[proofplan] We compute the [moment generating function](/page/Moment%20Generating%20Function) directly from the two-point law of a [Rademacher random variable](/page/Rademacher%20Random%20Variable). This gives the exact formula $M_\varepsilon(t)=(e^t+e^{-t})/2$. The sub-Gaussian bound is then proved by comparing the [power series](/page/Power%20Series) of $\cosh(t)$ and $e^{t^2/2}$ term by term, using the elementary factorial inequality $(2m)!\ge 2^m m!$. [/proofplan] [step:Evaluate the expectation using the two atoms of the Rademacher law] Fix $t\in\mathbb R$. Define the events $A_+\in\mathcal F$ and $A_-\in\mathcal F$ by \begin{align*} A_+=\{\omega\in\Omega:\varepsilon(\omega)=1\} \end{align*} and \begin{align*} A_-=\{\omega\in\Omega:\varepsilon(\omega)=-1\}. \end{align*} Since $\varepsilon$ is Rademacher, $\mathbb P(A_+)=\mathbb P(A_-)=1/2$, and $\mathbb P(A_+\cup A_-)=1$. The [random variable](/page/Random%20Variable) $e^{t\varepsilon}:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ takes the value $e^t$ on $A_+$ and the value $e^{-t}$ on $A_-$. Therefore, by the definition of expectation for a two-valued simple random variable, \begin{align*} M_\varepsilon(t)=\mathbb E[e^{t\varepsilon}]=e^t\mathbb P(A_+)+e^{-t}\mathbb P(A_-). \end{align*} Substituting the two probabilities gives \begin{align*} M_\varepsilon(t)=\frac{e^t+e^{-t}}{2}. \end{align*} [/step] [step:Compare the power series of $\cosh(t)$ and $e^{t^2/2}$] Define $c:\mathbb R\to\mathbb R$ by \begin{align*} c(t)=\frac{e^t+e^{-t}}{2}. \end{align*} The power series expansion of $c$ is \begin{align*} c(t)=\sum_{m=0}^{\infty}\frac{t^{2m}}{(2m)!}. \end{align*} The power series expansion of $e^{t^2/2}$ is \begin{align*} e^{t^2/2}=\sum_{m=0}^{\infty}\frac{t^{2m}}{2^m m!}. \end{align*} For each $m\in\mathbb N$ and also for $m=0$, the factorial inequality \begin{align*} (2m)!\ge 2^m m! \end{align*} holds, because the product defining $(2m)!$ contains the even factors $2,4,\ldots,2m$, whose product is $2^m m!$. Hence \begin{align*} \frac{t^{2m}}{(2m)!}\le \frac{t^{2m}}{2^m m!} \end{align*} for every $m\ge 0$, since $t^{2m}\ge 0$. Comparing the non-negative terms in the two convergent power series gives \begin{align*} c(t)\le e^{t^2/2}. \end{align*} [guided] We need a global inequality, valid for every real $t$, not only an estimate near $0$. The useful observation is that both functions have power series with only even powers of $t$, so all the terms being compared are non-negative. Define $c:\mathbb R\to\mathbb R$ by \begin{align*} c(t)=\frac{e^t+e^{-t}}{2}. \end{align*} Using the power series for the exponential function and averaging the series for $e^t$ and $e^{-t}$, the odd powers cancel and the even powers remain: \begin{align*} c(t)=\sum_{m=0}^{\infty}\frac{t^{2m}}{(2m)!}. \end{align*} On the other hand, substituting $t^2/2$ into the exponential series gives \begin{align*} e^{t^2/2}=\sum_{m=0}^{\infty}\frac{(t^2/2)^m}{m!}. \end{align*} Equivalently, \begin{align*} e^{t^2/2}=\sum_{m=0}^{\infty}\frac{t^{2m}}{2^m m!}. \end{align*} We now compare the coefficient of $t^{2m}$ in the two series. For each integer $m\ge 0$, the factorial $(2m)!$ contains the even factors $2,4,\ldots,2m$. Their product is \begin{align*} 2\cdot 4\cdots 2m=2^m m!. \end{align*} All remaining factors in $(2m)!$ are positive integers, so \begin{align*} (2m)!\ge 2^m m!. \end{align*} Since $t^{2m}\ge 0$, dividing by the positive denominators yields \begin{align*} \frac{t^{2m}}{(2m)!}\le \frac{t^{2m}}{2^m m!}. \end{align*} Thus every term in the power series for $c(t)$ is bounded above by the corresponding term in the power series for $e^{t^2/2}$. Because both series are convergent series with non-negative terms, termwise comparison gives \begin{align*} c(t)\le e^{t^2/2}. \end{align*} [/guided] [/step] [step:Combine the exact formula with the power series bound] From the first step, for every $t\in\mathbb R$, \begin{align*} M_\varepsilon(t)=\frac{e^t+e^{-t}}{2}=c(t). \end{align*} From the second step, \begin{align*} c(t)\le e^{t^2/2}. \end{align*} Therefore, for every $t\in\mathbb R$, \begin{align*} M_\varepsilon(t)\le e^{t^2/2}. \end{align*} This proves both the exact formula and the stated bound. [/step]