[proofplan] We prove the equivalence by passing through exponential integrability, tails, and moments with explicit constants. Exponential integrability gives the tail bound by applying [Markov's inequality](/theorems/514) to the non-negative [random variable](/page/Random%20Variable) $\exp(|X|/K_3)$. The tail bound gives moment growth by writing $|X|^p$ as an integral of indicators and estimating the resulting Gamma function integral. Moment growth gives exponential integrability by expanding the exponential series and choosing a numerical denominator large enough to make the series summable. The Orlicz norm condition is exactly the infimum formulation of the exponential integrability condition. [/proofplan] [step:Derive exponential tails from exponential integrability] Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space on which the real-valued random variable $X:(\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is defined. Define $A_1$, $A_2$, $A_3$, and $A_4$ as the infimal admissible constants in conditions $1$, $2$, $3$, and $4$, respectively; equivalently, $A_4:=\|X\|_{\psi_1}$ and $A_3$ is the infimum of all $K>0$ such that $\mathbb{E}[\exp(|X|/K)]\le 2$. Assume that $K_3>0$ satisfies \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right]\le 2. \end{align*} For each $t\ge 0$, define the event $E_t:=\{\omega\in\Omega: |X(\omega)|\ge t\}\in\mathcal{F}$. On $E_t$, \begin{align*} \exp\left(\frac{|X|}{K_3}\right)\ge \exp\left(\frac{t}{K_3}\right). \end{align*} Equivalently, this is [Markov's inequality](/theorems/514) applied to the non-negative random variable $\exp(|X|/K_3)$. In integral form, \begin{align*} \exp\left(\frac{t}{K_3}\right)\mathbb{P}(E_t) \le \int_{E_t}\exp\left(\frac{|X(\omega)|}{K_3}\right)\,d\mathbb{P}(\omega) \le \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 2. \end{align*} Multiplying by $\exp(-t/K_3)$ gives \begin{align*} \mathbb{P}(|X|\ge t)\le 2\exp\left(-\frac{t}{K_3}\right). \end{align*} Thus condition $3$ implies condition $1$, and the least constants satisfy $A_1\le A_3$. [guided] Assume that $K_3>0$ is an exponential-integrability constant, meaning \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right]\le 2. \end{align*} We want to convert this information about an average into information about a tail probability. Fix $t\ge 0$, and define \begin{align*} E_t:=\{\omega\in\Omega: |X(\omega)|\ge t\}. \end{align*} Since $X$ is measurable and $[t,\infty)$ is Borel after applying the measurable map $\omega\mapsto |X(\omega)|$, the set $E_t$ belongs to $\mathcal{F}$. On $E_t$, the inequality $|X|\ge t$ implies \begin{align*} \exp\left(\frac{|X|}{K_3}\right)\ge \exp\left(\frac{t}{K_3}\right). \end{align*} Integrating this lower bound over $E_t$ with respect to $\mathbb{P}$ gives \begin{align*} \int_{E_t}\exp\left(\frac{|X(\omega)|}{K_3}\right)\,d\mathbb{P}(\omega) \ge \int_{E_t}\exp\left(\frac{t}{K_3}\right)\,d\mathbb{P}(\omega) = \exp\left(\frac{t}{K_3}\right)\mathbb{P}(E_t). \end{align*} The same integral is bounded above by the integral over all of $\Omega$, so \begin{align*} \exp\left(\frac{t}{K_3}\right)\mathbb{P}(E_t) \le \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 2. \end{align*} Multiplying both sides by $\exp(-t/K_3)$ yields \begin{align*} \mathbb{P}(|X|\ge t)\le 2\exp\left(-\frac{t}{K_3}\right). \end{align*} This is exactly condition $1$ with $K_1=K_3$, so every admissible $K_3$ is also an admissible $K_1$. Taking infima gives $A_1\le A_3$. [/guided] [/step] [step:Convert the exponential tail bound into linear moment growth] Assume that $K_1>0$ satisfies \begin{align*} \mathbb{P}(|X|\ge t)\le 2\exp\left(-\frac{t}{K_1}\right) \end{align*} for every $t\ge 0$. Fix $p\ge 1$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,\infty)$ with its Borel $\sigma$-algebra. For each $\omega\in\Omega$, \begin{align*} |X(\omega)|^p=\int_0^\infty p t^{p-1}\mathbb{1}_{\{t\le |X(\omega)|\}}\,d\mathcal{L}^1(t). \end{align*} Integrating over $\Omega$ and applying Tonelli's theorem to the non-negative measurable function $(\omega,t)\mapsto p t^{p-1}\mathbb{1}_{\{t\le |X(\omega)|\}}$, \begin{align*} \mathbb{E}[|X|^p]=p\int_0^\infty t^{p-1}\mathbb{P}(|X|\ge t)\,d\mathcal{L}^1(t). \end{align*} Using the assumed tail bound inside this integral gives \begin{align*} \mathbb{E}[|X|^p]\le 2p\int_0^\infty t^{p-1}\exp\left(-\frac{t}{K_1}\right)\,d\mathcal{L}^1(t). \end{align*} Under the substitution $u=t/K_1$, the one-dimensional Lebesgue measure transforms as $d\mathcal{L}^1(t)=K_1\,d\mathcal{L}^1(u)$, and the interval $[0,\infty)$ maps onto $[0,\infty)$. Hence \begin{align*} \mathbb{E}[|X|^p] &\le 2pK_1^p\int_0^\infty u^{p-1}e^{-u}\,d\mathcal{L}^1(u). \end{align*} Define the Gamma function $\Gamma:[1,\infty)\to(0,\infty)$ as the map sending each $q\in[1,\infty)$ to \begin{align*} \Gamma(q):=\int_0^\infty