On Ergodic Secrecy Capacity for Gaussian MISO Wiretap Channels
aa r X i v : . [ c s . I T ] M a r On Ergodic Secrecy Capacity for GaussianMISO Wiretap Channels
Jiangyuan Li and Athina P. PetropuluDepartment of Electrical and Computer EngineeringDrexel University, Philadelphia, PA 19104Email: [email protected], [email protected]
Abstract A Gaussian multiple-input single-output (MISO) wiretap channel model is considered, where thereexists a transmitter equipped with multiple antennas, a legitimate receiver and an eavesdropper eachequipped with a single antenna. We study the problem of finding the optimal input covariance thatachieves ergodic secrecy capacity subject to a power constraint where only statistical information aboutthe eavesdropper channel is available at the transmitter. This is a non-convex optimization problem that isin general difficult to solve. Existing results address the case in which the eavesdropper or/and legitimatechannels have independent and identically distributed Gaussian entries with zero-mean and unit-variance,i.e., the channels have trivial covariances. This paper addresses the general case where eavesdropper andlegitimate channels have nontrivial covariances. A set of equations describing the optimal input covariancematrix are proposed along with an algorithm to obtain the solution. Based on this framework, we showthat when full information on the legitimate channel is available to the transmitter, the optimal inputcovariance has always rank one. We also show that when only statistical information on the legitimatechannel is available to the transmitter, the legitimate channel has some general non-trivial covariance, andthe eavesdropper channel has trivial covariance, the optimal input covariance has the same eigenvectorsas the legitimate channel covariance. Numerical results are presented to illustrate the algorithm.
Index Terms
Ergodic secrecy capacity, MISO wiretap channel, beamforming. Work supported by the Office of Naval Research under grant ONR-N-00010710500 and the National Science Foundationunder grant CNS-0905425.
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I. I
NTRODUCTION
Wireless physical (PHY) layer based security from a information-theoretic point of view has receivedconsiderable attention recently [1]. Such approaches exploit the physical characteristics of the wirelesschannel to enhance the security of communication systems. The wiretap channel, first introduced andstudied by Wyner [2], is the most basic physical layer model that captures the problem of communicationsecurity. Wyner showed that when an eavesdropper’s channel is a degraded version of the legitimatechannel, the source and destination can achieve a positive information rate (secrecy rate). The maximalsecrecy rate from the source to the destination is defined as the secrecy capacity ; for the degraded wiretapchannel the secrecy capacity is given as the largest between zero and the difference between the capacityat the legitimate receiver and the capacity at the eavesdropper. The Gaussian wiretap channel, in which theoutputs at the legitimate receiver and at the eavesdropper are corrupted by additive white Gaussian noise(AWGN), was studied in [3]. Along the same lines, the secrecy capacity of a deterministic Gaussian MIMOwiretap channel has been studied recently in [4]-[8]. In [9], the achievable rate in Gaussian MISO channelswas studied. In that context, the channel state information (CSI) of the legitimate channel was assumed tobe available, but only statistical information about the eavesdropper channel was assumed to be availableat the transmitter. In [9] it was shown that when the eavesdropper channel is a vector of independent andidentically distributed (i.i.d.) zero-mean complex circularly symmetric Gaussian random variables, i.e.,the channel has a trivial covariance matrix, the optimal communication strategy is beamforming, and thatthe beamforming direction depends on the CSI of the legitimate channel. In [10], the authors derived theergodic secrecy capacity of a Gaussian MIMO wiretap channel where only statistical information aboutthe legitimate and eavesdropper channels are available at the transmitter. It was shown that a circularlysymmetric Gaussian input is optimal. It was also shown in the same paper that when the eavesdropperand legitimate channels have i.i.d. Gaussian entries with zero-mean and unit-variance (trivial covariance),a circularly symmetric Gaussian input with diagonal covariance is optimal.In this paper, we consider a Gaussian multiple-input single-output (MISO) wiretap channel and assumethat only statistical information about the the eavesdropper channel is available at the transmitter. Re-garding the legitimate channel, we consider two scenarios: a) only statistical information of the legitimatechannel is available at the transmitter; b) full CSI on the legitimate channel is available at the transmitter.We extend the result of [9] and [10] proposed for the case of multiple-input single-output (MISO) wiretapchannel with trivial channel covariances to the case of nontrivial covariances. The non-trivial channelcovariance matrix corresponds to the case where there exists statistical correlation between the channel
June 6, 2018 DRAFT coefficients of different transmit-receive antenna pairs. Such cases arise when the transmit and receiveantennas are closely spaced relative to the signal wavelength. We address the problem of finding theoptimal input covariance that achieves ergodic secrecy capacity subject to a power constraint. This leadsto a non-convex optimization problem. The contributions of this paper are the following: • We derive a set of equations for the optimal input covariance matrix, and propose an algorithm toobtain the solution (please refer to Theorem 1 of Section IV). • We show that when the legitimate channel is completely known at the transmitter, in addition to theconditions of Theorem 1, the following hold: 1) the optimal input covariance matrix has rank one;2) the ergodic secrecy rate is increasing with the signal-to-noise ratio (SNR). • We show that when only statistical information on the legitimate channel is available to the transmit-ter, the legitimate channel has some general non-trivial covariance, and the eavesdropper channel hastrivial covariance, the optimal input covariance has the same eigenvectors as the legitimate channelcovariance. • We show that under high SNR, the optimal input covariance has rank one.The remainder of this paper is organized as follows. The mathematical model is introduced in § II.In § III, we give the explicit expression of ergodic secrecy rate, and in § IV, we derive the condition foroptimal input covariance. In § V, we analyze the dependence of ergodic secrecy rate on the SNR, and in § VI, we study the ergodic secrecy rate under high SNR. In § VII, an algorithm is proposed to search forthe solution. Numerical results are presented in § VIII to illustrate the proposed algorithm. Finally, § IXgives a brief conclusion. Several proofs appear in an Appendix.
