= Solution
Write $\psi(s)=\log\mathbb E e^{sX}$ for the <cumulant-generating function>, and let $D_+=\{s>0:\psi(s)<\infty\}$. This set is nonempty. In fact, finiteness of the <moment-generating function> at some $\eta>0$ and $-\eta$ gives
$$
\mathbb E e^{\eta|X|}\leq\mathbb E e^{\eta X}+\mathbb E e^{-\eta X}<\infty,
$$
so $X$ is an <integrable random variable>. Put $m=\mathbb EX$.
For the upper tail, use the following one-sided <convex conjugate> and its <strict generalized inverse of a nondecreasing function>:
$$
I(x)=\sup_{s\in D_+\cup\{0\}}\{sx-\psi(s)\},\qquad
\psi_X^{*-1}(t):=\inf\{x\in\mathbb R:I(x)>t\},\quad t>0.
$$
Here $\psi(0)=0$, and $I$ is nondecreasing because every slope in its defining <supremum> is nonnegative. For $x\geq m$, negative $s$ cannot increase the conjugate: $\psi(s)\geq sm$ gives $sx-\psi(s)\leq s(x-m)\leq0$, and $s=0$ already gives zero. Thus $I$ agrees with the full <Legendre transform of a cumulant-generating function> on its upper-tail branch. Using a <strict generalized inverse of a nondecreasing function> handles constant <random variables> and bounded supports without assuming a strictly increasing, everywhere finite conjugate. Directly from the definition,
$$
\{x:I(x)>t\}
=\bigcup_{s\in D_+}\left(\frac{t+\psi(s)}s,\infty\right),
\qquad
\boxed{\psi_X^{*-1}(t)=\inf_{s\in D_+}\frac{t+\psi(s)}s.}
$$
This threshold is finite: any fixed $s\in D_+$ gives a finite upper bound, while $e^u\geq1+u$ gives $\mathbb E e^{s(X-m)}\geq1$ and hence $\psi(s)\geq sm$, providing the lower bound $m$.
We now derive the needed <Chernoff bound>. For $s\in D_+$ and any real $x$, the pointwise inequality $e^{sX}\geq e^{sx}\mathbf1_{\{X>x\}}$ implies
$$
\mathbb P(X>x)\leq e^{\psi(s)-sx}.
$$
Let $q=\psi_X^{*-1}(t)$. For every $\varepsilon>0$, the <infimum> formula supplies an $s\in D_+$ such that $(t+\psi(s))/s<q+\varepsilon$. Thus $\mathbb P(X>q+\varepsilon)\leq e^{-t}$. As $\varepsilon\downarrow0$, these events increase to $\{X>q\}$; <continuity from below of a measure> proves
$$
\boxed{\mathbb P\{X>\psi_X^{*-1}(t)\}\leq e^{-t}.}
$$
The strict event and the <strict generalized inverse of a nondecreasing function> avoid any assumption that the optimizing $s$ exists. For example, if $X=m$ almost surely, the threshold is $m$ for every $t>0$, and the probability is zero.
Back to article page
