[proofplan] The scaling identity is obtained by comparing the defining admissible parameters in the Luxemburg-type definition of $\|\cdot\|_{\psi_2}$. For centering, we first record the triangle inequality for this Orlicz norm using the convexity of $\Phi(u)=e^{u^2}-1$. We then bound the deterministic constant $\mathbb E[X]$ by $\|X\|_{\psi_2}$ through the elementary inequality $u\le e^u-1$ for $u\ge 0$ and the Cauchy-Schwarz estimate on the probability space. [/proofplan] [step:Compare admissible parameters to prove the scaling identity] Let \begin{align*} K:=\|X\|_{\psi_2}. \end{align*} If $a=0$, then $aX=0$ $\mathbb P$-a.s. and the defining infimum gives $\|aX\|_{\psi_2}=0=|a|K$. Assume $a\ne 0$. For $t>0$, \begin{align*} \mathbb E\left[\exp\left(\frac{(aX)^2}{t^2}\right)\right]\le 2 \iff \mathbb E\left[\exp\left(\frac{X^2}{(t/|a|)^2}\right)\right]\le 2. \end{align*} Thus the admissible set for $aX$ is exactly \begin{align*} \left\{t>0:\frac{t}{|a|}\ \text{is admissible for }X\right\} = |a|\left\{s>0:s\ \text{is admissible for }X\right\}. \end{align*} Taking infima over these two sets gives \begin{align*} \|aX\|_{\psi_2}=|a|\,\|X\|_{\psi_2}. \end{align*} [/step] [step:Prove the triangle inequality for the $\psi_2$ Orlicz norm] Define the function $\Phi:[0,\infty)\to[0,\infty)$ by \begin{align*} \Phi(u):=e^{u^2}-1. \end{align*} Its first and second derivatives are \begin{align*} \Phi'(u)=2ue^{u^2}\ge 0, \qquad \Phi''(u)=2e^{u^2}(1+2u^2)\ge 0, \end{align*} for every $u\in[0,\infty)$, so $\Phi$ is increasing and convex on $[0,\infty)$. Let $Y,Z:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be random variables with finite $\psi_2$ norm. The admissible set for a [random variable](/page/Random%20Variable) is upward closed, because increasing the denominator decreases the non-negative ratio inside the exponential. Choose $r>\|Y\|_{\psi_2}$ and $s>\|Z\|_{\psi_2}$ such that \begin{align*} \mathbb E[\Phi(|Y|/r)]\le 1, \qquad \mathbb E[\Phi(|Z|/s)]\le 1. \end{align*} Set \begin{align*} \lambda:=\frac{r}{r+s}, \qquad \mu:=\frac{s}{r+s}. \end{align*} Then $\lambda,\mu\in(0,1)$ and $\lambda+\mu=1$. Since \begin{align*} \frac{|Y+Z|}{r+s} \le \lambda\frac{|Y|}{r}+\mu\frac{|Z|}{s}, \end{align*} convexity and monotonicity of $\Phi$ give \begin{align*} \Phi\left(\frac{|Y+Z|}{r+s}\right) \le \lambda\Phi\left(\frac{|Y|}{r}\right) + \mu\Phi\left(\frac{|Z|}{s}\right). \end{align*} Taking expectations yields \begin{align*} \mathbb E\left[\Phi\left(\frac{|Y+Z|}{r+s}\right)\right]\le 1. \end{align*} Therefore $\|Y+Z\|_{\psi_2}\le r+s$. Letting $r\downarrow\|Y\|_{\psi_2}$ and $s\downarrow\|Z\|_{\psi_2}$ gives \begin{align*} \|Y+Z\|_{\psi_2}\le \|Y\|_{\psi_2}+\|Z\|_{\psi_2}. \end{align*} [guided] The point of this step is that centering subtracts a constant, so we need a controlled way to estimate the norm of a sum. We prove the triangle inequality directly from the definition rather than citing it. Define the function $\Phi:[0,\infty)\to[0,\infty)$ by \begin{align*} \Phi(u):=e^{u^2}-1. \end{align*} The condition $\|Y\|_{\psi_2}\le r$ is the same as \begin{align*} \mathbb E[\Phi(|Y|/r)]\le 1. \end{align*} The useful structural facts are monotonicity and convexity. Indeed, \begin{align*} \Phi'(u)=2ue^{u^2}\ge 0, \qquad \Phi''(u)=2e^{u^2}(1+2u^2)\ge 0 \end{align*} for every $u\ge 0$, so $\Phi$ is increasing and convex on $[0,\infty)$. Let $Y,Z:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ be random