Wireless Communications and Networking Laboratory. Electrical Engineering Department. The Pennsylvania State University,
A New Outer Bound for the Gaussian Interference Channel with Confidential Messages Xiang He
Aylin Yener
Wireless Communications and Networking Laboratory Electrical Engineering Department The Pennsylvania State University, University Park, PA 16802
[email protected] [email protected]
Abstract-In this work, we derive new outer bounds for the two-user interference channel with confidential messages. An upper bound is found for the sum rate. When the interfering link of the first user is greater than 1, a new upper bound on 2R 1 + R2 is obtained by studying a special form of the threeuser interference channel. The bounds are then compared with known bounds for the symmetric interference channel under strong interference regime. In particular, examples are presented to showcase for channel parameters where positive secrecy rates are known to be achievable, the new bounds improve upon the known outer bounds on the secrecy capacity region. It is shown that, in some cases, the 2R 1 + R 2 bound also improves the bound on the sum rate.
I. INTRODUCTION
In a wireless environment, interference is always present. Traditionally, interference is viewed as a harmful physical phenomenon that should be avoided. Yet, from the secrecy perspective, if interference is more harmful to an eavesdropper, it can be a resource to protect confidential messages. To fully appreciate and evaluate the potential benefit of interference to secrecy, it is therefore important to find the fundamental transmission limits in an multi-user model with interference when secrecy constraints must be met. The simplest model of this type is the two-user interference channel, which has been studied under several different security constraints up to date. References [1 ]-[3] have studied the case where each receiver is an eavesdropper for the messages not intended for it. References [4], [5] have considered the case where the eavesdropper is external to both receivers and is interested in the secret messages of both users. Reference [6] has considered the case in the setting of cognitive radios, where both transmitters know one public message and try to send it to both receivers, and only one transmitter has a secret message intended only for its receiver. For a Gaussian two-user fully connected interference channel with the security constraints as defined in [1], several results have been derived up to date. On the achievability side, reference [1] has derived the achievable rate region when the channel has weak interference. Reference [2] derived the achievable rate region when only one user has secret messages to send. For the converse, reference [1] has derived an outer bound for the discrete memoryless case, which is difficult to evaluate for the Gaussian channel due to the presence of auxiliary random variables. In reference [2], a genie bound for
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individual rates is derived, which is close to achievable rates with weak interference when the transmission power is small. In this work, we focus on improving the converse results for this model. Our starting point is the converse proof in [3], which is a sum rate bound derived for a one-sided interference channel under weak interference. From the converse viewpoint, the technique in [3] essentially decomposes the Zchannel into a separate point to point link and a wire-tap channel. In this work, we verify that this technique actually does not rely on whether the channel is under weak interference or not. By converting the fully connected interference channel into the Z-channel and applying this technique, a sum rate bound is obtained. We next introduce a special form of the three-user interference channel, which we call the "Zigzag" channel, and show that an upper bound on 2R 1 + R 2 of the two-user case can be derived from the sum rate bound for this three-user channel. When the first user's interfering link channel gain is greater than 1, the sum rate upper bound of the corresponding threeuser channel can be derived by combining the techniques in [2], [3]. This leads to our new upper bound for the-two user case. These bounds are then compared for the symmetric twouser interference channel, in the strong but not very strong interference regime. In particular, we present numerical examples for channel parameters for which, using achievable results from [2], positive secrecy rates are achievable; and therefore deriving upper bounds is of interest. For these examples, the new bound derived from the Zigzag channel improves the previously known outer bounds on the secrecy capacity region. In one example, we observe that this bound even improves the equal rate point. The rest of the paper is organized as follows: Section II describes the model. Section III derives the upper bounds. Section IV compares these bounds and presents the numerical examples. Section V concludes the paper. The following notation is used throughout the paper: C (x) == ~ log2 (1 + x). Yj,i denotes the ith component of Yjn. Yji denotes its first i components. II. SYSTEM MODEL
The Gaussian two-user interference channel with confidential messages is shown in Figure 1. The received signals at the
318
WI
®Xl
Zl
81
va
@ Z2
W2
Fig. 1.
