Suppose $A\in \mathbb{R}^{n\times n}$ is a symmetric random matrix, with iid centered Bernoulli non-diagnonal entries $A_{ij} \sim \mathrm{Bernoulli}(p) - p$, for some $p \in (0,1)$, and zeros on its diagonal. For non-diagonal entries, We have $\mathbb{E}[A_{ij}] = 0$ and $\mathrm{Var}(A_{ij}) = p(1-p)$. If $p$ is fixed, then $A$ is a Wigner matrix up to some scaling.
I'm interested in bounding $|v_1^\top D v_1| = |\sum_{i=1}^n d_i v_{1i}^2|$ with high probability, where $v_1$ is the eigenvector of $A$ corresponding to its largest eigenvalue, and $D$ is a diagonal matrix with $D_{ii}=d_i = \sum_{j} A_{ij}$.
A trivial bound is $|\sum_{i=1}^n d_i v_{1i}^2 | \le \max |d_i| = O_{\mathbb{P}}(\sqrt{np(1-p)} \sqrt{\log n})$. This bound is far from optimal based on simulation (order of vDv assuming fixed p=0.05), which seems to suggest $|v_1^\top D v_1| = O_{\mathbb{P}}(1)$.
The major challenge is that $D$ and $v_1$ are not independent. Otherwise, one may apply contraction inequality like the following, $$\mathbb{E}\left[|\sum_{i=1}^n d_i v_{1i}^2|\right] \le \|v_1\|^2_{\infty} \mathbb{E}\left[|\sum_{i=1}^n d_i|\right] = O\left(n^{o(1)}\sqrt{p(1-p)}\right) , $$ which can then be easily bounded due to the delocalization of $v_1$, i.e., $\|v_1\|_{\infty} = n^{-\frac{1}{2}+o(1)} $
Alternatively, one may write $u^\top = (v_{11}^2, \ldots, v_{1n}^2)$, and then $$|\sum_{i=1}^n d_i v_{1i}^2| = |u^\top d| = \|u\|_2 \|d\|_2 |\cos(u, d)|.$$ But then we also have to show the angle between $u$ and $d$ is small, e.g., $|\cos(u, d)| \approx n^{-\frac{1}{2}}$
Some related questions:
- Will the problem be easier if we instead assume $A\sim\mathrm{GOE}$, namely $A_{ij}$ follows $\mathcal{N}(0,1)$?
- What if we allow $p$ to decrease with $n$? Simulation shows that if we let $p\asymp n^{-\frac{1}{2}}$ or $p\asymp n^{-\frac{1}{3}}$, the simulation still behaves like $|v_1^\top D v_1 | = O_{\mathbb{P}}(1)$. (order of vDv assuming p~n^{-1/2})
