Proving anti-concentration for the operator norm of a random matrix

Question

If $X$ is a random matrix then I would like to find $\theta >0$ and $\delta \in (0,1)$ s.t I can say, $$\mathbb{P} \Bigg [ \Big \vert \Vert X \Vert - \mathbb{E} [ \Vert X \Vert ] \Big \vert > \theta \Bigg ] > 1 - \delta $$

• I would like to know examples where such a thing is knowable.
• I am particularly interested in $X$ being PSD - best if there is as little as possible assumption of mutual independence among the entries.

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To be explicit we have, $\Vert X \Vert = \text{largest singular value of } X = \lambda_{\max}(X^\top X)$

Answer 1

Since the interest is in a PSD $X$, let me take $X=WW^{\rm T}$ with the elements of the $N\times M$ matrix $W$ i.i.d. with mean zero and variance $\sigma^2$. Note that the elements of $X$ itself are not independent. The distribution of the largest eigenvalue $x_{\rm max}$ of $X$ is known, see Distribution of the largest eigenvalue for real Wishart and Gaussian random matrices and a simple approximation for the Tracy-Widom distribution.

For $N,M\rightarrow\infty$ at fixed ratio $N/M$ the distribution $P(x_{\rm max})$ is narrowly peaked at $$\mu=(\sqrt{M-1/2}+\sqrt{N-1/2})^2\sigma^2,$$ with width $$\delta=\sqrt{\mu}\,\biggl(\frac{1}{\sqrt{N-1/2}}+\frac{1}{\sqrt{M-1/2}}\biggr)^{1/3}.$$ So $\mu$ is of order $N$ while $\delta$ is of order $N^{1/3}$, signifying a concentration of $x_{\rm max}$ at the average.