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Let $X_i$ be independent, mean zero, $n\times n$, symmetric random matrices. $\|X_i\|\leq K$ almost sure for $\forall I$.

We have matrix Bernstein's inequality for the tail probability as follows

$$\forall t\geq 0,P(\|\sum X_i\|\geq t)\leq 2n \cdot\text{exp}(-\frac{t^2}{\sigma^2+Kt/3})$$ where $\sigma^2=\|\sum \mathbb{E}X_i^2\|$.

My question is how to prove $$\|\sum X_i\|\lesssim\|\sum \mathbb{E}X_i^2\|^{1/2}\sqrt{1+\log n}+K(1+\log n)?$$

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  • $\begingroup$ for bounding the expectation you can use the layercake principle $E[Z]=\int_{0}^{\infty}P(Z>t)dt$ for nonnegative Z. $\endgroup$ Commented Jun 3, 2023 at 17:43
  • $\begingroup$ for getting an almost sure statement $|\sum X_i|\lessapprox t_{n}$, you could try a Borel-Cantelli argument for $A_{n}:=|\sum X_i|>t_{n}$ and then use the concentration inequality you mention. $\endgroup$ Commented Jun 3, 2023 at 17:44

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The problem is Exercise 5.4.11 in Vershynin's book High-Dimensional Probability. The book gives strong hints on how to solve it. It has also been discussed on math.stackexchange.com. Hope you can figure it out from here.

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