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Water-filling for noise variances one and four

Example

For the two parallel branches with noise variances $λ_{1} = 1$ and $λ_{2} = 4$ , under total power $P \geq 0$ , the optimal allocation is $(0, 0)$ and the capacity is $0$ at $P = 0$ . For $P > 0$ , the water level and allocations are $T = {\begin{cases} 1 + P, & 0 < P \leq 3, \\ (P + 5) / 2, & P \geq 3, \end{cases} (p_{1}, p_{2}) = {\begin{cases} (P, 0), & 0 \leq P \leq 3, \\ ((P + 3) / 2, (P - 3) / 2), & P \geq 3. \end{cases}$ The capacity is $C (P) = \frac{1}{2} \ln (1 + P)$ for $0 \leq P \leq 3$ and $C (P) = \ln ((P + 5) / 4)$ for $P \geq 3$ . At $P = 3$ , both formulas give $\ln 2$ .

Facts & Assumptions

Given: A two-dimensional parallel Gaussian channel with independent noise variances $1$ and $4$ , and a nonnegative total power $P$ .

[A1]

Finite-dimensional additive Gaussian channel with average power: the cost is squared Euclidean norm per vector use and rates are natural-log nats.

[F1]

Water-filling maximizes the finite parallel Gaussian allocation objective: the optimal allocation is $p_{j} = (T - λ_{j})^{+}$ with $T$ chosen so their sum is $P$ .

[F2]

Capacity of finite parallel Gaussian channels with and without feedback: this allocation's value is the ordinary and feedback capacity under the average power constraint.

Verification

1.1A1F1givencasesalgebra

At $P = 0$ , the only feasible allocation is $(0, 0)$ . If $0 < P \leq 3$ , choosing $T = 1 + P \leq 4$ makes $p_{1} = T - 1 = P$ and $p_{2} = (T - 4)^{+} = 0$ , whose sum is $P$ . If $P \geq 3$ , choosing $T = (P + 5) / 2 \geq 4$ gives $p_{1} = (P + 3) / 2$ , $p_{2} = (P - 3) / 2$ , and their sum is $P$ . At $P = 3$ the formulas agree. In each positive-power regime [F1] therefore confirms the displayed water level and nonnegative powers.

2.1F2step 1.1givenalgebra∎

Substituting the first allocation into the rate formula [F2] gives $\frac{1}{2} \ln (1 + P)$ ; substituting the second gives $\frac{1}{2} \ln (T) + \frac{1}{2} \ln (T / 4) = \ln (T / 2) = \ln ((P + 5) / 4)$ . At the common endpoint $P = 3$ , both are $\ln 2$ .

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Sources

Yury Polyanskiy and Yihong Wu, Information Theory: From Coding to Learning, Chapter 20 §20.4, Theorem 20.14 (standard reference, not scraped)
Yury Polyanskiy, MIT 6.441 Information Theory, Chapter 17 §17.3.2, Theorem 17.7 (standard reference, not scraped)