A. Notation
Upper case and lower case bold symbols denote matrices and vectors, respectively. Superscripts ∗ , T and † denote respectively conjugate, transposition and conjugate transposition. det( A ) and Tr( A ) denote the determinant and trace of matrix A , respectively. λ max ( A ) and λ min ( A ) denote the largest andsmallest eigenvalues of A , respectively. A (cid:23) means that A is Hermitian positive semi-definite, and A ≻ means that A is Hermitian positive definite. diag( a ) denotes a diagonal matrix with the elementsof the vector a along its diagonal. k a k denotes Euclidean norm of vector a . I n denotes the identity matrixof order n (the subscript is dropped when the dimension is obvious). E {·} denotes expectation operator.In this paper, log( · ) denotes base- e logarithm where e = 2 . · · · . June 6, 2018 DRAFT R h E h Fig. 1. System model.
II. S
YSTEM M ODEL AND P ROBLEM S TATEMENT
Consider a Gaussian MISO wiretap channel shown in Fig. 1, where the transmitter is equipped with n T antennas, while the legitimate receiver and an eavesdropper each have a single antenna. The receivedsignals at the legitimate receiver and the eavesdropper are respectively given by y R = h † R x + v R , (1) y E = h † E x + v E (2)where x is the n T × transmitted signal vector with zero mean and n T × n T covariance matrix R x (cid:23) ,i.e., x ∼ CN ( , R x ) ; h R , h E are respectively channel vectors between the transmitter and legitimatereceiver, and between the transmitter and eavesdropper; v R ∼ CN (0 , σ v ) , v E ∼ CN (0 , σ v ) are the noisesat the legitimate receiver and the eavesdropper, respectively. We can represent R x in terms of the averagesignal energy E s and normalized signal covariance matrix Q , so that R x = E s Q and Tr( Q ) = 1 . Thesignal-to-noise ratio (SNR) is defined as ρ , E s /σ v .We assume that full CSI is available at both the legitimate receiver and the eavesdropper, and onlythe statistical information on the eavesdropper channel is available at the transmitter. We consider twocases, depending on the type of information available at the transmitter on the legitimate channel:a) Only statistical information on the legitimate channel is available at the transmitter, i.e., the trans-mitter knows the distributions of h R and h E given by h R ∼ CN ( , Σ R ) , h E ∼ CN ( , Σ E ) withcovariances Σ R ≻ , and Σ E ≻ , respectively. The ergodic secrecy capacity of the Gaussian MISOwiretap system (2) equals [10] C s , max Q (cid:23) , Tr( Q )=1 C s ( Q ) (3) June 6, 2018 DRAFT where C s ( Q ) is the ergodic secrecy rate given by C s ( Q ) = E h R { log(1 + ρ h † R Qh R ) }− E h E { log(1 + ρ h † E Qh E ) } . (4)b) Full CSI on the legitimate channel is available at the transmitter. The ergodic secrecy rate is givenby [9] C s ( Q ) = log(1 + ρ h † R Qh R ) − E h E { log(1 + ρ h † E Qh E ) } . (5)The transmitter optimization problem is to find the optimal input covariance matrix Q to maximize C s ( Q ) for cases a) and b). We denote the feasible set as Ω = { Q | Q (cid:23) , Tr( Q ) = 1 } which is a convexset.The problem is of interest when a positive secrecy rate can be achieved, i.e., C s ( Q ) > for some Q .The conditions to ensure a positive ergodic capacity are provided in the following lemmas. Lemma 1:
For h R ∼ CN ( , Σ R ) , the sufficient and necessary condition under which C s ( Q ) > forsome Q is that Σ R − Σ E is non negative semi-definite. The proof is given in Appendix A.
Lemma 2:
When h R is completely known at the transmitter, a sufficient condition under which C s ( Q ) > for some Q is that h R h † R − Σ E is non negative semi-definite. The proof is given in Appendix B.III. C
ALCULATION OF E RGODIC S ECRECY R ATE
The calculation of the ergodic secrecy rate involves calculation of terms like E z { log(1 + ρ z † Qz ) } with z ∼ CN ( , R ) . To this end, following the analysis of [11, Eq. (64)], we give the following lemma. Theproof is given in Appendix C. Lemma 3:
Let R / QR / have a eigen-decomposition U D U † where D = diag( d , · · · , d M , , · · · , ,and d > · · · > d M > are the M non-zero eigenvalues. For z ∼ CN ( , R ) , it holds: E z { log(1 + ρ z † Qz ) } = M X j =1 F ( ρd j ) Q Mi = j (1 − d i /d j ) (6) where F ( x ) = e /x E (1 /x ) with E ( x ) = R ∞ x e − t t d t being the exponential integral. Based on (6), we can calculate C s ( Q ) by simply letting R = Σ R or R = Σ E . June 6, 2018 DRAFT
IV. C
ONDITIONS FOR O PTIMAL I NPUT C OVARIANCE
Next we obtain the necessary conditions for the optimal Q by using the Karush-Kuhn-Tucker (KKT)conditions. Let us construct the cost function L ( Q , θ, Ψ ) = C s ( Q ) − θ (Tr( Q ) −
