Exponential Markov Inequality

Question

I need to show that: $$ P[\alpha X\geq \epsilon]\leq\frac{E[e^{\alpha X}]}{e^\epsilon} , \epsilon>0 $$ Does this work the same way as the normal Markov-Inequality? Because with that way I couldn't really figure out the solution, I mean this way: $$ E[e^{\alpha X}]=\int_{-\infty}^{\infty}e^{\alpha X}f(x)dx=... $$

Answer 1

$\alpha X \geq \epsilon$ iff $e^{\alpha X} \geq e^{\epsilon}$. Hence $P(\alpha X \geq \epsilon)=P(Y \geq e^{\epsilon}) \leq \frac {EY} {e^{\epsilon}}$ where $Y=e^{\alpha X}$.