Alphabeta Math
LemmaStatement: AI-generatedProof: AI-generated
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Loop-free Bernoulli adjacency and its two centerings

Statement

Let A be the adjacency matrix of the simple graph G(n,p), with n≥1 and 0≤p≤1. Matrix expectation is taken entrywise: (EA)ij=E[Aij]. Write J for the all-ones matrix, I for the identity, X=A−EA, and Δ=A−pJ, the reference-degree centering at d∗=np. Then EA=p(J−I), Xii=0, Δii=−p, and Δ=X−pI. For p>0, the diagonal entries of Δ are not centered Bernoulli variables, since their mean is −p≠0. For p=0, they are the zero centered Bernoulli(0) variables. Here a centered Bernoulli(p) variable means B−p for a Bernoulli(p) variable B.

Facts & Assumptions

[F1]

The adjacency entry Aij is the indicator of the edge {i,j}. The simple graph has zero diagonal and mutually independent Bernoulli(p) entries above the diagonal. Adjacency, degrees, graph Laplacian, and spectral notation Erdős–Rényi random graph and high probability Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials

[F2]

The reference-degree centered adjacency matrix is A−(d∗/n)J. Centered graph matrices and reference degree

[F3]

For a real random variable Y on a finite probability space (Ω,w), E[Y]=∑ω∈ΩY(ω)w(ω); a real constant also denotes the corresponding constant random variable. Expectation of a real random variable on a finite probability space

[F4]

For p∈[0,1], a Bernoulli(p) variable B takes the value 1 with probability p and 0 with probability 1−p. Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials

Proof

Given: The data and hypotheses in the Statement.

Goal: Establish the stated conclusions, including the case p=0.

1.1givenF1F3algebra

Work on the finite probability space of simple graphs on [n] with the law in [F1]. For i≠j, [F1] and [F3] give E[Aij]=∑ω:Aij(ω)=1w(ω)=p, since that sum is the probability that the edge is present. For i=j, Aii=0 gives E[Aii]=0. Taking matrix expectation entrywise therefore gives EA=p(J−I) and Xii=0.

2.1givenF1F2step 1.1algebra

Substitute d∗=np in [F2]. Then Δ=A−pJ=A−p(J−I)−pI=X−pI. Since Aii=0 and Jii=1, this also gives Δii=−p.

3.1givenF1F3F4step 2.1algebra∎

By [F3], the deterministic variable Δii=−p has mean −p∑ω∈Ωw(ω)=−p. By [F4], for a Bernoulli(p) variable B, the centered variable B−p has mean p(1−p)+(1−p)(−p)=0 by the same weighted-sum definition. Thus for p>0 the diagonal entries cannot be centered Bernoulli variables. At p=0, B=0 almost surely and both B−p and Δii are zero. This proves the distinction with its degenerate exception throughout the stated parameter range.

Depends on

Used by · 1

Dependency tree · 2 levels

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