On Private Broadcasting over Independent Parallel Channels Ashish Khisti
Tie Liu
Electrical and Computer Engineering University of Toronto Toronto, ON Email:
[email protected]
Electrical and Computer Engineering Texas A & M College Station, TX Email:
[email protected]
Abstract—We study private broadcasting of two messages to two groups of users over reversely degraded parallel channels. Group 1 has two users, both interested in a common message whereas group 2 has only one user. The message for each group of users needs to be kept confidential from the other group. We characterize the capacity region for a special degradation structure of the channels and establish the optimality of a superposition codebook where codewords of a secure productcodebook form the cloud centers and codewords of a secure multicast codebook form the satellite codewords. An extension to Gaussian channels with a sum-power constraint is also obtained, where the optimality of Gaussian codebooks is established using an extremal inequality.
I. I NTRODUCTION The wiretap channel problem [1], [2] is an information theoretic model for secure communication at the physical layer. The compound-wiretap channel [3] model is an extension of the standard wiretap channel model where the channels of both the legitimate receiver and the eavesdropper belong to a set of possible channels. This setup also models the possibility of having multiple receivers in each of the two groups. The secrecy capacity of the compound wiretap channel remains open in general and only some special cases have been resolved [3]–[9]. In particular, the secrecy capacity has been resolved for the case of parallel independent channels when either there are multiple legitimate receivers and one eavesdropper [6] and when there are multiple eavesdroppers and one legitimate receiver [5], [8]. In the former setup, the capacity is achieved using a secure-multicast codebook, which consists of independent wiretap codebooks on each of the links that individually secure the message from an eavesdropper on that link. An alternate construction that involves indirect decoding is proposed for the case of two channels in [7]. In the other extreme, i.e., when there are multiple eavesdroppers and one legitimate receiver, a secure-product codebook construction is proposed in [5], [8] and shown to be optimal. A key property of this construction is that the output codeword on any given link is independent of the output codewords on other links. In both these cases the secrecy capacity coincides A. Khisti’s work was supported by an NSERC (Natural Sciences Engineering Research Council) Discovery Grant and an Ontario Early Researcher Award. T. Liu was supported by the National Science Foundation under Grants CCF-08-45848 and CCF-09-16867.
with a natural pairwise upper bound which is known to be not tight in general [6]. In this paper we propose a superposition coding technique that combines the secure multicast and secure product codebook construction discussed above. In our setup of interest, there are two groups of users. Group 1 consists of K users and group 2 consists of one user. There are two messages, one for each group and each message needs to be communicated to all users in the respective group and secured from the users in the other group. We refer to this setup as private broadcasting. It has been studied previously in [10]–[12] when there is a single user in each group (i.e., K = 1) and in reference [6] for a special compound channel. We establish the optimality of the superposition technique for the case when K = 2 and when a particular degradation structure is assumed across the parallel channels. II. P ROBLEM S TATEMENT AND M AIN R ESULTS Our setup involves two independent parallel channels, two groups and three receivers. Receiver 1 and 2 belong to group 1 and observe output y1 = (y11 , y12 ) and y2 = (y21 , y22 ) respectively whereas receiver 3 belongs to group 2 and observes the output symbol z = (z1 , z2 ). The channel input symbols are x1 and x2 . Throughout this paper we additionally assume the following degradation order x1 → y11 → z1 → y21 x2 → y22 → z2 → y12
(1) (2)
We intend to transmit message m1 to receiver 1 and 2 in group 1, while the message m2 must be transmitted to receiver 3 in group 2. A length n private broadcast code also satisfies 1 I(m1 ; zn ) ≤ εn , n
