Distribution of a product of two discrete i.i.d. variables The problem is to estimate the distribution of product of two $\textit{discretized Gaussian}$ random variables with zero means. The discretized Gaussian means that the p.m.f. looks like 
$D_s(x)=\rho_s(x)/\rho(\mathbb{Z}), x \in \mathbb{Z}$, 
where $\rho_s(x)=exp(-x^2/s^2)$ is Gaussian subject to discrete support and $\rho(\mathbb{Z})$ is the normalization.
The question is: how to show that the product of two independant discrete Gaussians $X_1 \leftarrow D_{s_1}, X_2\leftarrow D_{s_2}$ is a subgaussian (namely, the tail decays exponentially and how fast it decays) except the argument that the resulting variable is bounded since the st.dev=$s_1s_2$? Assume no relatively good p.m.f. is expected like in continuous case (Normal Product Distribution)?
Thanks in advance.  
 A: The term "sub-Gaussian" usually refers to random variables with tail decaying like that of a Gaussian, not exponential.
Now for a random variable $X$ being sub-Gaussian in this sense is equivalent to any of the following conditions:


*

*$\mathsf{P}\{|X| \ge C\} \le e^{-C^2/2(\sigma^2 + o(1))}, C \to +\infty$

*$\mathsf{E} \exp \lambda |X| \le e^{(\sigma^2 + o(1)) \lambda^2 /2}, \lambda \to +\infty$

*$(\mathsf{E} |X|^p)^{1/p} \le (\sigma e^{-1/2} + o(1))p^{1/2}, p \to +\infty$


A similar set of equivalent conditions characterize random variables with sub-exponential tails:


*

*$\mathsf{P}\{|X| \ge C\} \le e^{-C/(\sigma + o(1))}, C \to +\infty$

*$\mathsf{E} \exp \lambda |X| < \infty, |\lambda| < \sigma$

*$(\mathsf{E} |X|^p)^{1/p} \le (\sigma e^{-1} + o(1)) p, p \to +\infty$


For an explanation of that, see, e.g, the first chapter of Lugosi's lecture notes on measure concentration.
Now in your case the sub-Gaussian tail estimate for $X_i$ is easy to check. From there you get an $O(p^{1/2})$ moment estimate on them, therefore an $O(p)$ moment estimate on the product, therefore an exponential tail estimate on the product.
A: Changing the order of two operations: 
D) discretization of continuous probability distribution(s), 
P) calculating a product of two distributions,
shouldn’t change the results much. 
Thus, if by applying P) first and then D), we expect discretized normal product distribution, we shouldn’t expect something much different if applying D) first and P) after that.
E.g. in this paper, it was assumed that the noise term follows a discretised normal distribution and it lead to a conclusion that a product of two error terms follows a discretised Normal product distribution.
I don’t think we can assume that “no relatively good p.m.f. is expected like in continuous case (Normal Product Distribution)” as in fact we should expect some form of a discretised Normal product distribution.  