1) + Tr( ΨQ ) (7)where θ is the Lagrange multiplier associated with the constraint Tr( Q ) = 1 , and Ψ is the Lagrangemultiplier associated with the constraint Q (cid:23) . The KKT conditions enable us to write [20] Θ − θ I n T + Ψ = 0 , (8) Ψ (cid:23) , Tr( ΨQ ) = 0 , Q (cid:23) , Tr( Q ) = 1 , (9)where Θ = ( ∂C s ( Q ) ∂ Q ) T . By using the fact ∂ h † Qh ∂ Q = ( hh † ) T , we have: for case a) Θ = E h R (cid:26) ρ h R h † R ρ h † R Qh R (cid:27) − E h E (cid:26) ρ h E h † E ρ h † E Qh E (cid:27) (10)and for case b) Θ = ρ h R h † R ρ h † R Qh R − E h E (cid:26) ρ h E h † E ρ h † E Qh E (cid:27) . (11)From the KKT conditions (8) and (9), we obtain the equivalent (but without containing the Lagrangemultipliers) conditions for optimal Q consisting of a set of equations given in the following theorem.Please see Appendix D for details. Theorem 1:
The optimal Q (cid:23) satisfies QΘ = ΘQ = Tr( QΘ ) Q (12) λ max ( Θ ) = Tr( QΘ ) . (13)The above conditions imply that for the optimal Q , QΘ is a scaled version of Q . Any Q satisfying theconditions of Theorem 1 is called as KKT solution. In § VII, we propose an algorithm to search for theKKT solution. Θ is an important variable for the transmitter optimization problem. For the calculationof Θ , we give the following lemma. The proof is given in Appendix E. Lemma 4:
For z ∼ CN ( , R ) , it holds E z (cid:26) ρ zz † ρ z † Qz (cid:27) = ρ R / U YU † R / (14) June 6, 2018 DRAFT where Y is a diagonal matrix with ( k, k ) th entry given by Y kk = M X j =1 ,j = k F ( ρd j ) − F ( ρd k )( ρd j − ρd k ) Q Mi = j (1 − d i /d j )+ F ( ρd k ) Q Mi = k (1 − d i /d k ) , k ≤ M (15) Y kk = M X j =1 F ( ρd j ) ρd j Q Mi = j (1 − d i /d j ) , k > M. (16) with F ( x ) = x − x e /x E (1 /x ) , and F ( x ) , E ( x ) , U , d i , M are defined in Lemma 3. Based on (14), we can calculate Θ by simply letting R = Σ R or R = Σ E .In the following, we show that for some special cases, more information about Q than that of Theorem1 can be obtained. A. h R is completely known at the transmitter We will prove that if C s ( Q ) > for some Q , then the optimal Q always has rank one, i.e., beamformingis optimal. We put the proof in the second part of the subsection. In the first part of the subsection, weanalyze how this result reduces our problem to a problem of one variable.Based on this result, we let Q = uu † with k u k = 1 and the problem is reduced to C s ( Q ) = log(1 + ρ h † R uu † h R ) − E h E { log(1 + ρ h † E uu † h E ) } (17)which, by using (66) and (68), can be rewritten as C s ( Q ) = log(1 + ρ u † h R h † R u ) − F ( ρ u † Σ E u ) . (18)Let u † h R h † R u = z k h R k . Then ≤ z ≤ . Note that F ( x ) is an increasing function. Thus, for fixed z , u † Σ E u should be minimized. Define φ ( z ) = min u u † Σ E u (19) s . t . u † h R h † R u = z k h R k , and k u k = 1 . Then, our problem is reduced to C s ( z ) = log(1 + ρ k h R k z ) − F ( ρφ ( z )) , ≤ z ≤ . (20) June 6, 2018 DRAFT
Since the problem of (19) belongs to the class of quadratically constrained quadratic programming(QCQP) with two constraints, it can be exactly solved [17], and is equivalent to its semidefinite program-ming (SDP) relaxation, i.e., φ ( z ) = min X Tr( Σ E X ) (21) s . t . Tr( h R h † R X ) = z k h R k , and Tr( X ) = 1 , X (cid:23) . For any given z , the problem of (21) is an SDP and can be effectively solved via CVX software [25]. Lemma 5:
The function φ ( z ) is a convex function. The proof is given in Appendix F.Since φ ( z ) is a convex function, according to well-known properties of convex functions, we knowthat φ ( z ) is continuous and Lipschitz continuous [27, Corollary 2.3.1], and is differentiable at all but atmost countably many points (left and right derivatives always exists) [27, Theorem 2.3.4]. Further studyon φ ( z ) and proposing more effective method for the optimization of C s ( z ) can be our future work.For the special case Σ E = α I ( h E has a trivial covariance), (18) becomes C s ( Q ) = log(1 + ρ u † h R h † R u ) − F ( ρα ) . (22)Obviously, the optimal u = h R / k h R k , the optimal Q = h R h † R / k h R k and [max Q C s ( Q )] = log(1 + ρ k h R k ) − F ( ρα ) . (23)This case was considered in [9] and the above result is consistent with that in [9]. Remark : If the optimal u do not achieve C s ( Q ) > , then C s ( Q ) ≤ for any Q .In the remainder of the subsection, we give the proof for that if C s ( Q ) > for some Q , then theoptimal Q always has rank one. We first provide a lemma that will be helpful in the following. The proofis put in Appendix G. Lemma 6:
Let A be a positive definite matrix, a be a vector. If aa † − A has a positive eigenvalue,then it has all negative eigenvalues except for a positive eigenvalue. Via Lemma 6, we can show that, if C s ( Q ) > , then Θ has all negative eigenvalues except for apositive eigenvalue. To see why this is the case, recall that Θ = ρ h R h † R ρ h † R Qh R − E h E (cid:26) ρ h E h † E ρ h † E Qh E (cid:27) . (24) June 6, 2018 DRAFT
