007
1
A New Outer Bound and the
Noisy-Interference Sum-Rate Capacity for
Gaussian Interference Channels
Xiaohu Shang, Gerhard Kramer, and Biao Chen
Abstract
A new outer bound on the capacity region of Gaussian interference channels is developed. The bound
combines and improves existing genie-aided methods and is shown to give the sum-rate capacity for noisy
interference as defined in this paper. Specifically, it is shown that if the channel coefficients and power
constraints satisfy a simple condition then single-user detection at each receiver is sum-rate optimal, i.e.,
treating the interference as noise incurs no loss in performance. This is the first concrete (finite signal-
to-noise ratio) capacity result for the Gaussian interference channel with weak to moderate interference.
Furthermore, for certain mixed (weak and strong) interference scenarios, the new outer bounds give a
corner point of the capacity region.
Index terms — capacity, Gaussian noise, interference.
I. INTRODUCTION
The interference channel (IC) models communication systems where transmitters communicate with
their respective receivers while causing interference to all other receivers. For a two-user Gaussian IC,
the channel output can be written in the standard form [1]
Y1
=X1+√aX2+ Z1,
√
bX1+ X2+ Z2,
Y2
=
X. Shang and B. Chen are with Syracuse University, Department of EECS, 335 Link Hall, Syracuse, NY 13244. Phone:
(315)443-3332. Email: xshang@syr.ed and bichen@ecs.syr.edu. G. Kramer is with Bell Labs, Alcatel-Lucent, 600 Mountain
Ave. Murray Hill, NJ 07974-0636 Phone: (908)582-3964. Email: gkr@research.bell-labs.com.
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where√a and
√b are channel coefficients, Xiand Yiare the transmit and receive signals, and where the
user/channel input sequence Xi1,Xi2,··· ,Xinis subject to the power constraint?n
i = 1,2. The transmitted signals X1and X2are statistically independent. The channel noises Z1and Z2
j=1E(X2
ij) ≤ nPi,
are possibly correlated unit variance Gaussian random variables, and (Z1,Z2) is statistically independent
of (X1,X2). In the following, we denote this Gaussian IC as IC(a,b,P1,P2).
The capacity region of an IC is defined as the closure of the set of rate pairs (R1,R2) for which both
receivers can decode their own messages with arbitrarily small positive error probability. The capacity
region of a Gaussian IC is known only for three cases:
• a = 0, b = 0.
• a ≥ 1, b ≥ 1: see [2]–[4].
• a = 0, b ≥ 1; or a ≥ 1, b = 0: see [5]
For the second case both receivers can decode the messages of both transmitters. Thus this IC acts as two
multiple access channels (MACs), and the capacity region for the IC is the intersection of the capacity
region of the two MACs. However, when the interference is weak or moderate, the capacity region is still
unknown. The best inner bound of the capacity region is obtained in [4] by using superposition coding
and joint decoding. A simplified form of the Han-Kobayashi region was given by Chong-Motani-Garg
[6], [7]. Various outer bounds have been developed in [8]–[12]. Sato’s outer bound in [8] is derived by
allowing the receivers to cooperate. Carleial’s outer bound in [9] is derived by decreasing the noise power.
Kramer in [10] presented two outer bounds. The first is obtained by providing each receiver with just
enough information to decode both messages. The second outer bound is obtained by reducing the IC to
a degraded broadcast channel. Both of these two bounds dominate the bounds by Sato and Carleial. The
recent outer bounds by Etkin, Wang, and Tse in [11] are also based on genie-aided methods, and they
show that Han and Kobayashi’s inner bound is within one bit or a factor of two of the capacity region.
This result can also be established by the methods of Telatar and Tse [12]. We remark that neither of the
bounds of [10] and [11] implies each other. But as a rule of thumb, our numerical results show that the
bounds of [10] are better at low SNR while those of [11] are better at high SNR. The bounds of [12] are
not amenable to numerical evaluation since the optimal distributions of the auxiliary random variables
are unknown. None of the above outer bounds is known to be tight for the general Gaussian IC.
