I asked the following question on an early post:
Suppose $X$ is a random variable and $\phi:(-\infty,\infty) \to(0,\infty)$ satisfies $\phi(-t)=\phi(t)$. Assume that $\phi(\cdot)$ is an increasing function on $(0,\infty)$. Show that for each $t>0$, $P(|X| \ge t) \le E_\phi(X) / \phi(t)$.
However, after looking through the answer one more time, I realize that the answer that I accepted is a bit confusing for me.
The answer states that $P(|X| > t) \le 2E(\phi(X))/ \phi(t)$, but a comment implies that $P(|X| > t) \le E(\phi(X))/ \phi(t)$, which is what we are actually trying to show. Can anyone please help me make these two connections?