According to § V-1 (after Lemma 7), we know that, if C s ( Q ) > , then Tr( ΘQ ) > . Thus, Θ has atleast a positive eigenvalue. According to (14), we know that the second term in the right hand side of(41) is positive definite. Note that the first term in the right hand side of (41) has the form aa † . Thus,the desired result follows directly from Lemma 6.From Theorem 1, we know that the optimal Q and its associated Θ are commutable. Thus, there existsa unitary matrix U that simultaneously diagonalizes Q and Θ . Let Λ Q and Λ Θ be the correspondingdiagonal matrices. From (12) in Theorem 1, we know that Λ Q Λ Θ = Tr( ΘQ ) Λ Q (25)or equivalently, ( Λ Q ) kk ( Λ Θ ) kk = Tr( ΘQ )( Λ Q ) kk , k = 1 , · · · , n T . (26)Since Tr( ΘQ ) > , it follows from (26) that, if ( Λ Q ) kk > , then ( Λ Θ ) kk = Tr( ΘQ ) > . However, Λ Θ has all negative diagonal entries except for a positive one. Thus, Λ Q has only one nonzero diagonalentry, i.e, the optimal Q has rank one. B. Only statistical information on h R available at the transmitter and Σ E = α I During this subsection, we assume that Σ R has simple spectrum (all eigenvalues are distinct), since mul-tiple eigenvalues are rare for generic Hermitian matrices [19, § Σ R have the eigen-decomposition Σ R = V R Λ R V † R where Λ R = diag( η , · · · , η n T ) , η > η > · · · > η n T . We use Theorem 1 toshow that the optimal Q has the same eigenvectors as Σ R , i.e., V † R QV R is diagonal, denoted by Λ = diag( ζ , · · · , ζ n T ) . Using this result, our problem is reduced to C s ( Q ) = E h w { log(1 + ρ h † w Λ / R ΛΛ / R h w ) }− E h w { log(1 + αρ h † w Λh w ) } = E h w { log(1 + ρ X n T i =1 η i ζ i | h w,i | ) }− E h w { log(1 + αρ X n T i =1 ζ i | h w,i | ) } . (27)The power constraint is P n T i =1 ζ i = 1 . For the case n T = 2 , (27) becomes C s ( Q ) = E h w { log(1 + ρη ζ | h w, | + ρη (1 − ζ ) | h w, | ) }− E h w { log(1 + αρζ | h w, | + αρ (1 − ζ ) | h w, | ) } . (28)The constraint is ≤ ζ ≤ . Similarly to § III, the expectations in (27) and (28) can be expressed inexplicit form.
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In the remainder of the subsection, we give the proof. Let Q have the eigen-decomposition Q = VΛV † where Λ is diagonal. According to Appendix E, we express Θ as Θ = E h w (cid:26) ρ Σ R h w h † w Σ R ρ h † w Σ R QΣ R h w (cid:27) − E h w (cid:26) αρ h w h † w αρ h † w Qh w (cid:27) . (29)Let Σ R QΣ R = A and let A have the eigen-decomposition A = V A Λ A V † A where Λ A = diag( λ J , λ J , · · · , λ K J K ) , (30) λ > λ > · · · > λ K − > λ K ≥ are distinct eigenvalues, and J k ’s are identity matrices. By using thefact that Uh w and h w have the identical distributions for any unitary matrix U , we express Θ as Θ = Σ / R V A Y V † A Σ / R − VZ V † (31)where Y = E h w (cid:26) ρ h w h † w ρ h † w Λ A h w (cid:27) , (32) Z = E h w (cid:26) αρ h w h † w αρ h † w Λh w (cid:27) . (33)Similarly to Appendix E, it can be shown that Y and Z are both diagonal.Observe that VZ V † and Q are commutable. With this, from Theorem 1, we know that Θ and Q arecommutable which enables us to get Σ / R V A Y V † A Σ / R Q = QΣ / R V A Y V † A Σ / R . (34)By inserting Σ R QΣ R = V A Λ A V † A into (34), we get Σ R V A Y Λ A V † A = V A Λ A Y V † A Σ R . (35)Since Λ A and Y are both diagonal matrices, it holds that Y Λ A = Λ A Y . With this, by inserting Σ R = V R Λ R V † R into (35), we get Λ R V Λ A Y V † = V Λ A Y V † Λ R (36)where V = V † R V A . From (36) and the assumption that all diagonal entries of Λ R are distinct, we knowthat V Λ A Y V † is a diagonal matrix [22, Special matrices: diagonal]. On the other hand, it followsfrom Σ R QΣ R = V A Λ A V † A that Q = Σ − / R V A Λ A V † A Σ − / R which, when combined with the fact that Σ R = V R Λ R V † R , results in Q = V R Λ − / R ( V Λ A V † ) Λ − / R V † R . (37) June 6, 2018 DRAFT0
Next, we show that V Λ A V † is diagonal. Since V Λ A Y V † is diagonal, there exists a V , whichis the column-permuted version of V , such that V Λ A Y V † = Λ A Y , or equivalently, V Λ A Y = Λ A Y V . We aim to prove that V Λ A = Λ A V . According to (30), (32), (76), (77), it is not difficultto show that Λ A Y has the form of Λ A Y = diag( λ ′ J , λ ′ J , · · · , λ ′ K J K ) (38)where λ ′ > λ ′ > · · · > λ ′ K − > , λ ′ K ≥ , and λ ′ K = λ ′ k for k = 1 , · · · , K − . From (38) and the factthat V Λ A Y = Λ A Y V , we know that V has the form of V = diag( A , A , · · · , A K ) , where each A k is the same size as the corresponding J k [22, Special matrices: diagonal]. Thus, it is easy to verifythat V Λ A = Λ A V , or equivalently, V Λ A V † = Λ A . Therefore, since V is the column-permutedmatrix of V , it follows that V Λ A V † is diagonal. With this, from (37), we know that the optimal Q has the same eigenvectors as Σ R . V. D EPENDENCE OF C s ( Q ) ON ρ In this section we investigate how the SNR, ρ , impacts the ergodic secrecy rate.
1) Full CSI on h R at the transmitter: We first provide a lemma that will be helpful in the following.The proof is given in Appendix H.