~
III. UPPER BOUNDS WI
Vb X 2
:+
Y2@ D 2
A. Invariance property of the secrecy capacity region
W2
®
Two-user interference channel with confidential messages
two receivers can be expressed as
+ jaX2 + Zl
(1)
y 2 ==!bX I +X2 +Z2
(2)
YI == Xl
where Zi, i == 1,2 is a zero-mean Gaussian random variable with unit variance. Let n be the total number of channel uses. Node Si has the following power constraint:
Lemma 1: The secrecy capacity region of an interference channel with confidential messages is invariant under a joint channel noise distribution P(ZI, Z2), as long as it leads to the same marginal distributions P(ZI) and p(Z2)' Remark 1: The same property has been stated for the interference channel in [2] where only one user has secret message to send. Here, we verify this property when both users have secret messages to send. Proof' From Section II, we can see the secrecy capacity region is defined by the following four joint distributions: (10)
Without loss of generality, we examine p (WI, ¥.in). For this term, we have: (11)
p (YjIWI ) ==
(3)
L L
p(Xr X 2YjIWI )
(12)
p (XrIWI) p (X~) p (¥.inIXr X~)
(13)
Xf,X:;
==
where Xi,j is the jth component of Xf. Let Ii be the stochastic encoder used by the source node Si,i==1,2:
Xf,X:;
==
(4)
Let hi be the decoder used by the destination node D i , i == 1,2. Let Wi be the estimate of Wi computed by node D i , i == 1,2. (5)
We assume there is no common randomness shared by the encoders. This means, the input distribution to the channel is constrained to be
L
n
p(XrIWI)p(X~)IIp(}j,il¥.ii-IXrX~)
Xf,X:;
==
L
(14)
i=l n
p(XrIWI)p(X~) IIp(Zj,i)
Xf,X:;
(15)
i=l
Hence p(WI , ¥.in) only depends on the marginal distribution p (Zj). The same holds true for p(W2 , W jn ). Hence we have the lemma. • Remark 2: This invariance property also applies to K-user (K 2: 3) interference channel.
2
IIp (Wi) p (XfIWi )
(6)
i=l
Given (6), Xf, i == 1,2 are independent from each other. For D i to receive Wi reliably, we require
WI
®Xl
Zl
~
81
Vb
(7)
W2
In addition, since WI must be kept secret from D 2, and W 2 must be kept secret from D I , we require
=
Y2@ D 2
W2
®
Z-channel with a confidential message
~H (W2 IY/") n
(9)
The channel can be changed into a one-sided interference channel as follows:
Define R i == limn~oo ~H (Wi)' Then the equivocation capacity region of the two-user interference channel with confidential messages is then defined as any rate pair (R I, R 2 ) such that (7), (8) and (9) are met.
1) First we remove the eavesdropper at D I . 2) Then we reveal the signal X!j to node D I as genie information. The signals received by node D I essentially becomes Xl + Zr , X!j. Since X!j is independent from WI and Xl + Zr, node D I can discard X!j.
n~oo
~H (W2 ) n
:+
(8)
lim
=
;l
~H (Wi ly2n ) n
lim n~oo
~H (Wi) n
Fig. 2.
~X2
Z2
WI
lim n~oo
lim n~oo
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319
The resulting channel is a Z-channel shown in Figure 2. When b < 1, the outer bound on R I + R 2 for the Z-channel has been given in [3]:
WI
W3
Fig. 3.
