On the Gaussian Z-Interference channel

Transcription

1 On the Gaussian Z-Interference channel Max Costa, Chandra Nair, and David Ng University of Campinas Unicamp, The Chinese University of Hong Kong CUHK Abstract The optimality of the Han and Kobayashi achievable region with Gaussian signaling remains an open problem for Gaussian interference channels. In this paper we focus on the Gaussian Z-interference channel. We first show that using correlated over time Gaussian signals does not improve on the Han and Kobayashi achievable rate region. Secondly we compute the slope of the Han and Kobayashi achievable region with Gaussian signaling around the Sato s corner point. I. INTRODUCTION Gaussian interference channel is one of the most basic multiuser settings whose capacity region is as yet undetermined. The best known-achievable region for a two-receiver interference channel is due to Han and Kobayashi []. Recently it has been shown that there are two-receiver interference channels with discrete alphabets where the Han and Kobayashi region is strictly inside the capacity region []. However for the Gaussian setting, the optimality of the Han and Kobayashi region with Gaussian auxiliaries remains an open challenge. In the discrete memoryless setting, it was shown that a -letter extension coding in blocks of two symbols of the Han and Kobayashi scheme strictly outperforms the single-letter traditional scheme. There has been some attempts, for instance, see [3], at using correlated Gaussians to improve on the Han and Kobayashi scheme for the interference channel. Remark. It is worth mentioning that the authors in [3] claim that the rates they achieve outperform state-of-the-art coding schemes. This claim fails to hold if a comparison is made with the rates of the Han and Kobayashi scheme with Gaussian signals and the use of a time-sharing variable, Q, to do power control. A scheme employing the concept of noisebergs has been shown by one of the authors [4] to strictly improve the rate region with power control. The sub-optimality of the region without power control can also be shown by using perturbations along Hermite polynomials [5]. In this paper, the first result is a proof that correlated Gaussian signaling does not improve on the traditional single-letter scheme for the Gaussian interference channel. The second result concerns evaluation of the slope of the Han and Kobayashi region with Gaussian signaling around the corner point known as Sato s corner point. A. Preliminaries An interference channel is a model for communication where two point-to-point communications occur over a shared medium causing interference. The particular channel model that we study in this paper is called the Gaussian Z-interference setting, and the channel is described by: Y = X + Z Y = X + ax + Z. Here Z and Z are Gaussian variables, each distributed as N 0, and independent of X and X. We further assume power constraints P, P on X, X and that a 0,. Note that if a = 0 or a, then the capacity region is fully determined; hence this regime is the only open case. Z N 0, X a Y X Y Z N 0, Fig. : Gaussian Z interference channel The capacity region for this setting is defined in the usual sense see [6] for details and background work.

