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Capacity of finite parallel Gaussian channels with and without feedback

Statement

Assume AC and use the finite-dimensional channel of dimension $d$ and covariance $K$ from Finite-dimensional additive Gaussian channel with average power. Let $λ_{1}, \dots, λ_{d} > 0$ be the eigenvalues of $K$ and put $λ_{\min} : = \min_{j} λ_{j}$ . Both the ordinary no-feedback capacity $C_{n f} (P)$ and the causal-feedback capacity $C_{f b} (P)$ defined in Causal feedback codes for a finite-dimensional Gaussian channel equal $C_{K} (P) = \max_{p_{j} \geq 0, \sum_{j} p_{j} \leq P} \frac{1}{2} \sum_{j = 1}^{d} \ln (1 + \frac{p_{j}}{λ_{j}}) = \frac{1}{2} \sum_{j = 1}^{d} \ln^{+} ⁣ (\frac{T}{λ_{j}}),$ where $T$ is the unique water level solving $\sum_{j} (T - λ_{j})^{+} = P$ when $P > 0$ ; for $P = 0$ , $C_{K} (0) = 0$ and all $p_{j} = 0$ . Thus feedback does not increase capacity.

Facts & Assumptions

Given: AC, a finite-dimensional additive Gaussian channel with positive-definite covariance, a finite-message code, and an expected average-power budget.

[A1]

The Axiom of Choice and Finite-dimensional additive Gaussian channel with average power: the Gaussian channel has independent fresh $N_{d} (0, K)$ noises and uses squared Euclidean norm as cost under AC.

[A2]

Causal feedback codes for a finite-dimensional Gaussian channel: each causal encoder has the form $f_{t} (M, Y^{t - 1})$ , the decoder is measurable, and the power constraint is expected average input energy.

[F1]

Orthogonal coordinates diagonalize a finite-dimensional Gaussian noise law: orthogonal coordinates diagonalize $K$ and preserve power and code/error correspondence.

[F2]

Finite-input vector Gaussian mutual information is bounded by water-filling: for every finite-support vector input of second moment at most $u$ , its one-use mutual information is at most $C_{K} (u)$ .

[F3]

Water-filling maximizes the finite parallel Gaussian allocation objective: $C_{K} (u)$ is the maximum allocation value and is nondecreasing and concave.

[F4]

Jensen's inequality for an integrable cost random variable: for integrable $U \geq 0$ , the finite concave function $C_{K}$ satisfies $E C_{K} (U) \leq C_{K} (E U)$ .

[F5]

Fano's inequality for a finite message observed through a density: for an $N$ -message variable observed through conditional densities and decoded with error $e$ , $H (M ∣ Y) \leq \ln 2 + e \ln N$ .

[F6]

Strictly feasible Gaussian inputs yield cost-feasible Gaussian codes: every rate below the Gaussian-input mutual information of a strict allocation is achievable with no feedback under the average power budget.

[F7]

A positive-definite multivariate normal law has its Gaussian density: positive-definite Gaussian noise has a positive Lebesgue density.

[F8]

The natural logarithm as the inverse of the exponential function: $\ln$ is the natural logarithm.

[F9]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $\ln^{'} (x) = 1 / x$ for $x > 0$ , so integrating its derivative gives $\ln (1 + x) \leq x$ for $x \geq 0$ and continuity of the finite rate expression.

[F10]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere: a nonnegative measurable function of integral zero vanishes almost everywhere; on the probability space this says zero expected nonnegative energy is zero almost surely.

