[proofplan] We use the defining exponential integrability of the $\psi_1$ norm to obtain factorial moment bounds for $X$. After expanding the moment generating function, the centering condition removes the linear term, so only moments of order at least two remain. The factorials from the exponential series cancel against those moment bounds, leaving a geometric series when $|\lambda|$ is sufficiently small compared to the reciprocal scale set by $K$. Finally, the elementary inequality $1+u\le e^u$ turns the resulting quadratic bound into the claimed exponential estimate. [/proofplan] [step:Derive moment bounds from the $\psi_1$ norm] Let $(\Omega, \mathcal F, \mathbb P)$ denote the probability space on which the real-valued [random variable](/page/Random%20Variable) $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ is defined, where $\mathcal B(\mathbb R)$ denotes the Borel $\sigma$-algebra on $\mathbb R$. Since $\|X\|_{\psi_1}\le K$, the Luxemburg definition of the $\psi_1$ norm gives \begin{align*} \mathbb E\left[\exp\left(\frac{|X|}{K}\right)\right]\le 2. \end{align*} For each integer $p\ge 1$ and each $t\ge 0$, the exponential series gives \begin{align*} e^t \ge \frac{t^p}{p!}. \end{align*} Applying this with \begin{align*} t=\frac{|X|}{K} \end{align*} and integrating with respect to $\mathbb P$ gives \begin{align*} \frac{\mathbb E[|X|^p]}{K^p p!} \le \mathbb E\left[\exp\left(\frac{|X|}{K}\right)\right] \le 2. \end{align*} Thus, for every integer $p\ge 1$, \begin{align*} \mathbb E[|X|^p]\le 2K^p p!. \end{align*} [guided] Let $(\Omega, \mathcal F, \mathbb P)$ denote the probability space on which the real-valued random variable $X: (\Omega, \mathcal F) \to (\mathbb R, \mathcal B(\mathbb R))$ is defined, where $\mathcal B(\mathbb R)$ denotes the Borel $\sigma$-algebra on $\mathbb R$. The purpose of this step is to turn the Orlicz norm assumption into concrete bounds on ordinary moments. By the Luxemburg definition of the $\psi_1$ norm, the condition $\|X\|_{\psi_1}\le K$ means that the exponential moment at scale $K$ is controlled: \begin{align*} \mathbb E\left[\exp\left(\frac{|X|}{K}\right)\right]\le 2. \end{align*} Now fix an integer $p\ge 1$. For every real number $t\ge 0$, the exponential series has non-negative terms, so retaining only the $p$th term yields \begin{align*} e^t = \sum_{m=0}^{\infty}\frac{t^m}{m!} \ge \frac{t^p}{p!}. \end{align*} We apply this pointwise with \begin{align*} t=\frac{|X|}{K}, \end{align*} obtaining \begin{align*} \exp\left(\frac{|X|}{K}\right) \ge \frac{|X|^p}{K^p p!}. \end{align*} Both sides are non-negative measurable random variables, so integration with respect to $\mathbb P$ preserves the inequality: \begin{align*} \frac{\mathbb E[|X|^p]}{K^p p!} \le \mathbb E\left[\exp\left(\frac{|X|}{K}\right)\right] \le 2. \end{align*} Multiplying by $K^p p!$ gives the moment estimate \begin{align*} \mathbb E[|X|^p]\le 2K^p p!. \end{align*} This is the only place where the definition of the $\psi_1$ norm is used. [/guided] [/step] [step:Expand the moment generating function and remove the linear term] Let $\lambda\in\mathbb R$ satisfy \begin{align*} |\lambda|\le \frac{1}{2K}. \end{align*} Then, pointwise, \begin{align*} |\lambda X|\le \frac{|X|}{2K}, \end{align*} and therefore \begin{align*} e^{|\lambda X|}\le \exp\left(\frac{|X|}{2K}\right)\le \exp\left(\frac{|X|}{K}\right). \end{align*} The right-hand side is integrable by the previous step. For each integer $n\ge 0$, define the partial sum random variable $S_n: \Omega \to \mathbb R$ by \begin{align*} S_n(\omega)=\sum_{p=0}^{n}\frac{(\lambda X(\omega))^p}{p!