2 The Han and Kobayashi achievable region [] for the interference channel can be found in Section 6.5 of [6]. However when the interference is one-sided as in the Z-channel, the achievable region simplifies to the following. Theorem Han and Kobayashi Region. The union of rate pairs R, R satisfying the constraints R IX ; Y Q R IX ; Y U, Q R + R IU, X ; Y Q + IX ; Y U, Q over distributions p Q qp U,X Qu, x qp X Qx q is achievable for a discrete memoryless Z-interference channel. To achieve this region, it suffices to consider Q 3. The above region will also yield an achievable region in the Gaussian Z-interference channel defined by 7. Further, for every Q = q, if X = U + V, where U and V are zero-mean independent Gaussian random variables, and X is also an independent Gaussian random variable, then we call such a region as Han and Kobayashi region with Gaussian signaling. Theorem Han and Kobayashi region with Gaussian signaling. The union of rate pairs R, R satisfying the constraints R E Q log + P Q R E Q log P Q + + a α Q P Q R + R E Q log + α QP Q + log + P Q + a α Q P Q + a α Q P Q over α Q [0, ], P Q, P Q 0 satisfying E Q P Q P and E Q P Q P is achievable. By Bunt s extension of Caratheodory s theorem, it suffices to consider Q 5. The region described by Theorem will be referred to as R HK. By computing the Han and Kobayashi region with Gaussian signaling of the multi-letter extension of the Gaussian interference channel, the following region is achievable. Theorem 3 k-letter Han and Kobayashi region with Gaussian signaling. The union of rate pairs R, R satisfying the constraints R k E Q log I + K Q R k E Q log I + a K vq + K Q I + a K vq R + R k E Q log I + K vq + log I + a K Q + K Q I + a K vq over K q, K q 0, K vq K q satisfying k E QtrK Q P and k E QtrK Q P is achievable. As before it suffices to consider Q 5. The relation denotes the positive semi-definite partiarder among real symmetric matrices; while A and tra denotes the determinant and trace of matrix A, respectively. The region described by Theorem 3 will be referred to as R k HK. Clearly R HK R k HK C, the capacity region. Known results about the capacity region: In this section we summarize the previously known results about the capacity region of the Gaussian Z-interference channel. i It is known from [7], [8], [9] that the rate-pair R = log + P and R = log + P is a Pareto-optimal point on the boundary of the capacity region. Further it is also known that the above point maximizes the rate-sum R + R. Since C = log + P is the maximum achievable rate to receiver Y, the capacity region contains a +a P line-segment that starts at C, 0 and ends at C, log + P +a P. We call this extremal point corner point as Sato-point. The outer bounds in [9] and [0] shows that the Sato-point also maximizes R + R for any +a P a +P. This provides an outer bound to the slope of the capacity region around the Sato s corner point. ii It is known from [7], [], [] that the rate-pair R = log + a P +P and R = C = log + P, is another Pareto-optimal point on the boundary of the capacity region of the Gaussian Z interference channel. Hence the capacity region contains a line-segment that starts at 0, C and ends at log + a P +P, C. This recently established extremal point is the second corner point of the capacity region of the Z Gaussian interference channel. The outer bound to the capacity region of the interference channel does not yield any finite such that the maximum of R + R over

3 3 achievable rate pairs passes through this corner point. Two of the authors computed the slope of Han and Kobayashi region with Gaussian signaling around this new corner point [3]. B. Summary of our results In this article we establish the following results. Theorem 4. R k HK = R HK, k. We show that the k-letter extension of the Han and Kobayashi region with Gaussian signaling does not improve on the single-letter scheme. Theorem 5. The largest value of such that the maximum of R + R with R, R R HK occurs at the Sato point is given by a cr HK,s + P + a } P = min a, P + P where is the unique positive solution of γ = 0, where P γ := log + + a P + log a P + a P + a P + P. a P + P + a P + a P + P II. CORRELATED GAUSSIANS DO NOT IMPROVE THE REGION In this section we establish Theorem 4. Towards this end we make the following observations: both R k HK and R HK are convex regions, hence they can be characterized by the intersection of supporting hyperplanes. Further the hyperplane R + R to the capacity region passes through the Sato-point, which is present in both R k HK and R HK. Hence Theorem 4 is equivalent to showing that for every >, max R + R = max R + R. R,R R k R,R R HK HK The above condition can be re-expressed in terms of matrices as max E Q K q,k q 0,K vq K q: k E QtrK Q P, k E QtrK Q P + k log K vq + I k log a K vq + I = max E Q P q,p q 0,α q [0,]: log P Q + a P Q + + E Q trp Q P,E Q trp Q P + log α QP Q + log a α Q P Q +. The key to establishing is the following Theorem. Theorem 6. The maximum of the expression k log K Q + a K Q + I + log K Q + a K vq + I k log K + a K v + a K u + I + log K + a K v + I + log K v + I log a K v + I log P Q + a α Q P Q + where the k k Hermitian matrices satisfy the constraints: K, K u, K v 0, trk u + K v P, trk P can be attained by restricting K, K u, K v to be diagonal matrices. Remark. Identify K uq := K q K vq. It is immediate that equation will follow from Theorem 6 by the following reasoning: for every Q = q, Theorem 6 allows us to replace the matrices inside the expression by diagonal matrices, which then is just sum of terms of the form appearing on the right-hand-side. Note the extra factor k on the left hand-side changes this sum into a convex combination and the equality is immediate. The following notations will be used in the proof: i Given a k k matrix A, let λa denote the unordered set of eigenvalues of A, and let λ A λ A denote the k-tuple of eigenvalues of A arranged in decreasing increasing order respectively.

