$P[ e^{tX} > e^{ta} ] =?$

Question

Can anyone help me understand why given any random variable $X$, the following stands true?

$$ \forall t > 0, P( e^{tX} > e^{t\epsilon} ) \le e^{-t\epsilon} E[e^{tX}]. $$

I found it in the proof of Hoeffding's Inequality (Lemma 5).

Answer 1

We are given that $t > 0$. The event $\{e^{tX} > e^{t\epsilon}\}$ occurs exactly when $X > \epsilon$, and so $$\begin{align} P\{e^{tX} > e^{t\epsilon}\} &= P\{X > \epsilon\}\\ &= \int_\epsilon^\infty f_X(x)\,\mathrm dx\\ &= \int_{-\infty}^\infty \mathbf 1_{x\colon x > \epsilon}\, f_X(x)\,\mathrm dx\\ &\leq \int_{-\infty}^\infty e^{t(x-\epsilon)} f_X(x)\,\mathrm dx\\ &= e^{-t\epsilon} \int_{-\infty}^\infty e^{tx} f_X(x)\,\mathrm dx\\ &= e^{-t\epsilon} E[e^{tX}]. \end{align}$$

Answer 2

Since $P\left(e^{tX}\gt e^{t\xi}\right)$ is a decreasing function of $\xi$, $$ \begin{align} \mathrm{E}\left[e^{tX}\right] &=\int_{-\infty}^\infty P\left(e^{tX}\gt e^{t\xi}\right)\,\mathrm{d}e^{t\xi}\\ &\ge\int_{-\infty}^\epsilon P\left(e^{tX}\gt e^{t\xi}\right)\,\mathrm{d}e^{t\xi}\\ &\ge\int_{-\infty}^\epsilon P\left(e^{tX}\gt e^{t\epsilon}\right)\,\mathrm{d}e^{t\xi}\\[6pt] &= e^{t\epsilon}P\left(e^{tX}\gt e^{t\epsilon}\right) \end{align} $$