In this paper, we present a new outer bound on the capacity region of Gaussian ICs that improves on
the bounds of [10], [11]. The new bounds are based on a genie-aided approach and a recently proposed
extremal inequality [13]. Unlike the genie-aided method used in [10, Theorem 1], neither receiver is
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required to decode the messages from the other transmitter. Based on this outer bound, we obtain new
sum-rate capacity results (Theorem 2 and 3) for ICs satisfying some channel coefficient and power
constraint conditions. We show that the sum-rate capacity can be achieved by treating the interference
as noise when both the channel gain and the power are weak. We say that such channels have noisy
interference. For this kind of noisy interference, the simple single-user transmission and detection strategy
is sum-rate optimal. In Theorem 3, we show that for ICs with a > 1,0 < b < 1 and satisfying another
condition, the sum-rate capacity is achieved by letting user 1 fully recover messages from user 2 first
before decoding its own message, while user 2 only recovers its own messages.
This paper is organized as follows. In Section II, we present a new genie-aided outer bound and the
resulting sum-rate capacity for certain Gaussian ICs. We prove these results in Section III. Numerical
examples are given in Section IV, and Section V concludes the paper.
II. MAIN RESULTS
A. General outer bound
The following is a new outer bound on the capacity region of Gaussian ICs.
Theorem 1: If the rates (R1,R2) are achievable for IC(a,b,P1,P2) with 0 < a < 1,0 < b < 1, they
must satisfy the following constraints (1)-(3) for µ > 0,1+bP1
b+bP1≤ η1≤1
band a ≤ η2≤a+aP2
?+1
?+µ
?
+η2
2logaη2− a
1+aP2:
R1+ µR2
≤min
ρi∈[0,1]
(σ2
1,σ2
2)∈Σ
+µ
2log
1
2log
?
1 +P∗
1
σ2
1
?
−1
2log?aP∗
2+ 1 − ρ2
1
2log
?
1 + P1+ aP2−(P1+ ρ1σ1)2
P1+ σ2
1
?
?
1 +P∗
2
σ2
2
?
?
−µ
2log?bP∗
−η1
1+ 1 − ρ2
2
2log
?
1 + P2+ bP1−(P2+ ρ2σ2)2
P2+ σ2
2
?
,(1)
R1+ η1R2
≤
1
2log
1
2log(1 + P1+ aP2) −1
?
1 +bη1− 1
b − bη1
2log
?
1 +a − η2
η2− 1
1 +bη1− 1
1 − η1
+η1
2log(1 + bP1+ P2),
?
(2)
R1+ η2R2
≤
2log
?
?
1 +a − η2
?
,
(3)
where
Σ=



??σ2
??σ2



P2,
1,σ2
2
?
?
|
|
σ2
1> 0,0 < σ2
2≤1−ρ2
σ2
2> 0
1
a
?
?
,
if µ ≥ 1,
if µ < 1,
1,σ2
2
0 < σ2
1≤1−ρ2
2
b,,
(4)
and if µ ≥ 1 we have
P∗
1
=






P1,0 < σ2
1≤
??
P1+1−ρ2
1
µ− 1
?
P1+1−ρ2
?+
2
bµ
?+,
1−ρ2
2−bµσ2
bµ−b
1
,
??
σ2
1
µ− 1
1>1−ρ2
?
2
2
bµ
< σ2
1≤1−ρ2
2
bµ,
0,
bµ,
(5)
P∗
2
=0 < σ2
2≤1 − ρ2
1
a
,
(6)
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where (x)+? max{x,0}, and if 0 < µ < 1 we have
P∗
1
=P1,0 < σ2
1≤1 − ρ2
?
(µ − 1)P2+µ(1−ρ2
σ2
a
2
b
,
(7)
P∗
2
=









P2,0 < σ2
2≤(µ − 1)P2+µ(1−ρ2
1)
a
1)
a
?+,
µ(1−ρ2
1)−aσ2
a−aµ
2
,
??+
< σ2
2≤µ(1−ρ2
1)
a
,
0,
2>µ(1−ρ2
1)
.
(8)
Remark 1: The bounds (1)-(3) are obtained by providing different genie-aided signals to the receivers.
There is overlap of the range of µ, η1, and η2, and none of the bounds uniformly dominates the other
two bounds. Which one of them is active depends on the channel conditions and the rate pair.