4 4 ii Given two vectors v, w, we say v w if v majorizes w, i.e. if v [] v [] v [k] is a non-increasing arrangement of v and w [] w [] w [k] is a non-increasing arrangement of w; then m m v [l] w [l], m k with equality at m = k. l= l= Proof of Theorem 6. Suppose we fix the matrices K and K v satisfying the trace constraint; then we must choose K u so as to maximize K + a K v + a K u + I subject to trk u P trk v. If one further fixes the eigenvalues λk u then Fiedler s bound [4] says that K + a K v + a K u + I k i= λ i K + a K v + I + a λ i K u, and clearly equality is achieved if the matrices K + a K v + I and K u share the same eigenvectors with eigenvalues λ i K + a K v + I and λ i K u, respectively. Hence we seek to maximize k i= λ i K + a K v + I + a λ i K u, subject to k i= λ i K u P trk v and λ i K u 0. The optimal choice of this problem is the water-filling solution. Denote the optimal choice as K w u ; then it is immediate that λ i K + a K v + I + a K u = λ i K + a K v + I + a λ i Kw u, i =,.., k. Let K be a diagonal matrix with entries ordered as λ i K, and K v be a diagonal matrix with entries ordered as λ i K v. Applying Fiedler s bound [4] we see that K + a K v + I k i= λ i K + λ i K v + = K + a Kv + I. 3 We now invoke the celebrated Lidskii-Wielandt inequality see.6 in survey article [5] that establishes the majorization, Let K,w u Lemma yields λ K + a K v + I = λ K + λ a K v + I λ K + a K v + I. denotes the diagonal water filling matrix corresponding to K + a K v + I. Lemma implies that λ K + a K v + I + K,w u λ K + a K v + I + K w u. K + a K v + I + K,w u K + a K v + I + K w u. 4 Since K v + I = k i= + λ ik v and a K v + I = k i= + a λ i K v, we see that for a fixed choice of λk v and λk, the diagonal matrices K, Kv, and Ku,w maximize the expression term-by-term in Theorem 6. Varying over the choices of λk v and λk that satisfies the trace constraint establishes the theorem. A. Lemmas regarding majorization and its applications The following Lemma must be well-known but we cannot find an immediate reference, so we establish it below. Lemma. [Waterfilling preserves majorization] Let v, u be vectors such that v u. Let v and u denote the vectors obtained after water-filling operation with a quantity of water W > 0. Then v u. Proof. W.l.o.g. let v and u be arranged in decreasing order. After waterfilling operation note that v and u is also in decreasing order; and further they satisfy v i v i i m =, c m + i k u u i i n i =. n + i k Further, v m c v m+, u n c u n+, and W = k i=m+ c v i = k i=n+ c u i. c

5 5 We divide the proof into two cases: m < n and m n. Case : m < n. Note that v i + W = u i + W = i= i= m v i + k mc = i= n m u i + k nc u i + k mc i= i= m v i + k mc. i= The first inequality is due to u i c, m+ i n, and the second inequality is from v u. Hence c c. Thus to establish that v u, it suffices to show that l l v i u i, m + l n, as the rest of the choices of l are immediate from v u, c c, and that k i= v i = k We establish it by contradiction. Suppose is the first index in [m + : n] such that i= m v i + i= i=m+ i= c = i= v i > i= u i = i= u i. i= u i. Hence it must be that c > u lo = u, and since u i is decreasing, c u i, i. This would imply that v i = i= i= v i + a contradiction. This completes the proof in this case. Case : m n. Note that W = i=m+ c v i = i=n+ i=+ c > i= c u i u i + i=m+ i=+ u i = c u i u i, i= i=m+ c v i, where the first inequality is due to c u i 0, n + i m, and the second inequality is from v u, taif v has larger partial sum than taif u. Thus c c, as before. Similar to previous case, to establish that v u, it suffices to show that l v i i= l u i, n + l m, as the rest of the choices of l are immediate from v u, c c, and that k i= v i = k contradiction. Suppose is the first index in [n + : m] such that i= v i = i= i= v i > l 0 i= u i = n u i + i= l 0 i=n+ c. i= u i Hence it must be that v l0 > c, and since v i is decreasing, v i c, i l 0. This would imply that v i = i= i= v i + i=+ a contradiction. This completes the proof of the lemma. v i > i= u i + i=+ c = u i, i=. The proof follows again by Lemma. see A..d, page 66 in [6] Let A, B 0 and λa λb. Then k i= λ ia = A B = k i= λ ib. It basically follows from the concavity of log and the Hardy-Littlewood-Polya majorization inequality.

