[proofplan] We prove directly from the Orlicz norm definitions. The only estimate needed is the weighted arithmetic-geometric inequality, applied with weights chosen from the two $\psi_2$ scales. After converting $|XY|$ into a sum of $X^2$ and $Y^2$, we bound the exponential moment of $|XY|$ by separating the two factors through an elementary $L^2$ Cauchy-Schwarz estimate proved inside the argument. Finally we let the auxiliary scales decrease to the actual $\psi_2$ norms. [/proofplan] [step:Handle the degenerate case where one $\psi_2$ norm vanishes] Let \begin{align*} K_X:=\|X\|_{\psi_2}, \qquad K_Y:=\|Y\|_{\psi_2}. \end{align*} Suppose first that $K_X=0$. We show that $X=0$ $\mathbb P$-a.s. For every $\varepsilon>0$, the definition of the infimum gives a number $s_\varepsilon\in(0,\varepsilon)$ such that \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{s_\varepsilon^2}\right)\right]\le 2. \end{align*} Since $s_\varepsilon<\varepsilon$, pointwise monotonicity of the exponential gives \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{\varepsilon^2}\right)\right]\le 2. \end{align*} If $\mathbb P(|X|>\delta)>0$ for some $\delta>0$, then \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{\varepsilon^2}\right)\right] \ge \exp\left(\frac{\delta^2}{\varepsilon^2}\right)\mathbb P(|X|>\delta), \end{align*} which exceeds $2$ for all sufficiently small $\varepsilon>0$, a contradiction. Hence $\mathbb P(|X|>\delta)=0$ for every $\delta>0$, and therefore $X=0$ $\mathbb P$-a.s. Then $XY=0$ $\mathbb P$-a.s. and $\|XY\|_{\psi_1}=0$. The same argument applies if $K_Y=0$. Thus the desired estimate is immediate whenever $K_XK_Y=0$. For the rest of the proof assume \begin{align*} K_X>0, \qquad K_Y>0. \end{align*} [/step] [step:Choose admissible $\psi_2$ scales above the norms] Let $A$ and $B$ be [real numbers](/page/Real%20Numbers) satisfying \begin{align*} A>K_X, \qquad B>K_Y. \end{align*} By the definition of the infimum in the $\psi_2$ norm, there exist numbers $A_0\in(0,A)$ and $B_0\in(0,B)$ such that \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{A_0^2}\right)\right]\le 2, \qquad \mathbb E\left[\exp\left(\frac{|Y|^2}{B_0^2}\right)\right]\le 2. \end{align*} Since $A_0<A$ and $B_0<B$, pointwise monotonicity of $t\mapsto \exp(t)$ gives \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{A^2}\right)\right]\le 2, \qquad \mathbb E\left[\exp\left(\frac{|Y|^2}{B^2}\right)\right]\le 2. \end{align*} [/step] [step:Convert the product into a sum of squares at the chosen scales] For every $\omega\in\Omega$, apply the identity \begin{align*} 0\le \left(\sqrt{\frac{B}{A}}\,|X(\omega)|-\sqrt{\frac{A}{B}}\,|Y(\omega)|\right)^2 = \frac{B}{A}|X(\omega)|^2+\frac{A}{B}|Y(\omega)|^2-2|X(\omega)Y(\omega)|. \end{align*} Therefore \begin{align*} 2|X(\omega)Y(\omega)| \le \frac{B}{A}|X(\omega)|^2+\frac{A}{B}|Y(\omega)|^2. \end{align*} Dividing by $8AB$ yields the pointwise bound \begin{align*} \frac{|X(\omega)Y(\omega)|}{4AB} \le \frac{|X(\omega)|^2}{8A^2} + \frac{|Y(\omega)|^2}{8B^2}. \end{align*} [/step] [step:Bound the exponential moment of $XY$ by the two sub-Gaussian moments] Define $U:\Omega\to[0,\infty)$ by, for each $\omega\in\Omega$, \begin{align*} U(\omega)=\exp\left(\frac{|X(\omega)|^2}{8A^2}\right). \end{align*} Define $V:\Omega\to[0,\infty)$ by, for each $\omega\in\Omega$, \begin{align*} V(\omega)=\exp\left(\frac{|Y(\omega)|^2}{8B^2}\right). \end{align*} The pointwise estimate from the previous step gives \begin{align*} \exp\left(\frac{|XY|}{4AB}\right)\le UV. \end{align*} For non-negative real numbers $r$ and $s$, the inequality $2rs\le r^2+s^2$ follows from $(r-s)^2\ge0$. Applying this pointwise to \begin{align*} r=\lambda U(\omega), \qquad s=\lambda^{-1}V(\omega), \end{align*} where $\lambda>0$, taking expectations, and optimizing over $\lambda$, gives \begin{align*} \mathbb E[UV]\le \left(\mathbb E[U^2]\right)^{1/2}\left(\mathbb E[V^2]\right)^{1/2}. \end{align*} Since \begin{align*} U^2=\exp\left(\frac{|X|^2}{4A^2}\right) \le \exp\left(\frac{|X|^2}{A^2}\right), \qquad V^2=\exp\left(\frac{|Y|^2}{4B^2}\right) \le \exp\left(\frac{|Y|^2}{B^2}\right), \end{align*} the admissibility of $A$ and $B$ gives $\mathbb E[U^2]\le2$ and $\mathbb E[V^2]\le2$, so both second moments are finite. Therefore \begin{align*} \mathbb E\left[\exp\left(\frac{|XY|}{4AB}\right)\right] \le \mathbb E[UV] \le 2. \end{align*} [guided] Recall the setup used in this step: $A$ and $B$ are real numbers with $A>K_X=\|X\|_{\psi_2}$ and $B>K_Y=\|Y\|_{\psi_2}$, chosen so that \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{A^2}\right)\right]\le 2, \qquad \mathbb E\left[\exp\left(\frac{|Y|^2}{B^2}\right)\right]\le 2. \end{align*} The goal is to prove that $4AB$ is an admissible $\psi_1$ scale for the product $XY$. By definition, this means we must prove \begin{align*} \mathbb E\left[\exp\left(\frac{|XY|}{4AB}\right)\right]\le 2. \end{align*} For each $\omega\in\Omega$, the square \begin{align*} \left(\sqrt{\frac{B}{A}}\,|X(\omega)|-\sqrt{\frac{A}{B}}\,|Y(\omega)|\right)^2 \end{align*} is non-negative. Expanding it gives \begin{align*} 2|X(\omega)Y(\omega)| \le \frac{B}{A}|X(\omega)|^2+ \frac{A}{B}|Y(\omega)|^2. \end{align*} Dividing by $8AB$ gives the pointwise estimate \begin{align*} \frac{|X(\omega)Y(\omega)|}{4AB} \le \frac{|X(\omega)|^2}{8A^2} + \frac{|Y(\omega)|^2}{8B^2}. \end{align*} Exponentiating this inequality converts the sum into a product: \begin{align*} \exp\left(\frac{|XY|}{4AB}\right) \le \exp\left(\frac{|X|^2}{8A^2}\right) \exp\left(\frac{|Y|^2}{8B^2}\right). \end{align*} Define the non-negative [random variable](/page/Random%20Variable) $U:\Omega\to[0,\infty)$ by \begin{align*} U(\omega)=\exp\left(\frac{|X(\omega)|^2}{8A^2}\right). \end{align*} Define the non-negative random variable $V:\Omega\to[0,\infty)$ by \begin{align*} V(\omega)=\exp\left(\frac{|Y(\omega)|^2}{8B^2}\right). \end{align*} Then the estimate becomes \begin{align*} \exp\left(\frac{|XY|}{4AB}\right)\le UV. \end{align*} We now separate the expectation of the product $UV$. This does not require independence of $X$ and $Y$. The separation comes from the elementary $L^2$ Cauchy-Schwarz estimate, which we derive directly. For any $\lambda>0$ and any $\omega\in\Omega$, the inequality $2rs\le r^2+s^2$ applied to \begin{align*} r=\lambda U(\omega), \qquad s=\lambda^{-1}V(\omega) \end{align*} gives \begin{align*} 2U(\omega)V(\omega) \le \lambda^2 U(\omega)^2+\lambda^{-2}V(\omega)^2. \end{align*} Taking expectations, \begin{align*} 2\mathbb E[UV] \le \lambda^2\mathbb E[U^2]+\lambda^{-2}\mathbb E[V^2]. \end{align*} If either $\mathbb E[U^2]$ or $\mathbb E[V^2]$ is zero, then $UV=0$ $\mathbb P$-a.s. and the desired estimate is immediate. Otherwise choose \begin{align*} \lambda^2=\left(\frac{\mathbb E[V^2]}{\mathbb E[U^2]}\right)^{1/2}. \end{align*} This gives \begin{align*} \mathbb E[UV]\le \left(\mathbb E[U^2]\right)^{1/2}\left(\mathbb E[V^2]\right)^{1/2}. \end{align*} It remains to use the $\psi_2$ bounds for $X$ and $Y$. Since \begin{align*} U^2=\exp\left(\frac{|X|^2}{4A^2}\right) \le \exp\left(\frac{|X|^2}{A^2}\right), \end{align*} we get \begin{align*} \mathbb E[U^2]\le \mathbb E\left[\exp\left(\frac{|X|^2}{A^2}\right)\right]\le 2. \end{align*} Similarly, \begin{align*} V^2=\exp\left(\frac{|Y|^2}{4B^2}\right) \le \exp\left(\frac{|Y|^2}{B^2}\right), \end{align*} so \begin{align*} \mathbb E[V^2]\le \mathbb E\left[\exp\left(\frac{|Y|^2}{B^2}\right)\right]\le 2. \end{align*} Combining the estimates, \begin{align*} \mathbb E\left[\exp\left(\frac{|XY|}{4AB}\right)\right] \le \mathbb E[UV] \le \sqrt{2}\sqrt{2} = 2. \end{align*} Thus $4AB$ is an admissible $\psi_1$ scale for $XY$. [/guided] [/step] [step:Pass from admissible scales to the sharp norm estimate] The preceding step proves that, for every $A>K_X$ and every $B>K_Y$, \begin{align*} \mathbb E\left[\exp\left(\frac{|XY|}{4AB}\right)\right]\le 2. \end{align*} By the definition of the $\psi_1$ norm, \begin{align*} \|XY\|_{\psi_1}\le 4AB. \end{align*} Taking the infimum over all $A>K_X$ and all $B>K_Y$ gives \begin{align*} \|XY\|_{\psi_1}\le 4K_XK_Y = 4\|X\|_{\psi_2}\|Y\|_{\psi_2}. \end{align*} In particular $\|XY\|_{\psi_1}<\infty$, so $XY$ is sub-exponential. [/step]