u^{q-1}e^{-u}\,d\mathcal{L}^1(u). \end{align*} The elementary Gamma function estimate $\Gamma(p+1)^{1/p}\le 2p$ for $p\ge 1$ gives \begin{align*} \mathbb{E}[|X|^p]\le 2K_1^p\Gamma(p+1)\le 2(2pK_1)^p. \end{align*} Since $2^{1/p}\le 2$ for $p\ge 1$, \begin{align*} \left(\mathbb{E}[|X|^p]\right)^{1/p}\le 4K_1p. \end{align*} Thus condition $1$ implies condition $2$, and $A_2\le 4A_1$. [/step] [step:Obtain exponential integrability from moment growth] Assume that $K_2>0$ satisfies \begin{align*} \left(\mathbb{E}[|X|^p]\right)^{1/p}\le K_2p \end{align*} for every $p\ge 1$. Set \begin{align*} K_3:=4eK_2. \end{align*} For every integer $m\ge 1$, the moment hypothesis applied with $p=m$ gives \begin{align*} \mathbb{E}[|X|^m]\le (K_2m)^m. \end{align*} Using the [power series](/page/Power%20Series) for the exponential function, non-negativity of every term, and the [Monotone Convergence Theorem](/theorems/509) applied to the increasing sequence of partial sums, \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] = 1+\sum_{m=1}^{\infty}\frac{\mathbb{E}[|X|^m]}{m!K_3^m}. \end{align*} Substituting the moment bound and the definition $K_3=4eK_2$ gives \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 1+\sum_{m=1}^{\infty}\frac{(K_2m)^m}{m!(4eK_2)^m}. \end{align*} For each integer $m\ge 1$, the elementary factorial bound $m!\ge (m/e)^m$ gives \begin{align*} \frac{(K_2m)^m}{m!(4eK_2)^m} \le \frac{(K_2m)^m}{(m/e)^m(4eK_2)^m} = 4^{-m}. \end{align*} Therefore \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 1+\sum_{m=1}^{\infty}4^{-m} = 1+\frac{1}{3} \le 2. \end{align*} Thus condition $2$ implies condition $3$, and $A_3\le 4eA_2$. [guided] Assume the moment-growth condition: there is $K_2>0$ such that, for every real $p\ge 1$, \begin{align*} \left(\mathbb{E}[|X|^p]\right)^{1/p}\le K_2p. \end{align*} To prove exponential integrability, we need to control \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \end{align*} for a suitable constant $K_3$. The natural way to connect this expectation to the moment bounds is to expand the exponential into its power series. Choose \begin{align*} K_3:=4eK_2. \end{align*} For every integer $m\ge 1$, the hypothesis applies with $p=m$, so \begin{align*} \mathbb{E}[|X|^m]\le (K_2m)^m. \end{align*} Because all terms in the exponential power series are non-negative, the [Monotone Convergence Theorem](/theorems/509) permits integration to be interchanged with the increasing sequence of partial sums. Thus \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] = 1+\sum_{m=1}^{\infty}\frac{\mathbb{E}[|X|^m]}{m!K_3^m}. \end{align*} Substituting the moment estimate gives \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 1+\sum_{m=1}^{\infty}\frac{(K_2m)^m}{m!(4eK_2)^m}. \end{align*} The role of the numerical constant $4e$ is to dominate the factorial denominator uniformly in $m$. The elementary factorial estimate, which follows from the integral comparison underlying [Stirling's formula](/theorems/1109), is \begin{align*} m!\ge \left(\frac{m}{e}\right)^m. \end{align*} It implies \begin{align*} \frac{(K_2m)^m}{m!(4eK_2)^m} \le \frac{(K_2m)^m}{(m/e)^m(4eK_2)^m} = 4^{-m}. \end{align*} Therefore the exponential expectation is bounded by a geometric series: \begin{align*} \mathbb{E}\left[\exp\left(\frac{|X|}{K_3}\right)\right] \le 1+\sum_{m=1}^{\infty}4^{-m} = 1+\frac{1}{3} \le 2. \end{align*} This proves condition $3$ with $K_3=4eK_2$. Since every admissible $K_2$ gives the admissible exponential-integrability constant $4eK_2$, taking infima yields $A_3\le 4eA_2$. [/guided] [/step] [step:Identify the Orlicz norm condition with exponential integrability] By definition, \begin{align*} \|X\|_{\psi_1} = \inf\left\{K>0:\mathbb{E}\left[\exp\left(\frac{|X|}{K}\right)\right]\le 2\right\}. \end{align*} Hence $\|X\|_{\psi_1}<\infty$ if and only if the admissible set \begin{align*} \left\{K>0:\mathbb{E}\left[\exp\left(\frac{|X|}{K}\right)\right]\le 2\right\} \end{align*} is non-empty, which is exactly condition $3$. The infimal admissible constant in condition $3$ is the defining infimum of $\|X\|_{\psi_1}$, so $A_4=A_3$. This formulation uses infima rather than assuming that the infimum is attained; for example, when $X=0$ the infimum is $0$, while admissible constants are required to be positive. [/step] [step:Combine the implications and constants] The implications proved above give \begin{align*} 3\implies 1,\qquad 1\implies 2,\qquad 2\implies 3,\qquad 3\iff 4. \end{align*} Therefore all four conditions are equivalent. The accompanying constant estimates are \begin{align*} A_1\le A_3,\qquad A_2\le 4A_1,\qquad A_3\le 4eA_2,\qquad A_4=A_3. \end{align*} These inequalities show that the infimal admissible constants in the four characterizations are comparable by universal numerical constants, completing the proof. [/step]