# Is the normal product distribution sub-gaussian?

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.

• 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 $e^{-ax^2}, a>0$ – enthdegree Sep 28 '18 at 23:06
• 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) – enthdegree Sep 28 '18 at 23:15

## 1 Answer

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 $$\lambda \ge 1$$.

• 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. – Daniel López Sep 29 '18 at 11:10
• @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. – Nate Eldredge Sep 29 '18 at 14:20