TITLE: Prove an inequality given only the moment generating function
QUESTION [5 upvotes]: If given $X$ as a non-negative random variable that has mgf $M_X(s) = E[e^{sX}]$, how can I prove $\Pr(X \ge a) \le e^{-as} \centerdot M_X(s)$ where $a > 0$ and $s \ge 0$ ?

I figure the Markov inequality is a good start:
$\Pr(X \ge a) \le \dfrac{E[X]}{a}$
But this is where I'm stuck. Of course you can derive $M_x(s)$ at $s=0$ to get $E[X]$, but that doesn't really get you anywhere here.
Any suggestions?

REPLY [3 votes]: Markov's inequality gives, since $e^{sX}$ is always a positive r.v., and $e^{sX}>0,$
$$P[e^{sX} \ge e^{sa}] \le \dfrac{E(e^{sX})}{e^{sa}}=e^{-sa}\cdot M_X(s)$$
The result follows since $e^{sX} \ge e^{sa} \Leftrightarrow sX \ge sa \Leftrightarrow X\ge a $ .