}. \end{align*} For every $\omega\in\Omega$ one has $S_n(\omega)\to e^{\lambda X(\omega)}$ and \begin{align*} |S_n(\omega)|\le \sum_{p=0}^{n}\frac{|\lambda X(\omega)|^p}{p!}\le e^{|\lambda X(\omega)|}. \end{align*} The dominating random variable $e^{|\lambda X|}$ is integrable, so the [Dominated Convergence Theorem](/theorems/4) applied to the sequence $(S_n)_{n=0}^{\infty}$ justifies integrating the exponential series term by term. Hence \begin{align*} \mathbb E[e^{\lambda X}] = 1+\lambda\mathbb E[X]+\sum_{p=2}^{\infty}\frac{\lambda^p\mathbb E[X^p]}{p!}. \end{align*} Using $\mathbb E[X]=0$ and $|\mathbb E[X^p]|\le \mathbb E[|X|^p]$, we obtain \begin{align*} \mathbb E[e^{\lambda X}] \le 1+\sum_{p=2}^{\infty}\frac{|\lambda|^p\mathbb E[|X|^p]}{p!} \le 1+2\sum_{p=2}^{\infty}(K|\lambda|)^p. \end{align*} [guided] We now pass from moment bounds to a bound on the moment generating function. Fix $\lambda\in\mathbb R$ with \begin{align*} |\lambda|\le \frac{1}{2K}. \end{align*} The point of this restriction is to make the exponential $e^{|\lambda X|}$ dominated by the exponential integrable random variable already controlled by the $\psi_1$ assumption. Indeed, \begin{align*} |\lambda X|\le \frac{|X|}{2K}, \end{align*} and hence \begin{align*} e^{|\lambda X|}\le \exp\left(\frac{|X|}{2K}\right)\le \exp\left(\frac{|X|}{K}\right). \end{align*} The final random variable is integrable because the previous step proved \begin{align*} \mathbb E\left[\exp\left(\frac{|X|}{K}\right)\right]\le 2. \end{align*} For each integer $n\ge 0$, define the partial sum random variable $S_n: \Omega \to \mathbb R$ by \begin{align*} S_n(\omega)=\sum_{p=0}^{n}\frac{(\lambda X(\omega))^p}{p!}. \end{align*} Then $S_n(\omega)\to e^{\lambda X(\omega)}$ for every $\omega\in\Omega$. We also have the pointwise domination \begin{align*} |S_n(\omega)|\le \sum_{p=0}^{n}\frac{|\lambda X(\omega)|^p}{p!}\le e^{|\lambda X(\omega)|}. \end{align*} The dominating random variable $e^{|\lambda X|}$ is integrable by the estimate above. Hence the [Dominated Convergence Theorem](/theorems/4), applied to the sequence $(S_n)_{n=0}^{\infty}$, justifies integrating the exponential series term by term, giving \begin{align*} \mathbb E[e^{\lambda X}] = \mathbb E\left[\sum_{p=0}^{\infty}\frac{(\lambda X)^p}{p!}\right] = 1+\lambda\mathbb E[X]+\sum_{p=2}^{\infty}\frac{\lambda^p\mathbb E[X^p]}{p!}. \end{align*} The centering hypothesis $\mathbb E[X]=0$ removes the linear term. Taking absolute values in the remaining moments gives \begin{align*} \mathbb E[e^{\lambda X}] \le 1+\sum_{p=2}^{\infty}\frac{|\lambda|^p\mathbb E[|X|^p]}{p!}. \end{align*} Using the moment estimate from the first step, we get \begin{align*} \mathbb E[e^{\lambda X}] \le 1+\sum_{p=2}^{\infty}\frac{|\lambda|^p 2K^p p!}{p!}. \end{align*} Cancelling $p!