6 6 A. Han and Kobayashi region with Gaussian signaling III. SLOPE AT SATO S CORNER POINT For, the maximum of the weighted sum-rate R + R of the Han and Kobayashi region with Gaussian signaling for a Z-interference channel can be computed as max E Q P q,p q 0,α q [0,]: log P Q + a P Q + + log P Q + a α Q P Q + E Q trp Q P,E Q trp Q P + log α QP Q + log a α Q P Q +. 5 We know that when =, the maximum sum-rate for the Han and Kobayashi region as well as the capacity region is given by log + P + log P + a P + and is achieved at the rate pair R = log + P, and R = log + P a P +, which is referred to as Sato s corner point. We define } = sup : +, HK,s cr max R,R R HK R + R = log + P + log P a P + the largest such that the line R + R passing through Sato s corner point is a supporting hyperplane to R HK. Using the concave envelope interpretation see [7] for E Q we see that for any the value of max R,R R HK R +R is the upper concave envelope of f Q, Q evaluated at P, P, where f is defined by f Q, Q := log + a Q + Q + max α [0,] log + a αq + Q + a αq + log + αq + a αq + Q By taking derivative with respect to α, the optimal α = α satisfies: = a + Q a a + a Q + α Q We define the regions R = Q, Q : } a + Q a Q R = Q, Q : a + Q + a } Q a 6 Q + Q R 3 = Q, Q : a + Q + a Q a < < } a + Q Q + Q a Q where R, R, R 3 correspond to the cases α = 0, α = and 0 < α < respectively. This gives an explicit expression for f, log + a Q + Q + log + Q Q, Q R Q f Q, Q = log + +a Q + log + Q Q, Q R log +a Q +Q a Q + log log + log a +Q a + log a Q, Q R 3 Note that the hyperplane R + R passes through Sato s corner point if and only if P, P R and Cf P, P = f P, P. From Lemma in Appendix, this is equivalent to requiring that P, P R and g Q, Q attains global maximum at P, P, where g is defined by g Q, Q := f Q, Q a + a a P + P + a + P + P Q + a P + P } Q. Thus to establish Theorem 5, we need to show that g Q, Q attains a global maximum at P, P if and only if cr HK,s where HK,s cr where is the solution to h = 0, with h := log + P + a P a + P + a } P = min a, P + P a P + a P + a P + P } + log a } P + P + a P + a P + P. The proof of Theorem 5 is completed by analysing the local behaviour of g and isolating the local maxima.

