Products of variables with Gaussian tails are sub-exponential

Applying [F1] to $U^2,V^2$ gives $\mathbb{E}{U}^2,\mathbb{E}{V}^2\le 2\tau$. Since $2\mathrm{\mid }UV\mathrm{\mid }\le U^2+V^2$, $UV$ is integrable and $\mathrm{\mid }\mathbb{E}UV\mathrm{\mid }\le 2\tau$. Also, for $x>0$, $\mathrm{\mid }UV\mathrm{\mid }>x$ implies $\mathrm{\mid }U\mathrm{\mid }>\sqrt{x}$ or $\mathrm{\mid }V\mathrm{\mid }>\sqrt{x}$, so $\mathbb{P}(\mathrm{\mid }UV\mathrm{\mid }>x)\le 2e^{-x\mathrm{/}(2\tau )}$.

Put $W=UV-\mathbb{E}UV$. If $t\ge 4\tau$, then $\mathrm{\mid }W\mathrm{\mid }>t$ implies $\mathrm{\mid }UV\mathrm{\mid }>t-\mathrm{\mid }\mathbb{E}UV\mathrm{\mid }\ge t\mathrm{/}2$, so step 1.1 gives $\mathbb{P}(\mathrm{\mid }W\mathrm{\mid }>t)\le 2e^{-t\mathrm{/}(4\tau )}\le 4e^{-t\mathrm{/}(8\tau )}$. If $0<t<4\tau$, then $4e^{-t\mathrm{/}(8\tau )}>4e^{-1\mathrm{/}2}>1$, so the same final bound follows from $\mathbb{P}(\mathrm{\mid }W\mathrm{\mid }>t)\le 1$. Thus for all $t>0$, $\mathbb{P}(\mathrm{\mid }W\mathrm{\mid }>t)\le 4e^{-t\mathrm{/}(8\tau )}$.

By [F1] and step 2.1, for every integer $m\ge 2$, $\mathbb{E}\mathrm{\mid }W{\mathrm{\mid }}^m=m{\int }_0^{\mathrm{\infty }}{t}^{m-1}\mathbb{P}(\mathrm{\mid }W\mathrm{\mid }>t)dt\le 4(8\tau {)}^{m}m!\mathrm{.}$ The identity follows from substitution $u=t\mathrm{/}(8\tau )$ and ${\int }_0^{\mathrm{\infty }}{u}^{m-1}{e}^{-u}du=(m-1)!$.

Fix $\mathrm{\mid }\lambda \mathrm{\mid }\le 1\mathrm{/}(16\tau )$ and put $x=8\tau \mathrm{\mid }\lambda \mathrm{\mid }\le 1\mathrm{/}2$. The variable $W$ is integrable and centered by step 1.1. The nonnegative measurable partial sums $A_N={\sum }_{m=0}^N\mathrm{\mid }\lambda W{\mathrm{\mid }}^m\mathrm{/}m!$ increase pointwise to $e^{\mathrm{\mid }\lambda W\mathrm{\mid }}$. Finite linearity [F5] and monotone convergence [F3], applied only to these nonnegative partial sums, give $\mathbb{E}{e}^{\mathrm{\mid }\lambda W\mathrm{\mid }}=1+\mathrm{\mid }\lambda \mathrm{\mid }\mathbb{E}\mathrm{\mid }W\mathrm{\mid }+{\sum }_{m\ge 2}\frac{\mathrm{\mid }\lambda {\mathrm{\mid }}^m\mathbb{E}\mathrm{\mid }W{\mathrm{\mid }}^m}{m!}\le 1+\mathrm{\mid }\lambda \mathrm{\mid }\mathbb{E}\mathrm{\mid }W\mathrm{\mid }+4{\sum }_{m\ge 2}{x}^m<\mathrm{\infty }\mathrm{.}$ Thus $e^{\mathrm{\mid }\lambda W\mathrm{\mid }}$ is an integrable dominating variable.

The signed partial sums $S_N={\sum }_{m=0}^N(\lambda W{)}^m\mathrm{/}m!$ converge pointwise to $e^{\lambda W}$ and satisfy $\mathrm{\mid }{S}_N\mathrm{\mid }\le A_N\le e^{\mathrm{\mid }\lambda W\mathrm{\mid }}$. Dominated convergence [F4] and finite linearity [F5] justify taking expectations termwise; the scalar series of expectations is absolutely convergent by step 4.1. Since $\mathbb{E}W=0$, step 3.1 gives $\mathbb{E}{e}^{\lambda W}=1+{\sum }_{m\ge 2}\frac{{\lambda }^m\mathbb{E}{W}^m}{m!}\le 1+4{\sum }_{m\ge 2}{x}^m=1+\frac{4x^2}{1-x}\le 1+8x^2\le e^{8x^2}=e^{512{\tau }^2{\lambda }^2}\mathrm{.}$ By [F2] the parameters remain ${\nu }^2=1024{\tau }^2$ and $b=16\tau$.