Remark 2: Equations (2) and (3) are outer bounds for the capacity region of a Z-IC, and a Z-IC is
equivalent to a degraded IC [5]. For such channels, it can be shown that (2) and (3) are the same as the
outer bounds in [14]. For 0 ≤ η1≤1+bP1
Z-IC (or degraded IC) because there is no power sharing between the transmitters. Consequently,1+bP1
b+bP1and η2≥a+aP2
1+aP2, the bounds in (2) and (3) are tight for a
b+bP1
anda+aP2
1+aP2are the negative slopes of the tangent lines for the capacity region at the corner points.
Remark 3: The bounds in (2)-(3) turn out to be the same as the bounds in [10, Theorem 2]. We
show this by proving that (3) is equivalent to [10, page 584, (37)-(38)] but with equalities rather than
inequalities. Consider the rates
R1
=
1
2log
1
2log?1 + P′
P1
a
?
1 +
P′
1
P′
2+ 1/a
?
(9)
R2
=
2
?
(10)
P′
1+ P′
2
=
+ P2
(11)
for 0 ≤ P′
1≤ P1. We rewrite (9) and (10) in the form of the weighted sum
R1+ αR2=1
2log
?
1 +
P′
1
P′
2+ 1/a
?
+α
2log?1 + P′
2
?.
(12)
Observe that (12) represents a line with slope α where
α=−∂R1
∂R2
−∂R1
∂P′
=
2
?∂R2
?
∂P′
2
=−
a + aP′
1 + aP′
∂ log1 +P1/a+P2−P′
P′
∂P′
2
2
2+1/a
?
?∂ log(1 + P′
2)
∂P′
2
=
2
2
.
(13)
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We thus obtain
P′
2=
a − α
aα − a.
(14)
Substituting (14) into (12), we have
R1+ αR2=1
2log(1 + P1+ aP2) −1
2log
?
1 +a − α
α − 1
1+aP2follows from (13) and 0 ≤ P′
?
+α
2log
?
1 +a − α
aα − a
?
,
which is the same as (3). The relation a ≤ α ≤a+aP2
Remark 4: The bounds in [10, Theorem 2] are obtained by getting rid of one of the interference links
2≤ P2.
to reduce the IC into a Z interference channel (or Z-IC, see [5]). Next, the proof in [10] allowed the
transmitters to share their power, which further reduces the Z-IC into a degraded broadcast channel. Then
the capacity region of this degraded broadcast channel is an outer bound for the capacity region of the
original IC. The bounds in (2) and (3) are also obtained by reducing the IC to a Z-IC. Although we do not
explicitly allow the transmitters to share their power, it is interesting that these bounds are equivalent to the
bounds in [10, Theorem 2] with power sharing. In fact, a careful examination of our new derivation reveals
that power sharing is implicitly assumed. For example, for the term h(Xn
1+ Zn
1) −η1h
2) user 1 uses all the
?√bXn
1+ Zn
2
?
of (43) below, user 1 uses power P∗
1=bη1−1
b−bη1≤ P1, while for the term η1h(Yn
power P1. This is equivalent to letting user 1 use the power P∗
1for both terms, and letting user 2 use a
power that exceeds P2. To see this, consider (43) below and write
n(R1+ η1R2)≤
n
2log(P∗
n
2log(P∗
n
2log?P′
2? P2+ b(P1− P∗
1+ 1) −nη1
1+ 1) −nη1
1+ 1?−nη1
2
log(bP∗
1+ 1) +nη1
2
log(1 + bP1+ P2) + nǫ
=
2
log(bP∗
1+ 1) +nη1
2
log(1 + bP∗
1+ P2+ b(P1− P∗
?+ nǫ,
1)) + nǫ
=
2
log?bP′
1). Therefore, one can assume that user 2 uses extra power
1+ 1?+nη1
2
log?1 + bP′
1+ P′
2
where P′
1? P∗
1, and P′
provided by user 1.