1 I(m2 ; yjn ) ≤ εn , n
j = 1, 2 (3)
for some sequence εn that approaches zero as n → ∞. Theorem 1: Let auxiliary variables u1 and u2 satisfy the following Markov conditions: u1 → x1 → y11 → z1 → y21
(4)
u2 → x2 → y22 → z2 → y12
(5)
The capacity region is the union of all rate pairs (R1 , R2 ) that satisfy the following constraints: R1 ≤ min I(x1 ; y11 |u1 ) − I(x1 ; z1 |u1 ), (6) I(x2 ; y22 |u2 ) − I(x2 ; z2 |u2 ) R2 ≤ min {I(u1 ; z1 ) − I(u1 ; y21 ), I(u2 ; z2 ) − I(u2 ; y12 )} 2 (7)
m1=1 m1=2
code-words
Super-position Codebook – Channel 1
m2
u1n u2n
The proposed coding scheme is illustrated in Fig. 1. The message for group 2 is encoded using a product codebook and the corresponding codewords (u1n , u2n ) constitute the cloudcenters of a superposition codebook as shown. The codewords of the secure multicast codebook on each of the channels constitute the satellite codewords. For each message m1 , a total of 2nI(xi ;zi |ui ) satellite codewords are sampled conditionally given uin . Our secrecy analysis reveals that despite this structure in the codebook, it suffices for each user in group 1 to treat the satellite codeword on the weaker channel as a noisy version of the cloud center uin . The result in Theorem 1 can be extended to the case of Gaussian parallel channels. Consider a Gaussian Channel model described by y11 = x1 + w1 , y22 = x2 + w2 ,
z1 = y11 + n1 ,
y21 = z1 + s1
z2 = y22 + n2 ,
y12 = z2 + s2
(8)
where wj ∼ N (0, σj2 ) and nk ∼ N (0, δk2 ) and sk ∼ N (0, γk2 ) for j, k = 1, 2. A sum-power constraint E[x12 + x22 ] ≤ P is imposed on the input symbols. Theorem 2: The capacity region for private broadcasting over two-parallel Gaussian channels in (8) is the union of all rate tuple (R1 , R2 ) that satisfy R1 ≤ min(R11 , R12 ) R2 ≤ min(R21 , R22 ) where R11 = R12 = R21 =
R22 =
P11 P11 1 1 log 1 + 2 − log 1 + 2 2 σ1 2 σ1 + δ12 P12 P12 1 1 log 1 + 2 − log 1 + 2 2 σ2 2 σ2 + δ22 P21 1 log 1 + 2 P11 + σ12 + δ12 P21 1 − log 1 + 2 P11 + σ12 + δ12 + γ12 P22 1 log 1 + 2 P + σ22 + δ22 12 P22 1 − log 1 + 2 P12 + σ22 + δ22 + γ22
where Pij ≥ 0 satisfy P11 + P21 + P12 + P22 ≤ P .
(9) (10)
(11) (12)
(13)
Product-Codebook
m1=1 m1=2
2 nI ( x2 ; z2 |u2 )
code-words
Super-position Codebook – Channel 2
Fig. 1: Proposed Superposition Coding Scheme. The codewords of the product code (u1n , u2n ) constitute cloud-centers whereas the codewords of the multicast code constitute the sattelite codewords.
III. T HEOREM 1: C ODING S CHEME The coding scheme uses a product codebook for message m2 in order to guarantee that it is decodable by the user in group 2 and kept secure from all users in group 1. A multicast code is used for message m1 in order to guarantee that it is decodable by the users in group 1 and kept secure from the single user in group 2. On each channel, a superposition codebook is applied where the codeword of the product codebook is a cloud center and the codeword of the multicast codebook is the satellite codeword. The construction of the product codebook C2 is as follows. Consider the set M21 of all binary sequences of length N21 = n(I(u1 ; z1 ) − ε) and the set M22 of all binary sequences of length N22 = n(I(u2 ; z2 ) − ε). We partition the cartesian product M21 × M22 into 2nR2 bins so that each bin contains 2n(I(u1 ;z1 )+I(u2 ;z2 )−R2 ) elements of the productset. Each element m21 ∈ M21 is encoded using a codebook C21 whose codewords are sampled i.i.d. from the distribution pu1 (u1 ). The elements m22 ∈ M22 are encoded using a codebook C22 whose elements are sampled i.i.d. from the distribution pu2 (u2 ). Each message m2 ∈ [1, 2nR2 ] constitutes the bin-index of our product codebook. An element (m21 , m22 ) is selected uniformly at random from the bin associated with m2 . The corresponding codewords (u1n (m21 ), u2n (m22 )) constitute the cloud center of the superposition codebooks on each link. We note that by construction we have that each element (m21 , m22 ) is equally likely and, Pr(m21 = m21 , m22 = m22 ) =
(14)
2 nI ( x1 ; z1|u1 )
1 . |M21 | × |M22 |
(15)
For each codeword sequence u1n ∈ C21 , we generate codebook C11 (u1n ) consisting of a total of 2nR1 bins and 2n(I(x1 ;z1 |u1 )+ε) codewords in each bin. Each codeword is sampled i.i.d. from the conditional distribution px1 |u1 (x1i |u1i ).