$ in each summand yields \begin{align*} \mathbb E[e^{\lambda X}] \le 1+2\sum_{p=2}^{\infty}(K|\lambda|)^p. \end{align*} This is the key reduction: after centering, the MGF is bounded by a geometric tail beginning at quadratic order. [/guided] [/step] [step:Choose the local scale so the series is quadratic] Assume now that \begin{align*} |\lambda|\le \frac{1}{4K}. \end{align*} Then \begin{align*} K|\lambda|\le \frac{1}{4}, \end{align*} and \begin{align*} \sum_{p=2}^{\infty}(K|\lambda|)^p = K^2\lambda^2\sum_{q=0}^{\infty}(K|\lambda|)^q \le K^2\lambda^2\sum_{q=0}^{\infty}4^{-q} = \frac{4}{3}K^2\lambda^2. \end{align*} Therefore \begin{align*} \mathbb E[e^{\lambda X}] \le 1+\frac{8}{3}K^2\lambda^2. \end{align*} Using $1+u\le e^u$ for $u\ge 0$, with $u=\frac{8}{3}K^2\lambda^2$, gives \begin{align*} \mathbb E[e^{\lambda X}] \le \exp\left(\frac{8}{3}K^2\lambda^2\right). \end{align*} [guided] The previous step reduced the problem to estimating the geometric tail \begin{align*} \sum_{p=2}^{\infty}(K|\lambda|)^p. \end{align*} We now choose the local range of $\lambda$ so that the common ratio is uniformly less than $1$. Assume \begin{align*} |\lambda|\le \frac{1}{4K}. \end{align*} Then \begin{align*} K|\lambda|\le \frac{1}{4}. \end{align*} Factoring out the first nonzero power of the tail gives \begin{align*} \sum_{p=2}^{\infty}(K|\lambda|)^p = K^2\lambda^2\sum_{q=0}^{\infty}(K|\lambda|)^q. \end{align*} Since each term satisfies $(K|\lambda|)^q\le 4^{-q}$, comparison with the geometric series gives \begin{align*} \sum_{q=0}^{\infty}(K|\lambda|)^q \le \sum_{q=0}^{\infty}4^{-q} = \frac{1}{1-1/4} = \frac{4}{3}. \end{align*} Substituting this into the MGF estimate from the previous step yields \begin{align*} \mathbb E[e^{\lambda X}] \le 1+2\sum_{p=2}^{\infty}(K|\lambda|)^p \le 1+\frac{8}{3}K^2\lambda^2. \end{align*} Finally, the elementary inequality $1+u\le e^u$ for $u\ge 0$, applied with \begin{align*} u=\frac{8}{3}K^2\lambda^2, \end{align*} gives \begin{align*} \mathbb E[e^{\lambda X}] \le \exp\left(\frac{8}{3}K^2\lambda^2\right). \end{align*} This proves the desired quadratic MGF control on the local interval \begin{align*} |\lambda|\le \frac{1}{4K}. \end{align*} [/guided] [/step] [step:Record the radius and exponential constants separately] Define the universal radius constant and the universal exponential constant by \begin{align*} c:=\frac{1}{4}, \qquad C:=\frac{8}{3}. \end{align*} The preceding steps show that, whenever $\lambda\in\mathbb R$ satisfies \begin{align*} |\lambda|\le \frac{c}{K}, \end{align*} one has \begin{align*} \mathbb E[e^{\lambda X}] \le \exp(CK^2\lambda^2). \end{align*} These constants are numerical and independent of $X$, $K$, and $\lambda$, so they are universal. This proves the stated local MGF bound. [guided] The estimates have produced two different constants, and they serve different roles. The radius constant controls how small $\lambda$ must be, while the exponential constant controls the size of the quadratic bound. We therefore define \begin{align*} c:=\frac{1}{4}, \qquad C:=\frac{8}{3}. \end{align*} The previous step proves exactly that if $\lambda\in\mathbb R$ satisfies \begin{align*} |\lambda|\le \frac{c}{K}, \end{align*} then \begin{align*} \mathbb E[e^{\lambda X}] \le \exp(CK^2\lambda^2). \end{align*} It is important not to merge these two constants into one symbol: enlarging the exponential constant would also enlarge the permitted interval if the same symbol were used in both places. Keeping $c$ and $C$ separate records precisely what the proof establishes. Since both constants are fixed numerical constants, independent of the random variable $X$, the scale $K$, and the parameter $\lambda$, they are universal. This completes the proof. [/guided] [/step]