Are there any concentration inequalities (i.e. probability bounds on how a random variable deviates from its expectation) for the product of $n$ independent gaussian random variables with zero means and equal variances? What about different means and variances?

  • $\begingroup$ do you assume the independence of gaussians or no? $\endgroup$
    – tortue
    Nov 29 '19 at 12:13
  • $\begingroup$ @pointguard0 yes $\endgroup$
    – Sunny88
    Dec 2 '19 at 7:05

This paper computes the desired probability explicitly in Theorem 3 (page 4).

The theorem states that for independent standard normal Gaussians random variables $g_i$ one has $$ \mathbb{P}(X \le t) = \mathbb{P}(X \ge -t) = f_{0.5, 1}\bigg(\frac{2^n} t\bigg), $$ where $$ f_{\nu, \xi}(u) = \nu + \frac 1 {2^{\xi}} \cdot \frac 1 {\pi^{n/2}} \sum_{k=0}^\infty u^{-0.5 - k} \sum_{j=0}^{n-1} H_{kj} \cdot [\log u]^j, $$ for details and definition of $H_{kj}$ see the paper itself.

The expression seems to be complicate enough to argue to the concentration bound immediately from here, but I believe this is a good starting point.


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