7 7 Interior analysis: Lemma 3. HK,s cr min a + P + a P + a } P a, P + P a + P Proof. This is the condition that says P, P R and it is a local maximum. HK,s a P +P since P, P R ; else the optimizing α is not see 6. The second condition, cr HK,s +a P a +P follows from one of the second order conditions for P, P being a local maximum, namely det Hf P, P 0 see Hessian calculation in Appendix. +a Lemma 4. There is no local maximum of g in the interior of R for any P. a +P cr a +P +a P Proof. Since g is concave in R, there is at most one local maximum in the interior of R. The first order condition yields a + a = a Q + Q + a a P + P + a + P + P + a + = Q + Q + Q + a P + P Solving for Q gives But in R we have a +Q a Q. Substituting for Q yields Q = + a P + P + P + a + a P a 4 + P. +a But we also have P, a +P implying a. This gives a contradiction. Lemma 5. There are at most local maxima of g in the interior of R. The value of g at both points is bounded from +a above by g P, P, when P. a +P Proof. The first order condition for local-maximum yields The solutions are a + a Q + Q a + a Q + = + Q + a = Q + Q a + a a P + P + a + P + a P + P a Q = P or k Q = P + a P Q + P where k := +a P a +P. If k, there is only one solution in R + at P, P, so assume that k <. If Q, Q is any solution to above, then where ϕ x := log x + x. g Q, Q = f Q, Q a Q + Q + a + a Q Q + Q + a Q Q + Q = f Q, Q + + a Q + Q + a + Q + Q = ϕ + a Q + Q ϕ + a Q + ϕ + Q = ϕ + a P + P ϕ + a Q + ϕ + Q

8 8 Now let Q, Q to be the solution other than P, P. Then, g P, P g Q, Q = ϕ + a Q ϕ + a P ϕ + Q ϕ + P = ϕ a ϕ a k k k a k a ϕ a ϕ k a k Differentiating with respect to and simplifying gives g P, P g Q, Q = [ ] log + k k 0 k k k k +a since P a +P = k and x 0, log + x x. So g P, P g Q, Q g =k P, P g =k Q, Q = 0 and hence g P, P g Q, Q. +a Lemma 6. There is no local maximum of g in the interior of R 3 when < P. a +P Proof. The first order condition for g having a local maximum yields + a Q + Q Q + a + a Q + Q = a + Q = a + a a P + P + a + P + a P + P. By substituting the first equation into the second one, and then writing = a +Q a Q θ, where θ Q = a + P From the second order condition see Appendix det Hf Q, Q 0, or equivalently, a + Q +a which contradicts with that < P. a +P = Q + a P a + P + P +a Q +Q,, we get Thus the interior analysis yields that the value of g Q, Q at any interior local maximum is upper bounded by that of g P, P and P, P R if and only if a + P + a P + a } P min a, P + P a. + P The necessity comes from Lemmas 3, while the sufficiency comes from Lemmas 4, 5, and 6. Boundary analysis: The remaining cases are the boundaries Q = 0 and Q = 0. In this part, we first establish that g P, P g Q, Q for Q, Q on the boundaries if and only if is smaller than or equal to the upper bound in Lemma 3 and in Theorem 5. Then in Lemma 0 we reduce the minimum of three terms to that of two of them. This gives the critical in Theorem 5. Lemma 7. On the boundary Q = 0 we have that if and only if log+p+ +P log+a P + +a P. Proof. When Q = 0, we have g P, P max Q 0 g 0, Q f Q, Q = log + Q, g Q, Q = log + Q + a P + P Q.

9 9 Note that g 0, Q is concave in Q, and maximized at Q = a P + P. Since P, P R, min g P, P g 0, Q Q 0 = g P, P g 0, a P + P = log + a P + + a P 0 if and only if log+p+ +P log+a P + +a P. Lemma 8. + log + P + + P log + P + +P + a log + a P + +a P P a + P and hence, by Lemma 3 and Lemma 7, g P, P g Q, Q on the boundary Q = 0. Proof. This is equivalent to where ϕ x = + x log + x + xx. Let Note that Hence a 4 ϕ P ϕ a P 0 ψx = a 4 ϕx ϕa x. ϕ x = + x log + x x ϕ x = log + x + x ψx = a 4 ϕ x a ϕ a x x ψx = a 4 log + x + a x 0. So ψx is convex when x 0 and ψ 0 = 0, implying ψ is convex and increasing on x 0. Thus ψp ψ0 = 0. Lemma 9. On the boundary Q = 0 we have that if and only if, where as in Theorem 5. Proof. When Q = 0, we have f Q, Q = log + Q g Q, Q = log + Q g Q, 0 is concave in Q, maximized when g P, P max Q 0 g Q, 0 a + a a P + P + a + Q P + P a = + Q + a a P + P + a + P + P [0, ] since a + P + a P a P + P That is, there is always a maximizing Q 0. Note that, after some manipulations, we can express min g P, P g Q, 0 Q 0 [ log + = = h P + a P a } P + a P + a + log P + P a }] P + P + a P + a P + P which is concave in, equals 0 when = 0, the derivative with respect to is non-negative at = 0. Here h as in Theorem 5. Hence g P, P max Q 0 g Q, 0 if and only if. Thus combining the interior and boundary analysis we see that cr HK,s is the minimum of three quantities, two of them given by Lemma 3 and one given by Lemma 9. The proof is completed by showing that one of the three quantities is redundant.