Lemma 7:
For a positive constant x and a positive random variable Y , the following fact holds: log x > E (log Y ) = ⇒ x < E (cid:18) Y (cid:19) . (39) Here, = ⇒ means that the right side follows from the left side. By using Lemma 7, we can prove that, if C s ( Q ) > , then Tr( ΘQ ) > . To see why this is the case,we let x = 1 + ρ h † R Qh R and Y = 1 + ρ h † E Qh E which enables us to write C s ( Q ) = log x − E (log Y ) (40) Tr( ΘQ ) = E (cid:18) Y (cid:19) − x . (41)The desired result follows from Lemma 7.Taking the derivative of C s ( Q ) with respect to ρ , we get: ∂C s ( Q ) ∂ρ = h † R Qh R ρ h † R Qh R − E h E (cid:26) h † E Qh E ρ h † E Qh E (cid:27) = Tr( ΘQ ) ρ . (42) June 6, 2018 DRAFT1
Based on the fact that, if C s ( Q ) > , then Tr( ΘQ ) > , we get that if C s ( Q ) > , then ∂C s ( Q ) ∂ρ > .Thus, if C s ( Q ) > for some Q , then more power should achieve larger secrecy rate. In other words,we should use the maximum power.
2) Statistical information on h R at the transmitter: In this case, we deal with the situation Σ R (cid:23) Σ E .Taking the derivative of C s ( Q ) with respect to ρ , we get: ∂C s ( Q ) ∂ρ = 1 ρ E h w (cid:26)
11 + ρ h † w Σ / E QΣ / E h w (cid:27) − ρ E h w (cid:26)
11 + ρ h † w Σ / R QΣ / R h w (cid:27) . (43)Here, we have used (58) and (59) in Appendix A. According to Ostrowski theorem [24, p. 224], weknow that if A (cid:23) B and B ≻ , then λ k ( A / QA / ) ≥ λ k ( B / QB / ) , where λ k ( · ) denotes the k theigenvalue arranged in decreasing order. Since Σ R (cid:23) Σ E , similarly to the methodology in Appendix A,it is easy to prove that ∂C s ( Q ) ∂ρ > . Thus, more power should achieve larger secrecy rate. In other words,we should use the maximum power. Remarks : For the situation Σ R (cid:15) Σ E , whether or not C s ( Q ) > imply that ∂C s ( Q ) ∂ρ > has not beenproved. This can be our future work.VI. T HE O PTIMAL Q U NDER H IGH
SNRIn this subsection, we give an analysis for high SNR, i.e., ρ → ∞ . Our results show that for high SNR,the optimal Q has rank one, i.e., beamforming is optimal. The detailed analysis is given as follows. A. Full CSI about h R at the transmitter According to § IV-A, the optimal Q always has rank one. Let Q = uu † with k u k = 1 .For high SNR, by using the fact that log(1 + x ) ≈ log x for large x , we write C s ( Q ) ≈ log( ρ h † R uu † h R ) − E h E { log( ρ h † E uu † h E ) } = log( u † h R h † R u ) − E h w { log( h † w Σ / E uu † Σ / E h w ) } = log( u † h R h † R u ) − E h w, { log( u † Σ E u | h w, | ) } = log( u † h R h † R u ) − log( u † Σ E u ) − E { log | h w, | } = log u † h R h † R uu † Σ E u + γ ≤ log( h † R Σ − E h R ) + γ (44) June 6, 2018 DRAFT2 where we have used the fact that Uh w and h w have identical distributions for any unitary matrix U ,and E { log | h w, | } = − γ where γ = 0 . · · · is the Euler’s constant (since | h w, | ∼ χ (2) ,i.e., the chi-square distribution with degree of freedom ). In (44), the maximum is achieved when u = Σ − E h R / k Σ − / E h R k . B. Statistics information about h R at the transmitter For high SNR, similarly, we write C s ( Q ) ≈ E h R { log( ρ h † R Qh R ) } − E h E { log( ρ h † E Qh E ) } = E h w { log( h † w Σ R QΣ R h w ) } − E h w { log( h † w Σ E QΣ E h w ) } = E h w { log X n T k =1 a k | h w,k | } − E h w { log X n T k =1 b k | h w,k | } = E h w (cid:26) log P n T k =1 a k | h w,k | P n T k =1 b k | h w,k | (cid:27) (45)where a k ’s and b k ’s are the eigenvalues of Σ R QΣ R and Σ E QΣ E arranged in decreasing order, respec-tively, and we have used the fact that Uh w and h w have the identical distributions for any unitary matrix U . Let p and p n T be the largest and smallest eigenvalues of Σ R Σ − E Σ R . By writing Σ R QΣ R = Σ R Σ − E (cid:16) Σ E QΣ E (cid:17) Σ − E Σ R (46)and applying Ostrowski theorem [24, p. 224], we have a k = b k θ k , p n T ≤ θ k ≤ p ( k = 1 , · · · , n T ) . (47)Since rank( Σ R QΣ R ) = rank( Σ E QΣ E ) , the number of non-zero elements of a k ’ and b k ’s is the same,denoted by n . Thus, from (47), we have (cid:20) max k =1 , ··· ,n a k b k (cid:21) ≤ p = λ max ( Σ R Σ − E Σ R ) . (48)To proceed, we need the following lemma. Lemma 8:
For x k > , y k > , k = 1 , · · · , n , it holds x + · · · + x n y + · · · + y n ≤ max k =1 , ··· ,n x k y k . (49)The proof is simple: it is based on the following a + bc + d ≤ max (cid:8) ac , bd (cid:9) , ∀ a, b, c, d > . (50) June 6, 2018 DRAFT3
From (48), using Lemma 8, we get that: for any h w = 0 , P n T k =1 a k | h w,k | P n T k =1 b k | h w,k | = P nk =1 a k | h w,k | P nk =1 b k | h w,k | ≤ max k =1 , ··· ,n a k b k ≤ λ max ( Σ R Σ − E Σ R ) . (51)In (51), the maximum is achieved simultaneously for any h w = 0 when Q = u u † , u = Σ − E x / k Σ − E x k with x being the eigenvector associated with the largest eigenvalue of Σ − E Σ R Σ − E , and correspondingly, a = u † Σ R u , b = u † Σ E u , a = · · · = a n T = 0 , b = · · · = b n T = 0 , i.e., n = 1 . To see why this isthe case, noting that λ max ( Σ R Σ − E Σ R ) = λ max ( Σ − E Σ R Σ − E ) , it is easy to verify that a b = x † Σ − E Σ R Σ − E x x † x = λ max ( Σ − E Σ R Σ − E ) . (52)The desired result follows. Now, combining (45) and (51), we have C s ( Q ) ≈ E h w (cid:26) log P n T k =1 a k | h w,k | P n T k =1 b k | h w,k | (cid:27) ≤ log( λ max ( Σ R Σ − E Σ R )) . (53)In (53), the maximum is achieved when Q = u u † . Thus, the optimal Q has rank one.VII. F IXED P OINT I TERATION
