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Genie-aided outer bounds for the two-user Gaussian interference channel

Statement

For the complex Gaussian channel in Real and complex two-user Gaussian interference channels, the following are outer bounds in the listed regimes; all logarithms are base two. Write SNR and INR as defined there.

Weak regime: If ${I N R}_{1} < {S N R}_{2}$ and ${I N R}_{2} < {S N R}_{1}$ , every achievable pair satisfies $R_{1} \leq \log_{2} (1 + {S N R}_{1}), R_{2} \leq \log_{2} (1 + {S N R}_{2}),$ $R_{1} + R_{2} \leq \log_{2} (1 + {S N R}_{1}) + \log_{2} ⁣ (1 + \frac{{S N R}_{2}}{1 + {I N R}_{2}}),$ $R_{1} + R_{2} \leq \log_{2} (1 + {S N R}_{2}) + \log_{2} ⁣ (1 + \frac{{S N R}_{1}}{1 + {I N R}_{1}}),$ $R_{1} + R_{2} \leq \log_{2} ⁣ (1 + {I N R}_{1} + \frac{{S N R}_{1}}{1 + {I N R}_{2}}) + \log_{2} ⁣ (1 + {I N R}_{2} + \frac{{S N R}_{2}}{1 + {I N R}_{1}}),$ $2 R_{1} + R_{2} \leq \log_{2} (1 + {S N R}_{1} + {I N R}_{1}) + \log_{2} ⁣ (1 + {I N R}_{2} + \frac{{S N R}_{2}}{1 + {I N R}_{1}}) + \log_{2} ⁣ (\frac{1 + {S N R}_{1}}{1 + {I N R}_{2}}),$ $R_{1} + 2 R_{2} \leq \log_{2} (1 + {S N R}_{2} + {I N R}_{2}) + \log_{2} ⁣ (1 + {I N R}_{1} + \frac{{S N R}_{1}}{1 + {I N R}_{2}}) + \log_{2} ⁣ (\frac{1 + {S N R}_{2}}{1 + {I N R}_{1}}) .$

Mixed regime: If ${I N R}_{1} \geq {S N R}_{2}$ and ${I N R}_{2} < {S N R}_{1}$ , every achievable pair satisfies the two individual bounds above and $R_{1} + R_{2} \leq \log_{2} (1 + {S N R}_{1}) + \log_{2} ⁣ (1 + \frac{{S N R}_{2}}{1 + {I N R}_{2}}), R_{1} + R_{2} \leq \log_{2} (1 + {S N R}_{1} + {I N R}_{1}),$ $R_{1} + 2 R_{2} \leq \log_{2} (1 + {S N R}_{2} + {I N R}_{2}) + \log_{2} ⁣ (1 + {I N R}_{1} + \frac{{S N R}_{1}}{1 + {I N R}_{2}}) + \log_{2} ⁣ (1 + \frac{{S N R}_{2}}{1 + {I N R}_{1}}) .$ The other mixed regime is obtained by interchanging user indices 1 and 2.

Facts & Assumptions

Given: A complex Gaussian interference code with independent messages and the SNR/INR convention below; logarithms and entropies in the proof are base two.

[A1]

The Axiom of Choice: AC is inherited through the Gaussian-law and information suppliers.

[F1]

Real and complex two-user Gaussian interference channels: the channel has independent unit circular Gaussian receiver noises and the stated average powers.

[F2]

Fano bound for finite messages and density observations: a vanishing-error decoder gives a message-entropy bound by mutual information plus $o (n)$ .

[F3]

Mutual information tensorizes over a memoryless vector channel: vector mutual information on a memoryless channel is bounded by the sum of its coordinate values.

[F4]

Conditional mutual-information chain rule and data processing for standard-Borel variables: conditional chain rules and data processing identify deterministic-message and codeword information and allow revealing side information.

[F5]

Power bound for real and complex scalar Gaussian channels: the scalar Gaussian information power bound and its equality case.

[F6]

Entropy estimates for complex Gaussian interference converses: the entropy-difference bound and every unconditioned and genie-conditioned output entropy estimate used below, with only average power assumptions.

