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Orthogonal reduction of a correlated two-dimensional Gaussian channel

Example

Assume AC. Let $U = \frac{1}{\sqrt{2}} (\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array}), K = U^{T} (\begin{array}{cc} 1 & 0 \\ 0 & 4 \end{array}) U = (\begin{array}{cc} 5 / 2 & 3 / 2 \\ 3 / 2 & 5 / 2 \end{array}) .$ For the channel $Y = X + Z$ with $Z \sim N_{2} (0, K)$ and total power budget $P = 2$ , the transformed noise $U Z$ has independent variances $1$ and $4$ , the input energy is preserved, and both feedback and no-feedback capacities equal $\frac{1}{2} \ln 3$ nats per vector use.

Facts & Assumptions

Given: The matrix $U$ , covariance $K$ , AC, the Gaussian channel, and power budget $P = 2$ .

[A1]

The Axiom of Choice and Standard normal and normal laws: the scalar standard-normal law is used under AC and has density $(2 π)^{- 1 / 2} e^{- x^{2} / 2}$ .

[A2]

Finite-dimensional additive Gaussian channel with average power: the noise is additive with covariance $K$ and the power cost is $∥ X ∥_{2}^{2}$ .

[F1]

Orthogonal coordinates diagonalize a finite-dimensional Gaussian noise law: an orthogonal matrix diagonalizes the Gaussian covariance, the transformed components are independent, and squared norm is preserved.

[F2]

Orthogonal transformations preserve the standard normal product law: under AC, an orthogonal transformation preserves the independent standard-normal product law.

[F3]

Capacity of finite parallel Gaussian channels with and without feedback: a parallel Gaussian channel has feedback and no-feedback capacity equal to the water-filled allocation value.

Verification

1.1A1A2F1F2givenalgebra

Direct multiplication gives $U^{T} U = I_{2}$ and $U^{T} diag (1, 4) U = \frac{1}{2} (\begin{array}{cc} 5 & 3 \\ 3 & 5 \end{array}) = K$ , hence $U K U^{T} = diag (1, 4)$ , so $K$ is positive definite with eigenvalues $1, 4$ . The diagonalization argument of [F1], applying [F2] to the underlying independent standard-normal vector, gives that $U Z$ has independent coordinates with variances $1, 4$ . For every input, $∥ U X ∥_{2}^{2} = X^{T} U^{T} U X = ∥ X ∥_{2}^{2}$ and $U (X + Z) = U X + U Z$ .

2.1F1F3step 1.1givenalgebra∎

Since $P = 2 < 4 - 1 = 3$ , the water level is $T = 1 + P = 3$ : the first branch receives $p_{1} = T - 1 = 2$ and the second receives $p_{2} = 0$ . Substitution in [F3] gives capacity $\frac{1}{2} \ln (1 + 2 / 1) + \frac{1}{2} \ln (1 + 0 / 4) = \frac{1}{2} \ln 3$ . The orthogonal code correspondence from step 1.1 transfers this value to the original correlated channel.

Depends on

Used by

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 20 §20.4 (standard reference, not scraped)
Dmitry Panchenko, MIT 18.443 Statistics for Applications, Fall 2006, Lecture 4 (standard reference, not scraped)