10 0 Lemma 0. The following holds: a + P + a } P min a, a + P + a P + a } P = min P + P a, P + P a, + P where as in Theorem 5. Proof. It suffices to show that, if a +P +a P +a a P +P P, a +P or equivalently P a +a +P, then P a +P. +a That is, γ P a +P 0, where γ is defined in Theorem 5. Write P = + a P θ and k := +a P a +P. Then θ k and k. We would like to show hk 0, that is, k θ log + θ kk θ + + log kθ + θ 0 k log + θ + kθ + θ kθ + + log + θ + θ 0 The derivative of left hand side with respect to θ is equal to So k log + θ + +θ + kθ +θ This completes the proof of Theorem 5. B. Outerbound to the slope at Sato s point θ + θ + k kθ + θ kθ + θ = k θ + θ + kθ + θ = k θ k + θ + θ kθ 0 k + log kθ +θ is decreasing in θ, and = 0 when θ = 0. We are done. The outer bound the the capacity region by Sato [8] and Kramer [0] see also [] from which this form is taken states that any achievable rate-pair R, R must satisfy + a P e R + P e R a +. a A simple calculus argument shows that the largest such that the maximum of R + R occurs at Sato s corner point is given by Remark 3. Note that the inner bound for the slope HK,s cr OB,s cr := + a P a + P. coincides with the outer bound OB,s cr in the limit P. CONCLUSION In this paper we show that correlated Gaussians over time do not improve on the single-letter Han and Kobayashi achievable region with Gaussian signaling. We also computed the slope of the Han and Kobayashi region around the Sato s corner point. ACKNOWLEDGEMENT M. Costa would like to acknowledge the partial support by FAPESP. Chandra Nair wishes to acknowledge the support from the various GRF grants from the RGC, Hong Kong.