In this section we propose an algorithm to search for the KKT solution according to Theorem 1.When Θ and Q commute, Θ + γ I n T and Q commute for any real number γ , and vice versa. Let γ = (1 + β ) max { , − λ min ( Θ ) } , β > and let K = Θ + γ I n T . It holds that K ≻ . From (12), we get KQ = Tr( KQ ) Q . (54)Equation (54) looks like the eigenvalue equation Ax = λ x , where Tr( KQ ) is the eigenvalue and Q isthe corresponding eigenvector. Recall that the power iteration method is a classical method for computingthe eigenvector associated with the largest eigenvalue of a matrix [21, p. 533] x k +1 = Ax k k Ax k k , k = 0 , , · · · . (55)We can derive the similar algorithm. Note that there is a difference between (54) and the eigenvalueequation: Q is a Hermitian matrix. Thus, the iteration (55) cannot be used directly. From (54), since K and Q commute, thus, we have that KQ = K / QK / and hence Q = K / QK / Tr( K / QK / ) , f ( Q ) . (56) June 6, 2018 DRAFT4
Note that f ( Q ) (cid:23) and Tr( f ( Q )) = 1 for any Q ∈ Ω . The equation (56) defines a mapping from aconvex set to itself: Ω → Ω , Q f ( Q ) . The optimal Q corresponds to a fixed point of f ( Q ) , i.e., f ( Q ◦ ) = Q ◦ . To search for the KKT solution, the iterative expression is Q k +1 = f ( Q k ) , k = 0 , , · · · (57)The initial point Q can be set to I n T /n T , or any Q ∈ Ω . The iterations stop when the relative errorof C s ( Q ) in the successive iterations is less than a preset value, e.g., − or − . If the convergent Q satisfies (13), we obtain a KKT solution, otherwise, we choose a different initial point.VIII. N UMERICAL S IMULATIONS
In this section we provide some examples to illustrate the theoretical findings. We assume that Σ R isnormalized as Tr( Σ R ) = n T , and Σ E is multiplied correspondingly by a factor η . In simulations, weassume that the correlation matrices of legitimate and eavesdropper channels follow the Jakes’ correlationmodel [18], i.e., for p, q = 1 , · · · , n T Σ R ( p, q ) = J (cid:0) φ R | p − q | πd/λ (cid:1) , Σ E ( p, q ) = η J (cid:0) φ E | p − q | πd/λ (cid:1) where J ( · ) is the zero-order Bessel function of the first kind, d is the element spacing, λ is thewavelength, and φ R (or φ E ) is a parameter that controls the correlation among antennas and has itsvalue determined by the distance between the transmitter and receiver, and the incident angle of thewavefront. We set d/λ = 1 / . A. The transmitter has full information about the legitimate channel and only statistical informationabout the eavesdropper channel
We consider a MISO wiretap channel where n T = 4 , n R = n E = 1 . We set h R = [0 . . , . . , . . , . − . T and φ E = 0 . , η = 0 . . The eigenvaluesof h R h † R − Σ E are . , − . , − . , − . .Fig. 2 depicts the function C s ( z ) defined in (20) for z in [0 . , . with step . and SNR = 10 dB .Among these points, the optimal point is (0 . , . also depicted in Fig. 2.Fig. 3 depicts the ergodic secrecy rates during the iteration of the algorithm of Section § VII for
SNR = 10 dB and Q = I . The convergent ergodic secrecy rate is . . We can see that the June 6, 2018 DRAFT5 algorithm converges rapidly. If we do iterations for
SNR = 10 dB and Q = I , the convergentvalues are: λ ( Q ) = { . , . , . , . } ,λ ( Θ ) = { . , − . , − . , − . } Tr( ΘQ ) = 0 . . Fig. 4 plots the ergodic secrecy rates for different φ E from . to . . It can be seen from Fig. 4 thatthe ergodic secrecy rate decreases first, and then increases with φ E . Fig. 5 plots the ergodic secrecy ratesfor different SNR and φ E = 0 . . As is revealed in § V, the ergodic secrecy rate increases with
SNR . B. The transmitter has only statistical information about both the legitimate channel and the eavesdropperchannel
We set φ R = 0 . , φ E = 0 . , η = 0 . . The eigenvalues of Σ R − Σ E are . , . , . , . .Fig. 6 depicts the ergodic secrecy rates during the iteration for SNR = 10 dB and Q = I , whileFig. 7 depicts the ergodic secrecy rates for random Q ∈ Ω . We can see that the algorithm convergesrapidly. If we do iterations for SNR = 10 dB and Q = I , the convergent values are: λ ( Q ) = { . , . , . , . } ,λ ( Θ ) = { . , . , − . , − . } Tr( ΘQ ) = 0 . . We can see that the convergent Q has rank two.Fig. 8 plots the ergodic secrecy rates for different φ R from . to . . It can be seen from Fig. 8that the ergodic secrecy rate increases with φ R . Fig. 9 plots the ergodic secrecy rates for different φ E from . to . . Fig. 10 plots the ergodic secrecy rates for different SNR . As is revealed in § V, when Σ R ≻ Σ E , the ergodic secrecy rate increases with SNR .IX.
