I have been working on bounding $\rho_{\nu_1^{e_1} \nu_2^{e_2}}(u)$ for $\nu_i \sim \mathcal{N}(0,1)$ i.i.d. for large $u$, i.e. the density of a product of (positive integer) powers of independent Gaussians, as

$$ \rho_{\nu_1^{e_1} \nu_2^{e_2}}(u) = \Omega(exp(-u^c /2)) \tag{1} \label{eq1}$$ for some $c$. It is not too difficult to show that $$\rho_{\nu_i^{e_i}}(u) = \Omega(exp(-u^{2/e_i}/2).$$ I suspect that the tails of the product should be at most as large as the tails of the bigger of the two, i.e. of $\nu_1^{e_1}$, assuming $e_1 \geq e_2$. I have however only been able to show a statement in that direction for the probability, namely $$P(\lvert \nu_1^{e_1} \nu_2^{e_2} \rvert \geq u) \leq C exp(-x^{1/e_1}/2),$$ using the Markov inequality. My question is, is there a $c$ such that equation \eqref{eq1} holds, and if yes, how large can $c$ be?

Thank you for any suggestions.

I have been working on bounding $\rho_{\nu_1^{e_1} \nu_2^{e_2}}(u)$ for $\nu_i \sim \mathcal{N}(0,1)$ i.i.d. for large $u$, i.e. the density of a product of (positive integer) powers of independent Gaussians, as

$$ \rho_{\nu_1^{e_1} \nu_2^{e_2}}(u) = \Omega\big(\exp(-u^c /2)\big) \tag{1} \label{eq1}$$ for some $c$. It is not too difficult to show that $$\rho_{\nu_i^{e_i}}(u) = \Omega\big(\exp(-u^{2/e_i}/2)\big).$$ I suspect that the tails of the product should be at most as large as the tails of the bigger of the two, i.e. of $\nu_1^{e_1}$, assuming $e_1 \geq e_2$. I have however only been able to show a statement in that direction for the probability, namely $$P(\lvert \nu_1^{e_1} \nu_2^{e_2} \rvert \geq u) \leq C \exp(-x^{1/e_1}/2),$$ using the Markov inequality. My question is, is there a $c$ such that equation \eqref{eq1} holds, and if yes, how large can $c$ be?

Thank you for any suggestions.