+ R2) -
nc
1 (WI; y 2n ) + 1 (W2; y 2n ) ==1 (WI; YIn) - 1 (WI; v n ) - 1 (WI; Y;IV n ) + 1 (WI; v n ly2n ) + 1 (W2; y 2n ) n ~1 (WI; YIn) - 1 (WI; V ) ~1 (WI; YIn) -
+ 1 (WI; v n ly2n ) + 1 (W2; Y;)
n ~1 (WI; YIn) - 1 (WI; V ) n n + 1 (WI; v ly2 ) + 1 (X2'; y 2n )
(18)
(19) (20)
where limn~oo c == o. (18) follows from (7), (8) and Fano's inequality. Then, as in [3], we prove
1 (W1,. vnlyn) 2
n . yn) < 1 (X n . ynlx n ) + 1 (X 2' 2 2' 2 1
(22)
This is because:
1 (WI; v n ly2n ) + 1 (X2;Y;) n n ~1 (Xf; v ly2 ) + 1 (X 2;Y;) n ) + 1 (X n . X n ) n ==1 (X 2'. ynlx 2 1 2' 1
- 1 (X2; Xfly2n ) + 1 (Xf; v n ly2n ) ==1 (X 2;y 2n lXf) + 1 (Xf; v n ly2n ) - 1 (Xf; X 21y2n ) ==1 (X 2;y 2n lXf) + 1 (Xf; vnlv n + X 2)
- 1 (Xf; X 21V n + X~) ==1 (X 2;y 2n lXf) + 1 (Xf; X 21V n + X 2) - 1 (Xf; X2'IV n + X2') ==1 (X 2;y 2n lXf)
(23)
(26) (27)
(28)
(29)
Note that (26) follows from (6). Note also that to obtain (22), we do not rely on the joint distribution of Z 1, Z2. Finally, we prove, for b 2: 1, that: (30)
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:+
Y2@ D 2
W2
@
Z3
C0X
3
W3
@
The Zigzag channel with confidential messages
~vn == X n + ~zn
Vb
Y1n
Vb
1
2
1 zn n I -b Z- 1 = xn1+ zn1= X 1+ Vb 2+Vr:-T
n
(31) (32)
where 2r is a Gaussian random vector that has the same distribution as Z'2 but is independent from Z'2. According to Lemma 1, the secrecy capacity of the Z-channel in Figure 2 should remain unchanged. Under this correlation, we have the Markov chain WI - V n - YIn. Hence, we have 1 (WI; YIn) - 1 (WI; V n ) (WI; YIn) - 1 (WI; vnyln ) == -1 (WI; vnlyln ) ~ 0 (33)
The lemma follows from the fact that 1 (XJ]:; y 2n lXr) nC(P2 ) C. 2R I
+ R2
< •
when a > 1
When one of the interfering gains is greater or equal to 1, another bound can be derived. Here, we consider the case where a 2: 1, and derive an upper bound on 2R I + R 2 • To do so, we will first consider the Zigzag channel model shown in Figure 3. The channel outputs are:
(24)
(25)
€)X
Z2 2
WI
This is done by manipulating the correlation factor between ZI,Z2:
~1
(21)
~
va
Here we consider the case when b 2: 1. We verify that the techniques from [3] are still useful after combining it with Lemma 1. Lemma 2:
n(R I
Zl
81
Vb W2
Proof· We only prove the first term and focus on the case where b > 1. The second term can be proved in a similar fashion. Like in [3], we define V == Vbx I + Z2. Hence, we have y 2n == V n + X!j. Then the sum rate of the corresponding Z-channel can be bounded as:
®Xl
YI == Xl
+ ZI, Y2 == Vbx I + X 2 + Z2 + X 3 + Z3
Y3 == jax 2
(34)
(35)
where Zi, i == 1,2,3, is a zero-mean Gaussian random variable with unit variance. Let Pi be the power constraint of node Si. Let gi be the encoding function used by node Si and qi be the decoding function used by node D i . Node D i , i == 2, 3, is an eavesdropper for the secret message sent by node Si-I. At the same time, node D i , i == 1,2,3 is the intended receiver of the message sent by node Si. If the power constraint of the Zigzag channel is such that P3 == PI, then we can make the following statement: Lemma 3: If a secret rate pair of (R 1, R 2 ) is achievable in Figure 1, then the rate pair (R 1, R 2 , R 1) is also achievable in Figure 3. Proof' The rate pair (R I , R 2 , R I ) can be achieved as follows: Node Si, i == 1,2 uses the same encoder as node Si, i == 1, 2 in Figure 1: gi == ii, i == 1, 2.
320
Node 8 3 uses the same encoder as node 8 1 : 93 == fl. The input to the encoder of node 8 3 shares the same message set as {WI} but is independent from WI. Node D l receives Xr + Zr. At the same time, node D l can generate X!J:, that has the same distribution as X!J:, and feed YIn == Xr+ vaX!] +Zr to the decoder hI. Since WI, YIn has the same joint distribution as WI, YIn, node D l can decode WI reliably. Node D 2 uses the same decoder as node D 2 in Figure 1: q2 == h 2 . Node D 3 uses the same decoder as node D l in Figure 1: q3 == hI. Then it can be verified all reliability transmission conditions and secrecy constraints is met. Hence we proved the lemma.