In an analogous manner, for each u2n ∈ C22 , we generate a codebook C12 (u2n ) consisting of a total of 2nR1 bins and 2n(I(x2 ;z2 |u2 )+ε) codewords. Each codeword is sampled i.i.d. from the conditional distribution px2 |u2 (x2i |u2i ). Given a message m1 and a sequence pair (u1n , u2n ) we select a codeword pair (x1n , x2n ) uniformly at random from the bins corresponding to m1 in codebooks C11 (u1n ) and C12 (u2n ) respectively. Receiver 1 in group 1 searches for a codeword n ) that are jointly typical. This succeeds with pair (x1n , u1n , y11 high probability since our choice of R1 guarantees that the total number of codewords in the satellite codebook is no greater than 2nI(x1 ;y11 ) . In turn the receiver decodes m1 with high probability. Likewise user 2 in group 1 can decode n . Similarly by m1 with high probability using the output y22 construction of C21 and C22 , user 2 can decode sub-messages m21 and m22 with high probability and in turn find the corresponding message m2 . A. Secrecy-Analysis For message m1 we need to show that 1 I(m1 ; z1n , z2n |C) ≤ εn (16) n where C denotes the particular codebook selected in the ensemble. Note that from (15) the messages (m21 , m22 ) are mutually independent. From this it can be shown that outputs (z1n , z2n ) are conditionally independent given m1 . Thus we have, 1 1 1 I(m1 ; z1n , z2n |C) = H(z1n , z2n |C) − H(z1n , z2n |m1 , C) n n n (17) 1 1 1 = H(z1n , z2n |C) − H(z1n |m1 , C) − H(z2n |m1 , C) (18) n n n 1 1 n n ≤ I(z1 ; m1 |C) + I(z2 ; m1 |C) (19) n n By construction of each super-position codebook, there are 2nI(xi ;zi |ui ) codewords in each bin and hence through standard analysis we can show that n1 I(m1 ; zjn ) ≤ εn /2. Thus (16) follows. To establish the secrecy of message m2 with respect to the users in group 1 we show that provided R2 satisfies (7) we have that 1 n n H(m2 |yj1 , yj2 , C) ≥ R2 − εn , j = 1, 2. (20) n Consider the case when j = 1. 1 1 H(m2 |y1n , C) ≥ H(m2 |y1n , m1 , m21 , C) (21) n n 1 1 = H(m22 |y1n , m1 , m21 , C) − H(m22 |m2 , m21 , m1 , y1n , C) n n (22) 1 1 n = H(m22 |m1 , y12 , C) − H(m22 |m2 , m21 , m1 , y1n , C) n n (23) n ) → where (23) follows from the fact that m22 → (m1 , y12 n (m21 , y11 ) holds, since the two sub-messages (m21 , m22 ) are
mutually independent (c.f. (15)) and hence the knowledge of n ) do not help user 1 in group 1 to decode m22 . (m21 , y11 To lower bound the first term observe that from the symmetry of code construction 1 1 n n H(m22 |m1 , y12 , C) = H(m22 |m1 = 1, y12 , C) (24) n n and since m22 ∈ [1, 2n(I(u2 ;z2 )−ε) ], and for each u2n (m22 ) a total of 2nI(x2 ;z2 |u2 ) sequences are sampled in C12 in the bin m1 = 1 it follows from the analysis of a random wiretap code [13, Remark 22.2, pp. 555] that 1 n H(m22 |m1 = 1, y12 , C) ≥ I(u2 ; z2 ) − I(u2 ; y12 ) (25) n To bound the second term we note that from symmetry 1 H(m22 |m2 , m21 , m1 , y1n , C) n 1 = H(m22 |m2 = 1, m21 , m1 = 1, y1n , C). (26) n Furthermore in the product code construction, for each m21 ∈ M21 there are a total of |M22 | = 2n(I(u2 ;z2 )−ε) possible values for m22 . Due to random partitioning, which high probability the number of sequences in the bin with m2 = 1 is atmost |M22 |·2−n(R2 −ε) . Since R2 < I(u2 ; z2 )−I(u2 ; y22 )−2ε it can be shown along the lines of [13, Lemma 22.1, pp. 554] that 1 H(m22 |m2 = 1, m21 , m1 = 1, y1n , C) n ≤ I(u2 ; z2 ) − R2 − I(u2 ; y12 ) − 2ε.
(27)
Substituting (27) and (25) into (23) we recover (20) for j = 1. The result for j = 2 can be established in a similar fashion. IV. T HEOREM 1: C ONVERSE The intuition behind the converse is the following. Suppose we are not required to transmit any information to receiver 2 in group 1. Furthermore suppose that we are also not required to secure message m2 from user 1 in group 1. We only need to send message m1 to receiver 1 while securing it from receiver 3 in group 2 and transmit m2 while only securing it against user 2 in group 1. Looking at the degradation order on channel 2, it is apparent that no information must be transmitted on channel 2. Once we restrict our attention to channel 1 we have a degraded broadcast channel x1 → y11 → z1 → y21 where the message m1 needs to be protected from the two weaker receivers whereas the message m2 needs to be only protected from the weakest receiver. For this simplified problem we show that R1 ≤ I(x1 ; y11 |u1 , z1 ),
R2 ≤ I(u1 ; z1 |y21 ).