Remark 5: Theorem 1 improves [11, Theorem 3]. Specifically, for the three sum-rate bounds of [11,
Theorem 3], the first bound can be obtained from (43) with P∗
1= P1in (44). Therefore, the bound in
(2) is tighter than the first sum-rate bound of [11, Theorem 3]. Similarly, the bound in (3) is tighter than
the second sum-rate bound of [11, Theorem 3]. The third sum-rate bound in [11, Theorem 3] is a special
case of (1) with σ2
1=1
b,σ2
2=1
a,ρ1= ρ2= 0.
Remark 6: Our outer bound is not always tighter than that of [11] for all rate points. The reason is that
in [11, last two equations of (39)], different genie-aided signals are provided to the same receiver. Our
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outer bound can also be improved in a similar and more general way by providing different genie-aided
signals to the receivers. Specifically the starting point of the bound is
n(R1+ µR2) ≤
k
?
i=1
λiI (Xn
1;Yn
1,Ui) +
m
?
j=1
µiI (Xn
2;Yn
2,Wj) + nǫ,
(15)
where?k
B. Sum-rate capacity for noisy interference
i=1λi= 1,?m
j=1µj= µ,λi> 0,µj> 0.
The outer bound in Theorem 1 is in the form of an optimization problem. Four parameters ρ1,ρ2,σ2
1,σ2
2
need to be optimized for different choices of the weights µ,η1,η2. When µ = 1, Theorem 1 leads directly
to the following sum-rate capacity result.
Theorem 2: For the IC(a,b,P1,P2) satisfying
√a(bP1+ 1) +
√
b(aP2+ 1) ≤ 1,
(16)
the sum-rate capacity is
C =1
2log
?
1 +
P1
1 + aP2
?
+1
2log
?
1 +
P2
1 + bP1
?
.
(17)
Remark 7: The sum-rate capacity for a Z-IC with a = 0, 0 < b < 1 is a special case of Theorem 2
since (16) is satisfied. The sum capacity is therefore given by (17).
Theorem 2 follows directly from Theorem 1 with µ = 1. It is remarkable that a genie-aided bound
is tight if (16) is satisfied since the genie provides extra signals to the receivers without increasing the
rates. This situation is reminiscent of the recent capacity results for vector Gaussian broadcast channels
(see [15]). Furthermore, the sum-rate capacity (17) is achieved by treating the interference as noise. We
therefore refer to channels satisfying (16) as ICs with noisy interference. Note that (16) involves both
channel gains a,b and both powers P1and P2. The constraint (16) implies that
√a +
√b ≤ 1.
(18)
Moreover, as shown in Fig. 1, the powers P1and P2must be inside the triangle defined by:
P1
≥0,
P2
≥0,
b√aP1+ a
√
bP2
≤1 −√a −
√
b.
(19)
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These constraints can be considered as a counterpart of the IC with very strong interference [2] whose
powers should be inside the rectangle defined in Fig. 2:
a > 1,b > 1,
0 ≤ P1≤ a − 1,
0 ≤ P2≤ b − 1.
The ICs with noisy interference and ICs with very strong interference are two extreme cases in terms
of the decoding strategy to achieve the sum-rate capacity. In the former case, the sum-rate capacity is
achieved by treating interference as noise, while in the latter case, the interference is decoded before, or
together with, the intended messages.
For symmetric Gaussian ICs with a = b and P1= P2, the conditions in (18) and (19) become
a = b≤
1
4,
√a − 2a
2a2
(20)
P1= P2= P≤
.
(21)
“Noisy interference” is therefore “weaker” than “weak interference” as defined in [5] and [16], namely
√1+2P−1
2P
or
a ≤
P ≤1 − 2a
a2
.
(22)
Recall that [16] showed that for “weak interference” satisfying (22), treating interference as noise achieves
larger sum rate than time-or frequency-division multiplexing (TDM/FDM), and [5] claimed that in “weak
interference” the largest known achievable sum rate is achieved by treating the interference as noise.
C. Capacity region corner point
Theorem 3: For an IC(a,b,P1,P2) with a > 1, 0 < b < 1, the sum-rate capacity is
C =1
2log(1 + P1) +1
2log
?
1 +
P2
1 + bP1
?
(23)
when the following condition holds
(1 − ab)P1≤ a − 1.
(24)
A similar result follows by swapping a and b, and P1and P2.