Proof

1.1A1F1F2F3F4F5givenalgebra

Write $a = {S N R}_{1}$ , $b = {S N R}_{2}$ , $c = {I N R}_{1}$ , $d = {I N R}_{2}$ . Give $M_{2}$ to receiver 1 and subtract its deterministic codeword. Fano, conditional data processing in both directions, and tensorization give $n R_{1} \leq \sum_{j} I (X_{1, j}; h_{11} X_{1, j} + Z_{1, j}) + o (n)$ . The scalar Gaussian power bound followed by Jensen and the average input constraint gives $R_{1} \leq \log_{2} (1 + a)$ . Reflection gives $R_{2} \leq \log_{2} (1 + b)$ . The same argument applies when a direct gain vanishes and then forces the corresponding rate to zero. Throughout the proof independent messages give independent user input vectors, regardless of coordinate dependence inside a vector; the additive Fano remainders disappear on division by $n$ and passage to the coding limit.

2.1F1F2F4F6step 1.1givenconstructalgebra

Reveal $X_{2}^{n}$ to receiver 1. The chain rule and independent inputs give the following exact expansion of the two augmented mutual informations: $I (X_{1}^{n}; Y_{1}^{n} ∣ X_{2}^{n}) + I (X_{2}^{n}; Y_{2}^{n}) = h (Y_{2}^{n}) - h (Z_{1}^{n}) + h (h_{11} X_{1}^{n} + Z_{1}^{n}) - h (h_{12} X_{1}^{n} + Z_{2}^{n})$ . If $P_{1} = 0$ , the input vector is zero and this entropy difference and the claimed logarithmic ratio are both zero. Otherwise, if $a \geq d$ , the entropy-difference estimate [F6] applies with $s = ∣ h_{11} ∣^{2}$ , $t = ∣ h_{12} ∣^{2}$ , $P = P_{1}$ ; phases are removed by unit-modulus rotations. This is precisely the coupling condition $∣ h_{11} ∣ \geq ∣ h_{12} ∣$ : the lower-SNR observation can be formed by scaling the higher-SNR observation and adding circular Gaussian noise of variance $1 - t / s$ . The proof of [F6] covers zero gains without division by a cross gain. Its unconditioned-output bound therefore yields $R_{1} + R_{2} \leq \log_{2} (1 + b + d) + \log_{2} ((1 + a) / (1 + d)) = \log_{2} (1 + a) + \log_{2} (1 + b / (1 + d))$ . Only $a \geq d$ is needed. Reflecting the argument gives the second weak sum bound under $b \geq c$ , using the reflected noise coupling.

2.2F1F2F4F6step 1.1givenconstructalgebra

Define $S_{1}^{n} = h_{12} X_{1}^{n} + Z_{2}^{n}$ and $S_{2}^{n} = h_{21} X_{2}^{n} + Z_{1}^{n}$ using independent receiver noises. Give $S_{1}^{n}$ to receiver 1 and $S_{2}^{n}$ to receiver 2. Expanding both augmented informations gives $I (X_{1}^{n}; Y_{1}^{n}, S_{1}^{n}) + I (X_{2}^{n}; Y_{2}^{n}, S_{2}^{n}) = h (Y_{1}^{n} ∣ S_{1}^{n}) + h (Y_{2}^{n} ∣ S_{2}^{n}) - h (Z_{1}^{n}) - h (Z_{2}^{n})$ : the positive $h (S_{1}^{n})$ and $h (S_{2}^{n})$ cancel the interference-plus-noise entropies of the two conditional channels. Applying the two conditional-output estimates in [F6] and $h (Z_{i}^{n}) = n \log_{2} (π e)$ proves the third displayed weak sum bound. This calculation is valid in every regime and requires no degradation inequality.