Next we apply (42) to (39) and get:
n (R 1
+ R 2 + R 3 ) - n€ n n ~1 (Xr X 2; y l lY2 ) + 1 (X3; y 3 1X2) n (R 1
YIn) - 1 (WI; Y2 ) + 1 (W2; y 2n ) - 1 (W2; V3n ) + 1 (X3 ; y 3n IX2:) ==1 (WI; YIn) - 1 (WI; y 2n )
+ 1 (W2 ; Y2') n
y 2 == X2:
_l_~n _ va
3
-
2;v3n ly3n ) + 1 (X3 ;Y3 ) ==1 (X3; Y;IX2) + 1 (X3; X 2) - 1 (X3;X 21y3n ) + 1 (X 2;v3n ly3n ) ==1 (X3;Y;IX2)- 1 (X 2;X 31y3n ) + 1 (X 2;v3n ly3n ) n . y:nlx n ) - 1 (X n . xnlx n + ~n) ==1 (X3' 3 2 2' 3 3 3 + 1 (X2:; V3n lX3 + V3n ) ==1 (X3;y 3n 1X2) - 1 (X2;V3n lX3 + V3n ) + 1 (X2:; V3n lX3 + V3n ) == 1 (X3;y 3n IX2:) ~1 (X
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(43)
+ 1 (X!3; Y3n 1X 2)
x n + zr 2
If J Z!3
(50)
+
1-
(51)
~ 22
(52)
va
(53)
- I(W2 ;
=1 (W2 ; y 2n ) - I(W2 ; n
= - I(W2 ; y 2
llf
If If
V3n )
V3n ) V3n , y 2n )
:s: 0
(54)
(55)
Applying (55) to (50), we have
n(R l
+ R2 + R3) -
n€ n n ~1 (WI; YIn) - 1 (WI; y 2 ) + 1 (X 3;y 3 IX2:)
(56)
For the first two terms in (56), we have
1 (WI; YIn) - 1 (WI; y 2n ) n n ~1(Wl;YrIY2n) ~ 1 (XrX2; y l ly2 )
(57)
Applying (57) to (56) yields the lemma. • Next, we apply Lemma 1 to tighten the bound given by Lemma 4. Let p be the correlation factor between Zl and Z2. Note that (53) only determines the correlation factor between Z2 and Z3. Hence we still have the freedom to choose p. From [2], we have
lim !1 (Xr X 2;y ln lY2n ) n
n~oo
:s: f
(PI, P2, p, b, 0)
(58)
where 1
(44)
f(Pl ,P2,p,t,v) == 2"log2
(1
(45) (46)
(47)
Vbxr +
1 (W2 ; y 2n )
The proof is the same as the way we proved (22) and is repeated here for completeness: 1 (W2 ; V3n 1Y3 ) + 1 (X3;y 3n )
n
V3 )
where 2'2 is a n x 1 vector independent from Zr, with each component distributed as N(O, 1). Therefore
+ R2 + R3) -
Then, we apply the derivation from [3, Appendix] and prove 1 (W2; v3n ly3n ) + 1 (X3; y 3n ) ~ 1(X3; y 3n 1X2) (42)
If
+ VbXf + Z2:
= X 2+
(36)
n€ (37) n ==1 (WI; YIn) - 1 (WI; y 2 ) + 1 (W2; y 2n ) - 1 (W2; y 3n ) + 1 (W3 ; y 3n ) (38) n ~1 (WI; YIn) - 1 (WI; y 2 ) + 1 (W2; y 2n ) - 1 (W2; y 3n ) + 1 (X3 ;y 3n ) (39) Define V3n == vaX!] + Zr. Hence, y 3n == V3n + Xr. Then, for the last three terms in (39), we have: 1 (W2; y 2n ) - 1 (W2; y 3n ) + 1 (X3;y 3n ) n ~1 (W2; y2n ) - 1 (W2; V3 ) - 1 (W2; y3n lv3 ) + 1 (W2; v3n ly3n ) + 1 (X3;y3n ) (40) n ~1 (W2 ; Y2 ) - 1 (W2 ; V3 ) + 1 (W2; v3n ly3n ) + 1 (X3;y3n ) (41)
I(W2 ;
(49)
> 1, we have:
Since a
where limn~oo € == 0 Proof'
n(R l
n€
~1 (WI;
•
Next, we apply the technique from [3] to the Zigzag channel, and have the following lemma: Lemma 4: When a ~ 1,
+ R2 + R3) -
+ PI + vP2) (1 + tPl + P2) - (p + ViPl + VVP2) (1 - p2) (1 + tPl + P2)
2
(59)
The best p, as shown in [2], is given by
p = (1 (48)
321
+ t) PI + (1 + v)P2 + (JtV _1)2 PI P 2 -..;;s 2( ViPl + VVP2)
(60)
and
~
~
=
A. Genie Bound
is given by
[(Jt -1)2H + (JV -1)2P2 + (..;t;; -1)2PI P2] [(Jt + 1)2PI + (JV + 1)2 P2 + (..;t;; -
1)2 PI P2] (61)
We first examine the equal rate point R I == R 2 == R. For the genie bound given by (67), since the channel is symmetric, both I(X I ,X2;YI IY2) and I(X I ,X2;Y2IYI ) are tightest at the same p. Hence
R::; !(P,P,p,a,a)
Hence we have the following lemma: Lemma 5:
When P
it can be verified that
----* 00,
1
(62)
lim R - -2 log2
P~oo
This, along with Lemma 3, gives us the following upper bound for the 2-user interference channel with confidential messages: Theorem 1:
1 P 1 -log2 - - - -log2 P ::; 0 2 b+ 1 2
Proof· It can be verified that
limp~oo p ==
lim R I
1
+
lim C (P) - -log2 P == 0 2
P~oo
This yields the corollary.