Under the constraint (24), we have the following inequality:
1
2log
?
1 +
P2
1 + bP1
?
≤1
2log
?
1 +
aP2
1 + P1
?
.
(25)
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1 −√a −√b
√ab
1 −√a −√b
a√b
P1
P2
0
a > 0
b > 0
√a +√b ≤ 1
Fig. 1.Power region for the IC with noisy interference.
a−1
b−1
P1
P2
0
a ≥ 1
b ≥ 1
Fig. 2.Power region for the IC with very strong interference.
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Therefore, the sum-rate capacity is achieved by a simple scheme: user 1 transmits at the maximum rate
and user 2 transmits at the rate that both receivers can decode its message with single-user detection.
Observe further that this rate pair permits R2=1
2log
?
1 +
aP2
1+P1
?
when R1reaches its maximum. Such a
rate constraint was considered in [5, Theorem 1] which established a corner point of the capacity region.
However it was pointed out in [16] that the proof in [5] was flawed. Theorem 3 shows that the rate pair
of [16] is in fact a corner point of the capacity region when a > 1,0 < b < 1 and (24) is satisfied, and
this rate pair achieves the sum-rate capacity.
The sum-rate capacity of the degraded IC (ab = 1,0 < b < 1) is a special case of Theorem 3. Besides
this example, there are two other kinds of ICs to which Theorem 3 applies. The first case is ab > 1. In
this case, P1can be any positive value. The second case is ab < 1 and P1≤
signals from user 2 can be decoded first at both receivers.
a−1
1−ab. For both cases, the
D. State of the Art
We reiterate that both Theorems 2 and 3 are direct results of Theorem 1, and Theorem 1 is derived
by having a genie provide extra information to the receivers. We summarize the sum-rate capacity for
Gaussian ICs from Theorems 2 and 3 and previous results in [2]–[4]. In Fig. 3, four curves ab = 1,
a = 1, b = 1, and√a +√b ≤ 1 divide channel gain plane into 7 regimes. The sum-rate capacity for
each regime under certain power constraints is shown in Tab. I.
III. PROOFS OF THE MAIN RESULTS
We introduce some notation. We write vectors and matrices by using a bold font (e.g., X and S). When
useful we also write vectors with length n using the notation Xn. The ith entry of the vector X (or Xn)
is denoted as Xi. Random variables are written as uppercase letters (e.g., Xi) and their realizations as the
corresponding lowercase letter (e.g., xi). We usually write probability densities and distributions as p(x)
if the argument of p(·) is a lowercase version of the random variable corresponding to this density or
distribution. The notation h(X) and Cov(X) refers to the respective differential entropy and covariance
matrix of X. The notation of U|V = v and U|V denotes the random variable U conditioned on the event
V = v and the random variable V , respectively.
The proof utilizes the extremal inequalities introduced in [13]. We present them below for completeness.
Lemma 1:
[13, Theorem 1] For any µ ≥ 1 and any positive semi-definite S, a Gaussian X is an
optimal solution of the following optimization problem:
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01
a
2
0
1
2
ab = 1
√a +√b = 1
b = 1
a = 1
III
IV
III
V
VI
VII
b
Fig. 3.Gaussian IC channel coefficient regimes for Tab. I
TABLE I
SUM-RATE CAPACITY.
(a,b)(P1,P2)
sum-rate capacity
I
a ≥ 1,b ≥ 1P1> 0,P2> 0 min









1
2log(1 + P1) +1
2log(1 + P2)
1
2log(1 + P1+ aP2)
1
2log(1 + bP1+ P2)
?
same as above









II
ab ≥ 1,a ≤ 1
ab ≤ 1,b ≥ 1
ab ≥ 1,b ≤ 1
ab ≤ 1,a ≥ 1
√a +√b ≤ 1
√a +√b > 1,a < 1,b < 1
P1> 0,P2> 0
1
2log
?
1 +
P1
1+aP2
+1
2log(1 + P2)
III
P1> 0,P2≤
P1> 0,P2> 0
b−1
1−ab
IV
1
2log(1 + P1) +1
2log
?
1 +
P2
1+bP1
?