11 REFERENCES [] T. Han and K. Kobayashi, A new achievable rate region for the interference channel, Information Theory, IEEE Transactions on, vol. 7, no., pp , jan 98. [] C. Nair, L. Xia, and M. Yazdanpanah, Sub-optimality of the Han Kobayashi Achievable Region for Interference Channels, ArXiv e-prints, Feb. 05. [3] W. Huleihel and N. Merhav, Codewords With memory improve achievable rate regions of the memorylessgaussian interference channel, CoRR, vol. abs/ , 05. [Online]. Available: [4] M. H. M. Costa, Noisebergs in Z Gaussian interference channels, Information Theory and Applications Workshop ITA, 0. [5] E. Abbe and L. Zheng, A coordinate system for Gaussian networks, Information Theory, IEEE Transactions on, vol. 58, no., pp , Feb 0. [6] A. El Gamal and Y.-H. Kim, Network Information Theory. Cambridge University Press, 0. [7] M. H. M. Costa, On the Gaussian interference channel, Information Theory, IEEE Transactions on, vol. 3, no. 5, pp , Sept 985. [8] H. Sato, The capacity of the Gaussian interference channel under strong interference corresp., IEEE Transactions on Information Theory, vol. 7, no. 6, pp , Nov 98. [9], On degraded Gaussian two-user channels corresp., Information Theory, IEEE Transactions on, vol. 4, no. 5, pp , Sep 978. [0] G. Kramer, Outer bounds on the capacity of Gaussian interference channels, Information Theory, IEEE Transactions on, vol. 50, no. 3, pp , March 004. [] Y. Polyanskiy and Y. Wu, Wasserstein continuity of entropy and outer bounds for interference channels, CoRR, vol. abs/ , 05. [Online]. Available: [], Converse bounds for interference channels via coupling and proof of Costa s conjecture, in Proc. of IEEE ISIT, Barcelona, Spain, July 06. [3] M. H. M. Costa and C. Nair, Gaussian Z-interference channel: around the corner, Information Theory and Applications Workshop ITA, 06. [4] M. Fiedler, Bounds for the determinant of the sum of Hermitian matrices, Proceedings of the American Mathematical Society, vol. 30, no., pp. 7 3, 97. [Online]. Available: [5] T. Ando, Special issue honoring Ingram Olkin majorizations and inequalities in matrix theory, Linear Algebra and its Applications, vol. 99, pp. 7 67, 994. [Online]. Available: [6] A. Marshall, I. Olkin, and B. Arnold, Inequalities: Theory of Majorization and Its Applications, ser. Springer Series in Statistics. Springer New York, 00. [Online]. Available: [7] C. Nair, Upper concave envelopes and auxiliary random variables, International Journaf Advances in Engineering Sciences and Applied Mathematics, vol. 5, no., pp. 0, 03. [Online]. Available: [8] H. Sato, An outer bound to the capacity region of broadcast channels, IEEE Trans. Info. Theory, vol. IT-4, pp , May, 978. [9] M. H. M. Costa, A third critical point in the achievable region of the Z-Gaussian interference channel, Information Theory and Applications Workshop ITA, 04. A. Calculation via the Noiseberg region APPENDIX One of the authors proposed [4] that the Han and Kobayashi region with Gaussian signaling for the Z-interference channel is equivalent to the noiseberg region. This is based on the equivalence between the Gaussian Z-interference channel and the Gaussian degraded interference channel, demonstrated in [7]. The Gaussian degraded interference channel is described by Y = X + Z 7 Y = X + Y + Z where Z and Z are Gaussian variables distributed as N 0, and N 0, N, respectively, and independent of X and X. In addition we assume power constraints P, P on X, X. The equivalence between the Z and the degraded Gaussian interference channels holds if their three parameters are related by P = P, P = P a and N = a a. Applying the notation in Theorem, the noiseberg scheme a particular choice of parameters corresponds to Q = with: PQ = = λ, PQ = = λ, α =, P = 0. This leads to the constraints R λ R λ R + R λ log + P + λ log + P P log + + a α P log + α P + log + P + a α P + a + λ α P log + P. We have the constraints P = P λ, P = P λp λ. Therefore there are three free variables in the above expression: P, α, λ. Therefore the maximum weighted sum-rate of the noiseberg region is given by R + R = max λ α,p,λ log + α P + log + a α P + λ + a α P + log + P λp P λ λ P λ λ + log + + a. α P

12 Fig. : Slope of normal at Sato s point for P = P =. The noiseberg scheme applied to the degraded interference channel uses two parameters, namely the multiplex band λ and the noiseberg height h. The rates R and R associated with this scheme are given by R = λ log + P A λ + log + max0, h N + λ + N + P A log + P B λ R = λ log + P λ + N + P A λ where the partial powers P A and P A are given by λ + P λ P A = λ [P λ minh, N max0, h N ] λp P A = λ [P + P + λ minh, N ]. To simplify notation we restricted the primed power to P. The parameters λ and h are defined in a certain admissible region see details in [4]. The corner point that corresponds to Sato s point is R = log + P, R = log + P +P +N. To find the contour of the Gaussian Han and Kobayashi region at this point we evaluate the gradient of R h, λ + R h, λ, for λ close to zero and 0 h N. Equating the gradient to zero for the outermost rate contour we find Note that is the slope of the normal to the rate contour of the achievable region. As calculated in [9], the first of this derivative ratios can be expressed as P +h +P log + P +h +P dr /dλ dr /dλ = dr /dλ dr /dλ = dr /dh =. 8 dr /dh P +P +N log + P +P +N + 9 hp +P +P +P+P +N The second ratio of derivatives is dr /dh dr /dh = P + h + P + N + P + P + N P + P + P + P + h 0