(28)
Similarly by relaxing symmetric constraints on message transmission and secrecy we obtain the the remaining two constraints. A formal proof is discussed next. Consider a sequence of codes indexed by the length n that satisfy (3), as well as the Fano Inequality 1 1 H(m1 |yjn ) ≤ εn , H(m2 |zn ) ≤ εn , j = 1, 2. (29) n n
where we introduce yj = (yj1 , yj2 ) and z = (z1 , z2 ). By weakening user 2 on link 2 from y22 to z2 it follows from (3) that 1 n I(m2 ; y21 , z2n ) ≤ εn (30) n Now consider: n , z2n ) + 2nεn nR2 ≤ I(m2 ; zn ) − I(m2 ; y21 n n ≤ I(m2 ; z1n |y21 z2 ) + 2nεn n n ≤ I(m2 ; z1i |z1i−1 y21 , z2n ) + 2nεn
=
i=1 n
n I(m2 ; z1i |z1i−1 y21,i , z2n ) + 2nεn
≤ ≤
i=1 n
n I(m2 , z1i−1 y21,i+1 , z2n ; z1i |y2i )
(32) (33) (34)
(35)
i=1
= nI(u1 ; z1 |y21 , q) + 2nεn ≤ nI(u1 ; z1 |y21 ) + 2nεn
(36) (37)
i−1 → z1i−1 → where we use the Markov relation y21 n n (m2 , z1i , y21,i , z2 ) in (34) which follows from the structure of n the channel and in (35) we define u1i = (m2 , z1i−1 , y21,i+1 , z2n ) that satisfies u1i → xi → y11i → z1i → y21i and where (36) follows by defining a time-sharing random variable q to be uniformly distributed over {1, . . . , n} and independent of all other variables and letting u1 = (q, u1q ), x = xq , y21 = y21q and z1 = z1q respectively and finally (37) follows from the fact that q → u1 → (z1 , y21 ). The secrecy condition with respect to m1 is
1 1 1 I(m1 ; zn , m2 ) ≤ I(m1 ; zn ) + H(m2 |zn ) = 2εn , (38) n n n which follows by substituting (3) and (29). We use the following upper bound on the rate R1 , nR1 ≤ I(m1 ; y1n ) − I(m1 ; zn , m2 ) + 3nεn ≤ I(m1 ; y1n |zn , m2 ) + 3nεn = =
n n I(m1 ; y11 |z , m2 ) + 3nεn n n n H(y11 |z , m2 ) − H(y11 |m1 , m2 , zn )
(39) (40) (41) + 3nεn
(42)
where (41) follows from the fact that n n → z2n → (m1 , m2 , z1n , y11 ) holds from the structure y21 of the channel. We next bound each of the terms in (42). The second term can be lower bounded by: n n |m1 , m2 , zn ) ≥ H(y11 |m1 , m2 , zn , x1n ) H(y11 n H(y11i |x1i z1i ) =
it can be shown that
(43) (44)
i=1
where the last step follows from the memoryless property of the channel.
n
H(y11i |u1i , z1i ).,
(45)
i=1
(31)
+ 2nεn
I(u1i ; z1i |y21i ) + 2nεn ,
n n → y21,i+1 (y11i , z1i−1 , z1i , z2n m2 ) → z1,i+1
n n H(y11 |z , m2 ) ≤
i=1
n
Using the Markov relation
Substituting (44) and (45) into (42) we get n
R1 ≤
1 I(x1i ; y11i |z1i , u1i ) + 3εn n i=1
= I(x1 ; y11 |z1 , u1 ) + 3εn .