V
P1≤
a−1
1−ab,P2> 0
same as above
?
unknown
VI
√a(1 + bP1) +√b(1 + aP2) ≤ 1
P1> 0,P2> 0
1
2log
?
1 +
P1
1+aP2
+1
2log
?
1 +
P2
1+bP1
?
VII
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max
p(x)
h(X + U1) − µh(X + U2)
Cov(X) ? S,
subject to
where U1and U2are Gaussian vectors with strictly positive definite covariance matrices K1and K2,
respectively, and the maximization is over all X independent of U1and U2.
Lemma 2: [13, Corollary 4] For any real number µ and any positive semi-definite S, a Gaussian X
is an optimal solution of the following optimization problem:
max
p(x)
h(X + U1) − µh(X + U1+ U)
Cov(X) ? S,
subject to
where U1and U are two independent Gaussian vectors with strictly positive definite covariance matrices
K1and K, respectively, and the maximization is over all X independent of U1and U2.
For example, consider the following optimization problem
max
p(x)
h(X + U1) − µh(X + U2)
1
ntr(S) ≤ P,
subject to
S = E?XXT?,
(26)
and suppose that S∗is the optimal covariance matrix for X. When µ ≥ 1, the problem (26) is equivalent
to the problem of Lemma 1 with S replaced by S∗. Similarly, when µ < 1 the problem (26) is equivalent
to the problem of Lemma 2 with S replaced by S∗and U2 = U1+ U. Therefore a Gaussian X is
optimal for problem (26) in both cases. We further have the following two simple optimization results.
Corollary 1: The optimization problem of Lemma 1 with the matrix constraint replaced by the trace
constraint (or the problem (26) with µ ≥ 1) for the special case Cov(Ui) = σ2
Cov(X) = P∗I, where
iI, i = 1,2, has the solution
P∗=












0,0 < σ2
2< µσ2
1
σ2
2−µσ2
µ−1,
P,
1
µσ2
1≤ σ2
2≥ µσ2
2< µσ2
1+ (µ − 1)P
σ2
1+ (µ − 1)P
(27)
Alternatively, we can write (27) as
P∗=






P,0 < σ2
1≤
?σ2
µP
2
µ−µ−1
?+
µP
?+
σ2
2−µσ2
µ−1,
0,
1
?σ2
σ2
2
µ−µ−1
1>σ2
< σ2
1≤σ2
2
µ
2
µ
(28)
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12
Corollary 2: The optimization problem of Lemma 2 with the matrix constraint replaced by the trace
constraint (or the problem of (26) with µ < 1 and σ2
1≤ σ2
2) for the special case Cov(U1) = σ2
1I,
Cov(U) =?σ2
2− σ2
1
?I, where σ2
1≤ σ2
2, has the solution Cov(X) = P∗I where
P∗= P.
(29)
Proof: Suppose the eigenvalue decomposition of S is S = QΛQTand Λ = diag(λ1,...,λn). Since
Gaussian X is optimal, we have
h(X + U1) − µh(X + U2)
=1
2log?(2πe)n??S + σ2
2log??Λ + σ2
=1
2
i=1
? f(Λ)
1I???−µ
2log??Λ + σ2
?−µ
2log?(2πe)n??S + σ2
2I???
=1
1I??−µ
2I??+1 − µ
log?λi+ σ2
2
log(2πe)n
n
?
log?λi+ σ2
1
2
n
?
i=1
2
?+1 − µ
2
log(2πe)n
By using the Lagrangian of f(Λ) with the constraint?n
λiis λ∗
i=1λi= nP, it can be shown that the optimal
i= P∗with P∗defined in (27)-(29).
Finally we need another lemma to prove our main results.
Lemma 3: Suppose that (U,V ) is Gaussian with covariance matrix


σ2
1
ρσ1σ2
ρσ1σ2
σ2
2

, σ1 > 0,
σ2> 0, |ρ| < 1, and W is Gaussian with variance
variable X is independent of (U,V ) and X is independent of W, then we have
?1 − ρ2?σ2
1. If the discrete or continuous random
h(X + U|V ) = h(X + W)
(30)
Proof: We have
h(X + U|V )=
?
?
