[proofplan] The first assertion is a direct comparison of the $\psi_1$ norm of $X^2$ with the $\psi_2$ norm of $X$: the exponential appearing in one definition is exactly the exponential appearing in the other. To center $X^2$, we first control the deterministic quantity $\mathbb E[X^2]$ from the same exponential moment bound. Finally, we use the triangle inequality for the Luxemburg-Orlicz $\psi_1$ norm and compute the $\psi_1$ norm of a constant [random variable](/page/Random%20Variable). [/proofplan]

[step:Compare the defining exponential moments for $X^2$ and $X$]Let $K:=\|X\|_{\psi_2}$. If $K=0$, then for every $s>0$ there exists $r\in(0,s]$ such that \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{r^2}\right)\right]\le 2, \end{align*} because $0$ is the infimum of the admissible scales. Since $r\le s$, monotonicity of $u\mapsto \exp(|X|^2/u^2)$ on $(0,\infty)$ gives \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{s^2}\right)\right]\le \mathbb E\left[\exp\left(\frac{|X|^2}{r^2}\right)\right]\le 2. \end{align*} Hence $X=0$ $\mathbb P$-a.s.; therefore $X^2=0$ $\mathbb P$-a.s. and the asserted inequalities are immediate. Assume $K>0$. Let $\varepsilon>0$ be arbitrary. By the definition of the infimum, the number $K+\varepsilon$ is admissible in the definition of $\|X\|_{\psi_2}$, so \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{(K+\varepsilon)^2}\right)\right]\le 2. \end{align*} Since $|X^2|=|X|^2$, this is exactly \begin{align*} \mathbb E\left[\exp\left(\frac{|X^2|}{(K+\varepsilon)^2}\right)\right]\le 2. \end{align*} Thus $(K+\varepsilon)^2$ is admissible in the definition of $\|X^2\|_{\psi_1}$, and hence \begin{align*} \|X^2\|_{\psi_1}\le (K+\varepsilon)^2. \end{align*} Letting $\varepsilon\downarrow 0$ gives \begin{align*} \|X^2\|_{\psi_1}\le K^2=\|X\|_{\psi_2}^2. \end{align*}[/step]

[guided]The point of the first assertion is that the two Orlicz norms have been designed to match under squaring. Let \begin{align*} K:=\|X\|_{\psi_2}. \end{align*} If $K=0$, then the exponential moment condition holds at arbitrarily small scales. This forces $|X|=0$ $\mathbb P$-a.s., because otherwise $\exp(|X|^2/s^2)$ would become arbitrarily large on a set of positive probability as $s\downarrow 0$. Hence $X^2=0$ $\mathbb P$-a.s., and the result is immediate. Now suppose $K>0$. Because $K$ is an infimum, we do not assume that the defining inequality is attained exactly at $K$. Instead, fix an arbitrary $\varepsilon>0$. Then $K+\varepsilon$ is admissible in the definition of $\|X\|_{\psi_2}$, meaning \begin{align*} \mathbb E\left[\exp\left(\frac{|X|^2}{(K+\varepsilon)^2}\right)\right]\le 2. \end{align*} But the random variable whose $\psi_1$ norm we want to estimate is $X^2$, and \begin{align*} |X^2|=|X|^2. \end{align*} Therefore the preceding estimate can be rewritten as \begin{align*} \mathbb E\left[\exp\left(\frac{|X^2|}{(K+\varepsilon)^2}\right)\right]\le 2. \end{align*} This says precisely that $(K+\varepsilon)^2$ is an admissible scale in the definition of $\|X^2\|_{\psi_1}$. Hence \begin{align*} \|X^2\|_{\psi_1}\le (K+\varepsilon)^2. \end{align*} Since this holds for every $\varepsilon>0$, passing to the limit $\varepsilon\downarrow 0$ yields \begin{align*} \|X^2\|_{\psi_1}\le K^2=\|X\|_{\psi_2}^2. \end{align*}[/guided]

[step:Control the second moment by the sub-Gaussian norm] Let $\varepsilon>0$. Define the non-negative random variable $Z_\varepsilon:\Omega\to[0,\infty)$ by \begin{align*} Z_\varepsilon(\omega)=\frac{|X(\omega)|^2}{(K+\varepsilon)^2} \end{align*} for every $\omega\in\Omega$. From the preceding admissibility estimate and the elementary inequality $t\le e^t$ for $t\ge 0$, applied to $Z_\varepsilon$, we obtain \begin{align*} \frac{\mathbb E[X^2]}{(K+\varepsilon)^2} =\mathbb E[Z_\varepsilon] \le \mathbb E[e^{Z_\varepsilon}] \le 2. \end{align*} Thus \begin{align*} \mathbb E[X^2]\le 2(K+\varepsilon)^2. \end{align*} Letting $\varepsilon\downarrow 0$ gives \begin{align*} \mathbb E[X^2]\le 2K^2. \end{align*} [/step]

[step:Compute the sub-exponential norm of the centered square] Define the constant random variable $M:\Omega\to\mathbb R$ by \begin{align*} M(\omega)=\mathbb E[X^2] \end{align*} for every $\omega\in\Omega$. Then $X^2-\mathbb E[X^2]=X^2-M$. The Luxemburg-Orlicz norm $\|\cdot\|_{\psi_1}$ satisfies the triangle inequality: this follows from the convexity of $t\mapsto e^t$, since if $s_1,s_2>0$ are admissible for random variables $Y_1,Y_2:\Omega\to\mathbb R$, then \begin{align*} \exp\left(\frac{|Y_1+Y_2|}{s_1+s_2}\right) \le \frac{s_1}{s_1+s_2}\exp\left(\frac{|Y_1|}{s_1}\right)+\frac{s_2}{s_1+s_2}\exp\left(\frac{|Y_2|}{s_2}\right), \end{align*} and taking expectations gives admissibility of $s_1+s_2$ for $Y_1+Y_2$. Hence \begin{align*} \|X^2-M\|_{\psi_1}\le \|X^2\|_{\psi_1}+\|M\|_{\psi_1}. \end{align*} For a deterministic constant $a\in\mathbb R$, its $\psi_1$ norm is \begin{align*} \|a\|_{\psi_1}=\frac{|a|}{\log 2}, \end{align*} because $\mathbb E[\exp(|a|/s)]=\exp(|a|/s)\le 2$ is equivalent to $s\ge |a|/\log 2$. Therefore \begin{align*} \|M\|_{\psi_1}=\frac{\mathbb E[X^2]}{\log 2}\le \frac{2}{\log 2}K^2. \end{align*} Combining this with $\|X^2\|_{\psi_1}\le K^2$ gives \begin{align*} \|X^2-\mathbb E[X^2]\|_{\psi_1} \le \left(1+\frac{2}{\log 2}\right)K^2. \end{align*} Finally, \begin{align*} \mathbb E[X^2-\mathbb E[X^2]]=0, \end{align*} so $X^2-\mathbb E[X^2]$ is centered and sub-exponential with the asserted universal constant $C=1+2/\log 2$. [/step]