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Consider the normal product distribution, which is the distribution of the product of two or more independent normal variables. Particulary, focus in the case where the multiplied normal variables are $\mathcal{N}(0, 1)$.

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Considering the following definition of subgaussian tail (taken from [1]):

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Does the normal product distribution have subgaussian tail?

I have been doing some numerical experiments which suggest an affirmative answer ($a=0.1$):

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[1] Matoušek, J. (2008). On variants of the Johnson–Lindenstrauss lemma. Random Structures & Algorithms, 33(2), 142-156.

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  • $\begingroup$ did you know it is a difference of iid chi-squared distributions with one dof? this makes me guess that it is not sub-gaussian, as the tail of one of them is shaped like $\exp(-x/2-\ln(x)/2)$ which is eventually greater than anything like $e^{-ax^2}, a>0$ $\endgroup$
    – Christian Chapman
    Commented Sep 28, 2018 at 23:06
  • $\begingroup$ if you're just interested in the johnson-lindenstrauss lemma you can find a different component distribution for your matrices where it still holds, and have subgaussian tails (for instance, i think rademacher rvs work) $\endgroup$
    – Christian Chapman
    Commented Sep 28, 2018 at 23:15

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If it were subgaussian then we would have $E[e^{\lambda X Y}] < \infty$ for all $\lambda$. However, by conditioning and using independence we find $$E[e^{\lambda X Y}] = E[E[e^{\lambda X Y} \mid X]] = E[e^{\lambda^2 X^2/2}]$$ which is easily seen to be infinite for all $\lambda \ge 1$.

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  • $\begingroup$ I see, thank you for your answer. However, in my numerical simulations you can see that $P[x>\lambda] \le e^{-0.1\lambda^2}$ is true, except maybe for a very small amount at the end of the tail. Can this be interpreted as being "almost" subgaussian? In the sense that in practice such a small violation of the condition might not be problematic. $\endgroup$
    – Daniel López
    Commented Sep 29, 2018 at 11:10
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    $\begingroup$ @DanielLópez: The "end of the tail" is the whole point in the definition of subgaussian. If you choose a larger range for $\lambda$, things will look more dramatic. I suggest a log scale. I don't know what you mean by "not problematic"; I don't think you will be able to prove any of the usual helpful consequences of the subgaussian property if you assume $P(X > \lambda) \le e^{c \lambda^2}$ for small $\lambda$ only. $\endgroup$
    – Nate Eldredge
    Commented Sep 29, 2018 at 14:20

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