13 3 Example plots of these derivative ratios are given in [9] for P =, P = 4 and N = 3, i. e., P =, P = and a = 0.5. To find the point of intersection we can equate these two expressions and do some algebraic manipulation. Equivalently, we can take the derivative of Eq. 9 with respect to h and equate it to zero. We then get the following equation. + P + P P + N G + P + N N = + P + P + hg P + h + P, where the function Gx is defined as log + x/x. Solving this equation numerically or graphically for h we obtain a side product of this approach which is the optimal initial height of the noiseberg, as the rate point departs from Sato s point, if h N. Let this optimal value be denoted by h. Then we can use Eqs. 9 or 0 with h = h, and then Eq. 8 to get the normal slope at Sato s point. For example, if we take the case P =, P = 4 and N = 3, equivalent to P = P = and a = 0.5, we get h =.545, and = If the h solution to Eq. turns out to be greater than N, then there is no need for multiplexing with a noiseberg band, and the optimal strategy to exit the Sato s point is pure superposition, that is moving along the path with λ = 0 and h varying from N to N + P in the admissible λ, h parameter region. In this case the normal slope at Sato s point is easily found to be = P + N + P + N P + P. In Fig. we plot the slope of the normal to the achievable region at Sato s point for a in the interval [0, ] with P = P =. We show two curves which correspond to the optimal multiplexing strategy bottom curve and to the pure superposition scheme top curve. Note that for values of a below approximately 0.4, the two curves coincide, indicating that for these values of the interference gain, there is no advantage in multiplexing See To mux and not to mux in [9]. Remark 4. To illustrate the need for multiplexing for certain values of the channel parameters, we observe that the linear combination of rates R and R given by R + R = log + P + log + P +P +N is not concave in certain regions of the P, P plane and certain values of. In these cases the best rate combinations happen above the surface of the function, in its concave envelope, and require the multiplexing of two superposition schemes. As an example, in Fig.3 we show a surface plot of this function in the plane P, P with = and N = 3, for 0 P, P 8. The shading of the surface indicate the non-concavity. B. Gradient and Hessian of the function f In R, In R, In R 3, Q f = a + a Q + Q Q f = + a + Q + Q + Q Hf = [ a 4 +a Q +Q a a +a Q +Q +a Q +Q +a Q +Q Q f = a + a a Q + Q + a + Q + Q Q f = + a Q + Q Hf = a 4 a +a Q +Q + 4 +a Q +Q a +a Q +Q +Q ] a +a Q +Q +a Q +Q Q f = a + a Q + Q Q f = + a + Q + Q Q a + Q Hf = a 4 a +a Q +Q a +a Q +Q +a Q +Q +a Q +Q + Q a +Q

14 4 Fig. 3: Surface plot of R + R as a function of P and P with = and N = 3. By checking the values f and f at the boundaries, one can see that f is continuously differentiable on R >0. Lemma. Let f be a real-valued function differentiable at x. Then Cfx = fx if and only if f fx, attains global maximum at x. Here Cf and f denotes the concave envelope and gradient of f, respectively. Proof. It suffices to show that Cfx fx if and only if for all h we have fx fx + h fx, h. The if part is immediate, by taking concave envelope with respect to h and then putting h = 0. For the only if part, suppose on the contrary that there is ɛ > 0 and h 0 such that By differentiability of f at x, for ζ small enough. Now, for any δ 0,, fx + ɛ fx + h fx, h fx + ζ fx fx, ζ ɛ h ζ fx Cfx δ Cfx + h + δ Cfx δ δ h δfx + h + δfx δ δ h δɛ + δfx + fx, δh + δfx δ δ h

15 5 Rearranging gives for δ small enough. This gives a contradiction. fx δ δ ɛ + fx δ δ h fx, δ δ h δ δ ɛ + fx ɛ h δ δ h = fx + ɛ δ δ