CONCLUSION
We have investigated the problem of finding the optimal input covariance matrix that achieves ergodicsecrecy capacity subject to a power constraint. We extend the existing result to nontrivial covariances ofthe legitimate and eavesdropper channels. We have derived the necessary conditions for the optimal inputcovariance matrix in the form of a set of equations and propose an algorithm to solve the equations.
June 6, 2018 DRAFT6 A PPENDIX AP ROOF OF L EMMA Σ E − Σ R (cid:23) , then C s ≤ . Since h R ∼CN ( , Σ R ) , h E ∼ CN ( , Σ E ) , we can write h R = Σ / R h w , (58) h E = Σ / E h w . (59)By inserting (58) and (59) into (4), we get C s ( Q ) = E h w { log(1 + ρ h † w Σ / R QΣ / R h w ) }− E h w { log(1 + ρ h † w Σ / E QΣ / E h w ) } . (60)Let x ≥ x ≥ · · · ≥ x n T and y ≥ y ≥ · · · ≥ y n T be eigenvalues of Σ / R QΣ / R and Σ / E QΣ / E ,respectively. By using the fact that Uh w and h w have the identical distributions for any unitary matrix U , we have C s ( Q ) = E h w (cid:26) log (cid:18) ρ n T X i =1 x i | h w,i | (cid:19)(cid:27) − E h w (cid:26) log (cid:18) ρ n T X i =1 y i | h w,i | (cid:19)(cid:27) . (61)According to Ostrowski theorem [24, p. 224], we know that if A (cid:23) B and B ≻ , then λ k ( A / QA / ) ≥ λ k ( B / QB / ) , where λ k ( · ) denotes the k th eigenvalue arranged in decreasing order. With this, we knowthat x i ≤ y i , i = 1 , · · · , n T . On the other hand, it is easy to verify that the following function g ( z , · · · , z n T ) = E h w (cid:26) log (cid:18) ρ n T X i =1 z i | h w,i | (cid:19)(cid:27) (62)is strictly increasing with respect to z i , i = 1 , · · · , n T . Thus, we get that C s ( Q ) ≤ for any Q . Thiscompletes the first part.Second, we prove that if Σ R − Σ E is none negative semi-definite, then there exists a Q such that C s ( Q ) > . Let u be the eigenvector associated with the largest eigenvalue λ of Σ R − Σ E . Since λ > , we get that u † ( Σ R − Σ E ) u = λ > and u † Σ R u > u † Σ E u . We will prove that Q = uu † achieves C s ( Q ) > . In this case, we know that x = u † Σ R u , x = · · · = x n T = 0 , y = u † Σ E u and y = · · · = y n T = 0 . Since x > y and the function g ( z , · · · , z n T ) defined in (62) is strictly increasingwith respect to z , we get that C s ( Q ) > . This completes the proof. June 6, 2018 DRAFT7 A PPENDIX BP ROOF OF L EMMA C s ( Q ) = log(1 + ρ h † R Qh R ) − E h w { log(1 + ρ h † w Σ / E QΣ / E h w ) } . (63)It follows from the Jensen’s inequality [26, p. 25] that log E ( x ) ≥ E (log x ) . With this and the fact that E h w { h † w Ah w } = E h w { Tr( Ah w h † w ) } = Tr( A ) , we get C s ( Q ) ≥ log(1 + ρ h † R Qh R ) − log(1 + ρ Tr( QΣ E )) . (64)Note that h † R Qh R − Tr( QΣ E ) = Tr( Q ( h R h † R − Σ E )) . Let u be the eigenvector associated with the largesteigenvalue λ of h R h † R − Σ E . Since λ > , we know that Tr( uu † ( h R h † R − Σ E )) = u † ( h R h † R − Σ E ) u = λ > . Thus, C s ( Q ) > holds for Q = uu † . This completes the proof.A PPENDIX CP ROOF OF L EMMA E z { log(1 + ρ z † Qz ) } = Z ∞ e − t t (cid:18) − I + tρ R / QR / ) (cid:19) d t. (65)By inserting R / QR / = U D U † into (65), we get E z { log(1 + ρ z † Qz ) } = Z ∞ e − t (cid:18) t − t Q Mi =1 (1 + tρd i ) (cid:19) d t. (66)Performing partial fraction expansion, i.e., t − t Q Mi =1 (1 + tρd i ) = M X j =1 ρd j Q Mi = j (1 − d i /d j ) e − t tρd j (67)and using Z ∞ e − t ta d t = 1 a F ( a ) , (68)we get (6). This completes the proof. June 6, 2018 DRAFT8 A PPENDIX DP ROOF OF T HEOREM ΨQ = QΨ = 0 , that is, Ψ and Q commute and have the same eigenvectors [23,p.239] and their eigenvalue patterns are complementary in the sense that if λ i ( Q ) > , then λ i ( Ψ ) = 0 ,and vice versa [14]. This result, when combined with (8), implies that Θ and Q commute and have thesame eigenvectors. Further, we get ΘQ = QΘ = θ Q , which, when combined with Tr( Q ) = 1 and thefact Tr( QΘ ) is always real, leads to θ = Tr( QΘ ) and (12) (also see [15]).The condition (12) reveals that for the optimal Q , QΘ is a scaled version of Q . Further, the eigenvaluesof Θ corresponding to the positive eigenvalues of Q are all equal to Tr( QΘ ) , while the remainingeigenvalues of Θ are all less than or equal to Tr( QΘ ) , which follows from (8), (12) and Ψ (cid:23) . Basedon the above (13) follows. A PPENDIX EP ROOF OF L EMMA I . We write z = R / h w where h w =[ h w, , · · · , h w,n T ] T ∼ CN ( , I n T ) , and h w,k ’s follow i.i.d. CN (0 , . With this, we have I = E h w (cid:26) ρ R / h w h † w R / ρ h † w R / QR / h w (cid:27) . (69)By inserting R / QR / = U D U † into (69), we have I = ρ R / E h w (cid:26) h w h † w ρ h † w U D U † h w (cid:27) R / . (70)Then we use the fact that Uh w and h w have the identical distributions for any unitary matrix U to obtain I = ρ R / U YU † R / (71)where Y = E h w (cid:26) h w h † w ρ h † w D h w (cid:27) (72)with ( i, j ) th entries given by Y ij = E h w (cid:26) h w,i h ∗ w,j ρ P Mk =1 d k | h w,k | (cid:27) . (73)From the gamma integral [16] we have a z = 1Γ( z ) Z ∞ t z − e − ta d t, Re ( z ) > , a > (74) June 6, 2018 DRAFT9 where Γ( z ) = R ∞ u z − e − u d u , we let z = 1 to obtain a = R ∞ e − ta d t . With this identity, we can write Y ij = Z ∞ e − t E h w (cid:26) h w,i h ∗ w,j M Y k =1 e − tρd k | h w,k | (cid:27) d t. (75)Since h w,k ’s follow i.i.d. CN (0 , , we know Y ij = 0 for i = j , i.e., Y is a diagonal matrix with ( k, k ) thentries given by Y kk = Z ∞ e − t Q Mi =1 (1 + tρd i ) 11 + tρd k d t, k ≤ M (76) Y kk = Z ∞ e − t Q Mi =1 (1 + tρd i ) d t, k > M. (77)These integrals can be easily calculated. Performing partial fraction expansion Q Mi =1 (1 + tρd i ) = M X j =1 Q Mi = j (1 − d i /d j ) 11 + tρd j (78)and using (68) and Z ∞ e − t (1 + ta )(1 + tb ) d t = F ( a ) − F ( b ) a − b (79) Z ∞ e − t (1 + ta ) d t = F ( a ) , (80)we get (14). This completes the proof. A PPENDIX FP ROOF OF L EMMA φ ( tz + (1 − t ) z ) ≤ tφ ( z ) + (1 − t ) φ ( z ) , ∀ t ∈ [0 , . (81)Let X and X be the optimal X associated with z and z . Consider the problem associated with tz + (1 − t ) z , i.e., min X Tr( Σ E X ) (82) s . t . Tr( h R h † R X ) = ( tz + (1 − t ) z ) k h R k , Tr( X ) = 1 , X (cid:23) . It is easy to verify that t X + (1 − t ) X satisfies the constraints in the problem of (82) with thecorresponding objective value tφ ( z ) + (1 − t ) φ ( z ) . Thus, (81) holds and φ ( z ) is a convex function. June 6, 2018 DRAFT0 A PPENDIX GP ROOF OF L EMMA λ > be an eigenvalue of aa † − A , and we have det( aa † − A − λ I ) = 0 . (83)Note that A + λ I is positive definite. By using the identity det( B − aa † ) = (1 − a † B − a ) det( B ) for apositive definite matrix B , it follows from (83) that − a † ( A + λ I ) − a = 0 . (84)Denote ℓ ( λ ) , − a † ( A + λ I ) − a . It is easy to verify that ℓ ( λ ) is a strictly increasing function. Thus, ℓ ( λ ) has only one positive root, and ℓ (0) = 0 , i.e., is not a eigenvalue of aa † − A . Thus, all othereigenvalues are negative. This completes the proof.A PPENDIX HP ROOF OF L EMMA log x > E (log Y ) = ⇒ E (cid:16) log xY (cid:17) > . (85)By using Jensen’s inequality [26, p. 25], we have log E (cid:16) xY (cid:17) ≥ E (cid:16) log xY (cid:17) = ⇒ E (cid:16) xY (cid:17) > ⇒ x E (cid:18) Y (cid:19) > ⇒ x < E (cid:18) Y (cid:19) . (86)R EFERENCES [1] Y. Liang, H. V. Poor, and S. Shamai (Shitz),
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June 6, 2018 DRAFT2
0 0.2 0.4 0.6 0.8 1 0.51 1.52 2.53 z C s ( z ) ( b i t s / s / H z ) (0.55, 2.8143) Fig. 2. The function C s ( z ) , z = 0 .
01 : 0 .
01 : 0 . . Full CSI on h R is used, SNR = 10 dB .
0 2 4 6 8 10 12 14 161 1.21.41.61.82 2.22.42.62.83 k E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 3. Ergodic secrecy rate during the iteration. Full CSI on h R is used; Q = I , SNR = 10 dB . June 6, 2018 DRAFT3 φ E E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 4. Ergodic secrecy rate for different values of φ E . Full CSI on h R is used; Q = I , SNR = 10 dB .
0 5 10 15 201 1.52 2.53 3.54 4.5 SNR (dB) E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 5. Ergodic secrecy rate for different
SNR . Full CSI on h R is used. June 6, 2018 DRAFT4
0 2 4 6 8 10 12 14 16 181.41.61.82 2.22.4 k E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 6. Ergodic secrecy rate during the iteration. Only statistical info. on h R is used; Q = I , SNR = 10 dB .
0 5 10 15 20 25 30 35 400.81 1.21.41.61.82 2.22.4 k E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 7. Ergodic secrecy rate during the iteration. Only statistical info. on h R is used; random Q , SNR = 10 dB . June 6, 2018 DRAFT5 φ R E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 8. Ergodic secrecy rate for different values of φ R . Only statistical info. on h R is used; Q = I , SNR = 10 dB . φ E E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 9. Ergodic secrecy rate for different values of φ E . Only statistical info. on h R is used; Q = I , SNR = 10 dB . June 6, 2018 DRAFT6
0 5 10 15 200.51 1.52 2.53 3.54 SNR (dB) E r god i c S e c r e cy R a t e ( b i t s / s / H z ) Fig. 10. Ergodic secrecy rate for different
SNR . Only statistical info. on h R is used.is used.