Understanding $P[X \gt \alpha] = P[e^{\theta X} \gt e^{\theta \alpha}]$

Question

In Chernof Bound, it is expressed $P[X \gt \alpha] = P[e^{\theta X} \gt e^{\theta \alpha}]$ for given $M_X(\theta) = E[e^{\theta X}]$.

I am not getting how we can equal $P[X \gt \alpha] = P[e^{\theta X} \gt e^{\theta \alpha}]$.

Can you clarify this?

Answer 1

The fundamental observations are that \begin{align*} & X > a \\ \iff & \theta X > \theta a \qquad \text{(assuming $\theta > 0$}) \\ \iff & e^{\theta X} > e^{\theta a} \end{align*} because $e^x$ is an increasing function. Hence, the events $\{X > a\}$ and $\{e^{\theta X} > e^{\theta a}\}$ are the same event.

It may be instructive to compare a statement that would be proved similarly: $\mathbb P(X > 3) = \mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.