?
h?X + W′?
h(X + W)
fV(v)h(X + U |V = v)dv
?
fV(v)h?X + W′?dv
(a)
=fV(v)hX + W′+ρσ1
σ2
V
????V = v
?
dv
(b)
=
=
=
February 2, 2008DRAFT

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13
where W′is identically distributed as W but independent of (U,V ). (a) follows because
has the same joint distribution as (U,V ), (b) follows becauseρσ1
?
W′+ρσ1
σ2V,V
?
σ2V becomes a constant when conditioned
on V = v.
Since U|V = v is also Gaussian distributed with mean valueρσ1
3 shows that U|V can be replaced by an equivalent Gaussian random variable with the same variance.
σ2v and variance?1 − ρ2?σ2
1, Lemma
A. Proof of Theorem 1
Let N1and N2be two zero-mean Gaussian variables with variances σ2
1and σ2
2respectively, and set
E(N1Z1) = ρ1σ1and E(N2Z2) = ρ2σ2. We further define Nn
independent and identically distributed (i.i.d.) elements distributed as N1and N2, respectively.
1and Nn
2to be Gaussian vectors with n
Starting from Fano’s inequality, we have that reliable communication requires
n(R1+ µR2)
≤ I (Xn
≤ I (Xn
= I (Xn
1;Yn
1) + µI (Xn
2;Yn
2) + nǫ
1;Yn
1,Xn
1+ Nn
1) + µI (Xn
2;Yn
2,Xn
2+ Nn
2) + nǫ
1;Xn
1+ Nn
1) + I (Xn
1;Yn
1|Xn
1+ Nn
1) + µI (Xn
1) − h?√aXn
2|Nn
2;Xn
2+ Nn
2) + µI (Xn
2;Yn
2|Xn
2+ Nn
2) + nǫ
= h(Xn
1+ Nn
1) − h(Nn
1) + h(Yn
?√
1|Xn
bXn
1+ Nn
2+ Zn
1|Nn
1
?+ µh(Xn
2+ Zn
2) − µh(Nn
2)
+µh(Yn
2|Xn
2+ Nn
2) − µh
1+ Zn
2
?
+ nǫ
(31)
where ǫ → 0 as n → ∞. For h(Yn
have
1|Xn
1+ Nn
1), zero-mean Gaussian Xn
1and Xn
2are optimal, and we
1
nh(Yn
1|Xn
1+ Nn
1)≤
1
n
n
?
n
?
i=1
h(Y1i|X1i+ N1)
=
1
n
i=1
?h?X1i+√aX2i+ Z1,X1i+ N1
?
2i). Consider the function
f(p1,p2) = 1 + ap2+ p1−(p1+ ρ1σ1)2
?− h(X1i+ N1)?
=
1
2n
n
i=1
log
?
2πe
?
1 + aP2i+ P1i−(P1i+ ρ1σ1)2
P1i+ σ2
1
??
(32)
where P1i= E(X2
1i) and P2i= E(X2
p1+ σ2
1
(33)
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14
for which we compute
∂f
∂p1
∂f
∂p2
∂2f
∂p2
∂2f
∂p2
∂2f
∂p1∂p1
=σ2
1(σ1− ρ1)2
(p1+ σ2
1)2
≥ 0
(34)
= 1
(35)
1
= −2σ2
1(σ1− ρ1)2
(p1+ σ2
1)3
≤ 0
(36)
2
= 0
(37)
= 0.
(38)
Since log(x) is concave in x we have that the logarithm in (32) is concave in (P1i,P2i). We thus have
1
nh(Yn
1|Xn
1+ Nn
1)≤
1
2log
?
2πe
?
1 +a
n
n
?
i=1
P2i+1
n
n
?
i=1
P1i−
?1
??
n
?n
n
i=1P1i+ ρ1σ1
?n
?2
1
i=1P1i+ σ2
1
??
≤
1
2log
?
2πe
?
1 + aP2+ P1−(P1+ ρ1σ1)2
P1+ σ2
1
(39)
where the first inequality follows from Jensen’s inequality, and the second inequality follows from the
block power constraints1
n
?n
j=1Pij≤ Pi, i = 1,2, and (34).
