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Prove that $E[\max_{1\leq i\leq n}\vert X_i\vert]\leq C\sqrt{\log(2n)}\sigma$ for some constant $C \in R$, where $X_1, ..., X_n\sim \mathcal{N}(0,\sigma)$ are i.i.d.

Hint: Use Jensen's inequality: $f(E[Y])\leq E[f(Y)]$ for any convex function $f$.

I think that we can make use of the moment-generating function. $$E[e^{\lambda X}]=e^{\sigma^2\lambda^2/2}$$ where $\lambda\in R$ is a constant, and $X\sim\mathcal{N}(0, \sigma)$.

Moreover, i have tried to use $\max_{1\leq i\leq n}\vert X_i\vert\leq \sum_{i}^n \vert X_i\vert$, yet no good ideas.

Nick Y
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  • The question is "wrong" in the following sense. With the requested bound, one only needs the assumption of subgaussian, no normality or independence assumption is needed at all! In any case, you are on the right track, just combine what you have with Jensen's inequality (premultiplying by $\lambda$). – Andrew Mar 18 '24 at 04:45
  • You can get the proper font and spacing for $\max$ and $\log$ using \max and \log. For operators that don't have a command of their own, you can use \operatorname{name}. – joriki Mar 18 '24 at 05:01
  • @joriki Thank you! I have already edited the probelm. – Nick Y Mar 18 '24 at 06:58
  • @Andrew Thank you! – Nick Y Mar 18 '24 at 07:26

1 Answers1

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Assume $\lambda>0.$

$$ \begin{aligned} \mathbb{E}[\max_{1\leq i\leq N}X_i]&=\mathbb{E}[\frac{1}{\lambda}\log e^{\lambda\max_{1\leq i\leq N}X_i}]\\ &\leq \frac{1}{\lambda}\log\mathbb{E}[e^{\lambda\max_{1\leq i\leq N}X_i}]\\ &=\frac{1}{\lambda}\log\mathbb{E}[\max_{1\leq i\leq N}e^{\lambda X_i}]\\ &\leq \frac{1}{\lambda}\log\sum_{i=1}^N\mathbb{E}[e^{\lambda X_i}]\\ &\leq \frac{1}{\lambda}\log \sum_{i=1}^Ne^{\lambda^2\sigma^2/2}\\ &=\frac{\log n}{\lambda}+\frac{\lambda\sigma^2}{2} \end{aligned} $$

Let $\lambda=\sqrt{2(\log N)/\sigma^2}$, we can get $\mathbb{E}[\max_{1\leq i\leq N}X_i]\leq \sqrt{2\sigma^2\log n}$

Moreover, there is a fact that $\max_{1\leq i\leq N}\vert X_i\vert=\max_{1\leq i\leq 2N}X_i$, where $X_{N+i}=-X_i\sim\mathcal{N}(0,\sigma^2)$
Thus $\mathbb{E}[\max_{1\leq i\leq N}\vert X_i\vert]\leq \sqrt{2\log (2N)}\sigma=C\sqrt{\log(2 N)}\sigma$

Nick Y
  • 67