The Markov inequality is given as
$$ P(X \geq a) \leq \frac{\mathbb{E}(X)}{a} $$
The extended version for monotonically increasing function follows, if $\varphi$ is a monotonically increasing nonnegative function for the nonnegative reals
$$ P(|X| \geq a) \leq \frac{\mathbb{E}(\varphi(|X|))}{\varphi(a)} $$
The proof for this is given in Wikipedia article as
$$ P(|X| \geq a) = P(\varphi(|X|) \geq \varphi(a)) \leq \frac{\mathbb{E}(\varphi(|X|))}{\varphi(a)} $$
Why this statement $ P(|X| \geq a) = P(\varphi(|X|) \geq \varphi(a)) $ is true?
For e.g. let's say $\varphi$ is $e^x$, why is it that for any non-negative random variable $X$, then $P(X \geq a) = P(e^X \geq e^a) $, it doesn't seem obvious to me why this is true.
Best Answer
It is clearer if you go back to the very definition of the event. Say $X$ is a random variable from $(\Omega,\mathcal A)$ to $(\mathbb R,\mathcal B(\mathbb R))$ and recall that "$|X|\geq a$" is a notation for the set $\{\omega \in \Omega: |X(\omega)|\geq a\}$.
Since $\varphi$ is increasing, the following equality between sets holds: $$\{\omega \in \Omega: |X(\omega)|\geq a\} = \{\omega \in \Omega: \varphi(|X(\omega)|)\geq \varphi(a)\}.$$
Applying the measure $P$ on both sides and returning to the shorthand notation, $$P(|X|\geq a) = P\big(\varphi(|X|)\geq \varphi(a)\big).$$