Does the normal product distribution have subgaussian tail?

Question

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 $\mathcal{N}(0, 1)$.

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

Does the normal product distribution have subgaussian tail?

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

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

Answer 1

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.