For the same reason, we have
1
nh(Yn
2|Xn
2+ Nn
2) ≤1
2log
?
2πe
?
1 + bP1+ P2−(P2+ ρ2σ2)2
P2+ σ2
2
??
.
(40)
Let W′
2= Z2|N2, then W′
W2with variance 1 − ρ2
2is Gaussian distributed with variance 1−ρ2
2. From Lemma 3 and Corollaries 1 and 2 we have
?√
= h(Xn
1) − µh
?
≤n
2. Define a new Gaussian variable
h(Xn
1+ Nn
1) − µh
bXn
?√
Xn
1+ Zn
2|Nn
2
?
1+ Nn
bXn
1+ Wn
2
?
−nµ
= h(Xn
1+ Nn
1) − µh
1+Wn
2
√b
?
2
logb
2log?2πe?P∗
1+ σ2
1
??−nµ
2
log?2πe?bP∗
1+ 1 − ρ2
2
??,
(41)
where P∗
1is defined in (5) and (7). For the same reason, we have
µh(Xn
2+ Zn
2) − h?√aXn
2is defined in (6) and (8). From (31), (39)-(42) we obtain the rate constraint (1).
2+ Zn
1|Nn
1
?≤nµ
2
log?2πe?P∗
2+ σ2
2
??−n
2log?2πe?aP2+ 1 − ρ2
1
??, (42)
where P∗
February 2, 2008 DRAFT

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15
On the other hand, we have
n(R1+ η1R2)≤I (Xn
1;Yn
1) + η1I (Xn
2;Yn
2) + nǫ
≤I (Xn
1;Yn
1,Xn
2) + η1I (Xn
2;Yn
2) + nǫ
=I (Xn
1;Yn
1|Xn
2) + η1I (Xn
2;Yn
2) + nǫ
=h(Yn
1|Xn
2) − h(Yn
1|Xn
?√
2
1,Xn
2) + η1h(Yn
2) − η1h(Yn
− h(Zn
1+ 1) +nη1
2
2|Xn
2) + nǫ
=h(Xn
1+ Zn
1) − η1h
1+ 1) −nη1
bXn
1+ Zn
2
?
1) + η1h(Yn
2) + nǫ
≤
n
2log(P∗
log(bP∗
log(1 + bP1+ P2) + nǫ,
(43)
where the last step follows by Corollaries 1 and 2. We further have
P∗
1=









P1,η1≤1+bP1
1+bP1
b+bP1≤ η1≤1
η1≥1
1= 0 are redundant, we have
−η1
b+bP1
bη1−1
b−bη1,
0,
b
b.
(44)
Since the bounds in (43) when P∗
1= P1and P∗
1 +bη1− 1
b − bη1
R1+ η1R2
≤
1
2log
?
?
2log
?
1 +bη1− 1
1 − η1
?
+η1
2log(1 + bP1+ P2),
(45)
for1+bP1
b+bP1≤ η1≤1
b, which is (2). We similarly obtain (3).
B. Proof of Theorem 2
By choosing
σ2
1=
1
2b
1
2a
?
?
1 − aσ2
1 − bσ2
b(aP2+ 1)2− a(bP1+ 1)2+ 1 +
?
?
[b(aP2+ 1)2− a(bP1+ 1)2+ 1]2− 4b(aP2+ 1)2
?
?
(46)
σ2
2=
a(bP1+ 1)2− b(aP2+ 1)2+ 1 +
[a(bP1+ 1)2− b(aP2+ 1)2+ 1]2− 4a(bP1+ 1)2
(47)
ρ1=
?
?
2
(48)
ρ2=
1,
(49)
the bound (1) with µ = 1 is
R1+ R2≤1
2log
?
1 +
P1
1 + aP2
?
+1
2log
?
1 +
P2
1 + bP1
?
.
(50)
By one can achieve equality in (50) by treating the interference as noise at both receivers.
In order that the choice of σ2
1, σ2
2, ρ1and ρ2be feasible, there must exist at least one pair (σ2
1,σ2
2)
satisfying the following conditions:
σ2
1≥ 0,σ2
2≥ 0,ρ1≤ 1,ρ2≤ 1.
February 2, 2008 DRAFT