Proof

1.1A1A2F1F2F5F7givenalgebra

Use [F1] to express the channel in independent scalar branches. For a feedback code from [A2], conditional on each message $m$ the output vector has density $p (y^{n} ∣ m) = \prod_{t = 1}^{n} ϕ_{K} (y_{t} - f_{t} (m, y^{t - 1})),$ where $ϕ_{K} (z) = (2 π)^{- d / 2} (\det K)^{- 1 / 2} \exp (- z^{T} K^{- 1} z / 2)$ is positive by [F7]. Thus [F5] applies. Since $M$ is finite, define $I (M; Y^{n}) = H (M) - H (M ∣ Y^{n})$ using the posterior density of [F5]. Bayes' formula factors $\Pr (M = m ∣ Y^{n}) / \Pr (M = m)$ into the product over $t$ of $p (Y_{t} ∣ M = m, Y^{t - 1}) / p (Y_{t} ∣ Y^{t - 1})$ ; taking logarithms and expectations proves $I (M; Y^{n}) = \sum_{t} I (M; Y_{t} ∣ Y^{t - 1})$ . At a fixed history, $X_{t} = f_{t} (M, h)$ has finite support; grouping the finite mixture terms for messages with the same input shows $I (M; Y_{t} ∣ h) = I (X_{t}; Y_{t} ∣ h)$ because their conditional output densities are identical. Hence [F2] bounds this by $C_{K} (u_{t, h})$ , where $u_{t, h} = E [∥ X_{t} ∥^{2} ∣ h]$ .

2.1F3F4F8F9step 1.1givenalgebra

Let $T$ be uniform on $[n]$ , independent of the channel, and let $H = Y^{T - 1}$ and $U = u_{T, H}$ . Then $U \geq 0$ and $E U = n^{- 1} \sum_{t} E ∥ X_{t} ∥^{2} \leq P$ . For every allocation with total power at most $u$ , [F8]--[F9] give $\frac{1}{2} \sum_{j} \ln (1 + p_{j} / λ_{j}) \leq \frac{1}{2} \sum_{j} p_{j} / λ_{j} \leq u / (2 λ_{\min})$ ; hence $C_{K} (U)$ is integrable. Apply [F4] to step 1.1 and then monotonicity of $C_{K}$ to obtain $I (M; Y^{n}) \leq n C_{K} (E U) \leq n C_{K} (P)$ .

3.1F5step 2.1givenalgebracases

If $N = 1$ the rate is zero. If $N \geq 2$ , [F5] and $H (M) = \ln N$ give $I (M; Y^{n}) \geq (1 - e) \ln N - \ln 2$ . Combining with step 2.1 shows every reliable feedback rate is at most $C_{K} (P)$ . The no-feedback capacity is bounded by the same argument because no-feedback codes are a subclass.

4.1A1A2F3F6F9F10step 3.1givencasesalgebra∎

If $P = 0$ , put $V = \sum_{t = 1}^{n} ∥ X_{t} ∥^{2}$ . This is nonnegative and the expected-power constraint gives $E V \leq n P = 0$ , hence $E V = 0$ . By [F10], $V = 0$ almost surely, and every summand is therefore zero almost surely. Thus $X_{t} = 0$ for every $t$ outside one null set and $Y^{n} = Z^{n}$ almost surely. The fresh noise vector is independent of the message by [A1], so the outputs carry no message information and the Fano bound in step 3.1 gives zero reliable rate. One-message codes attain rate zero. If $P > 0$ , [F3] supplies the allocation $p_{j}^{*}$ with total power $P$ . For any $R < C_{K} (P)$ choose $ϵ > 0$ so the strict allocation $(1 - ϵ) p^{*}$ has Gaussian mutual information greater than $R$ ; continuity from [F9] gives such an $ϵ$ . Apply [F6] to obtain no-feedback codes at every such rate, and therefore also feedback codes. Letting $R ↑ C_{K} (P)$ proves both lower bounds, including the case $C_{K} (P) = 0$ by a one-message code.

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 20 §§20.3–20.4 and Chapter 23 §23.1 (standard reference, not scraped)
Yury Polyanskiy, MIT 6.441 Information Theory, Chapter 17 §§17.3.1–17.3.2 and Chapter 21 §21.1 (standard reference, not scraped)