(46) (47)
where the last step follows by since q is uniformly distributed over {1, . . . , n} and independent of all other variables and u1 = (q, u1,q ), x1 = x1q , y21 = y21q and z1 = z1q respectively. In an analogous manner we can define a random variable v for this problem such that R1 ≤ I(x2 ; y22 |u2 , z2 ),
R2 ≤ I(u2 ; z2 |y12 )
(48)
are satisfied. This completes the converse. V. P ROOF OF THE G AUSSIAN C ASE : T HEOREM 2 The achievability follows by selecting a Gaussian input distribution in Theorem 1. In particular let x1 ∼ N (0, P1 ) and x2 ∼ N (0, P2 ). Furthermore let u1 ∼ N (0, P21 ) and u2 ∼ N (0, P22 ). Let x1 = x11 + u1 , where x11 ∼ N (0, P11 ) is independent of u1 and similarly let x2 = x12 + u2 where x12 ∼ N (0, P12 ) is independent of u2 . Evaluating R11 = I(x1 ; y11 |u1 ) − I(x1 ; z1 |u1 ) R12 = I(x2 ; y22 |u2 ) − I(x2 ; z2 |u2 ) R21 = I(u1 ; z1 ) − I(u1 ; y21 )
(49)
R22 = I(u2 ; z2 ) − I(u2 ; y12 )
(52)
(50) (51)
with the above choice of input distribution results in (11)- (14). To establish that a Gaussian distribution in indeed optimal, consider, R11 = I(x1 ; y11 |u1 ) − I(x1 ; z1 |u1 ) = h(y11 |u1 ) − h(z1 |u1 ) − h(y11 |x1 ) + h(z1 |x1 )
(53)
= h(y11 |u1 ) − h(z1 |u1 ) − h(w1 ) + h(n1 )
(54)
where we use u1 → x1 → (y11 , z1 ) in (53) and use the channel model (8) in (54). Similarly, R21 = I(u1 ; z1 ) − I(u1 ; y21 ) = h(z1 ) − h(y21 ) + [h(y21 |u1 ) − h(z1 |u1 )]
(55) (56)
= h(z1 ) − h(z1 + s1 ) + [h(y21 |u1 ) − h(z1 |u1 )]. (57) 1 P1 + σ12 + δ12 ≤ log + [h(y21 |u1 ) − h(z1 |u1 )] 2 P1 + σ12 + δ12 + γ12 (58)
Δ
where P1 = E[x12 ] and the last step follows from the worstcase additive noise lemma [14]. Using this, it can also be shown that: 1 σ2 P1 + σ12 1 log 2 1 2 ≤ h(y11 |u1 ) − h(z1 |u1 ) ≤ log 2 σ1 + δ1 2 P1 + σ12 + δ12 (59) To establish (59) note that h(y11 |u) − h(z1 |u) = −I(z1 ; n1 |u) where the noise n1 ∼ N (0, δ12 ) is independent u. For the upper bound note that I(z1 ; n1 |u1 ) = I(z1 , u1 ; n1 ) ≥ I(z1 ; n) ≥
P1 + σ12 + δ12 1 log . 2 P1 + σ12
The last step follows from the worst-case additive noise lemma. For the lower bound, note that since u1 → x → z1 holds and n1 is independent of these random variables we have I(z1 ; n1 |u1 ) ≤ I(z1 ; n1 |x) = I(n1 + w1 ; n1 ) =
σ2 + δ2 1 log 1 2 1 . 2 σ1
Hence for some 0 ≤ P11 ≤ P1 we have that h(y11 |u1 ) − h(z1 |u1 ) =
P11 + σ12 1 log . 2 P11 + σ12 + δ12
(60)
Upon substituting (60) into (54) we establish (11). To establish the bound for R21 we use the following extremal inequality: max
{h(y11 |u1 ) + μh(y21 |u1 ) − (1 + μ)h(z1 |u1 )} 1 max log(t + σ12 ) + μ log(t + σ12 + δ12 + γ12 ) = 2 0≤t≤P1 2 2 (61) − (1 + μ) log(t + σ1 + δ1 )
px (·):E[x12 ]=P1
2 2 δ2 δ1 +γ1 By selecting μ = γ12 P11 + 1 it can be shown that 2 +σ1 1 the optimal t = P11 and through some straightforward computation it follows from (61) and (60) that h(y21 |u) − h(z1 |u) ≤
P11 + σ12 + δ12 + γ12 1 log . 2 P11 + σ12 + δ12
(62)
Substituting into (58) we have that R21 ≤
P1 + σ12 + δ12 P11 + σ12 + δ12 + γ12 1 1 log log + , 2 P1 + σ12 + δ12 + γ12 2 P11 + σ12 + δ12 (63)
from which we recover (13) by defining P21 = P1 − P11 . The upper bounds on R12 and R22 can be established in a similar fashion and will be skipped. VI. C ONCLUSION We study the problem of secure broadcasting of two messages to two groups of users over two independent and parallel channels. Group 1 has two receivers, both interested in a common message, whereas group 2 has only one receivers. Each message must be kept secure from user(s) in the other group. For a particular degradation structure we establish the optimality of a technique that combines a product code and a multicast code using a superposition codebook. For the case of
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