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Exact strong-interference region for the two-user Gaussian interference channel

Statement

For the real or complex Gaussian interference channel of Real and complex two-user Gaussian interference channels, assume the strong regime ∣h12∣≥∣h11∣,∣h21∣≥∣h22∣. Then the capacity region is the set of nonnegative rate pairs satisfying R1≤κlog⁡2(1+SNR1),R2≤κlog⁡2(1+SNR2), R1+R2≤κmin⁡{log⁡2(1+SNR1+INR1),log⁡2(1+SNR2+INR2)}, where κ=1/2 for real and 1 for complex channels.

Facts & Assumptions

Given: A length-n interference code with independent messages, the channel and power constraints of [F1], and the two gain inequalities in the statement.

[A1]

The Axiom of Choice: AC is inherited through the Gaussian random-coding and scalar-channel information suppliers.

[F1]

Real and complex two-user Gaussian interference channels: noises are independent unit-variance real or circular complex Gaussian; the input power constraints and SNR/INR conventions are as displayed there.

[F2]

Random coding achieves the finite-dimensional Gaussian multiple-access region: independent Gaussian codebooks achieve both receivers' multiple-access regions.

[F3]

Fano bound for finite messages and density observations: a finite message estimated from a density observation with error Pe has conditional entropy at most ln⁡2+Peln⁡∣M∣.

[F4]

Mutual information tensorizes over a memoryless vector channel: for a memoryless channel and any dependent input vector, vector mutual information is at most the sum of its coordinate mutual informations.

[F5]

Conditional mutual-information chain rule and data processing for standard-Borel variables: conditional mutual information obeys the chain rule and the data-processing inequality for standard-Borel variables, including finite message side information.

[F6]

Power bound for real and complex scalar Gaussian channels: for either scalar Gaussian model, I(X;aX+Z)≤κln⁡(1+∣a∣2E∣X−EX∣2) nats, with equality for the corresponding Gaussian inputs.

Proof

1.1A1F1F3F4F5F6givenalgebra

For the individual bound on R1, give M2 to receiver 1. Fano gives nR1ln⁡2≤I(M1;Y1n∣M2)+o(n) in nats. For each fixed M2, X2n is fixed; translating Y1n by the known vector −h21X2n is invertible, so [F5] in both directions preserves this conditional mutual information. The shifted channel from X1n is memoryless. The codeword map is deterministic and the channel law depends on M1 only through X1n, so [F5] in both Markov directions gives I(M1;Y1n∣M2)=I(X1n;h11X1n+Z1n∣M2). Apply [F4] for each M2 and average to bound this by ∑iI(X1,i;h11X1,i+Z1,i). By [F6], each term is at most κln⁡(1+∣h11∣2p1,i) nats, where p1,i=E∣X1,i∣2. Jensen's inequality and the power constraint give R1≤κlog⁡2(1+SNR1). Exchanging users gives the stated R2 bound.

1.2F1givenconstructalgebra

First suppose both cross gains are nonzero. Receiver 1 first forms its estimate M^1. On the event M^1=M1, it reconstructs X1n and subtracts h11X1n, obtaining h21X2n+Z1n. Since ∣h21∣≥∣h22∣, it scales this residual by h22/h21 and adds independent Gaussian noise of variance 1−∣h22/h21∣2; for complex channels use circular Gaussian noise. Adding back h12X1n gives a simulated Y~2n. For the error analysis, construct the same vector using the true M1: it then has exactly the law of receiver 2's Y2n, and on M^1=M1 it equals the receiver's simulation. Hence the joint error is at most Pe,1+Pe,2. Since the message pair is finite, the randomized decoder has a deterministic MAP decoder with no larger average error. Exchanging users and using ∣h12∣≥∣h11∣ gives the symmetric joint decoder at receiver 2.

2.1F1F3F4F5F6step 1.2algebra

In the nonzero-cross-gain case of step 1.2, apply the density-observation Fano bound [F3] to the joint decoder at receiver 1. Let M=(M1,M2) and X=(X1n,X2n). The deterministic codeword map and channel law give both Markov directions between M and X with observation Y1n, so [F5] yields I(M;Y1n)=I(X;Y1n). The two input vectors are independent because their messages are independent. By [F4], n(R1+R2)ln⁡2≤I(X1n,X2n;Y1n)+o(n)≤∑iI(X1,i,X2,i;Y1,i)+o(n). The output variance at time i is at most 1+∣h11∣2E∣X1,i∣2+∣h21∣2E∣X2,i∣2, because the independent inputs have zero cross covariance after centering. These single-coordinate mutual informations are finite by the power bound and [F6].

3.1F1F5F6step 1.1step 1.2step 2.1algebracases

For each coordinate put Si=h11X1,i+h21X2,i. The channel output depends on the input pair only through Si, so [F5] in both directions gives I(X1,i,X2,i;Y1,i)=I(Si;Si+Z1,i). Independence gives Var⁡(Si)≤∣h11∣2p1,i+∣h21∣2p2,i; [F6] therefore bounds this term by κln⁡(1+∣h11∣2p1,i+∣h21∣2p2,i) nats. Jensen's inequality and the average power constraints yield R1+R2≤κlog⁡2(1+SNR1+INR1). The joint decoder at receiver 2 analogously gives R1+R2≤κlog⁡2(1+SNR2+INR2). If h21=0, then h22=0, so R2=0 by step 1.1; the first sum bound is the R1 individual bound and the second is no tighter because INR2≥SNR1. If h12=0, then h11=0, so R1=0 and the second sum bound is the R2 individual bound while the first is no tighter because INR1≥SNR2. Thus both sum bounds hold also when a cross gain vanishes.

4.1F1F2step 1.1algebra∎

Use the same independent Gaussian codebook ensemble at the two receivers. The finite-receiver construction in the proof of [F2] sums the two ensemble-average errors and jointly selects codebooks meeting both deterministic power constraints, so one codebook pair achieves the intersection of the two Gaussian MAC regions. The strong gain inequalities make each unintended-receiver individual constraint at least as large as the corresponding intended-receiver constraint; the two sum constraints are exactly those displayed. Therefore every rate pair strictly inside the claimed region is achievable, and closure removes strictness.

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