[proofplan] We prove the one-sided upper tail by applying the exponential Markov argument to the non-negative [random variable](/page/Random%20Variable) $e^{\lambda X}$, where $\lambda>0$ is a free parameter. The assumed moment generating function bound reduces the estimate to minimizing a quadratic exponent in $\lambda$. The lower tail is obtained by applying the same argument with $-\mu X$, equivalently by using the given moment bound at the negative parameter $-\mu$. Finally, the two one-sided estimates are combined by the union bound. [/proofplan]

[step:Bound the upper tail by an optimized exponential moment]First let $t>0$ and let $\lambda>0$. Define the event $A_t\in\mathcal F$ by \begin{align*} A_t := \{\omega\in\Omega : X(\omega)\geq t\}. \end{align*} Since the exponential function is increasing on $\mathbb R$, we have \begin{align*} e^{\lambda X(\omega)}\geq e^{\lambda t}\mathbb 1_{A_t}(\omega) \end{align*} for every $\omega\in\Omega$. Taking expectations and using the assumed moment generating function bound gives \begin{align*} e^{\lambda t}\mathbb P(X\geq t) \leq \mathbb E[e^{\lambda X}] \leq e^{\sigma^2\lambda^2/2}. \end{align*} Dividing by $e^{\lambda t}>0$ yields \begin{align*} \mathbb P(X\geq t) \leq e^{-\lambda t+\sigma^2\lambda^2/2}. \end{align*} Choose \begin{align*} \lambda_t := \frac{t}{\sigma^2}. \end{align*} Because $t>0$ and $\sigma>0$, we have $\lambda_t>0$, so this choice is admissible. Substituting $\lambda=\lambda_t$ gives \begin{align*} -\lambda_t t+\frac{\sigma^2\lambda_t^2}{2} = -\frac{t^2}{\sigma^2}+\frac{t^2}{2\sigma^2} = -\frac{t^2}{2\sigma^2}. \end{align*} Therefore \begin{align*} \mathbb P(X\geq t)\leq e^{-t^2/(2\sigma^2)}. \end{align*}[/step]

[guided]We want to convert information about exponential moments into a probability estimate. The standard device is to compare the event $\{X\geq t\}$ with the size of $e^{\lambda X}$, where $\lambda>0$ is a parameter that will later be chosen optimally. Fix $t>0$ and $\lambda>0$. Define the event \begin{align*} A_t := \{\omega\in\Omega : X(\omega)\geq t\}. \end{align*} Since $X$ is a measurable real-valued random variable, $A_t\in\mathcal F$. On $A_t$, the inequality $X(\omega)\geq t$ implies $e^{\lambda X(\omega)}\geq e^{\lambda t}$ because the exponential function is increasing. Outside $A_t$, the indicator $\mathbb 1_{A_t}$ is zero. Hence the pointwise inequality \begin{align*} e^{\lambda X(\omega)}\geq e^{\lambda t}\mathbb 1_{A_t}(\omega) \end{align*} holds for every $\omega\in\Omega$. Taking expectations preserves the inequality because both sides are non-negative random variables. Thus \begin{align*} e^{\lambda t}\mathbb P(X\geq t) = \mathbb E[e^{\lambda t}\mathbb 1_{A_t}] \leq \mathbb E[e^{\lambda X}]. \end{align*} The hypothesis applies to this same positive parameter $\lambda\in\mathbb R$, so \begin{align*} \mathbb E[e^{\lambda X}] \leq e^{\sigma^2\lambda^2/2}. \end{align*} Combining the two inequalities and dividing by $e^{\lambda t}>0$ gives \begin{align*} \mathbb P(X\geq t) \leq e^{-\lambda t+\sigma^2\lambda^2/2}. \end{align*} The remaining task is to choose $\lambda$ so that the exponent is as small as possible. Define the quadratic function $q_t:(0,\infty)\to\mathbb R$ by \begin{align*} q_t(\lambda):=-\lambda t+\frac{\sigma^2\lambda^2}{2}. \end{align*} Its unconstrained minimizer is obtained from \begin{align*} q_t'(\lambda)=-t+\sigma^2\lambda=0, \end{align*} namely \begin{align*} \lambda_t:=\frac{t}{\sigma^2}. \end{align*} Because $t>0$ and $\sigma>0$, this value belongs to $(0,\infty)$ and is therefore allowed. Substitution gives \begin{align*} q_t(\lambda_t) = -\frac{t^2}{\sigma^2} + \frac{\sigma^2}{2}\frac{t^2}{\sigma^4} = -\frac{t^2}{2\sigma^2}. \end{align*} Therefore \begin{align*} \mathbb P(X\geq t)\leq e^{-t^2/(2\sigma^2)}. \end{align*}[/guided]

[step:Bound the lower tail using the negative moment parameter] Again let $t>0$ and let $\mu>0$. Define the event $B_t\in\mathcal F$ by \begin{align*} B_t := \{\omega\in\Omega : X(\omega)\leq -t\}. \end{align*} On $B_t$ we have $-\mu X(\omega)\geq \mu t$, so \begin{align*} e^{-\mu X(\omega)}\geq e^{\mu t}\mathbb 1_{B_t}(\omega). \end{align*} Taking expectations and applying the assumed moment bound with $\lambda=-\mu$ gives \begin{align*} e^{\mu t}\mathbb P(X\leq -t) \leq \mathbb E[e^{-\mu X}] \leq e^{\sigma^2\mu^2/2}. \end{align*} Hence \begin{align*} \mathbb P(X\leq -t) \leq e^{-\mu t+\sigma^2\mu^2/2}. \end{align*} Choosing \begin{align*} \mu_t := \frac{t}{\sigma^2} \end{align*} gives \begin{align*} \mathbb P(X\leq -t)\leq e^{-t^2/(2\sigma^2)}. \end{align*} [/step]

[step:Combine the two tails and include the endpoint $t=0$] For $t>0$, the event $\{|X|\geq t\}$ is contained in the union of the two events $\{X\geq t\}$ and $\{X\leq -t\}$. Therefore, by finite subadditivity of probability, \begin{align*} \mathbb P(|X|\geq t) \leq \mathbb P(X\geq t)+\mathbb P(X\leq -t) \leq 2e^{-t^2/(2\sigma^2)}. \end{align*} For $t=0$, \begin{align*} \mathbb P(|X|\geq 0)=1\leq 2=2e^{0}. \end{align*} Thus for every $t\geq 0$, \begin{align*} \mathbb P(|X|\geq t)\leq 2e^{-t^2/(2\sigma^2)}. \end{align*} [/step]