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zbh2047
  • Member for 6 years, 10 months
  • Last seen more than 2 years ago
  • Beijing, China
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The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian
Thank you for your suggestions. I am really sorry that my description for the question is not clear enough.
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The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian
Yes, I think so, too. But I cannot find an example that the norm depends on $n$. I tried high-demension ball and hypercube. I cannot come up with other isotropic high dimensional distributions which are easy to calculate the norm. You give an upper bound, but how to show the bound can be reached?
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The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian
In the book "high-dimensional probability" written by Roman Vershynin, page 60, 3.4.10.
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The norm of isotropic sub-Gaussian random vector may not be sub-Gaussian
Oh, it's not this meaning. It should mean that fix $\|X\|_{\psi_2}=K$, $\|\|X\|_2-\sqrt n\|_{\psi_2}$ can still be arbitraily large (that is, it depends on $n$). But when $X_i$ are independent, $\|\|X\|_2-\sqrt n\|_{\psi_2}$ can be bounded by constant.
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Prove that a sub-Gaussian random vector over a finite set $S \subset\mathbb R^n$ implies that $|S|$ is exponentially large
I am sorry that I didn't give the definition of $\|\cdot\|_{\psi_2}$ of a random vector. It has the same meaning as @Guillaume Aubrun said. I have added it in the question.
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Expectation of the norm of a random vector
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