3.1F1F2F4F6step 1.1step 2.1step 2.2givenalgebra

Add two copies of receiver 1's Fano bound and one of receiver 2's. Augment receiver 2 by $S_{2}^{n}$ and one receiver 1 copy by $X_{2}^{n}$ . The resulting information sum equals $h (Y_{1}^{n}) + h (Y_{2}^{n} ∣ S_{2}^{n}) - 2 h (Z_{1}^{n}) + h (h_{11} X_{1}^{n} + Z_{1}^{n}) - h (h_{12} X_{1}^{n} + Z_{2}^{n})$ . Indeed the three information terms contribute, respectively, $h (Y_{1}^{n}) - h (S_{2}^{n})$ , $h (S_{2}^{n}) + h (Y_{2}^{n} ∣ S_{2}^{n}) - h (h_{12} X_{1}^{n} + Z_{2}^{n}) - h (Z_{1}^{n})$ , and $h (h_{11} X_{1}^{n} + Z_{1}^{n}) - h (Z_{1}^{n})$ . The unconditioned, conditional, and entropy-difference estimates of [F6] give $2 R_{1} + R_{2} \leq \log_{2} (1 + a + c) + \log_{2} (1 + d + b / (1 + c)) + \log_{2} ((1 + a) / (1 + d))$ . The only degradation hypothesis is $a \geq d$ , exactly as in step 2.1. Reflection proves the displayed $R_{1} + 2 R_{2}$ bound under $b \geq c$ . Thus the stated weak regime supplies every required coupling.

3.2F1F2F4F6step 1.1step 2.1givenconstructalgebracases

In the mixed regime $c \geq b$ and $a > d$ , the individual bounds and first sum bound still follow from steps 1.1 and 2.1. For the MAC sum bound, receiver 1 first decodes $M_{1}$ . On the event of correct decoding it subtracts $h_{11} X_{1}^{n}$ and obtains $h_{21} X_{2}^{n} + Z_{1}^{n}$ . If $P_{2} = 0$ , step 1.1 gives $R_{2} = 0$ and the required MAC sum bound already follows from the individual bound. Otherwise, if $h_{21} \neq 0$ , scale by $h_{22} / h_{21}$ and add independent circular Gaussian noise of variance $1 - ∣ h_{22} / h_{21} ∣^{2}$ , which is nonnegative because $c \geq b$ and $P_{2} > 0$ . Add the known vector $h_{12} X_{1}^{n}$ and apply receiver 2's decoder. Coupling this simulation to the one using the true $M_{1}$ bounds its joint-message error by the sum of the two original errors; a deterministic MAP decoder has no larger average error. Joint-message Fano and the unconditioned-output estimate [F6] give $R_{1} + R_{2} \leq \log_{2} (1 + a + c)$ . If $h_{21} = 0$ , then $b = c = 0$ and step 1.1 forces $R_{2} = 0$ , so the same sum bound follows immediately. This establishes the exact mixed sum claim, including degenerate gains.

4.1F1F2F4F6step 1.1step 2.2step 3.2givenalgebra∎

For the mixed weighted bound add the three Fano information terms $I (X_{1}^{n}; Y_{1}^{n}, S_{1}^{n})$ , $I (X_{2}^{n}; Y_{2}^{n})$ , and $I (X_{2}^{n}; Y_{2}^{n}, S_{2}^{n} ∣ X_{1}^{n})$ . The first expands as $h (S_{1}^{n}) - h (Z_{2}^{n}) + h (Y_{1}^{n} ∣ S_{1}^{n}) - h (S_{2}^{n})$ , the second as $h (Y_{2}^{n}) - h (S_{1}^{n})$ , and the third as $h (S_{2}^{n}) - h (Z_{1}^{n}) + h (Y_{2}^{n} ∣ S_{2}^{n}, X_{1}^{n}) - h (Z_{2}^{n})$ . Their sum is $h (Y_{1}^{n} ∣ S_{1}^{n}) + h (Y_{2}^{n}) + h (Y_{2}^{n} ∣ S_{2}^{n}, X_{1}^{n}) - h (Z_{1}^{n}) - 2 h (Z_{2}^{n})$ . Apply the three corresponding bounds in [F6] to obtain exactly $R_{1} + 2 R_{2} \leq \log_{2} (1 + b + d) + \log_{2} (1 + c + a / (1 + d)) + \log_{2} (1 + b / (1 + c))$ . This particular weighted estimate is universal and uses no degradation condition. Reflection gives the other mixed regime. Combining the preceding steps proves every displayed inequality with precisely the hypotheses asserted in the statement.

Depends on

Used by

The two-user Gaussian interference capacity region to within one bit Theorem

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Raul H. Etkin, David N. C. Tse, and Hua Wang, Gaussian Interference Channel Capacity to Within One Bit (standard reference, not scraped)