C((a
+ l)P) 2: 0
+ 1, (70)
1
lim R - -log2((a + l)P) 2: 0 4
(64)
(71)
C. Comparison From Corollary 1, we have
lim R -
P~oo
(65) (66)
•
IV. COMPARISON TO KNOWN BOUNDS In this section, we compare Theorem 1 with several known outer bounds. We focus on the symmetric case PI == P2 == P, a == b, 1 < a < P + 1. Including the two bounds derived in this work by leveraging the technique from [2], [3], there are four bounds to compare 1) The genie bound from [2].
2) The bound on R I + R 2 derived in Section III-B. 3) Since a == b > 1, the bound derived in Section III-C can be used to bound 2R I + R 2 and R I + 2R 2 . 4) The bound for the strong interference channel without secrecy constraints.
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+ R2 -
P~oo
0 and
P 1 lim !(P,P,p,t,0)--2 Iog2 - 1t ==0
P~oo
(69)
+
When R I == R 2 == R this means
From this theorem, we have the following corollary: Corollary 1: Let PI == P2 == P. Then we have
+ R2 -
a
For the case without secrecy constraints, since a < P we can at least achieve the sum rate such that P~oo
lim 2R I
(a-1)2 p ( 1) ::; 0
B. Bound without Secrecy Constraints
(63)
P-H)()
(68)
~3 (~log2 ~ + ~ log2 2 b+ 1 2
p) : ;
0
(72)
Comparing (72) with (69), we observe when P is large, the bound on 2R I + R 2 is tighter than the genie bound (68) at the equal rate point. From Lemma 2, when R I == R 2 == R, a == b 2: 1, we have R ::; 0.5C(P)
(73)
Comparing (71) with (73), we notice (73) is always tighter. Comparing (73) with (72), we notice these two bounds may be close to each other when P is close to 1. When P is large, (73) is tighter than (72). These properties are demonstrated in Figure 4. Here a == 2, and P is chosen to be larger than 1, so 1 < a < P + 1. From [2], in this regime a positive secrecy rate is achievable, hence upper bounding the secrecy rate is of interest. It is shown in Figure 4 that at the equal rate point, the Zigzag channel bound is tighter than the genie bound in the shown range of P. Although not apparent in the figure, it is even tighter than the Z-Channel bound when P is close to 1. In Figure 5 and Figure 6, we present two numerical examples demonstrating that the new bounds improve previous outer bounds. In Figure 5, we demonstrate the bound 2R I + R 2 can indeed be dominant in bounding the sum rate when P is small. When P is large, as shown in Figure 6, it is looser than the Z channel bound in terms of sum rate, but it still helps to provide a tighter region. Also plotted in Figure 5 and Figure 6 are achievable rates obtained by time sharing between individual achievable rates
322
derived in [2]. Under the time sharing factor t, when a == b > 1, the following rates are achievable:
R1 = t ( C
2
R =
when
(H,a) - C C~~:,a) )
P 2 ,a
(74)
(1 - t) ( C (P2,/3) - C (1 a;~'/3))
P1,a::; a-I,
2: a-I,
Q)
0.8
Zigzag Channel Bound
0.7
R1+2R2
C/)
:J
(75)
Q)
c c
(76)
2: a-I
(77)
J' Z Channel Bound R +R
0.6