variables with finite $\psi_2$ norm. The admissible set is upward closed: if $q>0$ is admissible for $Y$ and $q'\ge q$, then $|Y|/q'\le |Y|/q$, so monotonicity of $\Phi$ gives $\mathbb E[\Phi(|Y|/q')]\le\mathbb E[\Phi(|Y|/q)]\le 1$; the same argument applies to $Z$. Choose parameters $r>\|Y\|_{\psi_2}$ and $s>\|Z\|_{\psi_2}$ for which \begin{align*} \mathbb E[\Phi(|Y|/r)]\le 1, \qquad \mathbb E[\Phi(|Z|/s)]\le 1. \end{align*} Define \begin{align*} \lambda:=\frac{r}{r+s}, \qquad \mu:=\frac{s}{r+s}. \end{align*} Then $\lambda+\mu=1$, and the ordinary triangle inequality in $\mathbb R$ gives \begin{align*} \frac{|Y+Z|}{r+s} \le \frac{|Y|}{r+s}+\frac{|Z|}{r+s} = \lambda\frac{|Y|}{r}+\mu\frac{|Z|}{s}. \end{align*} Because $\Phi$ is increasing on $[0,\infty)$ and convex, \begin{align*} \Phi\left(\frac{|Y+Z|}{r+s}\right) \le \Phi\left(\lambda\frac{|Y|}{r}+\mu\frac{|Z|}{s}\right) \le \lambda\Phi\left(\frac{|Y|}{r}\right) + \mu\Phi\left(\frac{|Z|}{s}\right). \end{align*} Taking expectation with respect to $\mathbb P$ gives \begin{align*} \mathbb E\left[\Phi\left(\frac{|Y+Z|}{r+s}\right)\right] \le \lambda\mathbb E[\Phi(|Y|/r)] + \mu\mathbb E[\Phi(|Z|/s)] \le \lambda+\mu = 1. \end{align*} Thus $r+s$ is admissible for $Y+Z$, so $\|Y+Z\|_{\psi_2}\le r+s$. Since $r$ and $s$ may be taken arbitrarily close to $\|Y\|_{\psi_2}$ and $\|Z\|_{\psi_2}$, respectively, \begin{align*} \|Y+Z\|_{\psi_2}\le \|Y\|_{\psi_2}+\|Z\|_{\psi_2}. \end{align*} [/guided] [/step] [step:Bound the norm of the mean by the norm of the random variable] Let \begin{align*} m:=\mathbb E[X]\in\mathbb R \end{align*} and regard $m$ as the constant random variable $\omega\mapsto m$ on $\Omega$. For a deterministic constant $c\in\mathbb R$, \begin{align*} \|c\|_{\psi_2} = \inf\left\{t>0:\exp\left(\frac{c^2}{t^2}\right)\le 2\right\} = \frac{|c|}{\sqrt{\log 2}}, \end{align*} with the value $0$ when $c=0$. Let $r>\|X\|_{\psi_2}$ be admissible; such an admissible $r$ exists because the admissible set is upward closed by monotonicity of $t\mapsto\exp(X^2/t^2)$ decreasing pointwise for $t>0$. Thus \begin{align*} \mathbb E\left[\exp\left(\frac{X^2}{r^2}\right)\right]\le 2. \end{align*} Since $u\le e^u-1$ for every $u\ge 0$, applied to $u=X^2/r^2$, we obtain \begin{align*} \mathbb E\left[\frac{X^2}{r^2}\right] \le \mathbb E\left[\exp\left(\frac{X^2}{r^2}\right)-1\right] \le 1. \end{align*} Hence \begin{align*} \mathbb E[X^2]\le r^2. \end{align*} Using \begin{align*} 0\le \mathbb E\left[(|X|-\mathbb E[|X|])^2\right] = \mathbb E[X^2]-(\mathbb E[|X|])^2, \end{align*} we get $\mathbb E[|X|]\le r$. Therefore \begin{align*} |m| = |\mathbb E[X]| \le \mathbb E[|X|] \le r. \end{align*} Letting $r\downarrow\|X\|_{\psi_2}$ gives \begin{align*} |\mathbb E[X]|\le \|X\|_{\psi_2}. \end{align*} Consequently, \begin{align*} \|\mathbb E[X]\|_{\psi_2} \le \frac{\|X\|_{\psi_2}}{\sqrt{\log 2}}. \end{align*} [/step] [step:Apply the triangle inequality to center the random variable] Using the triangle inequality from the previous step with $Y=X$ and $Z=-\mathbb E[X]$, and using the scaling identity for the constant random variable $\mathbb E[X]$, we obtain \begin{align*} \|X-\mathbb E[X]\|_{\psi_2} \le \|X\|_{\psi_2}+\|\mathbb E[X]\|_{\psi_2} \le \left(1+\frac{1}{\sqrt{\log 2}}\right)\|X\|_{\psi_2}. \end{align*} Thus the centering estimate holds with the universal constant \begin{align*} C:=1+\frac{1}{\sqrt{\log 2}}. \end{align*} This completes the proof. [/step]