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Let $v_1,...,v_N$ be positive real numbers. I would like a (preferably sharp up to a universal constant) bound on $\mathbb{E}(\max_i Z_i)$, where $Z_1,…,Z_N$ are real-valued random variables such that or every i=1,…,N, the logarithm of the moment-generating function of $Z_i$ satisfies $ψ_{Z_i}(λ)≤λ^2v_i/2 $ for all λ>0.

The following is a statement from Concentration Inequalities: A Nonasymptotic Theory of Independence(Stéphane Boucheron,Gábor Lugosi,Pascal Massart) and is similar to what I am looking for, but assumes that every random variable has the same variance factor.

Let $Z_1,…,Z_N$ be real-valued random variables where a v>0 exists such that for every i=1,…,N, the logarithm of the moment-generating function of $Z_i$ satisfies $ψ_{Z_i}(λ)≤λ^2v/2 $ for all λ>0 Then $\mathbb{E}(\max_i Z_i) \leq \sqrt{2v log(N)}$

mscj
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    The $Z_i$ are uniformly sub-Gaussian with parameter $v := \max{v_1,\dots, v_n}$. This won't be a sharp upper bound though. – Julius Aug 08 '23 at 15:31
  • Thank you. I had already worked this out, but it wasn't sharp enough – mscj Aug 08 '23 at 15:42

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Check out Exercise 2.5.10 of Vershynin, "High-Dimensional Probability". I believe this is what you are looking for.

To summarize the result, let $\Vert \cdot \|_\psi$ be the "sub-Gaussian norm" or "Orlicz norm", which takes the form $$ \Vert X\|_\psi = \inf_{t>0} \ t \ \ \text{subject to} \ \ \mathsf{E}\left\{ \exp\left(\dfrac{X^2}{t^2}\right) \right\} \leq 2. $$ This norm is essentially how Vershynin discusses the parameters (or as you say, variance factors) of sub-Gaussian random variables; check Definition 2.5.6. to see his treatment and how it relates to other characterizations of sub-Gaussian variables.

Now, for your problem, define $ K := \max_{1 \leq i \leq N} \Vert Z_i \Vert_\psi$. You can show (see the aforementioned Exercise and the below discussions) that for every $N\geq 2$ we have $$ \mathsf{E}\left\{ \max_{1 \leq i \leq N} |Z_i| \right\} \leq CK\sqrt{\log(N)}, $$ where $C$ is a universal constant factor.

See the below discussions you may find useful:

TShiong
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