[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*}
[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]
[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]
