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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.

Daniel López
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    +1. I think this may be easy to resolve by comparing the event $XY\gt \lambda$ to the event $\min(X,Y)\gt \sqrt{\lambda}.$ It may help to know [Mills' Ratio](https://stats.stackexchange.com/a/7206/919). – whuber Sep 26 '18 at 16:25
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    Re your experiments: these plots don't provide any information about subgaussianity because they don't have the resolution to let you see it. Use a probability plot of the simulated data instead. As an example, here's an exponential probability plot for simulated data in `R`: `n – whuber Sep 29 '18 at 15:36
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    the product of two iid standard Gaussians is not subgaussian: see e.g. here: https://math.stackexchange.com/questions/804093/product-of-sub-gaussian-rvs – jld Oct 02 '18 at 16:04

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Consider the case of the product of two independent standard Gaussian r.v's $X,Y$: $Z=XY$. Then, one can easily compute the MGF of $Z$: $$ \forall |\theta|<1,\qquad \mathbb{E}[e^{\theta Z}] = \mathbb{E}[\mathbb{E}[e^{\theta XY}\mid X]] = \mathbb{E}[e^{\frac{\theta^2}{2}X^2}] = \frac{1}{\sqrt{1-\theta^2}} $$ the last equality since $X^2$ is distributed as a chi-squared r.v.. In particular, the MGF of $Z$ is not defined for all of $\mathbb{R}$. But $\mathbb{E}[Z]=0$, so sub-gaussianity of $Z$ would be equivalent to its MGF being defined for all $\theta\in\mathbb{R}$ and satisfying $$ \phi(\theta) \leq e^{K\theta^2} $$ for some absolute constant $K>0$ (see e.g. this book, Proposition 2.5.2). Clearly, it does not.

Clement C.
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    (Note that however it is a **sub-exponential** r.v., as the product of two sub-gaussian r.v.'s is always subexponential.) – Clement C. Oct 30 '18 at 16:12