Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Gaussian feedback estimation-error covariance update

Statement

Let (X1,X2) be centered jointly Gaussian with variances P1,P2>0, correlation ρ∈[0,1), and let Z∼N(0,N) be independent, N>0. Put X=(X1,X2)T, a=(1,1)T, K=(P1ρP1P2ρP1P2P2), Y=aTX+Z, and D=N+aTKa. The residual X−E[X∣Y] is independent of Y and has covariance K−KaaTK/D. In particular, Var⁡(X1∣Y)=P1[N+P2(1−ρ2)]D,Var⁡(X2∣Y)=P2[N+P1(1−ρ2)]D, Cov⁡(X1−E[X1∣Y],X2−E[X2∣Y])=ρP1P2N−P1P2(1−ρ2)D. After scaling the two errors to variances P1,P2 and reversing the sign of the second, their correlation is ρnew=P1P2(1−ρ2)−ρN[N+P1(1−ρ2)][N+P2(1−ρ2)].

Facts & Assumptions

Given: A centered jointly Gaussian pair with positive variances, a correlation 0≤ρ<1, and independent Gaussian observation noise of variance N>0.

[F1]

Two-user real AWGN MAC with perfect causal output feedback: the output is the sum of the two real inputs and independent Gaussian noise; feedback encoders may normalize causal estimation errors.

Proof

1.1givenF1algebra

Write K=(P1ccP2) with c=ρP1P2, a=(1,1)T, and D=N+aTKa. Set R=X−(Ka/D)Y. Then Cov⁡(R,Y)=Ka−(Ka/D)D=0. Since (R,Y) is jointly Gaussian, this block-diagonal covariance makes its joint characteristic function factor, so R and Y are independent. Thus E[X∣Y]=KaY/D and the residual covariance is Cov⁡(R)=K−KaaTK/D.

2.1step 1.1algebra

Since Ka=(P1+c,P2+c)T and N+aTKa=D, subtracting the rank-one matrix gives diagonal entries P1[N+P2(1−ρ2)]/D, P2[N+P1(1−ρ2)]/D, and off-diagonal entry [cN+c2−P1P2]/D.

3.1step 2.1algebra∎

Dividing the off-diagonal covariance by the square root of the two diagonal variances and changing its sign yields the displayed ρnew; all denominators are positive because N>0.

Depends on

Used by

Dependency tree · two levels

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Sources